[
    {
        "id": "Biology_105",
        "problem": "The relationship between trichome density on leaves and ion uptake was plotted from a bromeliad plants treated with different concentrations of chloride during 1 week.\n\n[figure1]\n\nChloride concentration ( $\\mu \\mathrm{M}$ )\n\nThe relationship between chloride supply, trichome number and chlorin uptake.\nA: The results suggest that chloride is only absorbed by bromeliad trichoms.\nB: Induction of trichomes occurs at any increase in concentration of exogenously applied chloride.\nC: There is an exponential relationship between the chloride concentration and absorbed chloride by leaf trichomes.\nD: Chloride is actively excluded or secreted above $40 \\mu \\mathrm{M}$ concentration.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe relationship between trichome density on leaves and ion uptake was plotted from a bromeliad plants treated with different concentrations of chloride during 1 week.\n\n[figure1]\n\nChloride concentration ( $\\mu \\mathrm{M}$ )\n\nThe relationship between chloride supply, trichome number and chlorin uptake.\n\nA: The results suggest that chloride is only absorbed by bromeliad trichoms.\nB: Induction of trichomes occurs at any increase in concentration of exogenously applied chloride.\nC: There is an exponential relationship between the chloride concentration and absorbed chloride by leaf trichomes.\nD: Chloride is actively excluded or secreted above $40 \\mu \\mathrm{M}$ concentration.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-42.jpg?height=557&width=854&top_left_y=495&top_left_x=607"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1123",
        "problem": "Frogs and toads, as well as birds, use species-specific vocalisations as apparent mate-attraction signals. The figure below shows data from two closely-related frog species, Hyla ewingi and $H$. verreauxi, from southeastern Australia, whose ranges overlap partially.\n\n[figure1]\n\nWhich of the following is correct?\nA: The largely similar mating calls of the two species in their zones of exclusive distribution clearly indicate that the two species will hybridise with one another.\nB: The distinct differences in the mating calls of the two species in the zone of distributional overlap, however, suggests that both the calls may have evolved distinctive patterns in this zone by parallel evolution.\nC: The marked difference in the mating calls of $H$. ewingi between region $A$ and region of overlap is due to 'Founder effect'.\nD: In general, it is possible for two species from regions A and B to have never evolved reproductive isolation mechanisms.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFrogs and toads, as well as birds, use species-specific vocalisations as apparent mate-attraction signals. The figure below shows data from two closely-related frog species, Hyla ewingi and $H$. verreauxi, from southeastern Australia, whose ranges overlap partially.\n\n[figure1]\n\nWhich of the following is correct?\n\nA: The largely similar mating calls of the two species in their zones of exclusive distribution clearly indicate that the two species will hybridise with one another.\nB: The distinct differences in the mating calls of the two species in the zone of distributional overlap, however, suggests that both the calls may have evolved distinctive patterns in this zone by parallel evolution.\nC: The marked difference in the mating calls of $H$. ewingi between region $A$ and region of overlap is due to 'Founder effect'.\nD: In general, it is possible for two species from regions A and B to have never evolved reproductive isolation mechanisms.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-30.jpg?height=867&width=1523&top_left_y=594&top_left_x=317"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_874",
        "problem": "下图为人体某遗传病的家系图。不考虑染色体片段互换、基因突变和环境因素对该遗传病的影响。下列相关叙述正确的是\n\n[图1]\nA: 该遗传病为伴性遗传\nB: $\\mathrm{II}_{3}$ 的体细胞中可能存在 4 个致病基因\nC: $\\mathrm{III}_{2}$ 与 $\\mathrm{III}_{3}$ 生一个患病女孩的概率为 $1 / 4$ 或 $1 / 6$\nD: 若该遗传病的致病基因位于 $\\mathrm{X}$ 染色体上，则 $\\mathrm{I}_{1}$ 的基因型是唯一的\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为人体某遗传病的家系图。不考虑染色体片段互换、基因突变和环境因素对该遗传病的影响。下列相关叙述正确的是\n\n[图1]\n\nA: 该遗传病为伴性遗传\nB: $\\mathrm{II}_{3}$ 的体细胞中可能存在 4 个致病基因\nC: $\\mathrm{III}_{2}$ 与 $\\mathrm{III}_{3}$ 生一个患病女孩的概率为 $1 / 4$ 或 $1 / 6$\nD: 若该遗传病的致病基因位于 $\\mathrm{X}$ 染色体上，则 $\\mathrm{I}_{1}$ 的基因型是唯一的\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-043.jpg?height=500&width=991&top_left_y=1640&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_800",
        "problem": "如图为某家庭的遗传系谱图, 甲遗传病由等位基因 $\\mathrm{A} / \\mathrm{a}$ 控制, 乙遗传病由等位基因\n\n$\\mathrm{B} / \\mathrm{b}$ 控制, 其中一种遗传病的致病基因位于 $\\mathrm{X}$ 染色体上，下列相关叙述正确的是（ )\n\n[图1]\n\n## ㅇ正常男女\n\n四10患甲病男女\n\n患乙病女\n\n— 两病都患女\nA: 人群中甲病男性发病率高于女性, 乙病中男女发病率相等\nB: 若仅考虑乙病, $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$ 基因型为 $\\mathrm{BB}$ 或 $\\mathrm{Bb}$, 但两者不能同时为 $\\mathrm{Bb}$\nC: 若 $\\mathrm{III}_{2}$ 与 $\\mathrm{III}_{5}$ 为近亲结婚, 他们生一个患申, 乙两病孩子的概率为 $1 / 18$\nD: 图中 $\\mathrm{II}_{2}$ 和 $\\mathrm{II}_{6}$ 基因型相同, $\\mathrm{II}_{3} 、 \\mathrm{III}_{4}$ 和 $\\mathrm{III}_{5}$ 基因型相同的概率为 $5 / 12$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图为某家庭的遗传系谱图, 甲遗传病由等位基因 $\\mathrm{A} / \\mathrm{a}$ 控制, 乙遗传病由等位基因\n\n$\\mathrm{B} / \\mathrm{b}$ 控制, 其中一种遗传病的致病基因位于 $\\mathrm{X}$ 染色体上，下列相关叙述正确的是（ )\n\n[图1]\n\n## ㅇ正常男女\n\n四10患甲病男女\n\n患乙病女\n\n— 两病都患女\n\nA: 人群中甲病男性发病率高于女性, 乙病中男女发病率相等\nB: 若仅考虑乙病, $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$ 基因型为 $\\mathrm{BB}$ 或 $\\mathrm{Bb}$, 但两者不能同时为 $\\mathrm{Bb}$\nC: 若 $\\mathrm{III}_{2}$ 与 $\\mathrm{III}_{5}$ 为近亲结婚, 他们生一个患申, 乙两病孩子的概率为 $1 / 18$\nD: 图中 $\\mathrm{II}_{2}$ 和 $\\mathrm{II}_{6}$ 基因型相同, $\\mathrm{II}_{3} 、 \\mathrm{III}_{4}$ 和 $\\mathrm{III}_{5}$ 基因型相同的概率为 $5 / 12$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-83.jpg?height=474&width=879&top_left_y=180&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1181",
        "problem": "The graphs show the number of isolates of influenza (solid black line) and respiratory syncytial virus (dotted line), that were found in samples taken from people (mainly children) in Seattle, Washington.\n\nhttp://cid.oxfordjournals.org/content/37/2/201.full\n\n[figure1]\n[figure2]\n\nWhat can be concluded from these data?\nA: You cannot catch the flu in summer.\nB: Medical treatment works in individual patients after about two months.\nC: Influenza infection always peaks in February.\nD: Virus epidemics follow predictable patterns.\nE: Respiratory syncytial virus kills more people than the flu each year.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graphs show the number of isolates of influenza (solid black line) and respiratory syncytial virus (dotted line), that were found in samples taken from people (mainly children) in Seattle, Washington.\n\nhttp://cid.oxfordjournals.org/content/37/2/201.full\n\n[figure1]\n[figure2]\n\nWhat can be concluded from these data?\n\nA: You cannot catch the flu in summer.\nB: Medical treatment works in individual patients after about two months.\nC: Influenza infection always peaks in February.\nD: Virus epidemics follow predictable patterns.\nE: Respiratory syncytial virus kills more people than the flu each year.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-03.jpg?height=334&width=1002&top_left_y=1869&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-03.jpg?height=342&width=483&top_left_y=2236&top_left_x=364"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1527",
        "problem": "For the three different diseases, pedigree analysis was performed on family trees.\n\n[figure1]\n\nIf family C's disease is caused by a recessive allele, calculate the probability a new child of 1 and 2 gets the disease.\nA: 0.05\nB: 0.1\nC: 0.25\nD: 0.5\nE: 1\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor the three different diseases, pedigree analysis was performed on family trees.\n\n[figure1]\n\nIf family C's disease is caused by a recessive allele, calculate the probability a new child of 1 and 2 gets the disease.\n\nA: 0.05\nB: 0.1\nC: 0.25\nD: 0.5\nE: 1\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-43.jpg?height=1110&width=1636&top_left_y=500&top_left_x=240"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_639",
        "problem": "下列关于真核细胞中 DNA 分子复制的叙述, 正确的是 $(\\quad)$\nA: 是一种半保留复制\nB: 主要在细胞质中完成\nC: 以 DNA 分子的一条链为模板\nD: DNA 分子完成解旋后才开始碱基配对\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列关于真核细胞中 DNA 分子复制的叙述, 正确的是 $(\\quad)$\n\nA: 是一种半保留复制\nB: 主要在细胞质中完成\nC: 以 DNA 分子的一条链为模板\nD: DNA 分子完成解旋后才开始碱基配对\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_305",
        "problem": "为探究昆虫抗药性产生的原因, 将一雌一雄两只果蝇放在同一培养瓶内繁育获得同父同母的果蝇家系, 由此获得若干家系。将每一家系的果蝇均分至两个培养瓶，甲瓶中放有涂抹 DDT 的玻璃片, 乙瓶放有空白玻璃片。检测各家系甲瓶果蝇死亡率，若死亡率高, 则淘汰该家系 若死亡率低, 则按如图流程继续操作。重复十代后, 获得了抗 DDT 能力高于原家系几百倍的果蝇。以下关于此实验叙述错误的是（）\n\n[图1]\nA: 同一家系分至甲、乙瓶的果蝇具有相同的遗传背景\nB: 实验排除了 DDT 诱导果蝇产生抗药性变异的可能\nC: 乙瓶未放置 DDT, 无法实现 DDT 对果蝇家系的选择作用\nD: 本实验可以证明果蝇抗药性增强是选择的结果\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n为探究昆虫抗药性产生的原因, 将一雌一雄两只果蝇放在同一培养瓶内繁育获得同父同母的果蝇家系, 由此获得若干家系。将每一家系的果蝇均分至两个培养瓶，甲瓶中放有涂抹 DDT 的玻璃片, 乙瓶放有空白玻璃片。检测各家系甲瓶果蝇死亡率，若死亡率高, 则淘汰该家系 若死亡率低, 则按如图流程继续操作。重复十代后, 获得了抗 DDT 能力高于原家系几百倍的果蝇。以下关于此实验叙述错误的是（）\n\n[图1]\n\nA: 同一家系分至甲、乙瓶的果蝇具有相同的遗传背景\nB: 实验排除了 DDT 诱导果蝇产生抗药性变异的可能\nC: 乙瓶未放置 DDT, 无法实现 DDT 对果蝇家系的选择作用\nD: 本实验可以证明果蝇抗药性增强是选择的结果\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-67.jpg?height=489&width=414&top_left_y=181&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1422",
        "problem": "Hemoglobin and myoglobin are proteins that have oxygen-carrying capacity. Hemoglobin is found in red blood cells and myoglobin is found in muscles. Hemoglobin can bind to four oxygen molecules at any point in time. In the lungs, where there is a high oxygen concentration, hemoglobin binds to oxygen forming oxyhemoglobin. Oxyhemoglobin then circulates the body in blood and unloads oxygen to tissues with a low oxygen concentration, reforming hemoglobin. Myoglobin stores oxygen in the muscle tissue and only releases oxygen when the partial pressure of oxygen has fallen drastically. Myoglobin is only found in the bloodstream after muscle injury.\n\nThe oxygen dissociation curve, depicted below, describes the relationship between the partial pressure of oxygen ( $x$ axis) and the oxygen saturation of hemoglobin or myoglobin ( $y$ axis).\n\n[figure1]\n\nThe partial pressure of $\\mathrm{O}_{2}$ in the lungs is around $100 \\mathrm{mmHg}$. In the lungs, the $\\% \\mathrm{O}_{2}$ saturation of hemoglobin is:\nA: $90 \\%$\nB: $95 \\%$\nC: $100 \\%$\nD: $110 \\%$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHemoglobin and myoglobin are proteins that have oxygen-carrying capacity. Hemoglobin is found in red blood cells and myoglobin is found in muscles. Hemoglobin can bind to four oxygen molecules at any point in time. In the lungs, where there is a high oxygen concentration, hemoglobin binds to oxygen forming oxyhemoglobin. Oxyhemoglobin then circulates the body in blood and unloads oxygen to tissues with a low oxygen concentration, reforming hemoglobin. Myoglobin stores oxygen in the muscle tissue and only releases oxygen when the partial pressure of oxygen has fallen drastically. Myoglobin is only found in the bloodstream after muscle injury.\n\nThe oxygen dissociation curve, depicted below, describes the relationship between the partial pressure of oxygen ( $x$ axis) and the oxygen saturation of hemoglobin or myoglobin ( $y$ axis).\n\n[figure1]\n\nThe partial pressure of $\\mathrm{O}_{2}$ in the lungs is around $100 \\mathrm{mmHg}$. In the lungs, the $\\% \\mathrm{O}_{2}$ saturation of hemoglobin is:\n\nA: $90 \\%$\nB: $95 \\%$\nC: $100 \\%$\nD: $110 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-12.jpg?height=820&width=1014&top_left_y=892&top_left_x=538"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_130",
        "problem": "Scientists have isolated three different strains of bacteria $\\mathrm{ProA}^{-}, \\mathrm{ProB}^{-}$, and $\\mathrm{ProC}^{-}$that require added proline for growth. One is cold-sensitive, one is heat-sensitive, and one has a gene deleted. Cross-feeding experiments were carried out by streaking the strains out on agar plates containing minimal medium supplemented with a very low level of proline. In cross-feeding experiments, metabolite leaking from one strain can feed a neighbouring strain. After growth at three temperatures, the results were shown in Figure below.\n[figure1]\n\nFig.Q.3. Results of cross-feeding experiments with three strains defective in proline biosynthesis. Dark areas show high cell growth rate; grey areas show low cell growth\n\n$$\n\\text { rate; wt, wild type. }\n$$\nA: The intermediate that accumulates in the $\\mathrm{ProC}^{-}$strain comes after the block in the ProA ${ }^{-}$strain.\nB: The intermediate that accumulates in the $\\mathrm{ProB}^{-}$strain comes after the block in the ProA ${ }^{-}$strain.\nC: There are at least three different genes that affect proline biosynthesis.\nD: Under at least one condition, the proline that is produced is rapidly used for protein synthesis and is prevented from being synthesized in excess of needs.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nScientists have isolated three different strains of bacteria $\\mathrm{ProA}^{-}, \\mathrm{ProB}^{-}$, and $\\mathrm{ProC}^{-}$that require added proline for growth. One is cold-sensitive, one is heat-sensitive, and one has a gene deleted. Cross-feeding experiments were carried out by streaking the strains out on agar plates containing minimal medium supplemented with a very low level of proline. In cross-feeding experiments, metabolite leaking from one strain can feed a neighbouring strain. After growth at three temperatures, the results were shown in Figure below.\n[figure1]\n\nFig.Q.3. Results of cross-feeding experiments with three strains defective in proline biosynthesis. Dark areas show high cell growth rate; grey areas show low cell growth\n\n$$\n\\text { rate; wt, wild type. }\n$$\n\nA: The intermediate that accumulates in the $\\mathrm{ProC}^{-}$strain comes after the block in the ProA ${ }^{-}$strain.\nB: The intermediate that accumulates in the $\\mathrm{ProB}^{-}$strain comes after the block in the ProA ${ }^{-}$strain.\nC: There are at least three different genes that affect proline biosynthesis.\nD: Under at least one condition, the proline that is produced is rapidly used for protein synthesis and is prevented from being synthesized in excess of needs.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-009.jpg?height=468&width=1578&top_left_y=888&top_left_x=246"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1078",
        "problem": "Which one of the following plant traits is NOT associated with colonization on land from aquatic habitat?\nA: Presence of apical meristems.\nB: Absence of alternation of generations.\nC: Formation of multicellular, dependent embryos.\nD: Production of gametes within multicellular organs.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following plant traits is NOT associated with colonization on land from aquatic habitat?\n\nA: Presence of apical meristems.\nB: Absence of alternation of generations.\nC: Formation of multicellular, dependent embryos.\nD: Production of gametes within multicellular organs.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_747",
        "problem": "果蝇体细胞中 $\\mathrm{X}$ 染色体的数目 (X) 与常染色体染色体组数目 (A) 之比决定果蝇的性别, 若 $\\mathrm{X}: \\mathrm{A} \\leq 0.5$, 果蝇表现为雄性; 若 $\\mathrm{X}: \\mathrm{A} \\geq 1$, 果蝇表现为雌性。果蝇的 $\\mathrm{Y}$ 染色体只决定果蝇的育性。已知性染色体组成为 $X X X 、 Y Y$ 的果蝇致死。让基因型为: $X^{b} X^{b} Y$的白眼雌果蝇和基因型为 $X^{B} Y$ 的红眼雄果蝇交配, 子代的表型及比例为红眼雌: 白眼雄：白眼雌：红眼雄=46:46:4:4.下列叙述错误的是 （ ）\nA: 性染色体组成为 XYY 的果蝇表现为雄性可育\nB: 子代红眼雌果蝇的性染色体组成有 XX 和 XXY 两种\nC: 子代红眼雄果蝇的性染色体组成和基因型与亲代雄果蝇完全相同\nD: 亲代白眼雌果蝇产生的配子的种类及比例为: $X^{b}: X^{b} Y: X^{b} X^{b}: Y=2: 2: 1: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇体细胞中 $\\mathrm{X}$ 染色体的数目 (X) 与常染色体染色体组数目 (A) 之比决定果蝇的性别, 若 $\\mathrm{X}: \\mathrm{A} \\leq 0.5$, 果蝇表现为雄性; 若 $\\mathrm{X}: \\mathrm{A} \\geq 1$, 果蝇表现为雌性。果蝇的 $\\mathrm{Y}$ 染色体只决定果蝇的育性。已知性染色体组成为 $X X X 、 Y Y$ 的果蝇致死。让基因型为: $X^{b} X^{b} Y$的白眼雌果蝇和基因型为 $X^{B} Y$ 的红眼雄果蝇交配, 子代的表型及比例为红眼雌: 白眼雄：白眼雌：红眼雄=46:46:4:4.下列叙述错误的是 （ ）\n\nA: 性染色体组成为 XYY 的果蝇表现为雄性可育\nB: 子代红眼雌果蝇的性染色体组成有 XX 和 XXY 两种\nC: 子代红眼雄果蝇的性染色体组成和基因型与亲代雄果蝇完全相同\nD: 亲代白眼雌果蝇产生的配子的种类及比例为: $X^{b}: X^{b} Y: X^{b} X^{b}: Y=2: 2: 1: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_70",
        "problem": "The diagram below illustrates a neural pathway and the features of the associated neurotransmitters are described in the table.\n\nIons concentrations inside and outside of the cells are same as their physiological normal values in body and inhibition of an inhibitory neuron results in the stimulation of postsynaptic neuron.\n\n\"+\" Symbol in table indicate the activation of the ion channels which increases the ion permeability across the membrane.\n[figure1]\nA: The function of N1 neurotransmitter is same as acetylcholine.\nB: Neuron $\\mathrm{G}$ is not depolarized and neuron $\\mathrm{F}$ is stimulated so muscle M1 contracts.\nC: Neuron $\\mathrm{G}$ gets depolarized as a result of $\\mathrm{Na}$ ions influx.\nD: Neuron $\\mathrm{F}$ gets depolarized as a result of $\\mathrm{Cl}$ ions efflux.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe diagram below illustrates a neural pathway and the features of the associated neurotransmitters are described in the table.\n\nIons concentrations inside and outside of the cells are same as their physiological normal values in body and inhibition of an inhibitory neuron results in the stimulation of postsynaptic neuron.\n\n\"+\" Symbol in table indicate the activation of the ion channels which increases the ion permeability across the membrane.\n[figure1]\n\nA: The function of N1 neurotransmitter is same as acetylcholine.\nB: Neuron $\\mathrm{G}$ is not depolarized and neuron $\\mathrm{F}$ is stimulated so muscle M1 contracts.\nC: Neuron $\\mathrm{G}$ gets depolarized as a result of $\\mathrm{Na}$ ions influx.\nD: Neuron $\\mathrm{F}$ gets depolarized as a result of $\\mathrm{Cl}$ ions efflux.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-31.jpg?height=1312&width=1146&top_left_y=714&top_left_x=454"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_926",
        "problem": "某种植物的粉花和白花是一对相对性状, 现利用该植物进行了两组杂交实验, 实验\n结果如下：表所示，相关基因用 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b} 、 \\mathrm{C} / \\mathrm{c} \\ldots$ 表示，下列相关叙述正确的是（ ）\n\n| 组别 | 亲本 | $F_{1}$ |\n| :--- | :--- | :--- |\n| (1) | 白花×白花 | 白花: 粉花=3: 1 |\n| (2) | 粉花×粉花 | 粉花: 白花=9: 7 |\nA: 该植物的花色由 1 对等位基因控制, 且遵循分离定律\nB: 两个组别中所有的亲本都是杂合子\nC: 某白花植株自交不会出现性状分离, 则该植株为双隐性个体\nD: 组(1)的 $F_{1}$ 个体随机授粉, $F_{2}$ 中纯合粉花植株的比例是 $1 / 128$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种植物的粉花和白花是一对相对性状, 现利用该植物进行了两组杂交实验, 实验\n结果如下：表所示，相关基因用 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b} 、 \\mathrm{C} / \\mathrm{c} \\ldots$ 表示，下列相关叙述正确的是（ ）\n\n| 组别 | 亲本 | $F_{1}$ |\n| :--- | :--- | :--- |\n| (1) | 白花×白花 | 白花: 粉花=3: 1 |\n| (2) | 粉花×粉花 | 粉花: 白花=9: 7 |\n\nA: 该植物的花色由 1 对等位基因控制, 且遵循分离定律\nB: 两个组别中所有的亲本都是杂合子\nC: 某白花植株自交不会出现性状分离, 则该植株为双隐性个体\nD: 组(1)的 $F_{1}$ 个体随机授粉, $F_{2}$ 中纯合粉花植株的比例是 $1 / 128$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_426",
        "problem": "某生物的卵原细胞在培养液中既能进行有丝分裂也能进行减数分裂。研究人员在该生物卵原细胞进行减数分裂过程中, 发现了“逆反”减数分裂现象。将一个双链均被 ${ }^{14} \\mathrm{C}$标记的基因 $\\mathrm{A}$ 和一个双链均被 ${ }^{13} \\mathrm{C}$ 标记的基因 $\\mathrm{A}$ 插入一个卵原细胞的一条染色体的两端。将此卵原细胞在普通 ${ }^{12} \\mathrm{C}$ 培养液中培养, 先完成一次有丝分裂, 再发生如图所示的 “逆反”减数分裂, 共产生 8 个子细胞。下列叙述错误的是（）\n\n[图1]\nA: “逆反”减数分裂时, 同源染色体在减数分裂I分离, 姐妹染色单体在减数分裂II 分离\nB: 8 个子细胞中, 最多有 4 个卵细胞同时含有 ${ }^{13} \\mathrm{C}$ 标记和 ${ }^{14} \\mathrm{C}$ 标记\nC: 8 个子细胞中, 可能有 1 个卵细胞同时含有 ${ }^{13} \\mathrm{C}$ 标记和 ${ }^{14} \\mathrm{C}$ 标记、 1 个卵细胞含 ${ }^{13} \\mathrm{C}$标记\nD: 8 个子细胞中, 可能有 2 个卵细胞同时含有 ${ }^{13} \\mathrm{C}$ 标记和 ${ }^{14} \\mathrm{C}$ 标记、 6 个极体都不含 ${ }^{14} \\mathrm{C}$ 标记和 ${ }^{13} \\mathrm{C}$ 标记\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某生物的卵原细胞在培养液中既能进行有丝分裂也能进行减数分裂。研究人员在该生物卵原细胞进行减数分裂过程中, 发现了“逆反”减数分裂现象。将一个双链均被 ${ }^{14} \\mathrm{C}$标记的基因 $\\mathrm{A}$ 和一个双链均被 ${ }^{13} \\mathrm{C}$ 标记的基因 $\\mathrm{A}$ 插入一个卵原细胞的一条染色体的两端。将此卵原细胞在普通 ${ }^{12} \\mathrm{C}$ 培养液中培养, 先完成一次有丝分裂, 再发生如图所示的 “逆反”减数分裂, 共产生 8 个子细胞。下列叙述错误的是（）\n\n[图1]\n\nA: “逆反”减数分裂时, 同源染色体在减数分裂I分离, 姐妹染色单体在减数分裂II 分离\nB: 8 个子细胞中, 最多有 4 个卵细胞同时含有 ${ }^{13} \\mathrm{C}$ 标记和 ${ }^{14} \\mathrm{C}$ 标记\nC: 8 个子细胞中, 可能有 1 个卵细胞同时含有 ${ }^{13} \\mathrm{C}$ 标记和 ${ }^{14} \\mathrm{C}$ 标记、 1 个卵细胞含 ${ }^{13} \\mathrm{C}$标记\nD: 8 个子细胞中, 可能有 2 个卵细胞同时含有 ${ }^{13} \\mathrm{C}$ 标记和 ${ }^{14} \\mathrm{C}$ 标记、 6 个极体都不含 ${ }^{14} \\mathrm{C}$ 标记和 ${ }^{13} \\mathrm{C}$ 标记\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-073.jpg?height=429&width=1036&top_left_y=2107&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_835",
        "problem": "正常人的一条 16 号染色体含有两个 $\\alpha$ 基因, 如下图甲所示。若 4 个 $\\alpha$ 基因均缺失,则患 Barts 病, 若缺失 3 个 $\\alpha$ 基因, 则患 $\\mathrm{HbH}$ 病, 其余情况均正常。某家系 $\\alpha$-地中海贫血症（Barts 病和 $\\mathrm{HbH}$ 病）的遗传情况如下图乙所示，若不发生其他变异，下列分析正确的是（）\n\n第16号染色体\n\n甲\n\n[图1]\n\n$\\square \\bigcirc$ 正常男女患Barts病女患 $\\mathrm{HbH}$ 病男女\nA: II-3 缺失 2 个 $\\alpha$ 基因的染色体来源于 I-2\nB: I-1 和 II-5 的两条 16 号染色体上的 $\\alpha$ 基因组成不相同\nC: II-5 和 II-6 可能会生出患 $\\mathrm{HbH}$ 病的孩子\nD: III-3 和 III-4 生出一个患病女孩的概率为 $1 / 12$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n正常人的一条 16 号染色体含有两个 $\\alpha$ 基因, 如下图甲所示。若 4 个 $\\alpha$ 基因均缺失,则患 Barts 病, 若缺失 3 个 $\\alpha$ 基因, 则患 $\\mathrm{HbH}$ 病, 其余情况均正常。某家系 $\\alpha$-地中海贫血症（Barts 病和 $\\mathrm{HbH}$ 病）的遗传情况如下图乙所示，若不发生其他变异，下列分析正确的是（）\n\n第16号染色体\n\n甲\n\n[图1]\n\n$\\square \\bigcirc$ 正常男女患Barts病女患 $\\mathrm{HbH}$ 病男女\n\nA: II-3 缺失 2 个 $\\alpha$ 基因的染色体来源于 I-2\nB: I-1 和 II-5 的两条 16 号染色体上的 $\\alpha$ 基因组成不相同\nC: II-5 和 II-6 可能会生出患 $\\mathrm{HbH}$ 病的孩子\nD: III-3 和 III-4 生出一个患病女孩的概率为 $1 / 12$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-11.jpg?height=323&width=738&top_left_y=1455&top_left_x=659"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_39",
        "problem": "The skeletal muscle fibers are divided into three types depending on the speed and energy sources of muscle contraction:\n\n- Type I: Slow twitch, oxidative muscle fibers.\n- Type IIa: Fast twitch, oxidative muscle fibers.\n- Type IIb: Fast twitch, glycolytic muscle fibers.\n\nFigure Q. 29 shows the correlation of mRNA expression levels of the genes Myh7, Myh2, and Myh1 specific for muscle fiber types I, IIa, and IIb, respectively, in human skeletal muscles of legs: quadriceps, gastrocnemius, and soleus muscles.\n\n[figure1]\nA: The soleus muscles of the sprinter athletes are prone to be more developed than those in the marathon athletes.\nB: The ratio of mitochondria number per muscle mass in the quadriceps muscle is less than that in the gastrocnemius muscle.\nC: The soleus muscle contains less sarcoplasmic reticulum than the gastrocnemius muscle does.\nD: Regularly doing the endurance exercise for a long period of time can increase the number of glycolytic muscle fibers in the gastrocnemius muscles.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe skeletal muscle fibers are divided into three types depending on the speed and energy sources of muscle contraction:\n\n- Type I: Slow twitch, oxidative muscle fibers.\n- Type IIa: Fast twitch, oxidative muscle fibers.\n- Type IIb: Fast twitch, glycolytic muscle fibers.\n\nFigure Q. 29 shows the correlation of mRNA expression levels of the genes Myh7, Myh2, and Myh1 specific for muscle fiber types I, IIa, and IIb, respectively, in human skeletal muscles of legs: quadriceps, gastrocnemius, and soleus muscles.\n\n[figure1]\n\nA: The soleus muscles of the sprinter athletes are prone to be more developed than those in the marathon athletes.\nB: The ratio of mitochondria number per muscle mass in the quadriceps muscle is less than that in the gastrocnemius muscle.\nC: The soleus muscle contains less sarcoplasmic reticulum than the gastrocnemius muscle does.\nD: Regularly doing the endurance exercise for a long period of time can increase the number of glycolytic muscle fibers in the gastrocnemius muscles.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-065.jpg?height=928&width=1299&top_left_y=1061&top_left_x=390"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1445",
        "problem": "The graph below shows the concentration of methane $\\left(\\mathrm{CH}_{4}\\right)$ in the Earth's atmosphere alongside the intensity of solar radiation over the past few millennia.\n\n[figure1]\n\nAccording to the graph, if the trend of $\\mathrm{CH}_{4}$ concentration had continued to match the trend on solar radiation intensity, the concentration at present would most likely be:\nA: Less than $500 \\mathrm{ppb}$.\nB: Between 500 and $550 \\mathrm{ppb}$.\nC: Between 550 and $600 \\mathrm{ppb}$.\nD: Between 600 and $700 \\mathrm{ppb}$.\nE: Above 700 ppb.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph below shows the concentration of methane $\\left(\\mathrm{CH}_{4}\\right)$ in the Earth's atmosphere alongside the intensity of solar radiation over the past few millennia.\n\n[figure1]\n\nAccording to the graph, if the trend of $\\mathrm{CH}_{4}$ concentration had continued to match the trend on solar radiation intensity, the concentration at present would most likely be:\n\nA: Less than $500 \\mathrm{ppb}$.\nB: Between 500 and $550 \\mathrm{ppb}$.\nC: Between 550 and $600 \\mathrm{ppb}$.\nD: Between 600 and $700 \\mathrm{ppb}$.\nE: Above 700 ppb.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-24.jpg?height=1080&width=1051&top_left_y=451&top_left_x=525"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_354",
        "problem": "果蝇 X 染色体上有 $\\mathrm{B}$ （棒眼）、W（杏色眼）基因, 欲检测某雄果蝇 $\\mathrm{X}$ 染色体上是否发生了基因突变（非 $\\mathrm{B} 、 \\mathrm{~W}$ 基因），用 $\\mathrm{B} 、 \\mathrm{~W}$ 基因纯合雌果蝇与之杂交， $\\mathrm{F}_{1}$ 雌雄个体相互交配得到 $\\mathrm{F}_{2}$ 。下列相关叙述正确的是 ( )\nA: 若 $F_{2}$ 中被检测基因所控制的性状在雌雄果蝇中相同, 说明发生了隐性突变\nB: 若 $F_{2}$ 中被检测基因所控制的性状只有雄性果蝇表现，说明发生了显性突变\nC: 若 $F_{2}$ 中性别比例为雌蝇: 雄蝇=2: 1 , 说明发生了显性致死突变\nD: 若 $F_{2}$ 表现型比例为 1: 1: 1: 1 , 说明待测 $X$ 染色体上没有隐性致死突变\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇 X 染色体上有 $\\mathrm{B}$ （棒眼）、W（杏色眼）基因, 欲检测某雄果蝇 $\\mathrm{X}$ 染色体上是否发生了基因突变（非 $\\mathrm{B} 、 \\mathrm{~W}$ 基因），用 $\\mathrm{B} 、 \\mathrm{~W}$ 基因纯合雌果蝇与之杂交， $\\mathrm{F}_{1}$ 雌雄个体相互交配得到 $\\mathrm{F}_{2}$ 。下列相关叙述正确的是 ( )\n\nA: 若 $F_{2}$ 中被检测基因所控制的性状在雌雄果蝇中相同, 说明发生了隐性突变\nB: 若 $F_{2}$ 中被检测基因所控制的性状只有雄性果蝇表现，说明发生了显性突变\nC: 若 $F_{2}$ 中性别比例为雌蝇: 雄蝇=2: 1 , 说明发生了显性致死突变\nD: 若 $F_{2}$ 表现型比例为 1: 1: 1: 1 , 说明待测 $X$ 染色体上没有隐性致死突变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1469",
        "problem": "The Great Barrier Reef is among the world's most awe-inspiring and unique ecosystems. It spans more than 2,000 kilometres, has more than 600 types of coral, 1,600 types of fish and is of immense cultural significance, especially for traditional owners of the land.\n\nThe species of coral Acropora digitifera is unable to produce cysteine as it lacks the gene for an essential enzyme. However, the same organisms that give the coral its colour (dinoflagellates) also provide the organism with cysteine. This is an example of what kind of relationship?\nA: Commensalism\nB: Predator-prey\nC: Mutualism\nD: Parasitism\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe Great Barrier Reef is among the world's most awe-inspiring and unique ecosystems. It spans more than 2,000 kilometres, has more than 600 types of coral, 1,600 types of fish and is of immense cultural significance, especially for traditional owners of the land.\n\nThe species of coral Acropora digitifera is unable to produce cysteine as it lacks the gene for an essential enzyme. However, the same organisms that give the coral its colour (dinoflagellates) also provide the organism with cysteine. This is an example of what kind of relationship?\n\nA: Commensalism\nB: Predator-prey\nC: Mutualism\nD: Parasitism\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_659",
        "problem": "家族性高胆固醇血症 (FH) 是一种遗传病, 纯合子患者在人群中出现的频率约 1/10000。如图是某 FH 家系的系谱图, 下列叙述正确的是（ ）\n\n[图1]\n\nII\n\n[图2]\nA: FH 为常染色体显性遗传病\nB: FH 患者双亲中至少有一人为 FH 患者\nC: 杂合子患者在人群中出现的频率为 $99 / 5000$\nD: $\\mathrm{III}_{6}$ 的患病基因由父母双方共同提供\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n家族性高胆固醇血症 (FH) 是一种遗传病, 纯合子患者在人群中出现的频率约 1/10000。如图是某 FH 家系的系谱图, 下列叙述正确的是（ ）\n\n[图1]\n\nII\n\n[图2]\n\nA: FH 为常染色体显性遗传病\nB: FH 患者双亲中至少有一人为 FH 患者\nC: 杂合子患者在人群中出现的频率为 $99 / 5000$\nD: $\\mathrm{III}_{6}$ 的患病基因由父母双方共同提供\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-090.jpg?height=58&width=45&top_left_y=228&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-090.jpg?height=480&width=762&top_left_y=211&top_left_x=447"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_293",
        "problem": "Select the chemical property that is shared by all types of lipids forming the plasma membrane.\nA: Polar head\nB: Sugar component\nC: Glycerol backbone\nD: Phosphate group\nE: Hydrophobic region\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSelect the chemical property that is shared by all types of lipids forming the plasma membrane.\n\nA: Polar head\nB: Sugar component\nC: Glycerol backbone\nD: Phosphate group\nE: Hydrophobic region\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1302",
        "problem": "## RATITE EVOLUTION\n\nRatites are named for their flat \"raft-like\" sternum that lacks a keel and thus cannot provide sufficient anchorage for flight muscles to effectively generate the power required to fly. Cassowary (Australia and Papua New Guinea), Emu (Australia), Rhea (South America), Ostrich (Africa), Kiwi (New Zealand), and extinct Moa (New Zealand) are all examples of ground dwelling, flightless ratites. Given their large, flightless nature, biologists have proposed that they have all descended from a common ancestor present in ancient Gondwana before it broke apart. The exact timing and evolutionary origin of the Moa and Kiwi in New Zealand have been greatly debated in the scientific literature.\n\n[figure1]\nfrom Antarctica and India\n\n- 130 mya South America breaks free of Africa\n- 80 mya Zealandia breaks free\n- 60 mya New Zealand separates from Australia\n- 40 mya Australia separates from Antarctica\n- 30 mya South America breaks from West Antarctica (Antarctica freezes over)\n\n[figure2]\n\nFuture Zealandia\n\n## Flying Cousins\n\nRatites are one of only two groups of birds belonging to the \"old jaw\" Paleognaths originating in Gondwana, the other comprises the 47 living Tinamou species of Central and South America. Tinamou species are generally ground dwelling, though they do have wings that allow for limited flight.\n\nIt is generally accepted that loss of flight in birds is due to the development of successful foraging behaviours and diminished predation on eggs and nests.\n\n## Fossil Evidence\n\nExamination of the oldest known Moa fossils, dating from 19 million years ago, determined they had thickened leg bones and no wing structures at all. At least two species of flightless Moa were present in New Zealand at this time.\n\nThe oldest Tinamou fossils in South America are 10 million years old.\n\nThe oldest Kiwi fossil dates back 1 million years. Though diminished in size, kiwis have wings and flight feathers. Kiwis are the only known bird to have nostrils at the end of their beaks and the only ratite with two functioning ovaries.Tinamou are the only group of Paleognath birds that have the ability to fly. Which of the following is LEAST likely to account for their flying ability?\nA: A common ancestor became isolated from other Paleognaths prior to them becoming flightless.\nB: The rich biodiversity of South America provided a selective advantage for flying.\nC: Possessing wings provide benefits for Tinamou that are unrelated to flying.\nD: Possessing wings allows Tinamou to escape from ground dwelling predators.\nE: Tinamou evolved from a flightless bird that regained the ability to fly.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## RATITE EVOLUTION\n\nRatites are named for their flat \"raft-like\" sternum that lacks a keel and thus cannot provide sufficient anchorage for flight muscles to effectively generate the power required to fly. Cassowary (Australia and Papua New Guinea), Emu (Australia), Rhea (South America), Ostrich (Africa), Kiwi (New Zealand), and extinct Moa (New Zealand) are all examples of ground dwelling, flightless ratites. Given their large, flightless nature, biologists have proposed that they have all descended from a common ancestor present in ancient Gondwana before it broke apart. The exact timing and evolutionary origin of the Moa and Kiwi in New Zealand have been greatly debated in the scientific literature.\n\n[figure1]\nfrom Antarctica and India\n\n- 130 mya South America breaks free of Africa\n- 80 mya Zealandia breaks free\n- 60 mya New Zealand separates from Australia\n- 40 mya Australia separates from Antarctica\n- 30 mya South America breaks from West Antarctica (Antarctica freezes over)\n\n[figure2]\n\nFuture Zealandia\n\n## Flying Cousins\n\nRatites are one of only two groups of birds belonging to the \"old jaw\" Paleognaths originating in Gondwana, the other comprises the 47 living Tinamou species of Central and South America. Tinamou species are generally ground dwelling, though they do have wings that allow for limited flight.\n\nIt is generally accepted that loss of flight in birds is due to the development of successful foraging behaviours and diminished predation on eggs and nests.\n\n## Fossil Evidence\n\nExamination of the oldest known Moa fossils, dating from 19 million years ago, determined they had thickened leg bones and no wing structures at all. At least two species of flightless Moa were present in New Zealand at this time.\n\nThe oldest Tinamou fossils in South America are 10 million years old.\n\nThe oldest Kiwi fossil dates back 1 million years. Though diminished in size, kiwis have wings and flight feathers. Kiwis are the only known bird to have nostrils at the end of their beaks and the only ratite with two functioning ovaries.\n\nproblem:\nTinamou are the only group of Paleognath birds that have the ability to fly. Which of the following is LEAST likely to account for their flying ability?\n\nA: A common ancestor became isolated from other Paleognaths prior to them becoming flightless.\nB: The rich biodiversity of South America provided a selective advantage for flying.\nC: Possessing wings provide benefits for Tinamou that are unrelated to flying.\nD: Possessing wings allows Tinamou to escape from ground dwelling predators.\nE: Tinamou evolved from a flightless bird that regained the ability to fly.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-24.jpg?height=711&width=585&top_left_y=689&top_left_x=135",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-24.jpg?height=708&width=1259&top_left_y=688&top_left_x=747"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_345",
        "problem": "潜伏感染是指 HIV 病毒 DNA 整合到宿主细胞基因组中, 但不产生游离的病毒颗粒, 整合了 HIV 基因组的宿主细胞就成为了“病毒储存库”。HIV 病毒储存库通常存在于记忆 $\\mathrm{CD} 4+\\mathrm{T}$ 细胞中, 也可以存在于单核巨噬细胞和星形胶质细胞中。\n\n82. HIV 是具有包膜的 RNA 病毒, 下列叙述正确的是（ ）\nA: 包膜是 HIV 的细胞膜, 其结构基础为磷脂双分子层\nB: HIV 通过包膜与宿主细胞质膜融合, 只释放 RNA 进入细胞\nC: 宿主细胞成为了“病毒储存库”的过程仅需逆转录酶\nD: HIV 病毒 DNA 与宿主细胞基因组整合需在酶的催化下进行\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n潜伏感染是指 HIV 病毒 DNA 整合到宿主细胞基因组中, 但不产生游离的病毒颗粒, 整合了 HIV 基因组的宿主细胞就成为了“病毒储存库”。HIV 病毒储存库通常存在于记忆 $\\mathrm{CD} 4+\\mathrm{T}$ 细胞中, 也可以存在于单核巨噬细胞和星形胶质细胞中。\n\n82. HIV 是具有包膜的 RNA 病毒, 下列叙述正确的是（ ）\n\nA: 包膜是 HIV 的细胞膜, 其结构基础为磷脂双分子层\nB: HIV 通过包膜与宿主细胞质膜融合, 只释放 RNA 进入细胞\nC: 宿主细胞成为了“病毒储存库”的过程仅需逆转录酶\nD: HIV 病毒 DNA 与宿主细胞基因组整合需在酶的催化下进行\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1318",
        "problem": "Measurements were made on the growth of young spruce trees in a pure stand and in a mixture with pine trees. The density of trees in both stands was the same. The effect of providing the plants with a calcium and nitrogen source (basic slag) was also investigated at the same time. The results of these investigations on twelve-year old trees are shown in the table.\n\n|  | Mean height of spruce trees in centimeters in: |  |\n| :--- | :---: | :---: |\n|  | Pure stand of spruce | Mixed stand of spruce and pine |\n| Not treated | 60 | 100 |\n| Basic slag applied after planting | 95 | 130 |\n\nUsing only these data, which is the BEST interpretation?\nA: Spruce and pine are in competition with each other.\nB: The growth of pine is decreased by the presence of spruce.\nC: Calcium ions stimulate the growth of spruce more in mixed stands.\nD: The growth of spruce is increased by the presence of pine\nE: Pine trees grow faster with the addition of basic slag.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMeasurements were made on the growth of young spruce trees in a pure stand and in a mixture with pine trees. The density of trees in both stands was the same. The effect of providing the plants with a calcium and nitrogen source (basic slag) was also investigated at the same time. The results of these investigations on twelve-year old trees are shown in the table.\n\n|  | Mean height of spruce trees in centimeters in: |  |\n| :--- | :---: | :---: |\n|  | Pure stand of spruce | Mixed stand of spruce and pine |\n| Not treated | 60 | 100 |\n| Basic slag applied after planting | 95 | 130 |\n\nUsing only these data, which is the BEST interpretation?\n\nA: Spruce and pine are in competition with each other.\nB: The growth of pine is decreased by the presence of spruce.\nC: Calcium ions stimulate the growth of spruce more in mixed stands.\nD: The growth of spruce is increased by the presence of pine\nE: Pine trees grow faster with the addition of basic slag.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1501",
        "problem": "The RECOVERY trial run by the NHS is the most successful trial to identify proven treatments for COVID-19, and has disproven several popular candidates.\n\nWhat features should a randomised control trial like RECOVERY NOT have?\nA: Doctors do not know what drug they are administering until after the trial.\nB: Patients do not know which drug they received until after the trial.\nC: Hypothesis fixed at start of the trial.\nD: Number of patients used in trial maximised.\nE: Include patients without COVID-19\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe RECOVERY trial run by the NHS is the most successful trial to identify proven treatments for COVID-19, and has disproven several popular candidates.\n\nWhat features should a randomised control trial like RECOVERY NOT have?\n\nA: Doctors do not know what drug they are administering until after the trial.\nB: Patients do not know which drug they received until after the trial.\nC: Hypothesis fixed at start of the trial.\nD: Number of patients used in trial maximised.\nE: Include patients without COVID-19\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1214",
        "problem": "Two cultures of algae were exposed to ${ }^{14} \\mathrm{CO}_{2}$ for 5 seconds and 15 seconds respectively. They were then killed and the soluble products of photosynthesis extracted and used to produce the two chromatograms shown below.\n\n[figure1]\n\nWhich one of the following statements is a correct inference from the data?\nA: Glycerate 3-phosphate is the first product of $\\mathrm{CO}_{2}$ fixation.\nB: Malate is the last product of $\\mathrm{CO}_{2}$ fixation.\nC: Hexose monophosphates are unstable and break down into glycerate 3-phosphate.\nD: Glycerate 3-phosphate is converted to triose phosphate.\nE: Triose phosphates are formed after glycerate 3-phosphate.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTwo cultures of algae were exposed to ${ }^{14} \\mathrm{CO}_{2}$ for 5 seconds and 15 seconds respectively. They were then killed and the soluble products of photosynthesis extracted and used to produce the two chromatograms shown below.\n\n[figure1]\n\nWhich one of the following statements is a correct inference from the data?\n\nA: Glycerate 3-phosphate is the first product of $\\mathrm{CO}_{2}$ fixation.\nB: Malate is the last product of $\\mathrm{CO}_{2}$ fixation.\nC: Hexose monophosphates are unstable and break down into glycerate 3-phosphate.\nD: Glycerate 3-phosphate is converted to triose phosphate.\nE: Triose phosphates are formed after glycerate 3-phosphate.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-09.jpg?height=534&width=1259&top_left_y=1298&top_left_x=410"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_294",
        "problem": "Many disease causing mutations in enzyme coding genes affect parameters related to interactions of the enzymes with their coenzyme, which in turn lowers the rate of the reaction. The cofactor of a group of enzymes called TPP- or thiamine-dependent enzymes such as transketolase is thiamine pyrophosphate (TPP).\n\nThe kinetic properties of a TPP-dependent enzyme isolated from a patient's tissue and a normal individual's tissue were compared under condition of substrate saturation. The TPP-free forms of the enzyme was prepared and used for enzyme kinetic measurements. The Lineweaver-Burk like plots are shown below.\n\n[figure1]\nA: A is related to the patient's enzyme.\nB: Maximum velocity of $\\mathrm{B}$ is higher than the maximum velocity of $\\mathrm{A}$.\nC: It can be concluded that the enzyme in A has a lower affinity for its substrate(s).\nD: Administration of thiamine to patients is expected to restore enzymatic activity.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nMany disease causing mutations in enzyme coding genes affect parameters related to interactions of the enzymes with their coenzyme, which in turn lowers the rate of the reaction. The cofactor of a group of enzymes called TPP- or thiamine-dependent enzymes such as transketolase is thiamine pyrophosphate (TPP).\n\nThe kinetic properties of a TPP-dependent enzyme isolated from a patient's tissue and a normal individual's tissue were compared under condition of substrate saturation. The TPP-free forms of the enzyme was prepared and used for enzyme kinetic measurements. The Lineweaver-Burk like plots are shown below.\n\n[figure1]\n\nA: A is related to the patient's enzyme.\nB: Maximum velocity of $\\mathrm{B}$ is higher than the maximum velocity of $\\mathrm{A}$.\nC: It can be concluded that the enzyme in A has a lower affinity for its substrate(s).\nD: Administration of thiamine to patients is expected to restore enzymatic activity.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-18.jpg?height=614&width=1022&top_left_y=778&top_left_x=517"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_871",
        "problem": "某真核生物 DNA 片段的结构示意图如下, 下列叙述正确的是( )\n\n[图1]\nA: (1)的形成需要 DNA 聚合酶催化\nB: (2)表示腺嘌呤脱氧核苷酸\nC: (3)的形成只能发生在细胞核\nD: 若 a 链中 $\\mathrm{A}+\\mathrm{T}$ 占 $48 \\%$, 则 DNA 分子中 $\\mathrm{G}$ 占 $26 \\%$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某真核生物 DNA 片段的结构示意图如下, 下列叙述正确的是( )\n\n[图1]\n\nA: (1)的形成需要 DNA 聚合酶催化\nB: (2)表示腺嘌呤脱氧核苷酸\nC: (3)的形成只能发生在细胞核\nD: 若 a 链中 $\\mathrm{A}+\\mathrm{T}$ 占 $48 \\%$, 则 DNA 分子中 $\\mathrm{G}$ 占 $26 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-18.jpg?height=428&width=351&top_left_y=1802&top_left_x=173"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_933",
        "problem": "应用基因工程技术将抗虫基因 A 和抗除草剂基因 B 成功导入植株 W（2n=40）的染色体组中。植株 W 自交, 子代中既不抗虫也不抗除草剂的植株所占比例为 $1 / 16$ 。取植\n株 $\\mathrm{W}$ 某部位的一个细胞放在适宜条件下培养，产生 4 个子细胞。用荧光分子检测基因 $\\mathrm{A}$ 和基因 B（基因 A、基因 B 均能被苂光标记）。下列叙述正确的是\nA: 植株 W 获得抗虫和抗除草剂变异性状，其原理是染色体变异\nB: 若 4 个子细胞都只含有一个荧光点, 则子细胞中的染色体数是 40\nC: 若有的子细胞不含荧光点, 则细胞分裂过程中发生过基因重组\nD: 若 4 个子细胞都含有两个荧光点, 则细胞分裂过程中出现过 20 个四分体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n应用基因工程技术将抗虫基因 A 和抗除草剂基因 B 成功导入植株 W（2n=40）的染色体组中。植株 W 自交, 子代中既不抗虫也不抗除草剂的植株所占比例为 $1 / 16$ 。取植\n株 $\\mathrm{W}$ 某部位的一个细胞放在适宜条件下培养，产生 4 个子细胞。用荧光分子检测基因 $\\mathrm{A}$ 和基因 B（基因 A、基因 B 均能被苂光标记）。下列叙述正确的是\n\nA: 植株 W 获得抗虫和抗除草剂变异性状，其原理是染色体变异\nB: 若 4 个子细胞都只含有一个荧光点, 则子细胞中的染色体数是 40\nC: 若有的子细胞不含荧光点, 则细胞分裂过程中发生过基因重组\nD: 若 4 个子细胞都含有两个荧光点, 则细胞分裂过程中出现过 20 个四分体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_703",
        "problem": "苯丙酮尿症是一种常见的氨基酸代谢障碍病, 绝大多数患者病因是苯丙氨酸羟化酶\n\n（PAH）基因突变导致苯丙氨酸羟化酶缺乏。已知人群中染色体上 PAH 基因两侧限制性核酸内切酶 MspI 酶切位点的分布存在两种形式 (图 1)。图 2 是某患者的家族系谱图,其中部分成员 $\\mathrm{I}_{1} 、 \\mathrm{I}_{2} 、 \\mathrm{II}_{1}$ 和 $\\mathrm{II}_{2}$ 的 DNA 经限制性内切酶 $\\mathrm{MspI}$ 酶切后进行电泳分离, 并利用苂光标记的 PAH 基因片段与酶切片段杂交, 得到 DNA 条带分布情况如图 3, 在电泳过程中，分子量较小的条带会因电泳速度过快导致难以观察。下列分析错误的是（）\n\n[图1]\n\n图 1\n\n[图2]\n\n图 2\n\n[图3]\n\n图3\nA: 图 2 中 $\\mathrm{II}_{1}$ 与 $\\mathrm{II}_{4}$ 基因型相同的概率为 $4 / 9$\nB: $\\mathrm{I}_{2}$ 个体 $19 \\mathrm{~kb}$ 的 DNA 条带中一定含有正常 PAH 基因\nC: $\\mathrm{II}_{3}$ 个体一定为 PAH 基因纯合体，可能有 1 个探针杂交条带\nD: 若 $\\mathrm{II}_{4}$ 电泳只有 $19 \\mathrm{~kb}$ 的 DNA 条带, 则为 PAH 基因杂合体的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n苯丙酮尿症是一种常见的氨基酸代谢障碍病, 绝大多数患者病因是苯丙氨酸羟化酶\n\n（PAH）基因突变导致苯丙氨酸羟化酶缺乏。已知人群中染色体上 PAH 基因两侧限制性核酸内切酶 MspI 酶切位点的分布存在两种形式 (图 1)。图 2 是某患者的家族系谱图,其中部分成员 $\\mathrm{I}_{1} 、 \\mathrm{I}_{2} 、 \\mathrm{II}_{1}$ 和 $\\mathrm{II}_{2}$ 的 DNA 经限制性内切酶 $\\mathrm{MspI}$ 酶切后进行电泳分离, 并利用苂光标记的 PAH 基因片段与酶切片段杂交, 得到 DNA 条带分布情况如图 3, 在电泳过程中，分子量较小的条带会因电泳速度过快导致难以观察。下列分析错误的是（）\n\n[图1]\n\n图 1\n\n[图2]\n\n图 2\n\n[图3]\n\n图3\n\nA: 图 2 中 $\\mathrm{II}_{1}$ 与 $\\mathrm{II}_{4}$ 基因型相同的概率为 $4 / 9$\nB: $\\mathrm{I}_{2}$ 个体 $19 \\mathrm{~kb}$ 的 DNA 条带中一定含有正常 PAH 基因\nC: $\\mathrm{II}_{3}$ 个体一定为 PAH 基因纯合体，可能有 1 个探针杂交条带\nD: 若 $\\mathrm{II}_{4}$ 电泳只有 $19 \\mathrm{~kb}$ 的 DNA 条带, 则为 PAH 基因杂合体的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-098.jpg?height=289&width=417&top_left_y=1783&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-098.jpg?height=231&width=514&top_left_y=1821&top_left_x=791",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-098.jpg?height=211&width=465&top_left_y=1825&top_left_x=1321"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_521",
        "problem": "在红眼野生型果蝇中发现了极少数紫眼突变体。已知 Henna 基因是苯丙氨酸羟化酶的编码基因。为证明 Henna 基因突变是紫眼出现的原因, 进行了如下实验。\n\n实验 1：突变体和野生型果蝇正反交, 后代均为紫红眼\n\n实验 2: 将野生型果蝇的 1 个 Henna 基因转给突变体甲, 发现甲的眼色变紫红; 将野生型果蝇的 2 个 Henna 基因转给突变体乙，发现乙的眼色恢复至野生型\n\n实验 3: 检测果蝇眼部蝶呤含量, 发现甲的含量仅为野生型的 79\\%, 乙的含量升至 98\\%，与野生型差异不显著\n\n下列说法正确的是（）\nA: 果蝇眼色性状受常染色体基因控制\nB: 苯丙氨酸差化酶催化果蝇眼部蝶呤的合成\nC: Henna 基因突变导致眼部蝶呤含量降低\nD: 完全敲除野生型的 Henna 基因, 果蝇眼色会变紫色\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n在红眼野生型果蝇中发现了极少数紫眼突变体。已知 Henna 基因是苯丙氨酸羟化酶的编码基因。为证明 Henna 基因突变是紫眼出现的原因, 进行了如下实验。\n\n实验 1：突变体和野生型果蝇正反交, 后代均为紫红眼\n\n实验 2: 将野生型果蝇的 1 个 Henna 基因转给突变体甲, 发现甲的眼色变紫红; 将野生型果蝇的 2 个 Henna 基因转给突变体乙，发现乙的眼色恢复至野生型\n\n实验 3: 检测果蝇眼部蝶呤含量, 发现甲的含量仅为野生型的 79\\%, 乙的含量升至 98\\%，与野生型差异不显著\n\n下列说法正确的是（）\n\nA: 果蝇眼色性状受常染色体基因控制\nB: 苯丙氨酸差化酶催化果蝇眼部蝶呤的合成\nC: Henna 基因突变导致眼部蝶呤含量降低\nD: 完全敲除野生型的 Henna 基因, 果蝇眼色会变紫色\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_551",
        "problem": "下图为甲、乙两种单基因遗传病的遗传家系图，其中一种遗传病为伴性遗传。人群中乙病的发病率为 $1 / 256$ 。\n[图1]\n\n下列叙述正确的是(\nA: 甲病是伴 X 染色体隐性遗传病\nB: $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{6}$ 的基因型不同\nC: 若 $\\mathrm{II}_{1}$ 与某正常男性结婚, 所生正常孩子的概率为 $25 / 51$\nD: 若 $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生一个孩子, 同时患两种病的概率为 $1 / 17$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为甲、乙两种单基因遗传病的遗传家系图，其中一种遗传病为伴性遗传。人群中乙病的发病率为 $1 / 256$ 。\n[图1]\n\n下列叙述正确的是(\n\nA: 甲病是伴 X 染色体隐性遗传病\nB: $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{6}$ 的基因型不同\nC: 若 $\\mathrm{II}_{1}$ 与某正常男性结婚, 所生正常孩子的概率为 $25 / 51$\nD: 若 $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生一个孩子, 同时患两种病的概率为 $1 / 17$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-08.jpg?height=322&width=850&top_left_y=678&top_left_x=204"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1512",
        "problem": "Peaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nIn actuality, nectarines are the result of a recessive mutation in a single gene. A farmer grows only peaches. She crosses her peaches and gives the seeds to a neighbour to set-up his own farm. However, $9 \\%$ of the plants on his new farm turn into nectarines.\n\nWhat is the frequency of the mutation on the original farm?\nA: 0.03\nB: 0.09\nC: 0.3\nD: 0.9\nE: 0.45\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPeaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nIn actuality, nectarines are the result of a recessive mutation in a single gene. A farmer grows only peaches. She crosses her peaches and gives the seeds to a neighbour to set-up his own farm. However, $9 \\%$ of the plants on his new farm turn into nectarines.\n\nWhat is the frequency of the mutation on the original farm?\n\nA: 0.03\nB: 0.09\nC: 0.3\nD: 0.9\nE: 0.45\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-45.jpg?height=782&width=1231&top_left_y=474&top_left_x=241"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_98",
        "problem": "Zinc $(\\mathrm{Zn})$ and iron $(\\mathrm{Fe})$ are both micronutrients for plants. Plants obtain $\\mathrm{Zn}$ and $\\mathrm{Fe}$ ions from soil through the root system, and transport them to the shoot. Plant culture media usually contains low concentrations of these micronutrients. Half-strength MS medium, a typical plant culture medium, contains $15 \\mu \\mathrm{M} \\mathrm{Zn}^{2+}$ and $50 \\mu \\mathrm{M}$ $\\mathrm{Fe}^{2+}$.\n\nAlthough micronutrients are essential for plant growth, at excess concentrations they inhibit plant growth. To examine the inhibitory effects of excess micronutrients, Arabidopsis thaliana plants were grown on halfstrength MS media, supplemented with additional $\\mathrm{Zn}^{2+}$ and/or $\\mathrm{Fe}^{2+}$.\n[figure1]\n\nFigure 1. Effects of additional $\\mathrm{Zn}$ and $\\mathrm{Fe}$ ions on plant growth\n\nPlants that had been grown on half-strength MS media, supplemented with additional $\\mathrm{Zn}^{2+}$ and/or $\\mathrm{Fe}^{2+}$ at the indicated concentrations, were pictured from above (a), measured for the dry weight of the shoot (b), and analyzed for the $\\mathrm{Zn}$ contents in the shoot and root (c).\nA: Zn accumulation in the shoot shows a stronger correlation to growth defects than the correlation shown by $\\mathrm{Zn}$ accumulation in the root.\nB: The growth defect caused by excess $\\mathrm{Zn}^{2+}$ in the culture medium is mitigated by the addition of $\\mathrm{Fe}^{2+}$.\nC: High concentrations of $\\mathrm{Fe}^{2+}$ in the culture medium suppresses $\\mathrm{Zn}^{2+}$ uptake by the root.\nD: Total $\\mathrm{Zn}$ amount in the shoot is not affected by the addition of $\\mathrm{Fe}^{2+}$ in the medium.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nZinc $(\\mathrm{Zn})$ and iron $(\\mathrm{Fe})$ are both micronutrients for plants. Plants obtain $\\mathrm{Zn}$ and $\\mathrm{Fe}$ ions from soil through the root system, and transport them to the shoot. Plant culture media usually contains low concentrations of these micronutrients. Half-strength MS medium, a typical plant culture medium, contains $15 \\mu \\mathrm{M} \\mathrm{Zn}^{2+}$ and $50 \\mu \\mathrm{M}$ $\\mathrm{Fe}^{2+}$.\n\nAlthough micronutrients are essential for plant growth, at excess concentrations they inhibit plant growth. To examine the inhibitory effects of excess micronutrients, Arabidopsis thaliana plants were grown on halfstrength MS media, supplemented with additional $\\mathrm{Zn}^{2+}$ and/or $\\mathrm{Fe}^{2+}$.\n[figure1]\n\nFigure 1. Effects of additional $\\mathrm{Zn}$ and $\\mathrm{Fe}$ ions on plant growth\n\nPlants that had been grown on half-strength MS media, supplemented with additional $\\mathrm{Zn}^{2+}$ and/or $\\mathrm{Fe}^{2+}$ at the indicated concentrations, were pictured from above (a), measured for the dry weight of the shoot (b), and analyzed for the $\\mathrm{Zn}$ contents in the shoot and root (c).\n\nA: Zn accumulation in the shoot shows a stronger correlation to growth defects than the correlation shown by $\\mathrm{Zn}$ accumulation in the root.\nB: The growth defect caused by excess $\\mathrm{Zn}^{2+}$ in the culture medium is mitigated by the addition of $\\mathrm{Fe}^{2+}$.\nC: High concentrations of $\\mathrm{Fe}^{2+}$ in the culture medium suppresses $\\mathrm{Zn}^{2+}$ uptake by the root.\nD: Total $\\mathrm{Zn}$ amount in the shoot is not affected by the addition of $\\mathrm{Fe}^{2+}$ in the medium.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-45.jpg?height=706&width=1736&top_left_y=954&top_left_x=161"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_425",
        "problem": "鼠的灰色（A）对褐色（a）是一对相对性状，被甲基化修饰的基因不能表达。下列相关说法错误的是（）\n\n[图1]\nA: 图示过程是由于基因突变导致配子发生改变\nB: 基因印记均在亲代细胞减数分裂过程中建立\nC: 亲代雌鼠与雄鼠的基因型和表现型均相同\nD: 子代小鼠的表现型及比例为灰色: 褐色 $=1: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n鼠的灰色（A）对褐色（a）是一对相对性状，被甲基化修饰的基因不能表达。下列相关说法错误的是（）\n\n[图1]\n\nA: 图示过程是由于基因突变导致配子发生改变\nB: 基因印记均在亲代细胞减数分裂过程中建立\nC: 亲代雌鼠与雄鼠的基因型和表现型均相同\nD: 子代小鼠的表现型及比例为灰色: 褐色 $=1: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-24.jpg?height=503&width=1059&top_left_y=745&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1529",
        "problem": "The way groups of organisms live depends on fundamental aspects of their biology. For each of these characteristics, state whether they belong to plants, animals, fungi and/or bacteria.\n\nMost can reproduce sexually.\nA: Plants, animals, fungi\nB: Plants, bacteria, fungi\nC: Animals, fungi, bacteria\nD: Plants, animals, bacteria\nE: Plants, animals, bacteria, fungi\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe way groups of organisms live depends on fundamental aspects of their biology. For each of these characteristics, state whether they belong to plants, animals, fungi and/or bacteria.\n\nMost can reproduce sexually.\n\nA: Plants, animals, fungi\nB: Plants, bacteria, fungi\nC: Animals, fungi, bacteria\nD: Plants, animals, bacteria\nE: Plants, animals, bacteria, fungi\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_737",
        "problem": "已知小麦抗病对感病为显性, 无芒对有芒为显性, 两对性状独立遗传。用纯合的抗病无芒与感病有芒杂交, $\\mathrm{F}_{1}$ 自交, 播种所有的 $\\mathrm{F}_{2}$, 假定所有的 $\\mathrm{F}_{2}$ 植珠都能成活, 在 $\\mathrm{F}_{2}$ 植株开花前, 拔掉所有的有芒植株, 并对剩余植株套袋。假定剩余的每株 $F_{2}$ 收获的种子数量相等, 且 $F_{3}$ 的表现型符合遗传定律。从理论上讲 $F_{3}$ 中表现感病植株的比例为 ( )\nA: $1 / 8$\nB: $3 / 8$\nC: $1 / 16$\nD: $3 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知小麦抗病对感病为显性, 无芒对有芒为显性, 两对性状独立遗传。用纯合的抗病无芒与感病有芒杂交, $\\mathrm{F}_{1}$ 自交, 播种所有的 $\\mathrm{F}_{2}$, 假定所有的 $\\mathrm{F}_{2}$ 植珠都能成活, 在 $\\mathrm{F}_{2}$ 植株开花前, 拔掉所有的有芒植株, 并对剩余植株套袋。假定剩余的每株 $F_{2}$ 收获的种子数量相等, 且 $F_{3}$ 的表现型符合遗传定律。从理论上讲 $F_{3}$ 中表现感病植株的比例为 ( )\n\nA: $1 / 8$\nB: $3 / 8$\nC: $1 / 16$\nD: $3 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_13",
        "problem": "The translational rate of an mRNA can be estimated from sodium dodecyl sulfate polyacrylamide gel electrophoresis (SDS-PAGE). In this experiment, a tobacco mosaic virus (TMV) mRNA, that encodes a $116 \\mathrm{kDa}$ protein, was translated in a rabbit reticulocyte lysate in the presence of ${ }^{35} \\mathrm{~S}$-methionine. The lysate contained all the components of rabbit reticulocyte translational machinery. Samples were removed at 1-minute intervals and subjected to SDS-PAGE. The separated translation products were visualized by autoradiography. As can be seen in the figure below, the polypeptides get larger with time, until the full-length protein appears at about 25 minutes.\n\n[figure1]\n\nFig.Q.2. Time course of synthesis of a TMV protein in a rabbit-reticulocyte lysate.\nA: The rate of TMV protein synthesis is exponentially proportional to time.\nB: With an average molecular mass of an amino acid of 110 daltons, the rate of protein synthesis is approximately 35 to 40 amino acids per minute.\nC: The speed of ribosome movement along the mRNA is constant.\nD: The mRNA may contain more than two rare codons in its sequence.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe translational rate of an mRNA can be estimated from sodium dodecyl sulfate polyacrylamide gel electrophoresis (SDS-PAGE). In this experiment, a tobacco mosaic virus (TMV) mRNA, that encodes a $116 \\mathrm{kDa}$ protein, was translated in a rabbit reticulocyte lysate in the presence of ${ }^{35} \\mathrm{~S}$-methionine. The lysate contained all the components of rabbit reticulocyte translational machinery. Samples were removed at 1-minute intervals and subjected to SDS-PAGE. The separated translation products were visualized by autoradiography. As can be seen in the figure below, the polypeptides get larger with time, until the full-length protein appears at about 25 minutes.\n\n[figure1]\n\nFig.Q.2. Time course of synthesis of a TMV protein in a rabbit-reticulocyte lysate.\n\nA: The rate of TMV protein synthesis is exponentially proportional to time.\nB: With an average molecular mass of an amino acid of 110 daltons, the rate of protein synthesis is approximately 35 to 40 amino acids per minute.\nC: The speed of ribosome movement along the mRNA is constant.\nD: The mRNA may contain more than two rare codons in its sequence.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-007.jpg?height=800&width=1555&top_left_y=1064&top_left_x=271",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-008.jpg?height=611&width=664&top_left_y=282&top_left_x=364",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-008.jpg?height=616&width=647&top_left_y=1321&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1408",
        "problem": "[figure1]\n\nThis cell is best described as what type of cell?\nA: white blood cell\nB: prokaryotic cell\nC: plant cell\nD: animal cell\nE: epithelial cell\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nThis cell is best described as what type of cell?\n\nA: white blood cell\nB: prokaryotic cell\nC: plant cell\nD: animal cell\nE: epithelial cell\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-03.jpg?height=434&width=485&top_left_y=314&top_left_x=777"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1094",
        "problem": "In corn, three enzymes catalyze the same reaction. Corresponding genes $\\left(\\mathrm{a}^{+}\\right.$, $\\mathrm{b}^{+} \\& \\mathrm{c}^{+}$) are located on three different chromosomes. The reaction is as follows:\n\nColourless compound $\\xrightarrow{\\mathrm{a}^{+} \\text {or } \\mathrm{b}^{+} \\text {or } \\mathrm{c}^{+}}$red compound\n\nThe normal functioning of any one of these genes is sufficient to convert colourless compound to the red compound. The mutant alleles of $\\mathrm{a}^{+}, \\mathrm{b}^{+} \\& \\mathrm{c}^{+}$are $\\mathrm{a}, \\mathrm{b}$ and $\\mathrm{c}$ respectively.\n\nIf red $\\left(\\mathrm{a}^{+} / \\mathrm{a}, \\mathrm{b}^{+} / \\mathrm{b}, \\mathrm{c}^{+} / \\mathrm{c}\\right)$ plants are selfed, what proportion of progeny will be colourless? (Express your answer as a fraction.)",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIn corn, three enzymes catalyze the same reaction. Corresponding genes $\\left(\\mathrm{a}^{+}\\right.$, $\\mathrm{b}^{+} \\& \\mathrm{c}^{+}$) are located on three different chromosomes. The reaction is as follows:\n\nColourless compound $\\xrightarrow{\\mathrm{a}^{+} \\text {or } \\mathrm{b}^{+} \\text {or } \\mathrm{c}^{+}}$red compound\n\nThe normal functioning of any one of these genes is sufficient to convert colourless compound to the red compound. The mutant alleles of $\\mathrm{a}^{+}, \\mathrm{b}^{+} \\& \\mathrm{c}^{+}$are $\\mathrm{a}, \\mathrm{b}$ and $\\mathrm{c}$ respectively.\n\nIf red $\\left(\\mathrm{a}^{+} / \\mathrm{a}, \\mathrm{b}^{+} / \\mathrm{b}, \\mathrm{c}^{+} / \\mathrm{c}\\right)$ plants are selfed, what proportion of progeny will be colourless? (Express your answer as a fraction.)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_113",
        "problem": "In an experiment, a researcher isolated a neuron and placed it in the standard Ringer solution (isotonic physiological solution). He measured the resting membrane potential of the axon, then stimulated the axon and measured its action potential (Record A, Figure Q.22).\n\nSubsequently, he repeated the experiment several times, each with a different modified Ringer solution. Figure Q. 22 shows some of the results (Records B to E).\n\n[figure1]\n\nRecord A\n\n[figure2]\n\n[figure3]\n\nRecord B\n[figure4]\n\nRecord E\n\nFigure Q. 22\nA: If the modified Ringer solution contained a lower concentration of $\\mathrm{Na}^{+}$than the standard Ringer solution, the action potential is Record B.\nB: If the modified Ringer solution contained a lower concentration of $\\mathrm{K}^{+}$than the standard Ringer solution, the action potential is Record C.\nC: If the modified Ringer solution contained a substance that increased membrane permeability to $\\mathrm{K}^{+}$, the action potential is Record $\\mathrm{D}$.\nD: If the modified Ringer solution contained a substance that increased membrane permeability to $\\mathrm{Cl}^{-}$, the action potential is Record $\\mathrm{E}$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn an experiment, a researcher isolated a neuron and placed it in the standard Ringer solution (isotonic physiological solution). He measured the resting membrane potential of the axon, then stimulated the axon and measured its action potential (Record A, Figure Q.22).\n\nSubsequently, he repeated the experiment several times, each with a different modified Ringer solution. Figure Q. 22 shows some of the results (Records B to E).\n\n[figure1]\n\nRecord A\n\n[figure2]\n\n[figure3]\n\nRecord B\n[figure4]\n\nRecord E\n\nFigure Q. 22\n\nA: If the modified Ringer solution contained a lower concentration of $\\mathrm{Na}^{+}$than the standard Ringer solution, the action potential is Record B.\nB: If the modified Ringer solution contained a lower concentration of $\\mathrm{K}^{+}$than the standard Ringer solution, the action potential is Record C.\nC: If the modified Ringer solution contained a substance that increased membrane permeability to $\\mathrm{K}^{+}$, the action potential is Record $\\mathrm{D}$.\nD: If the modified Ringer solution contained a substance that increased membrane permeability to $\\mathrm{Cl}^{-}$, the action potential is Record $\\mathrm{E}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-051.jpg?height=536&width=397&top_left_y=873&top_left_x=218",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-051.jpg?height=590&width=393&top_left_y=1508&top_left_x=220",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-051.jpg?height=510&width=397&top_left_y=886&top_left_x=834",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-051.jpg?height=604&width=1024&top_left_y=1492&top_left_x=834"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1470",
        "problem": "On the left is corn yield, on the right is seal reproduction. Interpret these and answer the questions below:\n[figure1]\n\nThese two figures are evidence of:\nA: density dependent competition\nB: keystone predators\nC: density independent population regulation\nD: resource scarcity increases as the density of a population decreases\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOn the left is corn yield, on the right is seal reproduction. Interpret these and answer the questions below:\n[figure1]\n\nThese two figures are evidence of:\n\nA: density dependent competition\nB: keystone predators\nC: density independent population regulation\nD: resource scarcity increases as the density of a population decreases\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-06.jpg?height=520&width=1200&top_left_y=1599&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1296",
        "problem": "[figure1] AS \n[figure2]\nA: ![]([figure3])\nB: ![]([figure4])\nC: ![]([figure5])\nD: ![]([figure6])\nE: ![]([figure7])\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1] AS \n[figure2]\n\nA: ![]([figure3])\nB: ![]([figure4])\nC: ![]([figure5])\nD: ![]([figure6])\nE: ![]([figure7])\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-15.jpg?height=300&width=668&top_left_y=2029&top_left_x=116",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-15.jpg?height=297&width=674&top_left_y=2030&top_left_x=925",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-15.jpg?height=177&width=186&top_left_y=2501&top_left_x=227",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-15.jpg?height=175&width=175&top_left_y=2511&top_left_x=495",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-15.jpg?height=171&width=168&top_left_y=2513&top_left_x=767",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-15.jpg?height=171&width=154&top_left_y=2519&top_left_x=1042",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-15.jpg?height=183&width=166&top_left_y=2504&top_left_x=1299"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_533",
        "problem": "中国科学家通过基因编辑技术改变小鼠未受精卵细胞的基因甲基化, 并将该种细胞植入雌性小鼠子宫中, 获得的小鼠幼崽不仅能够成活, 有一只小鼠还可以正常生殖并产生后代，下列叙述正确的是（ ）\nA: 甲基化改变了基因的碱基排列顺序\nB: 小鼠幼崽体细胞中含 2 个染色体组\nC: 基因甲基化情况不能够遗传给后代\nD: 该技术改变了关键基因在胚胎中的表达\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n中国科学家通过基因编辑技术改变小鼠未受精卵细胞的基因甲基化, 并将该种细胞植入雌性小鼠子宫中, 获得的小鼠幼崽不仅能够成活, 有一只小鼠还可以正常生殖并产生后代，下列叙述正确的是（ ）\n\nA: 甲基化改变了基因的碱基排列顺序\nB: 小鼠幼崽体细胞中含 2 个染色体组\nC: 基因甲基化情况不能够遗传给后代\nD: 该技术改变了关键基因在胚胎中的表达\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1444",
        "problem": "[figure1]\n\nIf a mutation occurred which caused a protein's primary structure to be half the original length, what is the most likely effect this would have on the cell?\nA: Mutated proteins would fold, resulting in impaired cellular function.\nB: The mutated amino acid chain would be hydrolysed, resulting in no net effect on the cell.\nC: The defect in the protein would spread to other nearby cells, forming a plaque of malformed proteins.\nD: The protein would still be able to work normally if it had a quaternary structure, since each individual polypeptide chain comprises only a small part of the overall protein.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nIf a mutation occurred which caused a protein's primary structure to be half the original length, what is the most likely effect this would have on the cell?\n\nA: Mutated proteins would fold, resulting in impaired cellular function.\nB: The mutated amino acid chain would be hydrolysed, resulting in no net effect on the cell.\nC: The defect in the protein would spread to other nearby cells, forming a plaque of malformed proteins.\nD: The protein would still be able to work normally if it had a quaternary structure, since each individual polypeptide chain comprises only a small part of the overall protein.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-36.jpg?height=560&width=1394&top_left_y=388&top_left_x=248"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1227",
        "problem": "The diagram below shows the velocity of nerve impulse conduction as a function of fiber diameter in a variety of animals.\n\n[figure1]\n\nModified from Bullock and Horridge, 1965, Structure and function of the Nervous System of Invertebrates. W. H. Freeman and Company.\n\nConsidering the information in the diagram, which of the statements below is NOT correct?\nA: Giant squid have thick nerve fibers\nB: Myelinated fibers conduct nerve impulses faster than unmyelinated fibers.\nC: The giant fibers of the earthworm are thicker than those of polychaete worms.\nD: Teleost myelinated fibers are thicker than frog myelinated fibers.\nE: The thicker mammalian myelinated fibers show the fastest conduction rate.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram below shows the velocity of nerve impulse conduction as a function of fiber diameter in a variety of animals.\n\n[figure1]\n\nModified from Bullock and Horridge, 1965, Structure and function of the Nervous System of Invertebrates. W. H. Freeman and Company.\n\nConsidering the information in the diagram, which of the statements below is NOT correct?\n\nA: Giant squid have thick nerve fibers\nB: Myelinated fibers conduct nerve impulses faster than unmyelinated fibers.\nC: The giant fibers of the earthworm are thicker than those of polychaete worms.\nD: Teleost myelinated fibers are thicker than frog myelinated fibers.\nE: The thicker mammalian myelinated fibers show the fastest conduction rate.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-05.jpg?height=837&width=1282&top_left_y=364&top_left_x=153"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1451",
        "problem": "Turmeric is a flowering plant, Curcuma longa of the ginger family, Zingiberaceae. It has a very long history of medicinal use, dating back nearly 4000 years. In the last 25 years, over 3000 new publications have been released regarding the health effects of turmeric. The primary compound in turmeric is curcumin, which has been purported to have antiinflammatory and antioxidant effects. Scientists conducted a study investigating the effect of varying concentrations and length of exposure of curcumin on adipocytes from a human cell line. Figure A shows the effect of curcumin concentration on the viability of a population of adipocytes.\n\n[figure1]\n\nFurther investigation showed that the effect on viability was due to changes in the regulation of apoptosis. Caspases are known to play essential roles in apoptosis and are an important regulator of this process is the Bax protein, which causes the release of cytochrome $c$.\n\n$\\mathrm{Bcl}-2$ is a protein that prevents the release of cytochrome $\\mathrm{c}$. The Bax/Bcl-2 ratio determines the balance of apoptosis. Figure B shows the effect of $0,10,20$ and $40 \\mu \\mathrm{M}$ curcumin for a $24 \\mathrm{~h}$ period on the Bax/Bcl- 2 ratio. An asterisk ( ${ }^{*}$ ) denotes a statistically significant result. Figure $\\mathrm{C}$ shows the result of a western blot.\n\nB\n\n[figure2]\n\n[figure3]\n\n$\\mathrm{Bcl}-2$\n\n[figure4]\n\nCurcumin:\nA: stimulates the growth of fat cells by increasing the production of $\\mathrm{Bcl}-2$\nB: inhibits the growth of fat cells by increasing the expression of Bax relative to $\\mathrm{Bcl}-2$\nC: inhibits the growth of fat cells by decreasing the synthesis of Bax\nD: inhibits the growth of fat cells by increasing the expression of $\\mathrm{Bcl}-2$ relative to Bax\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTurmeric is a flowering plant, Curcuma longa of the ginger family, Zingiberaceae. It has a very long history of medicinal use, dating back nearly 4000 years. In the last 25 years, over 3000 new publications have been released regarding the health effects of turmeric. The primary compound in turmeric is curcumin, which has been purported to have antiinflammatory and antioxidant effects. Scientists conducted a study investigating the effect of varying concentrations and length of exposure of curcumin on adipocytes from a human cell line. Figure A shows the effect of curcumin concentration on the viability of a population of adipocytes.\n\n[figure1]\n\nFurther investigation showed that the effect on viability was due to changes in the regulation of apoptosis. Caspases are known to play essential roles in apoptosis and are an important regulator of this process is the Bax protein, which causes the release of cytochrome $c$.\n\n$\\mathrm{Bcl}-2$ is a protein that prevents the release of cytochrome $\\mathrm{c}$. The Bax/Bcl-2 ratio determines the balance of apoptosis. Figure B shows the effect of $0,10,20$ and $40 \\mu \\mathrm{M}$ curcumin for a $24 \\mathrm{~h}$ period on the Bax/Bcl- 2 ratio. An asterisk ( ${ }^{*}$ ) denotes a statistically significant result. Figure $\\mathrm{C}$ shows the result of a western blot.\n\nB\n\n[figure2]\n\n[figure3]\n\n$\\mathrm{Bcl}-2$\n\n[figure4]\n\nCurcumin:\n\nA: stimulates the growth of fat cells by increasing the production of $\\mathrm{Bcl}-2$\nB: inhibits the growth of fat cells by increasing the expression of Bax relative to $\\mathrm{Bcl}-2$\nC: inhibits the growth of fat cells by decreasing the synthesis of Bax\nD: inhibits the growth of fat cells by increasing the expression of $\\mathrm{Bcl}-2$ relative to Bax\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-08.jpg?height=560&width=797&top_left_y=774&top_left_x=618",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-09.jpg?height=411&width=557&top_left_y=1128&top_left_x=407",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-09.jpg?height=168&width=640&top_left_y=1138&top_left_x=1048",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-09.jpg?height=140&width=572&top_left_y=1346&top_left_x=1116"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1535",
        "problem": "Blue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nWhich factors are NOT important for getting reliable data from these population surveys?\nA: Measure the same number of whales each year.\nB: Carry out measurements in a range of areas representing different parts of the ocean.\nC: Maximise the number of samples taken in each survey.\nD: Sample at the same places each year.\nE: Sample in the same season each year.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBlue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nWhich factors are NOT important for getting reliable data from these population surveys?\n\nA: Measure the same number of whales each year.\nB: Carry out measurements in a range of areas representing different parts of the ocean.\nC: Maximise the number of samples taken in each survey.\nD: Sample at the same places each year.\nE: Sample in the same season each year.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-36.jpg?height=948&width=1673&top_left_y=657&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1329",
        "problem": "The stellate barnacle, Elminius modestus, is found in the midlittoral zone of New Zealand's estuaries and sheltered rocky shores. The barnacle is hermaphroditic, though it requires cross-fertilization. It reaches sexual maturity about 8 weeks after settling and an average sized animal produces about 1800-4000 eggs (Barnes \\& Barnes, 1968). The generation time is short and up to 12 successive broods are produced, 2 weeks apart. The nauplius larvae hatch and develop into cypris larvae which are attracted to places where adult conspecifics are present, settling and maturing into the adult barnacle.\n\nThe dispersion pattern and type of survivorship curve shown by this species is\nA: Clumped, I\nB: Uniform, I\nC: Clumped, II\nD: Clumped, III\nE: Uniform, III\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe stellate barnacle, Elminius modestus, is found in the midlittoral zone of New Zealand's estuaries and sheltered rocky shores. The barnacle is hermaphroditic, though it requires cross-fertilization. It reaches sexual maturity about 8 weeks after settling and an average sized animal produces about 1800-4000 eggs (Barnes \\& Barnes, 1968). The generation time is short and up to 12 successive broods are produced, 2 weeks apart. The nauplius larvae hatch and develop into cypris larvae which are attracted to places where adult conspecifics are present, settling and maturing into the adult barnacle.\n\nThe dispersion pattern and type of survivorship curve shown by this species is\n\nA: Clumped, I\nB: Uniform, I\nC: Clumped, II\nD: Clumped, III\nE: Uniform, III\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1314",
        "problem": "## BIRDS OF A FEATHER - EVOLUTIONARY RELATIONSHIPS AMONGST THE RATITES\n\n[figure1]\n\nOur national bird, the kiwi, is a ratite, a group of flightless birds that includes the emu and cassowary in Australia and New Guinea, the ostrich in Africa, and the rhea in South America. There are also two recently extinct groups that include the largest birds ever known: our own moa and the elephant birds from Madagascar who reached heights of up to $3 \\mathrm{~m}$. Ratites and tinamous (found in South America and weak fliers) belong to an ancestral group of birds called \"palaeognaths\" and are the sister group (closest relatives) to all other living birds, the \"neognaths\".\n\nThe evolutionary relationships within the ratites have been the subject of considerable research as these birds are believed to have originated through vicariant speciation driven by the continental breakup of the supercontinent Gondwana. Vicariant speciation is the process by which new species are formed from the separation of the original population into two or more populations by a geographic barrier. Researchers at the Australian Centre for Ancient DNA, and the Allan Wilson Centre for Molecular Ecology in New Zealand have recently published a study in Science which examines ancient DNA and clarifies ratite evolution.\n\nThe maps at right show the position of continents during the Late Cretaceous and Tertiary. Continental landmasses are coloured according to the order in which they broke away from the remaining Gondwanan landmass: Africa and Madagascar (dark gray) split 100 to 130 Million years ago (Ma), followed by New Zealand (red; 60 to $80 \\mathrm{Ma}$ ), then finally Australia, Antarctica, and South America (green; 30 to $50 \\mathrm{Ma})$.\n\nSource: Mitchell et al. Science 344, 898 (2014).\n\n[figure2]Which relationship amongst the palaeognaths might be expected if they are the result of vicariant speciation only?\nA: Ostrich and elephant birds are sister groups.\nB: The moa and emu are sister groups.\nC: Tinamous are most closely related to the cassowary.\nD: Rhea and cassowary are sister groups.\nE: None of the above are consistent with vicariant speciation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## BIRDS OF A FEATHER - EVOLUTIONARY RELATIONSHIPS AMONGST THE RATITES\n\n[figure1]\n\nOur national bird, the kiwi, is a ratite, a group of flightless birds that includes the emu and cassowary in Australia and New Guinea, the ostrich in Africa, and the rhea in South America. There are also two recently extinct groups that include the largest birds ever known: our own moa and the elephant birds from Madagascar who reached heights of up to $3 \\mathrm{~m}$. Ratites and tinamous (found in South America and weak fliers) belong to an ancestral group of birds called \"palaeognaths\" and are the sister group (closest relatives) to all other living birds, the \"neognaths\".\n\nThe evolutionary relationships within the ratites have been the subject of considerable research as these birds are believed to have originated through vicariant speciation driven by the continental breakup of the supercontinent Gondwana. Vicariant speciation is the process by which new species are formed from the separation of the original population into two or more populations by a geographic barrier. Researchers at the Australian Centre for Ancient DNA, and the Allan Wilson Centre for Molecular Ecology in New Zealand have recently published a study in Science which examines ancient DNA and clarifies ratite evolution.\n\nThe maps at right show the position of continents during the Late Cretaceous and Tertiary. Continental landmasses are coloured according to the order in which they broke away from the remaining Gondwanan landmass: Africa and Madagascar (dark gray) split 100 to 130 Million years ago (Ma), followed by New Zealand (red; 60 to $80 \\mathrm{Ma}$ ), then finally Australia, Antarctica, and South America (green; 30 to $50 \\mathrm{Ma})$.\n\nSource: Mitchell et al. Science 344, 898 (2014).\n\n[figure2]\n\nproblem:\nWhich relationship amongst the palaeognaths might be expected if they are the result of vicariant speciation only?\n\nA: Ostrich and elephant birds are sister groups.\nB: The moa and emu are sister groups.\nC: Tinamous are most closely related to the cassowary.\nD: Rhea and cassowary are sister groups.\nE: None of the above are consistent with vicariant speciation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-27.jpg?height=759&width=559&top_left_y=246&top_left_x=126",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-27.jpg?height=1477&width=802&top_left_y=1072&top_left_x=1118"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_948",
        "problem": "果蝇棒眼和圆眼、正常翅和翅外展这两对相对性状分别由两对等位基因控制, 其位于常染色体或 X 染色体上未知。现用一只棒眼正常翅雌果蝇与一只圆眼正常翅雄果蝇杂交, $\\mathrm{F}_{1}$ 中棒眼正常翅:棒眼翅外展: 圆眼正常翅: 圆眼翅外展 $=3: 1: 3: 1$ 。对亲本果蝇进行基因检测，检测过程用限制酶 MstII仅处理果蝇眼形（棒眼和圆眼）相关基因，得到大小不同的片段后进行电泳, 电泳结果如图。下列说法错误的是（）\n\n[图1]\nA: 可判断出圆眼对棒眼为显性\nB: 可判断出正常翅和翅外展位于常染色体上\nC: 棒眼基因可能比圆眼基因多 1 个 MstII识别位点\nD: $F_{1}$ 随机交配两代后， $F_{3}$ 中棒眼的基因频率是 $1 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇棒眼和圆眼、正常翅和翅外展这两对相对性状分别由两对等位基因控制, 其位于常染色体或 X 染色体上未知。现用一只棒眼正常翅雌果蝇与一只圆眼正常翅雄果蝇杂交, $\\mathrm{F}_{1}$ 中棒眼正常翅:棒眼翅外展: 圆眼正常翅: 圆眼翅外展 $=3: 1: 3: 1$ 。对亲本果蝇进行基因检测，检测过程用限制酶 MstII仅处理果蝇眼形（棒眼和圆眼）相关基因，得到大小不同的片段后进行电泳, 电泳结果如图。下列说法错误的是（）\n\n[图1]\n\nA: 可判断出圆眼对棒眼为显性\nB: 可判断出正常翅和翅外展位于常染色体上\nC: 棒眼基因可能比圆眼基因多 1 个 MstII识别位点\nD: $F_{1}$ 随机交配两代后， $F_{3}$ 中棒眼的基因频率是 $1 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-50.jpg?height=265&width=594&top_left_y=1164&top_left_x=363"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_700",
        "problem": "果蝇唾液腺细胞染色体上有许多宽窄不一的横纹, 下列对横纹的描述中正确的是 ( )\nA: 宽窄不一是 DNA 复制次数不一致所致\nB: 每条横纹各代表一个基因\nC: 横纹便于在染色体上进行基因定位\nD: 横纹的数目和位置在同种果蝇中都不同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇唾液腺细胞染色体上有许多宽窄不一的横纹, 下列对横纹的描述中正确的是 ( )\n\nA: 宽窄不一是 DNA 复制次数不一致所致\nB: 每条横纹各代表一个基因\nC: 横纹便于在染色体上进行基因定位\nD: 横纹的数目和位置在同种果蝇中都不同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_751",
        "problem": "某些植物的着丝粒组蛋白 3(CENH3) 发生某种突变后, 会导致合子发育过程中产生单倍体，相关过程如图所示。下列叙述错误的是（ ）\n\n[图1]\nA: 这种突变可能不影响减数分裂但却影响有丝分裂\nB: 这种产生单倍体的方法不需要进行植物组织培养\nC: 所得到的全部单倍体胚胎的基因型均与父本保持一致\nD: 单倍体胚胎也含有来自 CENH3 突变体母本的遗传物质\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某些植物的着丝粒组蛋白 3(CENH3) 发生某种突变后, 会导致合子发育过程中产生单倍体，相关过程如图所示。下列叙述错误的是（ ）\n\n[图1]\n\nA: 这种突变可能不影响减数分裂但却影响有丝分裂\nB: 这种产生单倍体的方法不需要进行植物组织培养\nC: 所得到的全部单倍体胚胎的基因型均与父本保持一致\nD: 单倍体胚胎也含有来自 CENH3 突变体母本的遗传物质\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-45.jpg?height=425&width=1304&top_left_y=1735&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_337",
        "problem": "某种 $X Y$ 型性别决定的二倍体动物, 其控制毛色的等位基因 $G 、 g$ 只位于 $X$ 染色体上, 仅 $\\mathrm{G}$ 表达时为黑色, 仅 $\\mathrm{g}$ 表达时为灰色, 二者均不表达时为白色。受表观遗传的影响, $G 、 g$ 来自父本时才表达, 来自母本时不表达。一对亲本杂交, 有关子代的分析错误的是 ( )\nA: 雌雄表型不同\nB: 白色只出现在雄性中\nC: 雌性只有一种表型\nD: 最多出现三种表型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种 $X Y$ 型性别决定的二倍体动物, 其控制毛色的等位基因 $G 、 g$ 只位于 $X$ 染色体上, 仅 $\\mathrm{G}$ 表达时为黑色, 仅 $\\mathrm{g}$ 表达时为灰色, 二者均不表达时为白色。受表观遗传的影响, $G 、 g$ 来自父本时才表达, 来自母本时不表达。一对亲本杂交, 有关子代的分析错误的是 ( )\n\nA: 雌雄表型不同\nB: 白色只出现在雄性中\nC: 雌性只有一种表型\nD: 最多出现三种表型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1546",
        "problem": "The way groups of organisms live depends on fundamental aspects of their biology. For each of these characteristics, state whether they belong to plants, animals, fungi and/or bacteria.\n\nHave cell walls.\nA: Plants, animals, fungi\nB: Plants, bacteria, fungi\nC: Animals, fungi, bacteria\nD: Plants, animals, bacteria\nE: Plants, animals, bacteria, fungi\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe way groups of organisms live depends on fundamental aspects of their biology. For each of these characteristics, state whether they belong to plants, animals, fungi and/or bacteria.\n\nHave cell walls.\n\nA: Plants, animals, fungi\nB: Plants, bacteria, fungi\nC: Animals, fungi, bacteria\nD: Plants, animals, bacteria\nE: Plants, animals, bacteria, fungi\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1228",
        "problem": "Data for the distribution and morphology of the Chatham Island (Leucocarbo onslowi) and Stewart Island (L. chalconotus) shags are shown in the figure below.\n\n[figure1]\n\nRawlence NJ, Till CE, Scofield RP, Tennyson AJD, Collins CJ, et al. (2014) Strong Phylogeographic Structure in a Sedentary Seabird, the Stewart Island Shag (Leucocarbo chalconotus). PLoS ONE 9(3): e90769. doi:10.1371/journal.pone. 0090769\n\nThe pie charts in Section $\\mathrm{C}$ indicate the proportion of the different morphological forms at each locality. White sectors indicate the proportion of pied morph expected and black the bronze morph. Grey sectors indicate the proportion of shags with scattered papillae and orange, small caruncles in breeding plumage.\n\nWhat features would you expect to see in a shag population in Foveaux Strait?\nA: $20 \\%$ pied morphs and $80 \\%$ dark-bronze morphs.\nB: $30 \\%$ pied morphs and $80 \\%$ dark-bronze morphs AND $50 \\%: 50 \\%$ small bright orange caruncles: dark to dull orange scattered papillae in breeding plumage.\nC: $50 \\%$ pied and dark-bronze morphs AND $50 \\%: 50 \\%$ small bright orange caruncles: dark to dull orange scattered papillae in breeding plumage.\nD: $60 \\%$ pied morphs and $40 \\%$ dark-bronze morphs AND all with small bright orange caruncles.\nE: $60 \\%$ pied morphs and $40 \\%$ dark-bronze morphs all with dark to dull orange scattered papillae in breeding plumage.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nData for the distribution and morphology of the Chatham Island (Leucocarbo onslowi) and Stewart Island (L. chalconotus) shags are shown in the figure below.\n\n[figure1]\n\nRawlence NJ, Till CE, Scofield RP, Tennyson AJD, Collins CJ, et al. (2014) Strong Phylogeographic Structure in a Sedentary Seabird, the Stewart Island Shag (Leucocarbo chalconotus). PLoS ONE 9(3): e90769. doi:10.1371/journal.pone. 0090769\n\nThe pie charts in Section $\\mathrm{C}$ indicate the proportion of the different morphological forms at each locality. White sectors indicate the proportion of pied morph expected and black the bronze morph. Grey sectors indicate the proportion of shags with scattered papillae and orange, small caruncles in breeding plumage.\n\nWhat features would you expect to see in a shag population in Foveaux Strait?\n\nA: $20 \\%$ pied morphs and $80 \\%$ dark-bronze morphs.\nB: $30 \\%$ pied morphs and $80 \\%$ dark-bronze morphs AND $50 \\%: 50 \\%$ small bright orange caruncles: dark to dull orange scattered papillae in breeding plumage.\nC: $50 \\%$ pied and dark-bronze morphs AND $50 \\%: 50 \\%$ small bright orange caruncles: dark to dull orange scattered papillae in breeding plumage.\nD: $60 \\%$ pied morphs and $40 \\%$ dark-bronze morphs AND all with small bright orange caruncles.\nE: $60 \\%$ pied morphs and $40 \\%$ dark-bronze morphs all with dark to dull orange scattered papillae in breeding plumage.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-14.jpg?height=1450&width=1562&top_left_y=343&top_left_x=247"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_11",
        "problem": "In Drosophila melanogaster, yellow body and white eye are both X-linked recessive genes.\n\nWild-type males were crossed with yellow females with white eyes, and $\\mathrm{F}_{1}$ progenies were produced in the numbers and phenotypes shown in the table below.\n\n| Progeny group | Progeny phenotype and sex | Progeny number |\n| :---: | :---: | :---: |\n| $(a)$ | wild-type female | 3,996 |\n| $(b)$ | yellow males with white eyes | 3,997 |\n| $(c)$ | yellow females with white eyes | 4 |\n| $(d)$ | wild-type male | 3 |\n\nWhich of the following is the best explanation for how progeny groups (c) and (d) were produced?\nA: Genetic recombination during meiosis I.\nB: Genetic recombination during meiosis II.\nC: Somatic mutations in the eye and body of wild-type flies.\nD: Nondisjunction of sex chromosome.\nE: Dosage compensation for X-linked genes.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn Drosophila melanogaster, yellow body and white eye are both X-linked recessive genes.\n\nWild-type males were crossed with yellow females with white eyes, and $\\mathrm{F}_{1}$ progenies were produced in the numbers and phenotypes shown in the table below.\n\n| Progeny group | Progeny phenotype and sex | Progeny number |\n| :---: | :---: | :---: |\n| $(a)$ | wild-type female | 3,996 |\n| $(b)$ | yellow males with white eyes | 3,997 |\n| $(c)$ | yellow females with white eyes | 4 |\n| $(d)$ | wild-type male | 3 |\n\nWhich of the following is the best explanation for how progeny groups (c) and (d) were produced?\n\nA: Genetic recombination during meiosis I.\nB: Genetic recombination during meiosis II.\nC: Somatic mutations in the eye and body of wild-type flies.\nD: Nondisjunction of sex chromosome.\nE: Dosage compensation for X-linked genes.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_688",
        "problem": "某野生型蝴蝶的性别决定方式是 ZW 型, 体色是深紫色。研究人员让白色纯合品系 $M$ 与纯合野生型蝴蝶进行正反交实验, 所得 $F_{1}$ 的体色均为深紫色。实验一: 让 $F_{1}$ 的一只雌蝶与 $F_{1}$ 的多只雄蝶交配, 后代为深紫色和白色, 且比例为 3: 1. 实验二: 让 $F_{1}$ 的一只雄蝶与多只品系 $\\mathrm{M}$ 的雌蝶交配, 后代表型及比例为深紫色: 紫色: 黑色: 白色=4:\n\n1：1：4. 假设个体每次产生的配子比例不变, 下列相关叙述正确的是（ ）\nA: 蝴蝶体色至少由常染色体上的两对等位基因控制\nB: 实验二中雄蝶产生的配子有 4 种, 比例为 $4: 1: 1: 4$\nC: 实验一中 $\\mathrm{F}_{1}$ 的雌蝶与实验二中 $\\mathrm{F}_{1}$ 的雄蝶相互交配, 子代紫色个体占 $1 / 10$\nD: $F_{1}$ 的任意雌蝶与品系 $M$ 杂交, 后代表型比例均为深紫色: 白色 $=1: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某野生型蝴蝶的性别决定方式是 ZW 型, 体色是深紫色。研究人员让白色纯合品系 $M$ 与纯合野生型蝴蝶进行正反交实验, 所得 $F_{1}$ 的体色均为深紫色。实验一: 让 $F_{1}$ 的一只雌蝶与 $F_{1}$ 的多只雄蝶交配, 后代为深紫色和白色, 且比例为 3: 1. 实验二: 让 $F_{1}$ 的一只雄蝶与多只品系 $\\mathrm{M}$ 的雌蝶交配, 后代表型及比例为深紫色: 紫色: 黑色: 白色=4:\n\n1：1：4. 假设个体每次产生的配子比例不变, 下列相关叙述正确的是（ ）\n\nA: 蝴蝶体色至少由常染色体上的两对等位基因控制\nB: 实验二中雄蝶产生的配子有 4 种, 比例为 $4: 1: 1: 4$\nC: 实验一中 $\\mathrm{F}_{1}$ 的雌蝶与实验二中 $\\mathrm{F}_{1}$ 的雄蝶相互交配, 子代紫色个体占 $1 / 10$\nD: $F_{1}$ 的任意雌蝶与品系 $M$ 杂交, 后代表型比例均为深紫色: 白色 $=1: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_623",
        "problem": "果蝇眼色由常染色体上的（A/a）和 $\\mathrm{X}$ 染色体上的（B/b）两对等位基因共同控制,具体关系如图所示。下列说法不正确的是（不考虑突变）（）\n\n[图1]\nA: 红眼雄果蝇的后代中雌性个体不会出现白色\nB: 据图分析，红眼果蝇应该有 6 种基因型\nC: 白眼雄果蝇与纯合粉眼雌果蝇杂交。 $F_{1}$ 有红眼、粉眼, 则亲代雄果蝇基因型为 $\\operatorname{Aax}^{b} \\mathrm{Y}$\nD: 将 $\\mathrm{C}$ 选项杂交组合的 $\\mathrm{F}_{1}$ 代果蝇随机进行交配, 则 $\\mathrm{F}_{2}$ 代果蝇中, 粉眼果蝇所占比例为 $20 / 64$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇眼色由常染色体上的（A/a）和 $\\mathrm{X}$ 染色体上的（B/b）两对等位基因共同控制,具体关系如图所示。下列说法不正确的是（不考虑突变）（）\n\n[图1]\n\nA: 红眼雄果蝇的后代中雌性个体不会出现白色\nB: 据图分析，红眼果蝇应该有 6 种基因型\nC: 白眼雄果蝇与纯合粉眼雌果蝇杂交。 $F_{1}$ 有红眼、粉眼, 则亲代雄果蝇基因型为 $\\operatorname{Aax}^{b} \\mathrm{Y}$\nD: 将 $\\mathrm{C}$ 选项杂交组合的 $\\mathrm{F}_{1}$ 代果蝇随机进行交配, 则 $\\mathrm{F}_{2}$ 代果蝇中, 粉眼果蝇所占比例为 $20 / 64$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-66.jpg?height=140&width=597&top_left_y=978&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_161",
        "problem": "Among 1290 individuals of a population under study, the numbers of individuals with blood group types $\\mathrm{M}, \\mathrm{MN}$, and $\\mathrm{N}$ are, respectively, 340,880 , and 70 .\nA: Based on the data, the frequencies of $\\mathrm{M}$ and $\\mathrm{N}$ alleles in the population are, respectively, 0.7 and 0.3 .\nB: The population describe above is in Hardy Weinberg equilibrium with respect to alleles of the MN blood group.\nC: If there is random mating in a population in which frequency of $M$ alleles is 0.6 and frequency of $\\mathrm{N}$ alleles is 0.4 , the frequency of offspring with the $\\mathrm{NN}$genotype will be 0.16 .\nD: In a start generation with frequency of $\\mathrm{M}=0.6$ and frequency of $\\mathrm{N}=0.4$, after three generations of random mating the frequency of MM individuals in thepopulation of the fourth generation who are offspring of $\\mathrm{MN*} \\mathrm{MN}$ matings will be $<10 \\%$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAmong 1290 individuals of a population under study, the numbers of individuals with blood group types $\\mathrm{M}, \\mathrm{MN}$, and $\\mathrm{N}$ are, respectively, 340,880 , and 70 .\n\nA: Based on the data, the frequencies of $\\mathrm{M}$ and $\\mathrm{N}$ alleles in the population are, respectively, 0.7 and 0.3 .\nB: The population describe above is in Hardy Weinberg equilibrium with respect to alleles of the MN blood group.\nC: If there is random mating in a population in which frequency of $M$ alleles is 0.6 and frequency of $\\mathrm{N}$ alleles is 0.4 , the frequency of offspring with the $\\mathrm{NN}$genotype will be 0.16 .\nD: In a start generation with frequency of $\\mathrm{M}=0.6$ and frequency of $\\mathrm{N}=0.4$, after three generations of random mating the frequency of MM individuals in thepopulation of the fourth generation who are offspring of $\\mathrm{MN*} \\mathrm{MN}$ matings will be $<10 \\%$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_103",
        "problem": "Amino acids resulting from the degradation of proteins can be further metabolized by conversion to intermediates of the citric acid cycle. The following labelled amino acids are obtained by degradation of a labelled protein.\n[figure1]\nThe Krebs cycle is below. Note that most $\\alpha$-amino acids can be directly converted into their corresponding $\\alpha$-keto acid by transamination reaction as also shown below.\n[figure2]\nA: Upon introduction of the aspartate into the Krebs cycle, label will first appear in oxaloacetate.\nB: Upon introduction of the alanine into the Krebs cycle, label will first appear in citrate.\nC: Labelled alanine will yield $14-\\mathrm{CO} 2$ during the first turn of the cycle.\nD: Labelled glutamate will yield $14-\\mathrm{CO} 2$ in the second turn of the cycle.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAmino acids resulting from the degradation of proteins can be further metabolized by conversion to intermediates of the citric acid cycle. The following labelled amino acids are obtained by degradation of a labelled protein.\n[figure1]\nThe Krebs cycle is below. Note that most $\\alpha$-amino acids can be directly converted into their corresponding $\\alpha$-keto acid by transamination reaction as also shown below.\n[figure2]\n\nA: Upon introduction of the aspartate into the Krebs cycle, label will first appear in oxaloacetate.\nB: Upon introduction of the alanine into the Krebs cycle, label will first appear in citrate.\nC: Labelled alanine will yield $14-\\mathrm{CO} 2$ during the first turn of the cycle.\nD: Labelled glutamate will yield $14-\\mathrm{CO} 2$ in the second turn of the cycle.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://i.postimg.cc/pTRm02qq/image.png",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-18.jpg?height=846&width=1470&top_left_y=1119&top_left_x=293"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1229",
        "problem": "External Anatomy of Tambja verconis\n\n[figure1]\n\nhttp://scientificillustration.tumblr.com/post/42376170639/wadeangeli art-this-dorid-nudibranch-sea-slug\nAt left is a diagram of the beautiful nudibranch (sea slug), Tambja verconis. This sea slug is found in northeastern waters of New Zealand. What is the approximate maximum length of this sea slug in $\\mathrm{cm}$ ?\nA: $20 \\mathrm{~mm}$\nB: $2 \\mathrm{~cm}$\nC: $4.8 \\mathrm{~mm}$\nD: $48 \\mathrm{~mm}$\nE: $4.8 \\mathrm{~cm}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nExternal Anatomy of Tambja verconis\n\n[figure1]\n\nhttp://scientificillustration.tumblr.com/post/42376170639/wadeangeli art-this-dorid-nudibranch-sea-slug\nAt left is a diagram of the beautiful nudibranch (sea slug), Tambja verconis. This sea slug is found in northeastern waters of New Zealand. What is the approximate maximum length of this sea slug in $\\mathrm{cm}$ ?\n\nA: $20 \\mathrm{~mm}$\nB: $2 \\mathrm{~cm}$\nC: $4.8 \\mathrm{~mm}$\nD: $48 \\mathrm{~mm}$\nE: $4.8 \\mathrm{~cm}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-16.jpg?height=668&width=756&top_left_y=366&top_left_x=156"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_177",
        "problem": "The menstrual cycle is the regular natural change that occurs in the female reproductive system. In the ovary, where oogenesis take place, each cycle can be divided into three phases consisting of the follicular phase, ovulation and luteal phase. The menstrual cycle is controlled by hormones of hypothalamus-pituitary-ovary axis. The figure shows alternations in body temperature and hormonal changes during the menstrual cycle.\n\n[figure1]\nA: The increase shown at point $\\mathrm{A}$ is caused by the effect of oestrogen on the anterior pituitary.\nB: Curve B shows changes in progesterone level during menstrual cycle.\nC: The source of the increase in concentrations indicated at point C and D are granulosa cells and corpus luteum, respectively.\nD: Substance $\\mathrm{E}$ is secreted by follicular cells.\nE: The cause of the sudden increase shown at point $\\mathrm{E}$ is positive feedback of oestrogen on the anterior pituitary and absence of progesterone.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe menstrual cycle is the regular natural change that occurs in the female reproductive system. In the ovary, where oogenesis take place, each cycle can be divided into three phases consisting of the follicular phase, ovulation and luteal phase. The menstrual cycle is controlled by hormones of hypothalamus-pituitary-ovary axis. The figure shows alternations in body temperature and hormonal changes during the menstrual cycle.\n\n[figure1]\n\nA: The increase shown at point $\\mathrm{A}$ is caused by the effect of oestrogen on the anterior pituitary.\nB: Curve B shows changes in progesterone level during menstrual cycle.\nC: The source of the increase in concentrations indicated at point C and D are granulosa cells and corpus luteum, respectively.\nD: Substance $\\mathrm{E}$ is secreted by follicular cells.\nE: The cause of the sudden increase shown at point $\\mathrm{E}$ is positive feedback of oestrogen on the anterior pituitary and absence of progesterone.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-29.jpg?height=1168&width=808&top_left_y=638&top_left_x=630"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_944",
        "problem": "亨廷顿症(HD)是由于异常 Htt 蛋白堆积, 从而损伤神经系统功能的一种单基因遗传病,相关基因不位于 $\\mathrm{Y}$ 染色体上，平均发病年龄为 45 岁左右。已知组蛋白乙酰化可促进转录, 异常 Htt 蛋白的积累会抑制组蛋白的乙酰化, 从而引起细胞凋亡。下列叙述错误的是（ ）\nA: 组蛋白乙酰化不会引起 DNA 的核苷酸序列发生改变\nB: 异常 Htt 蛋白积累可能会影响染色质的状态和解旋程度\nC: 异常 Htt 蛋白的积累会抑制 RNA 聚合酶与启动部位的结合\nD: 抑制组蛋白乙酰化引起的异常性状可通过配子直接遗传给子代\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n亨廷顿症(HD)是由于异常 Htt 蛋白堆积, 从而损伤神经系统功能的一种单基因遗传病,相关基因不位于 $\\mathrm{Y}$ 染色体上，平均发病年龄为 45 岁左右。已知组蛋白乙酰化可促进转录, 异常 Htt 蛋白的积累会抑制组蛋白的乙酰化, 从而引起细胞凋亡。下列叙述错误的是（ ）\n\nA: 组蛋白乙酰化不会引起 DNA 的核苷酸序列发生改变\nB: 异常 Htt 蛋白积累可能会影响染色质的状态和解旋程度\nC: 异常 Htt 蛋白的积累会抑制 RNA 聚合酶与启动部位的结合\nD: 抑制组蛋白乙酰化引起的异常性状可通过配子直接遗传给子代\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_173",
        "problem": "F1 subunit (a peripheral membrane protein) of the ATP synthase catalyses ATP synthesis using proton motive force responsible for the rotation of $F_{0}$ subunit (integral membrane protein complex) in one direction. $F_{1}$ is composed of three $a$ and three $b$ subunits arranged in alternating manner around a central shaft, the $\\gamma$ subunit.\n\nTo study the rotation, Masasuke Yoshida and his team attached a fluorescently labelled actin filament to $\\gamma$ and watched its movement.\n\n[figure1]\n\nFig.Q.55A. Attachment of labelled actin filament to ATP synthase.\n\nRotating actin filaments were observed by an inverted fluorescence microscope after addition of $2 \\mathrm{mM}$ ATP into a chamber containing actin-taged $a_{3} b_{3} \\gamma$ complex immobilized on the bottom side) as a mirror image formed on a camera. The time interval between images was $220 \\mathrm{~ms}$. A series of 13 images were taken and is shown in Fig. Q.55B.\n[figure2]\n\nFig.Q.55B. Sequential image of rotating filament attatched to the sunbunit in the $a_{3} b_{3} \\gamma$ subcomplex. The number indicate the shot image\nA: Hydrolysis of ATP by $F_{1}$ leads to the conformational change of $a$ and $b$ subunits.\nB: From the set of figures, the filament rotated anticlockwise (looking from the membrane side).\nC: Rotary rate is around 0.4 rounds per second.\nD: Rotating the actin filament in the opposite direction is coupled with ATP synthesis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nF1 subunit (a peripheral membrane protein) of the ATP synthase catalyses ATP synthesis using proton motive force responsible for the rotation of $F_{0}$ subunit (integral membrane protein complex) in one direction. $F_{1}$ is composed of three $a$ and three $b$ subunits arranged in alternating manner around a central shaft, the $\\gamma$ subunit.\n\nTo study the rotation, Masasuke Yoshida and his team attached a fluorescently labelled actin filament to $\\gamma$ and watched its movement.\n\n[figure1]\n\nFig.Q.55A. Attachment of labelled actin filament to ATP synthase.\n\nRotating actin filaments were observed by an inverted fluorescence microscope after addition of $2 \\mathrm{mM}$ ATP into a chamber containing actin-taged $a_{3} b_{3} \\gamma$ complex immobilized on the bottom side) as a mirror image formed on a camera. The time interval between images was $220 \\mathrm{~ms}$. A series of 13 images were taken and is shown in Fig. Q.55B.\n[figure2]\n\nFig.Q.55B. Sequential image of rotating filament attatched to the sunbunit in the $a_{3} b_{3} \\gamma$ subcomplex. The number indicate the shot image\n\nA: Hydrolysis of ATP by $F_{1}$ leads to the conformational change of $a$ and $b$ subunits.\nB: From the set of figures, the filament rotated anticlockwise (looking from the membrane side).\nC: Rotary rate is around 0.4 rounds per second.\nD: Rotating the actin filament in the opposite direction is coupled with ATP synthesis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-013.jpg?height=598&width=1184&top_left_y=728&top_left_x=430",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-013.jpg?height=524&width=1370&top_left_y=1884&top_left_x=362"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_369",
        "problem": "转座子是染色体上一段能够自主复制和移位的 DNA 序列。复制型转座子的 DNA 通过转录 RNA 来合成相应 DNA 片段, 然后插入新位点。非复制型转座子直接将自身 DNA 从原来位置切除并插入新的位点。澳洲野生稻在过去近 300 万年内, 3 种独立的转座子增加了约 $4 \\times 10^{8}$ 个碱基对，总量约占整个基因组的 $60 \\%$ 以上。下列叙述，错误的是\nA: 复制型转座子转座过程需要逆转录酶、DNA 连接酶等参与\nB: 澳洲野生稻的基因组大小的变化主要是复制型转座子引起的\nC: 转座子可能会造成基因和染色体的改变, 给基因组造成潜在危害\nD: 转座子可能造成基因组不稳定, 降低遗传多样性, 不利于生物进化\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n转座子是染色体上一段能够自主复制和移位的 DNA 序列。复制型转座子的 DNA 通过转录 RNA 来合成相应 DNA 片段, 然后插入新位点。非复制型转座子直接将自身 DNA 从原来位置切除并插入新的位点。澳洲野生稻在过去近 300 万年内, 3 种独立的转座子增加了约 $4 \\times 10^{8}$ 个碱基对，总量约占整个基因组的 $60 \\%$ 以上。下列叙述，错误的是\n\nA: 复制型转座子转座过程需要逆转录酶、DNA 连接酶等参与\nB: 澳洲野生稻的基因组大小的变化主要是复制型转座子引起的\nC: 转座子可能会造成基因和染色体的改变, 给基因组造成潜在危害\nD: 转座子可能造成基因组不稳定, 降低遗传多样性, 不利于生物进化\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1343",
        "problem": "The diagram shows protein filaments in a muscle myofibril.\n\n[figure1]\n\nWhich one of the following, $A$ to $D$, correctly describes what would happen to the $A$ band and the $\\mathrm{H}$ zone when the muscle contracts?\nA: A band: shorter, H zone: shorter\nB: A band: same, H zone: same\nC: A band: shorter, H zone: same\nD: A band: same, H zone: longer\nE: A band: same, H zone: shorter\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram shows protein filaments in a muscle myofibril.\n\n[figure1]\n\nWhich one of the following, $A$ to $D$, correctly describes what would happen to the $A$ band and the $\\mathrm{H}$ zone when the muscle contracts?\n\nA: A band: shorter, H zone: shorter\nB: A band: same, H zone: same\nC: A band: shorter, H zone: same\nD: A band: same, H zone: longer\nE: A band: same, H zone: shorter\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-09.jpg?height=457&width=1054&top_left_y=337&top_left_x=501"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_989",
        "problem": "Which of the following statements about skeletal muscle is NOT correct?\nA: The length (distance) of a single muscle contraction depends on the concentration of $\\mathrm{Ca}^{2+}$ ions in the sarcoplasmic reticulum\nB: Muscles with short sarcomeres contract faster than muscles with long sarcomeres\nC: The velocity of muscle contractions is determined by myosin-ATPase activity\nD: Tetanus is the effect of repeated stimulations within a very short interval\nE: Rigor mortis (death rigidity) appears when the concentration of $\\mathrm{Ca}^{2+}$ in cytoplasm is high, but ATP is lacking\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following statements about skeletal muscle is NOT correct?\n\nA: The length (distance) of a single muscle contraction depends on the concentration of $\\mathrm{Ca}^{2+}$ ions in the sarcoplasmic reticulum\nB: Muscles with short sarcomeres contract faster than muscles with long sarcomeres\nC: The velocity of muscle contractions is determined by myosin-ATPase activity\nD: Tetanus is the effect of repeated stimulations within a very short interval\nE: Rigor mortis (death rigidity) appears when the concentration of $\\mathrm{Ca}^{2+}$ in cytoplasm is high, but ATP is lacking\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_578",
        "problem": "某动物的基因型为 $\\mathrm{AaBb}$, 其一个原始生殖细胞的 DNA 都被 ${ }^{32} \\mathrm{P}$ 标记, 放在不含 ${ }^{32} \\mathrm{P}$的培养液中培养, 在分裂过程中形成了图中的甲和乙（图中示部分染色体）, 图丙为该生物体内的细胞某段时期内核 DNA 含量变化, 下列叙述正确的是（ ）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\nA: 图甲细胞正在发生基因重组, 对应图丙的 $\\mathrm{CD}$ 或 $\\mathrm{HK}$ 段\nB: 图乙细胞名称可能为次级精母细胞或第一极体\nC: 若是经过两次胞质分裂形成了图乙细胞, 则该细胞中含 ${ }^{32} \\mathrm{P}$ 的核 DNA 有 $\\mathrm{a}$ 个或 $(a+1)$ 个\nD: 图甲中细胞两极的 DNA 分子数可能不同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某动物的基因型为 $\\mathrm{AaBb}$, 其一个原始生殖细胞的 DNA 都被 ${ }^{32} \\mathrm{P}$ 标记, 放在不含 ${ }^{32} \\mathrm{P}$的培养液中培养, 在分裂过程中形成了图中的甲和乙（图中示部分染色体）, 图丙为该生物体内的细胞某段时期内核 DNA 含量变化, 下列叙述正确的是（ ）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\n\nA: 图甲细胞正在发生基因重组, 对应图丙的 $\\mathrm{CD}$ 或 $\\mathrm{HK}$ 段\nB: 图乙细胞名称可能为次级精母细胞或第一极体\nC: 若是经过两次胞质分裂形成了图乙细胞, 则该细胞中含 ${ }^{32} \\mathrm{P}$ 的核 DNA 有 $\\mathrm{a}$ 个或 $(a+1)$ 个\nD: 图甲中细胞两极的 DNA 分子数可能不同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-20.jpg?height=437&width=374&top_left_y=1392&top_left_x=356",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-20.jpg?height=434&width=305&top_left_y=1388&top_left_x=767",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-20.jpg?height=372&width=643&top_left_y=1433&top_left_x=1135"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1395",
        "problem": "Anna's father has five daughters: 1. Nana, 2. Nemu, 3. Nino, 4. Nomi. What is the name of the fifth daughter?\nA: Nune\nB: Nume\nC: Nunu\nD: Numu\nE: Anna\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAnna's father has five daughters: 1. Nana, 2. Nemu, 3. Nino, 4. Nomi. What is the name of the fifth daughter?\n\nA: Nune\nB: Nume\nC: Nunu\nD: Numu\nE: Anna\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1461",
        "problem": "Male guppies (Poecilia reticulata) show a complex colour pattern polymorphism that varies with predation pressure, reflecting a balance between selection for camouflage by predators and selection for attention by sexual selection. Three experimental ponds were used to study this phenomenon, mimicking the real condition on the native habitat of the guppies and its predators, Rivulus hartii and Crenicichla alta. One pond has the control group while the other two ponds were had one of the two predators added. In the field, $C$. alta was observed to be more dangerous than $R$. hartii.\n\n[figure1]\n\nFigure: Changes in the number of spots per fish during the course of the experiment. Line ' $\\mathrm{K}$ ' stands for pond without predator, ' $\\mathrm{R}$ ' for pond with $R$. hartii, and ' $\\mathrm{C}$ ' for pond with $C$. alta. In the X-axis, ' $F$ ' stands for the time when the foundation population was started, ' $S$ ' stands for the time when the predators were added, then 'l' and 'Il' stands for the numbering of the following censuses.\n\nWhich of the following is correct?\nA: The colour pattern is responsible for the reduced fitness of $P$. reticulata.\nB: Sexual selection for colour pattern of the $P$. reticulata cannot be inferred from the data.\nC: The colour pattern of the $P$. reticulata may be advantageous in escaping $R$. hartii.\nD: The two predators possibly use two different mechanisms to detect $P$. reticulata.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMale guppies (Poecilia reticulata) show a complex colour pattern polymorphism that varies with predation pressure, reflecting a balance between selection for camouflage by predators and selection for attention by sexual selection. Three experimental ponds were used to study this phenomenon, mimicking the real condition on the native habitat of the guppies and its predators, Rivulus hartii and Crenicichla alta. One pond has the control group while the other two ponds were had one of the two predators added. In the field, $C$. alta was observed to be more dangerous than $R$. hartii.\n\n[figure1]\n\nFigure: Changes in the number of spots per fish during the course of the experiment. Line ' $\\mathrm{K}$ ' stands for pond without predator, ' $\\mathrm{R}$ ' for pond with $R$. hartii, and ' $\\mathrm{C}$ ' for pond with $C$. alta. In the X-axis, ' $F$ ' stands for the time when the foundation population was started, ' $S$ ' stands for the time when the predators were added, then 'l' and 'Il' stands for the numbering of the following censuses.\n\nWhich of the following is correct?\n\nA: The colour pattern is responsible for the reduced fitness of $P$. reticulata.\nB: Sexual selection for colour pattern of the $P$. reticulata cannot be inferred from the data.\nC: The colour pattern of the $P$. reticulata may be advantageous in escaping $R$. hartii.\nD: The two predators possibly use two different mechanisms to detect $P$. reticulata.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-13.jpg?height=480&width=517&top_left_y=685&top_left_x=244"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_736",
        "problem": "下图为某伴 $\\mathrm{X}$ 染色体隐性遗传病的系谱, 基因检测发现致病基因 $\\mathrm{d}$ 有两种突变形式, 记作 $\\mathrm{d}_{\\mathrm{A}}$ 与 $\\mathrm{d}_{\\mathrm{B}}$ 。 $\\mathrm{II}_{1}$ 还患有先天性睪丸发育不全综合征 (性染色体组成为 XXY)。不考虑新的基因突变和染色体变异，下列分析正确的是（ ）\n[图1]\nA: 正常情况下, $\\mathrm{II}_{1}$ 体细胞中染色体数和猫叫综合征患者相同\nB: 致病基因 $\\mathrm{d}$ 有两种突变形式 $\\mathrm{d}_{\\mathrm{A}}$ 与 $\\mathrm{d}_{\\mathrm{B}}$, 说明基因突变具有随机性\nC: $\\mathrm{II}_{1}$ 性染色体异常, 可能因为 $\\mathrm{I}_{2}$ 的减数分裂 I 或减数分裂II异常\nD: $\\mathrm{II}_{3}$ 与正常男性婚配, 所生子女患有该伴 X 染色体隐性遗传病的概率是 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为某伴 $\\mathrm{X}$ 染色体隐性遗传病的系谱, 基因检测发现致病基因 $\\mathrm{d}$ 有两种突变形式, 记作 $\\mathrm{d}_{\\mathrm{A}}$ 与 $\\mathrm{d}_{\\mathrm{B}}$ 。 $\\mathrm{II}_{1}$ 还患有先天性睪丸发育不全综合征 (性染色体组成为 XXY)。不考虑新的基因突变和染色体变异，下列分析正确的是（ ）\n[图1]\n\nA: 正常情况下, $\\mathrm{II}_{1}$ 体细胞中染色体数和猫叫综合征患者相同\nB: 致病基因 $\\mathrm{d}$ 有两种突变形式 $\\mathrm{d}_{\\mathrm{A}}$ 与 $\\mathrm{d}_{\\mathrm{B}}$, 说明基因突变具有随机性\nC: $\\mathrm{II}_{1}$ 性染色体异常, 可能因为 $\\mathrm{I}_{2}$ 的减数分裂 I 或减数分裂II异常\nD: $\\mathrm{II}_{3}$ 与正常男性婚配, 所生子女患有该伴 X 染色体隐性遗传病的概率是 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-08.jpg?height=370&width=1450&top_left_y=1088&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_131",
        "problem": "Figure 1 shows the evolution of cetaceans. Studies using stable isotopes indicate that Pakicetus and Amburocetus ate freshwater fish, while Remingtonocetus, Miacetus and Basirosaurus ate seawater fish. Indhyus was, like most extant artiodactyls, a terrestrial herbivorous animal.\n\n[figure1]\n\n[figure2]\n\nFigure 1. A phylogenetic tree of cetaceans and artiodactyls. Extinct fossil species are shown with ${ }^{\\dagger}$. All extant cetaceans are classified into two subfamilies: Mysticeti (baleen whales, which do not possess enamel-based teeth but possess baleen plates) and Odontoceti (toothed whales, which possess teeth). The lifestyle (AQ: aquatic, AM: amphibious, TE: terrestrial) of each group is also shown. Lifestyles of fossil species are inferred based on morphological characteristics. MYA: million years ago\nA: Under the maximum-parsimony criterion, the most recent common ancestor of the hippopotamus and modern cetaceans is amphibious.\nB: There were no fully-aquatic cetaceans $50 \\mathrm{MYA}$.\nC: Based on these studies, the evolutionary scenario of cetaceans is speculated as follows: 1 . becoming carnivorous, 2 . becoming fully aquatic, 3 . invasion to oceanic environments, 4 . divergence into baleen whales and toothed whales.\nD: The enamelin gene, which encodes an essential protein for the formation of teeth enamel, was lost during the evolution of cetaceans before 35 MYA.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFigure 1 shows the evolution of cetaceans. Studies using stable isotopes indicate that Pakicetus and Amburocetus ate freshwater fish, while Remingtonocetus, Miacetus and Basirosaurus ate seawater fish. Indhyus was, like most extant artiodactyls, a terrestrial herbivorous animal.\n\n[figure1]\n\n[figure2]\n\nFigure 1. A phylogenetic tree of cetaceans and artiodactyls. Extinct fossil species are shown with ${ }^{\\dagger}$. All extant cetaceans are classified into two subfamilies: Mysticeti (baleen whales, which do not possess enamel-based teeth but possess baleen plates) and Odontoceti (toothed whales, which possess teeth). The lifestyle (AQ: aquatic, AM: amphibious, TE: terrestrial) of each group is also shown. Lifestyles of fossil species are inferred based on morphological characteristics. MYA: million years ago\n\nA: Under the maximum-parsimony criterion, the most recent common ancestor of the hippopotamus and modern cetaceans is amphibious.\nB: There were no fully-aquatic cetaceans $50 \\mathrm{MYA}$.\nC: Based on these studies, the evolutionary scenario of cetaceans is speculated as follows: 1 . becoming carnivorous, 2 . becoming fully aquatic, 3 . invasion to oceanic environments, 4 . divergence into baleen whales and toothed whales.\nD: The enamelin gene, which encodes an essential protein for the formation of teeth enamel, was lost during the evolution of cetaceans before 35 MYA.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-55.jpg?height=92&width=1330&top_left_y=688&top_left_x=183",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-55.jpg?height=939&width=1631&top_left_y=781&top_left_x=241"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_377",
        "problem": "某 XY 型性别决定的昆虫的眼色有紫眼、红眼和白眼之分, 分别受等位基因 A1、\n\nA2、A3 控制。为研究眼色的遗传规律, 某小组用一只红眼雌性个体与一只紫眼雄性个体杂交, $F_{1}$ 中雌性全为紫眼, 雄性既有红眼, 也有白眼。不考虑突变和 $X 、 Y$ 染色体的同源区段。下列有关叙述错误的是（ ）\nA: 该昆虫种群中与眼色有关的基因型有 9 种\nB: 亲本红眼雌性个体的细胞中最多有四个 3 种眼色基因出现\nC: 选择白眼雌性与紫眼雄性个体杂交, 根据眼色可快速判断子代的性别\nD: $F_{1}$ 中的紫眼雌性与红眼雄性个体杂交, 子代出现红眼个体的比例为 $3 / 8$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某 XY 型性别决定的昆虫的眼色有紫眼、红眼和白眼之分, 分别受等位基因 A1、\n\nA2、A3 控制。为研究眼色的遗传规律, 某小组用一只红眼雌性个体与一只紫眼雄性个体杂交, $F_{1}$ 中雌性全为紫眼, 雄性既有红眼, 也有白眼。不考虑突变和 $X 、 Y$ 染色体的同源区段。下列有关叙述错误的是（ ）\n\nA: 该昆虫种群中与眼色有关的基因型有 9 种\nB: 亲本红眼雌性个体的细胞中最多有四个 3 种眼色基因出现\nC: 选择白眼雌性与紫眼雄性个体杂交, 根据眼色可快速判断子代的性别\nD: $F_{1}$ 中的紫眼雌性与红眼雄性个体杂交, 子代出现红眼个体的比例为 $3 / 8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_435",
        "problem": "栽培番茄含有来自野生番茄的 Mi1 抗虫基因, 它使番茄产生对根结线虫（侵染番茄的根部)、长管蚜和烟粉風三种害虫的抗性。相关叙述正确的是（ ）\nA: 长管蚜和番茄之间是捕食关系，两者协同进化\nB: Mi1 抗虫基因的产生是野生番茄长期适应环境的结果\nC: 在含 Mi1 基因的番茄植株上生长的烟粉風种群基因频率会发生变化\nD: 长期种植含 Mi1 基因的番茄, 土壤中根结线虫的数量会越来越少\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n栽培番茄含有来自野生番茄的 Mi1 抗虫基因, 它使番茄产生对根结线虫（侵染番茄的根部)、长管蚜和烟粉風三种害虫的抗性。相关叙述正确的是（ ）\n\nA: 长管蚜和番茄之间是捕食关系，两者协同进化\nB: Mi1 抗虫基因的产生是野生番茄长期适应环境的结果\nC: 在含 Mi1 基因的番茄植株上生长的烟粉風种群基因频率会发生变化\nD: 长期种植含 Mi1 基因的番茄, 土壤中根结线虫的数量会越来越少\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1414",
        "problem": "The following graph shows relative oxygen concentrations in an airtight container containing one spinach plant exposed to varying light intensities. The y axis indicates oxygen concentration and the $x$ axis indicates increasing light intensity from left to right. The units of both axes are arbitrary; they do not matter.\n\n[figure1]\n\nConsider the stages A, B, C and D marked at the top of the graph. At which stage is light limiting the rate of photosynthesis?\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph shows relative oxygen concentrations in an airtight container containing one spinach plant exposed to varying light intensities. The y axis indicates oxygen concentration and the $x$ axis indicates increasing light intensity from left to right. The units of both axes are arbitrary; they do not matter.\n\n[figure1]\n\nConsider the stages A, B, C and D marked at the top of the graph. At which stage is light limiting the rate of photosynthesis?\n\nA: A\nB: B\nC: C\nD: D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-18.jpg?height=765&width=1354&top_left_y=563&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1373",
        "problem": "The following graph illustrates the change in membrane potential $(\\mathrm{mV})$ during an action potential.\n\n[figure1]\n\nDuring the resting state, the membrane potential is $-70 \\mathrm{mV}$. Which of the following statements is true?\nA: In the resting state, there is no movement of ions across the plasma membrane.\nB: The membrane potential is maintained by the balance between active transport and simple diffusion of ions across the membrane.\nC: In the resting state, the concentration of potassium ions is greater inside the cell than outside the cell.\nD: If the cell is not in the resting state, an action potential cannot be generated.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph illustrates the change in membrane potential $(\\mathrm{mV})$ during an action potential.\n\n[figure1]\n\nDuring the resting state, the membrane potential is $-70 \\mathrm{mV}$. Which of the following statements is true?\n\nA: In the resting state, there is no movement of ions across the plasma membrane.\nB: The membrane potential is maintained by the balance between active transport and simple diffusion of ions across the membrane.\nC: In the resting state, the concentration of potassium ions is greater inside the cell than outside the cell.\nD: If the cell is not in the resting state, an action potential cannot be generated.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-14.jpg?height=651&width=694&top_left_y=443&top_left_x=270"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_419",
        "problem": "某家系的遗传系谱图如下，1号个体不携带乙病的致病基因。据图分析下列有关说法不正确的是（ ）[图1]\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某家系的遗传系谱图如下，1号个体不携带乙病的致病基因。据图分析下列有关说法不正确的是（ ）[图1]\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-54.jpg?height=659&width=938&top_left_y=1161&top_left_x=362"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1111",
        "problem": "[figure1]\n\nThe diagram depicts the spatial distribution of two plant species, Acacia tortillis (a tree) and Euclea divinorum (a shrub) inhabiting the Savannah grasslands in Southern Africa.\n\nBased on it, which of the following is correct?\nA: Acacia tortillis shows a uniform distribution.\nB: Euclea divinorum shows a clumped distribution.\nC: The clumps of Eculea divinorum show a uniform distribution.\nD: A. tortillis and E. divinorum form the dominant vegetation in the landscape.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\n[figure1]\n\nThe diagram depicts the spatial distribution of two plant species, Acacia tortillis (a tree) and Euclea divinorum (a shrub) inhabiting the Savannah grasslands in Southern Africa.\n\nBased on it, which of the following is correct?\n\nA: Acacia tortillis shows a uniform distribution.\nB: Euclea divinorum shows a clumped distribution.\nC: The clumps of Eculea divinorum show a uniform distribution.\nD: A. tortillis and E. divinorum form the dominant vegetation in the landscape.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-32.jpg?height=520&width=857&top_left_y=778&top_left_x=645"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1413",
        "problem": "Bacteriophages, also known as phages, are viruses that can infect and kill bacteria. The susceptibility of a bacterial strain to infection by phages can be measured using a plaque assay.\n\nThe process of performing a plaque assay is: Phages are added to a lawn of bacterial cells growing on a plate. When one phage infects a cell, this infected cell is lysed and new phages are released. These newly released phages infect and lyse neighboring cells (the phages cannot travel to distant cells). The infection process proceeds until the area of cell lysis becomes so large that it is visible by the naked eye. This area of visible cell lysis, caused initially by one phage, is called a plaque. Therefore, the number of plaques is an estimate of the number of phages that were initially added to the lawn of bacterial cells. This is the plaque count.\n\nLet the efficiency of plaquing (EOP) be the ratio of a mutant bacterial strain's plaque count to that of the wild type.\n\nUsing the above definition of EOP, which of the following is CORRECT?\nA: An EOP value equal to 1 is impossible.\nB: An EOP value over 1 is impossible.\nC: An EOP value of 10-5 represents greater susceptibility to phage infection than a value of $10-1$\nD: An EOP value of 10-5 represents greater resistance to phage infection than a value of $10-1$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBacteriophages, also known as phages, are viruses that can infect and kill bacteria. The susceptibility of a bacterial strain to infection by phages can be measured using a plaque assay.\n\nThe process of performing a plaque assay is: Phages are added to a lawn of bacterial cells growing on a plate. When one phage infects a cell, this infected cell is lysed and new phages are released. These newly released phages infect and lyse neighboring cells (the phages cannot travel to distant cells). The infection process proceeds until the area of cell lysis becomes so large that it is visible by the naked eye. This area of visible cell lysis, caused initially by one phage, is called a plaque. Therefore, the number of plaques is an estimate of the number of phages that were initially added to the lawn of bacterial cells. This is the plaque count.\n\nLet the efficiency of plaquing (EOP) be the ratio of a mutant bacterial strain's plaque count to that of the wild type.\n\nUsing the above definition of EOP, which of the following is CORRECT?\n\nA: An EOP value equal to 1 is impossible.\nB: An EOP value over 1 is impossible.\nC: An EOP value of 10-5 represents greater susceptibility to phage infection than a value of $10-1$\nD: An EOP value of 10-5 represents greater resistance to phage infection than a value of $10-1$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_609",
        "problem": "某雌性动物的 2 对等位基因 $\\mathrm{A} 、 \\mathrm{a}$ 和 $\\mathrm{B} 、 \\mathrm{~b}$ 位于 1 对同源染色体上, 其卵巢中的一个卵原细胞减数分裂产生的一个卵细胞和三个极体的基因组成分别是 $\\mathrm{AB} 、 \\mathrm{Ab} 、 \\mathrm{aB}$ 、 $\\mathrm{ab}$ ，原因是\nA: 减数第一次分裂前的间期发生了基因突变 \nB: 减数第一次分裂前期四分体的非姐妹染色单体之间发生了交叉互换\nC: 减数第二次分裂的后期发生了非等位基因的自由组合\nD: 减数第一次分裂的后期发生了非等位基因的自由组合\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某雌性动物的 2 对等位基因 $\\mathrm{A} 、 \\mathrm{a}$ 和 $\\mathrm{B} 、 \\mathrm{~b}$ 位于 1 对同源染色体上, 其卵巢中的一个卵原细胞减数分裂产生的一个卵细胞和三个极体的基因组成分别是 $\\mathrm{AB} 、 \\mathrm{Ab} 、 \\mathrm{aB}$ 、 $\\mathrm{ab}$ ，原因是\n\nA: 减数第一次分裂前的间期发生了基因突变 \nB: 减数第一次分裂前期四分体的非姐妹染色单体之间发生了交叉互换\nC: 减数第二次分裂的后期发生了非等位基因的自由组合\nD: 减数第一次分裂的后期发生了非等位基因的自由组合\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1249",
        "problem": "Conservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.Use the Lincoln-Petersen estimator to calculate the population size of the Maui's dolphin.\nA: 57\nB: 55\nC: 61\nD: 104\nE: 107\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nConservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.\n\nproblem:\nUse the Lincoln-Petersen estimator to calculate the population size of the Maui's dolphin.\n\nA: 57\nB: 55\nC: 61\nD: 104\nE: 107\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1325",
        "problem": "In photosynthesis, light is most directly involved in\nA: reduction of $\\mathrm{CO}_{2}$ to phosphoglycerate\nB: synthesis of starch\nC: release of oxygen from $\\mathrm{CO}_{2}$\nD: splitting of water\nE: regeneration of ribulose bisphosphate\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn photosynthesis, light is most directly involved in\n\nA: reduction of $\\mathrm{CO}_{2}$ to phosphoglycerate\nB: synthesis of starch\nC: release of oxygen from $\\mathrm{CO}_{2}$\nD: splitting of water\nE: regeneration of ribulose bisphosphate\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1514",
        "problem": "The complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nHuman cells have precise human-complement inhibitors on their surface so they are not harmed. Which two pathways must they inhibit?\nA: Pathway A and B\nB: Pathway B and C\nC: Pathway C and A\nD: Pathway C and D\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nHuman cells have precise human-complement inhibitors on their surface so they are not harmed. Which two pathways must they inhibit?\n\nA: Pathway A and B\nB: Pathway B and C\nC: Pathway C and A\nD: Pathway C and D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=790&width=1714&top_left_y=486&top_left_x=228",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=930&width=1806&top_left_y=1335&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1135",
        "problem": "Evolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).This phylogeny suggests the closest relative of the kiwi is?\nA: Moa\nB: Tinamou\nC: Elephant bird\nD: Emu\nE: Rhea\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nEvolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).\n\nproblem:\nThis phylogeny suggests the closest relative of the kiwi is?\n\nA: Moa\nB: Tinamou\nC: Elephant bird\nD: Emu\nE: Rhea\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-28.jpg?height=1868&width=1627&top_left_y=748&top_left_x=206"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_850",
        "problem": "某高等雄性动物基因型为 $\\mathrm{AaBbCc}$, 基因 $\\mathrm{A}$ 和 $\\mathrm{b}$ 位于同一条染色体上。将该动物的一个精原细胞（DNA 被 ${ }^{32} \\mathrm{P}$ 全部标记）置于不含 ${ }^{32} \\mathrm{P}$ 的培养液中培养。该细胞进行一次有丝分裂后, 其中的一个子细胞又进行减数分裂后得到的精细胞如图所示 (不考虑基因突变和另一对染色体的互换), 其中仅 1 号染色体上不带有 ${ }^{32} \\mathrm{P}$ 标记。只考虑图中所示染色体，关于该精原细胞产生的另外三个精细胞的叙述正确的是（ ）\n\n[图1]\nA: 精细胞的基因型可能为 Abc, 1 号和 2 号染色体的 DNA 均不含 ${ }^{32} \\mathrm{p}$\nB: 精细胞的基因型可能为 $\\mathrm{ABC}, 1$ 号和 2 号染色体的 DNA 可能均含 ${ }^{32} \\mathrm{p}$\nC: 精细胞的基因型可能为 $\\mathrm{abC}, 1$ 号和 2 号染色体的 DNA 可能均含 ${ }^{32} \\mathrm{p}$\nD: 精细胞的基因型可能为 $\\mathrm{aBC}, 1$ 号和 2 号染色体的 DNA 可能均含 ${ }^{32} \\mathrm{P}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某高等雄性动物基因型为 $\\mathrm{AaBbCc}$, 基因 $\\mathrm{A}$ 和 $\\mathrm{b}$ 位于同一条染色体上。将该动物的一个精原细胞（DNA 被 ${ }^{32} \\mathrm{P}$ 全部标记）置于不含 ${ }^{32} \\mathrm{P}$ 的培养液中培养。该细胞进行一次有丝分裂后, 其中的一个子细胞又进行减数分裂后得到的精细胞如图所示 (不考虑基因突变和另一对染色体的互换), 其中仅 1 号染色体上不带有 ${ }^{32} \\mathrm{P}$ 标记。只考虑图中所示染色体，关于该精原细胞产生的另外三个精细胞的叙述正确的是（ ）\n\n[图1]\n\nA: 精细胞的基因型可能为 Abc, 1 号和 2 号染色体的 DNA 均不含 ${ }^{32} \\mathrm{p}$\nB: 精细胞的基因型可能为 $\\mathrm{ABC}, 1$ 号和 2 号染色体的 DNA 可能均含 ${ }^{32} \\mathrm{p}$\nC: 精细胞的基因型可能为 $\\mathrm{abC}, 1$ 号和 2 号染色体的 DNA 可能均含 ${ }^{32} \\mathrm{p}$\nD: 精细胞的基因型可能为 $\\mathrm{aBC}, 1$ 号和 2 号染色体的 DNA 可能均含 ${ }^{32} \\mathrm{P}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-008.jpg?height=409&width=437&top_left_y=161&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_666",
        "problem": "下图为某家族的遗传家系图, 已知甲、乙两种遗传病均为单基因遗传病, 且乙病为伴性遗传病, II-1 无致病基因, II-2 为纯合子, 经调查, 人群中甲病的患病概率为\n75. $5 \\%$ 。（假设无变异发生且相关基因不位于 X、Y 染色体的同源区段, 男女比例为 1 :\n\n1)。下列说法正确的是（ ）\n\n[图1]\nA: 控制甲病和乙病的致病基因在遗传过程中遵循自由组合定律\nB: III-2 的致病基因可能来自于I-1 或I-2 内\nC: II-3 和II-4 再生一个正常男孩的概率为 1/4\nD: III-4 携带甲病基因的概率是 $1 / 11$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下图为某家族的遗传家系图, 已知甲、乙两种遗传病均为单基因遗传病, 且乙病为伴性遗传病, II-1 无致病基因, II-2 为纯合子, 经调查, 人群中甲病的患病概率为\n75. $5 \\%$ 。（假设无变异发生且相关基因不位于 X、Y 染色体的同源区段, 男女比例为 1 :\n\n1)。下列说法正确的是（ ）\n\n[图1]\n\nA: 控制甲病和乙病的致病基因在遗传过程中遵循自由组合定律\nB: III-2 的致病基因可能来自于I-1 或I-2 内\nC: II-3 和II-4 再生一个正常男孩的概率为 1/4\nD: III-4 携带甲病基因的概率是 $1 / 11$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-072.jpg?height=448&width=1357&top_left_y=1638&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_826",
        "problem": "果蝇的 X、Y 染色体（如图所示）有同源区段（I片段）和非同源区段（II-1、II-2 片段)。有关杂交实验结果如下表。下列对结果的分析错误的是（）\n\n[图1]\nA: I片段和II-1 上的基因在遗传上均与性别相关联\nB: 在II-1 与II-2 中, 基因结构不同与碱基对的数目及排列顺序有关\nC: 通过杂交组合一，直接判断刚毛为显性性状\nD: 通过杂交组合二, 可以判断控制该性状的基因位于II-1 片段\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的 X、Y 染色体（如图所示）有同源区段（I片段）和非同源区段（II-1、II-2 片段)。有关杂交实验结果如下表。下列对结果的分析错误的是（）\n\n[图1]\n\nA: I片段和II-1 上的基因在遗传上均与性别相关联\nB: 在II-1 与II-2 中, 基因结构不同与碱基对的数目及排列顺序有关\nC: 通过杂交组合一，直接判断刚毛为显性性状\nD: 通过杂交组合二, 可以判断控制该性状的基因位于II-1 片段\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-88.jpg?height=874&width=1219&top_left_y=571&top_left_x=316"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_847",
        "problem": "若某二倍体高等动物 $(2 n=4)$ 的基因型为 DdEe, 其 1 个精原细胞（其中一对同源染色体的 DNA 被 ${ }^{32} \\mathrm{P}$ 全部标记）在不含 ${ }^{32} \\mathrm{P}$ 的培养液中培养一段时间, 分裂过程中形成的其中 1 个次级精母细胞如图所示, 图中细胞有 2 个染色体 DNA 含有 ${ }^{32} \\mathrm{P}$ 。下列叙述正确的是 ( )\n\n[图1]\nA: 形成图中细胞的过程中未发生变异\nB: 图中细胞含有 ${ }^{32} \\mathrm{P}$ 的 DNA 位于姐妹染色单体上\nC: 该精原细胞形成图中细胞的过程中至少经历了两次胞质分裂\nD: 与图中细胞同时产生的另一次级精母细胞中有 1 或 2 个染色体 DNA 含有 ${ }^{32} \\mathrm{P}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n若某二倍体高等动物 $(2 n=4)$ 的基因型为 DdEe, 其 1 个精原细胞（其中一对同源染色体的 DNA 被 ${ }^{32} \\mathrm{P}$ 全部标记）在不含 ${ }^{32} \\mathrm{P}$ 的培养液中培养一段时间, 分裂过程中形成的其中 1 个次级精母细胞如图所示, 图中细胞有 2 个染色体 DNA 含有 ${ }^{32} \\mathrm{P}$ 。下列叙述正确的是 ( )\n\n[图1]\n\nA: 形成图中细胞的过程中未发生变异\nB: 图中细胞含有 ${ }^{32} \\mathrm{P}$ 的 DNA 位于姐妹染色单体上\nC: 该精原细胞形成图中细胞的过程中至少经历了两次胞质分裂\nD: 与图中细胞同时产生的另一次级精母细胞中有 1 或 2 个染色体 DNA 含有 ${ }^{32} \\mathrm{P}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-36.jpg?height=756&width=485&top_left_y=153&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1340",
        "problem": "Plasmolysis occurs when living plant cells such as those of the onion bulb scale epidermis, are placed in strong sugar solution. After plasmolysis, which one of the following occupies the region between the plasma membrane and the cell wall?\nA: sugar solution\nB: pure vacuolar sap\nC: diluted vacuolar sap\nD: water\nE: air\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPlasmolysis occurs when living plant cells such as those of the onion bulb scale epidermis, are placed in strong sugar solution. After plasmolysis, which one of the following occupies the region between the plasma membrane and the cell wall?\n\nA: sugar solution\nB: pure vacuolar sap\nC: diluted vacuolar sap\nD: water\nE: air\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_40",
        "problem": "Some plants in the Bromeliaceae essentially lack stems and absorptive roots. They take up water from reservoirs known as \"tanks\" that are formed by overlaps of bases of different leaves (Fig. P-Q). All Bromeliads have profuse trichome coatings on both sides of their leaves. The coatings consist of tiny sliver white cells that are able to take up moisture and nutrients and transfer these into the plant.\n\nMorphology and anatomy of leaf of a bromeliad is shown. The leaf was cut at the base of the tank region, sealed, and immersed in aqueous solution of a fluorescent dye, for $2 \\mathrm{~h}$. Cross-sections were made from the blade (Fig. R) and tank (Fig. S) before and after staining, respectively. Red line indicates pathway for water between vein and trichome (tri).\n[figure1]\nA: \"a\" in Figure $\\mathrm{R}$ denotes aerenchyma\nB: \"b\" in Figure $\\mathrm{S}$ denotes hydrenchyma (aquatic parenchymal cells)\nC: The leaf anatomy suggests a C4 or CAM photosynthetic type\nD: Red line indicates pathway for water between vein and trichome on the upper side of leaf (adaxial)\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSome plants in the Bromeliaceae essentially lack stems and absorptive roots. They take up water from reservoirs known as \"tanks\" that are formed by overlaps of bases of different leaves (Fig. P-Q). All Bromeliads have profuse trichome coatings on both sides of their leaves. The coatings consist of tiny sliver white cells that are able to take up moisture and nutrients and transfer these into the plant.\n\nMorphology and anatomy of leaf of a bromeliad is shown. The leaf was cut at the base of the tank region, sealed, and immersed in aqueous solution of a fluorescent dye, for $2 \\mathrm{~h}$. Cross-sections were made from the blade (Fig. R) and tank (Fig. S) before and after staining, respectively. Red line indicates pathway for water between vein and trichome (tri).\n[figure1]\n\nA: \"a\" in Figure $\\mathrm{R}$ denotes aerenchyma\nB: \"b\" in Figure $\\mathrm{S}$ denotes hydrenchyma (aquatic parenchymal cells)\nC: The leaf anatomy suggests a C4 or CAM photosynthetic type\nD: Red line indicates pathway for water between vein and trichome on the upper side of leaf (adaxial)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-40.jpg?height=1224&width=1060&top_left_y=780&top_left_x=495"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_720",
        "problem": "下图为某种遗传病的家系图，该病与两对独立遗传的基因有关，且两对基因均可单独致病。两个联姻的家庭都不携带对方家庭的致病基因, 所生后代患病的概率均为 0 。调查对象中没有发现基因突变和染色体变异的个体。下列叙述错误的是（ ）\n\n[图1]\nA: 两对基因均位于常染色体上\nB: 该病属于多基因隐性遗传病\nC: I-1 与I-2, I-3 与I-4, III-1 与III-2 的基因型两两相同\nD: 若II-2 与II-5 婚配, 其后代携带致病基因的概率为 $5 / 9$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为某种遗传病的家系图，该病与两对独立遗传的基因有关，且两对基因均可单独致病。两个联姻的家庭都不携带对方家庭的致病基因, 所生后代患病的概率均为 0 。调查对象中没有发现基因突变和染色体变异的个体。下列叙述错误的是（ ）\n\n[图1]\n\nA: 两对基因均位于常染色体上\nB: 该病属于多基因隐性遗传病\nC: I-1 与I-2, I-3 与I-4, III-1 与III-2 的基因型两两相同\nD: 若II-2 与II-5 婚配, 其后代携带致病基因的概率为 $5 / 9$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-75.jpg?height=391&width=811&top_left_y=1255&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1521",
        "problem": "A section of DNA is shown.\n\n[figure1]\n\nA cysteine base is included in the image, what else is in the image?\nA: Thymine\nB: Glycine\nC: Guanosine\nD: Glucose\nE: Ribonucleotide\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA section of DNA is shown.\n\n[figure1]\n\nA cysteine base is included in the image, what else is in the image?\n\nA: Thymine\nB: Glycine\nC: Guanosine\nD: Glucose\nE: Ribonucleotide\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-08.jpg?height=845&width=1382&top_left_y=431&top_left_x=226"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_828",
        "problem": "利用转基因技术将抗病基因和抗除草剂基因转入豌豆, 获得抗病抗除草剂的单株A1、 A2 和 A3, 分别自然繁殖一代, 子代性状如下表所示。已知目的基因能一次或多次插入并整合到受体细胞染色体上。下列叙述正确的是（ ）\n\n| 子代个数 | 抗病抗除草剂 | 抗病不抗除草剂 | 不抗病抗除草剂 | 不抗病不抗除草剂 |\n| :--- | :--- | :--- | :--- | :--- |\n| A1 子代 | 83 | 41 | 43 | 0 |\n| A2 子代 | 160 | 7 | 8 | 0 |\n| A3 子代 | 93 | 31 | 28 | 10 |\nA: 单株 A1 有 1 个抗病基因和 1 个抗除草剂基因转入一条染色体上\nB: 单株 A2 子代抗病抗除草剂个体的一对同源染色体的每条染色体上各含有 1 个抗病基因和 1 个抗除草剂基因\nC: 给 A1 后代中植株喷施除草剂, 让存活植株自然繁殖一代, 子代中抗病不抗除草剂植株比例为 $1 / 2$\nD: 用 A3 后代纯合抗病不抗除草剂与纯合不抗病抗除草剂单株杂交, 子二代中纯合抗病抗除草剂的个体占 $1 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n利用转基因技术将抗病基因和抗除草剂基因转入豌豆, 获得抗病抗除草剂的单株A1、 A2 和 A3, 分别自然繁殖一代, 子代性状如下表所示。已知目的基因能一次或多次插入并整合到受体细胞染色体上。下列叙述正确的是（ ）\n\n| 子代个数 | 抗病抗除草剂 | 抗病不抗除草剂 | 不抗病抗除草剂 | 不抗病不抗除草剂 |\n| :--- | :--- | :--- | :--- | :--- |\n| A1 子代 | 83 | 41 | 43 | 0 |\n| A2 子代 | 160 | 7 | 8 | 0 |\n| A3 子代 | 93 | 31 | 28 | 10 |\n\nA: 单株 A1 有 1 个抗病基因和 1 个抗除草剂基因转入一条染色体上\nB: 单株 A2 子代抗病抗除草剂个体的一对同源染色体的每条染色体上各含有 1 个抗病基因和 1 个抗除草剂基因\nC: 给 A1 后代中植株喷施除草剂, 让存活植株自然繁殖一代, 子代中抗病不抗除草剂植株比例为 $1 / 2$\nD: 用 A3 后代纯合抗病不抗除草剂与纯合不抗病抗除草剂单株杂交, 子二代中纯合抗病抗除草剂的个体占 $1 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_907",
        "problem": "大部分龟类没有性染色体，其受精卵在较低温度下发育为雄性，在较高温度下发育为雌性; 剑尾鱼在 $\\mathrm{pH}$ 为 7.2 时全部发育成雄性, 而 $\\mathrm{pH}$ 为 7.8 时绝大多数发育成雌性;鳗鲰在高密度养殖时雄性个体占有较高比例。三类个体完成发育后, 不再发生性别转变。下列叙述错误的是（\nA: 温度、酸碱度、种群密度等因素能影响细胞的分化, 进而影响生物的个体发育\nB: 性别被确定后不再发生变化，体现了细胞分化的稳定性和不可逆性\nC: 随着细胞分化程度的提高, 细胞的全能性往往会逐渐降低, 但遗传物质通常不发生改变\nD: 以上三种生物的性别由环境决定，细胞内不含有性染色体和与性别形成有关的基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n大部分龟类没有性染色体，其受精卵在较低温度下发育为雄性，在较高温度下发育为雌性; 剑尾鱼在 $\\mathrm{pH}$ 为 7.2 时全部发育成雄性, 而 $\\mathrm{pH}$ 为 7.8 时绝大多数发育成雌性;鳗鲰在高密度养殖时雄性个体占有较高比例。三类个体完成发育后, 不再发生性别转变。下列叙述错误的是（\n\nA: 温度、酸碱度、种群密度等因素能影响细胞的分化, 进而影响生物的个体发育\nB: 性别被确定后不再发生变化，体现了细胞分化的稳定性和不可逆性\nC: 随着细胞分化程度的提高, 细胞的全能性往往会逐渐降低, 但遗传物质通常不发生改变\nD: 以上三种生物的性别由环境决定，细胞内不含有性染色体和与性别形成有关的基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_525",
        "problem": "某兴趣小组进行以下三组实验, 并分别完成数据分析: (1)某封闭实验田里的野生植株, 花色有红色和白色两种, 随机选取 30 株红花植株让其自由交配, $F_{1}$ 中红花: 白花 $=63: 1$; (2)将一株杂合紫花踠豆连续自交繁殖三代, $\\mathrm{F}_{3}$ 中紫花:白花 $=9: 7$; (3)将两株纯合的蓝花植株杂交 $\\rightarrow F_{1}$ 全为紫花, 让 $F_{1}$ 自交 $\\rightarrow F_{2}$ 紫花: 蓝花 $=9: 7$ 。不考虑致死等特殊情况,下列推断正确的是（ ）\nA: (1)数据说明花色的遗传至少由 2 对独立遗传的等位基因决定\nB: (2)数据反映出踠豆花色的遗传符合基因的自由组合定律\nC: 如果逐代淘汰(2)中白花植株，则 $\\mathrm{F}_{3}$ 中白花的基因频率为 10\\%\nD: 如果选取(3)中 $F_{1}$ 紫花植株进行测交, 子代紫花:蓝花为 $1: 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某兴趣小组进行以下三组实验, 并分别完成数据分析: (1)某封闭实验田里的野生植株, 花色有红色和白色两种, 随机选取 30 株红花植株让其自由交配, $F_{1}$ 中红花: 白花 $=63: 1$; (2)将一株杂合紫花踠豆连续自交繁殖三代, $\\mathrm{F}_{3}$ 中紫花:白花 $=9: 7$; (3)将两株纯合的蓝花植株杂交 $\\rightarrow F_{1}$ 全为紫花, 让 $F_{1}$ 自交 $\\rightarrow F_{2}$ 紫花: 蓝花 $=9: 7$ 。不考虑致死等特殊情况,下列推断正确的是（ ）\n\nA: (1)数据说明花色的遗传至少由 2 对独立遗传的等位基因决定\nB: (2)数据反映出踠豆花色的遗传符合基因的自由组合定律\nC: 如果逐代淘汰(2)中白花植株，则 $\\mathrm{F}_{3}$ 中白花的基因频率为 10\\%\nD: 如果选取(3)中 $F_{1}$ 紫花植株进行测交, 子代紫花:蓝花为 $1: 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_330",
        "problem": "某精原细胞 ( $2 \\mathrm{~N}=8$ ) 的 DNA 分子双链均用 ${ }^{15} \\mathrm{~N}$ 标记后置于含 ${ }^{14} \\mathrm{~N}$ 的培养基中培养,经过连续两次细胞分裂后，检测子细胞中的情况。下列推断，错误的是（ ）\nA: 若进行有丝分裂, 则含 ${ }^{15} \\mathrm{~N}$ 染色体的子细胞所占比例不唯一, 至少占 $50 \\%$\nB: 若进行减数分裂, 则第二次分裂后期每个细胞中含 ${ }^{15} \\mathrm{~N}$ 的染色体有 8 条\nC: 若子细胞中部分染色体含 ${ }^{15} \\mathrm{~N}$ ，则分裂过程中可能会发生同源染色体的分离\nD: 若子细胞中染色体都含 ${ }^{15} \\mathrm{~N}$, 则分裂过程中可能会发生非同源染色体自由组合\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某精原细胞 ( $2 \\mathrm{~N}=8$ ) 的 DNA 分子双链均用 ${ }^{15} \\mathrm{~N}$ 标记后置于含 ${ }^{14} \\mathrm{~N}$ 的培养基中培养,经过连续两次细胞分裂后，检测子细胞中的情况。下列推断，错误的是（ ）\n\nA: 若进行有丝分裂, 则含 ${ }^{15} \\mathrm{~N}$ 染色体的子细胞所占比例不唯一, 至少占 $50 \\%$\nB: 若进行减数分裂, 则第二次分裂后期每个细胞中含 ${ }^{15} \\mathrm{~N}$ 的染色体有 8 条\nC: 若子细胞中部分染色体含 ${ }^{15} \\mathrm{~N}$ ，则分裂过程中可能会发生同源染色体的分离\nD: 若子细胞中染色体都含 ${ }^{15} \\mathrm{~N}$, 则分裂过程中可能会发生非同源染色体自由组合\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-060.jpg?height=279&width=1034&top_left_y=483&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-060.jpg?height=303&width=874&top_left_y=865&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-060.jpg?height=294&width=214&top_left_y=867&top_left_x=1224",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-060.jpg?height=157&width=1336&top_left_y=1401&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-060.jpg?height=157&width=229&top_left_y=1566&top_left_x=728",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-060.jpg?height=151&width=209&top_left_y=1569&top_left_x=1209"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1274",
        "problem": "The table shows the results of an investigation into survival in relation to clutch size in Swiss starlings.\n\n| Number of young in clutch | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Number of young marked | 65 | 328 | 127 <br> 8 | 395 <br> 6 | 617 <br> 5 | 315 <br> 6 | 651 | 120 |\n| Number of marked birds recaptured after 3 <br> months | 0.0 | 1.8 | 2.0 | 2.1 | 2.1 | 1.7 | 1.5 | 0.8 |\n\nThe information suggests that clutch size in Swiss starlings is a result of\nA: disruptive selection\nB: stabilising selection\nC: directional selection\nD: artificial selection\nE: recurrent mutation\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe table shows the results of an investigation into survival in relation to clutch size in Swiss starlings.\n\n| Number of young in clutch | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Number of young marked | 65 | 328 | 127 <br> 8 | 395 <br> 6 | 617 <br> 5 | 315 <br> 6 | 651 | 120 |\n| Number of marked birds recaptured after 3 <br> months | 0.0 | 1.8 | 2.0 | 2.1 | 2.1 | 1.7 | 1.5 | 0.8 |\n\nThe information suggests that clutch size in Swiss starlings is a result of\n\nA: disruptive selection\nB: stabilising selection\nC: directional selection\nD: artificial selection\nE: recurrent mutation\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1221",
        "problem": "When plant cells from the zone of elongation are compared with those from the apical meristem, they have\nA: a larger surface area to volume ratio\nB: a higher proportion undergoing mitosis\nC: a larger nuclear to cytoplasmic ratio\nD: more prominent vacuoles\nE: a larger number of ribosomes\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen plant cells from the zone of elongation are compared with those from the apical meristem, they have\n\nA: a larger surface area to volume ratio\nB: a higher proportion undergoing mitosis\nC: a larger nuclear to cytoplasmic ratio\nD: more prominent vacuoles\nE: a larger number of ribosomes\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1075",
        "problem": "A glucose-fed yeast cell growing in an aerobic environment consumes $1.8 \\mathrm{~g}$ of glucose per unit time. If these cells are moved to anaerobic environment, how many moles of glucose the cells will consume to generate ATP at the same rate? (Write the answer in decimal form.)",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\n\nproblem:\nA glucose-fed yeast cell growing in an aerobic environment consumes $1.8 \\mathrm{~g}$ of glucose per unit time. If these cells are moved to anaerobic environment, how many moles of glucose the cells will consume to generate ATP at the same rate? (Write the answer in decimal form.)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir units are, in order, [moles, moles], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            "moles",
            "moles"
        ],
        "answer_sequence": null,
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1074",
        "problem": "For organisms, whose genomes are completely sequenced, the variation between the genomes can be measured from the temperature at which single-stranded DNA molecules anneal to form a double helix.\n$P$ represents a melting curve when both DNA strands were taken from a mocking bird. Q represents a melting curve when one DNA strand was from a mocking bird and another from starling.\n\nI to IV represent melting curves obtained when DNA strands of different bird species shown in the cladogram were paired.\n\n[figure1]\n\nTEMPERATURE $\\left({ }^{\\circ} \\mathrm{C}\\right)$\n\n[figure2]\n\nwhich of the following is correct?\nA: Curve I is likely to represent DNA duplex formed by one strand from a mocking bird and another from a thrasher.\nB: Common starlings, mynahs and catbirds form a monophyletic clade where the most recent common ancestor evolved around 20 mya.\nC: Curves II and III are likely to represent hybrids between DNA of mocking bird with family Sylvoidea and family Passeroidea respectively.\nD: The greater the curve to the left of $P$, greater is the complementarity between the sequences.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor organisms, whose genomes are completely sequenced, the variation between the genomes can be measured from the temperature at which single-stranded DNA molecules anneal to form a double helix.\n$P$ represents a melting curve when both DNA strands were taken from a mocking bird. Q represents a melting curve when one DNA strand was from a mocking bird and another from starling.\n\nI to IV represent melting curves obtained when DNA strands of different bird species shown in the cladogram were paired.\n\n[figure1]\n\nTEMPERATURE $\\left({ }^{\\circ} \\mathrm{C}\\right)$\n\n[figure2]\n\nwhich of the following is correct?\n\nA: Curve I is likely to represent DNA duplex formed by one strand from a mocking bird and another from a thrasher.\nB: Common starlings, mynahs and catbirds form a monophyletic clade where the most recent common ancestor evolved around 20 mya.\nC: Curves II and III are likely to represent hybrids between DNA of mocking bird with family Sylvoidea and family Passeroidea respectively.\nD: The greater the curve to the left of $P$, greater is the complementarity between the sequences.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-35.jpg?height=651&width=721&top_left_y=688&top_left_x=298",
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-35.jpg?height=683&width=894&top_left_y=726&top_left_x=1144"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_142",
        "problem": "The figures 1-4 illustrate the schematic cross-sectional anatomy of bones and flying muscles of a bird and a bat.\n\n(1)\n\n[figure1]\n\n(2)\n\n[figure2]\n\n(3)\n\n[figure3]\n\n(4)\n\n[figure4]\nA: Figure 1 and 2 are the bat structures.\nB: The animal of figure 1 and 2 has seven cervical vertebrae.\nC: In figure 1 and 2 the only mobile joint, which is involved in wing's flapping is scapula-humoral.\nD: The muscle supracoracoideus in figure 3 , is responsible for downstroke movement in flapping cycle.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe figures 1-4 illustrate the schematic cross-sectional anatomy of bones and flying muscles of a bird and a bat.\n\n(1)\n\n[figure1]\n\n(2)\n\n[figure2]\n\n(3)\n\n[figure3]\n\n(4)\n\n[figure4]\n\nA: Figure 1 and 2 are the bat structures.\nB: The animal of figure 1 and 2 has seven cervical vertebrae.\nC: In figure 1 and 2 the only mobile joint, which is involved in wing's flapping is scapula-humoral.\nD: The muscle supracoracoideus in figure 3 , is responsible for downstroke movement in flapping cycle.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-30.jpg?height=479&width=617&top_left_y=503&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-30.jpg?height=494&width=603&top_left_y=1004&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-30.jpg?height=408&width=754&top_left_y=550&top_left_x=1025",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-30.jpg?height=434&width=668&top_left_y=1045&top_left_x=1025"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_19",
        "problem": "When a light beam enters another medium in an oblique angle, its direction will change. This phenomenon is called light refraction. The amount of this change in beam direction can be calculated by Snell's law, where \" $n$ \" is the refraction index.\n[figure1]\n\nThis phenomenon is also the principle of conversion and diversion of beams in lenses. Power of a lens can be calculated by the following formula. Optical power of a lens demonstrates the degree of convergence or divergence of the light beam.\n\n( $D$ : Power of lens, $r$ : radius of curvature)\n\n$$\nD=\\left(n_{1}-n_{2}\\right) / r\n$$\n\nThe figure below shows the eye structure of Anableps $s p$. which can simultaneously see objects in both aquatic and terrestrial scenes.\n\n[figure2]\n\n(a)\n\n[figure3]\n\n(b)\n\n[figure4]\n\n(c)\n\nA) Anableps sp. B) Eye of Anablebs sp. C) Schematic diagram of Anableps. sp. eye\nA: The largest refraction index belongs to the lens.\nB: The largest amount of refraction belongs to the lens.\nC: The light beam would be more divergent when passing through the aqueous humour than the cornea.\nD: Considering the different diameters of the lens of Anableps fish, light ray 1 and 2 belong to terrestrial and aquatic spaces, respectively.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nWhen a light beam enters another medium in an oblique angle, its direction will change. This phenomenon is called light refraction. The amount of this change in beam direction can be calculated by Snell's law, where \" $n$ \" is the refraction index.\n[figure1]\n\nThis phenomenon is also the principle of conversion and diversion of beams in lenses. Power of a lens can be calculated by the following formula. Optical power of a lens demonstrates the degree of convergence or divergence of the light beam.\n\n( $D$ : Power of lens, $r$ : radius of curvature)\n\n$$\nD=\\left(n_{1}-n_{2}\\right) / r\n$$\n\nThe figure below shows the eye structure of Anableps $s p$. which can simultaneously see objects in both aquatic and terrestrial scenes.\n\n[figure2]\n\n(a)\n\n[figure3]\n\n(b)\n\n[figure4]\n\n(c)\n\nA) Anableps sp. B) Eye of Anablebs sp. C) Schematic diagram of Anableps. sp. eye\n\nA: The largest refraction index belongs to the lens.\nB: The largest amount of refraction belongs to the lens.\nC: The light beam would be more divergent when passing through the aqueous humour than the cornea.\nD: Considering the different diameters of the lens of Anableps fish, light ray 1 and 2 belong to terrestrial and aquatic spaces, respectively.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-21.jpg?height=494&width=1428&top_left_y=530&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-21.jpg?height=180&width=737&top_left_y=1458&top_left_x=662",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-21.jpg?height=294&width=308&top_left_y=1686&top_left_x=671",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-21.jpg?height=297&width=305&top_left_y=1685&top_left_x=1047"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_510",
        "problem": "家蚕 $(2 \\mathrm{n}=28)$ 为 $\\mathrm{ZW}$ 型性别决定生物。雌家蚕的一个卵原细胞中, 其染色体上的 DNA 分子双链用放射性 ${ }^{32} \\mathrm{P}$ 标记, 将其置于不含 ${ }^{32} \\mathrm{P}$ 的培养基中培养进行分裂, 形成了 4 个子细胞。下列叙述错误的（）\nA: 若进行减数分裂, 且未发生交换现象, 则产生的子细胞均含有放射性\nB: 若进行有丝分裂, 则子细胞中可能有 2 个子细胞的 Z 染色体含放射性\nC: 若子细胞为卵细胞, 且发生交换现象, 子细胞可能无 Z 染色体也无放射性\nD: 若子细胞为卵原细胞, 在分裂过程中细胞内可能出现 2 个或 4 个染色体组\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n家蚕 $(2 \\mathrm{n}=28)$ 为 $\\mathrm{ZW}$ 型性别决定生物。雌家蚕的一个卵原细胞中, 其染色体上的 DNA 分子双链用放射性 ${ }^{32} \\mathrm{P}$ 标记, 将其置于不含 ${ }^{32} \\mathrm{P}$ 的培养基中培养进行分裂, 形成了 4 个子细胞。下列叙述错误的（）\n\nA: 若进行减数分裂, 且未发生交换现象, 则产生的子细胞均含有放射性\nB: 若进行有丝分裂, 则子细胞中可能有 2 个子细胞的 Z 染色体含放射性\nC: 若子细胞为卵细胞, 且发生交换现象, 子细胞可能无 Z 染色体也无放射性\nD: 若子细胞为卵原细胞, 在分裂过程中细胞内可能出现 2 个或 4 个染色体组\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1433",
        "problem": "[figure1]\n\nA protein's primary structure consists of a series of amino acids held together by peptide (covalent) bonds. Which of the following would alter the primary protein structure?\nA: Boiling the sample\nB: Increasing $\\mathrm{pH}$ of the solution\nC: Decreasing pH of the Osolution\nD: Agitating the solution with a stirring rod\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nA protein's primary structure consists of a series of amino acids held together by peptide (covalent) bonds. Which of the following would alter the primary protein structure?\n\nA: Boiling the sample\nB: Increasing $\\mathrm{pH}$ of the solution\nC: Decreasing pH of the Osolution\nD: Agitating the solution with a stirring rod\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-36.jpg?height=560&width=1394&top_left_y=388&top_left_x=248"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_355",
        "problem": "易位是指发生在非同源染色体之间的部分染色体片段交换。雄果蝇第 $2 、 3$ 号染色体发生易位形成易位杂合子 (两条新的染色体记为 $2^{\\prime} 、 3^{\\prime}$ )。它减数分裂时会形成“十字形”构型, 在随后分离时会出现邻近分离或交互分离, 形成不同染色体组成的配子（如图甲); 果蝇的紫眼（b）和灰体（e）基因分别位于 $2 、 3$ 号染色体上。现以易位野生型雄果蝇（基因组成及位置如图乙）与紫眼灰体雌果蝇为亲本杂交, 结果 $F_{1}$ 中仅出现野生型及紫眼灰体, 没有紫眼体色野生型 $\\left(\\mathrm{bbe}^{+} \\mathrm{e}\\right)$ 和眼色野生型灰体 $\\left(\\mathrm{b}^{+} \\mathrm{bee}\\right)$ 。下列相关分析正确的是 ( )\n[图1]\nA: 据图推测, 2、3 号染色体间的易位可能发生在四分体中的染色单体之间\nB: $F_{1}$ 中没有 $b e^{+} e$ 和 $b^{+} b c e$ 个体可能是染色体组成为 $2^{\\prime} 3^{\\prime}$ 和 23 的配子无活性\nC: 若要产生与图乙相同的 $\\mathrm{F}_{1}$ 个体, 雄配子需由亲代雄果蝇通过邻近分离产生\nD: 不考虑互换, 让 $F_{1}$ 中野生型个体间随机交配, 则后代中易位纯合子占 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n易位是指发生在非同源染色体之间的部分染色体片段交换。雄果蝇第 $2 、 3$ 号染色体发生易位形成易位杂合子 (两条新的染色体记为 $2^{\\prime} 、 3^{\\prime}$ )。它减数分裂时会形成“十字形”构型, 在随后分离时会出现邻近分离或交互分离, 形成不同染色体组成的配子（如图甲); 果蝇的紫眼（b）和灰体（e）基因分别位于 $2 、 3$ 号染色体上。现以易位野生型雄果蝇（基因组成及位置如图乙）与紫眼灰体雌果蝇为亲本杂交, 结果 $F_{1}$ 中仅出现野生型及紫眼灰体, 没有紫眼体色野生型 $\\left(\\mathrm{bbe}^{+} \\mathrm{e}\\right)$ 和眼色野生型灰体 $\\left(\\mathrm{b}^{+} \\mathrm{bee}\\right)$ 。下列相关分析正确的是 ( )\n[图1]\n\nA: 据图推测, 2、3 号染色体间的易位可能发生在四分体中的染色单体之间\nB: $F_{1}$ 中没有 $b e^{+} e$ 和 $b^{+} b c e$ 个体可能是染色体组成为 $2^{\\prime} 3^{\\prime}$ 和 23 的配子无活性\nC: 若要产生与图乙相同的 $\\mathrm{F}_{1}$ 个体, 雄配子需由亲代雄果蝇通过邻近分离产生\nD: 不考虑互换, 让 $F_{1}$ 中野生型个体间随机交配, 则后代中易位纯合子占 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-49.jpg?height=434&width=1438&top_left_y=1142&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1496",
        "problem": "A section of DNA is shown.\n\n[figure1]\n\nBond 1 is...\nA: ionic.\nB: covalent.\nC: hydrogen.\nD: peptide.\nE: phosphate.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA section of DNA is shown.\n\n[figure1]\n\nBond 1 is...\n\nA: ionic.\nB: covalent.\nC: hydrogen.\nD: peptide.\nE: phosphate.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-08.jpg?height=845&width=1382&top_left_y=431&top_left_x=226"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_527",
        "problem": "某 XY 型动物的红眼和白眼分别由等位基因 A、a 控制，灰体、黑体分别由等位基\n\n因 B、b 控制显性基因对隐性基因为完全显性。现有双杂合的红眼灰体雌性个体与白眼黑体雄性个体交配; 欲根据子代性状判断基因在染色体上的相对位置，不考虑突变及其\n他变异, 下列说法错误的是\nA: 若两对基因均位于常染色体上且独立遗传, 则雌雄后代中均有 4 种表型\nB: 若有一对基因位于 $\\mathrm{X}$ 染色体上, 在考虑性别的情况下则后代共有 8 种表型\nC: 若两对基因均位于 X 染色体上, 则雌雄后代中白眼黑体的个体均占 $1 / 2$\nD: 若后代有 2 种表型、且与亲本相同、则两对基因的遗传不遵循自由组合定律\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某 XY 型动物的红眼和白眼分别由等位基因 A、a 控制，灰体、黑体分别由等位基\n\n因 B、b 控制显性基因对隐性基因为完全显性。现有双杂合的红眼灰体雌性个体与白眼黑体雄性个体交配; 欲根据子代性状判断基因在染色体上的相对位置，不考虑突变及其\n他变异, 下列说法错误的是\n\nA: 若两对基因均位于常染色体上且独立遗传, 则雌雄后代中均有 4 种表型\nB: 若有一对基因位于 $\\mathrm{X}$ 染色体上, 在考虑性别的情况下则后代共有 8 种表型\nC: 若两对基因均位于 X 染色体上, 则雌雄后代中白眼黑体的个体均占 $1 / 2$\nD: 若后代有 2 种表型、且与亲本相同、则两对基因的遗传不遵循自由组合定律\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1055",
        "problem": "One of the classical ways to study the function of any unknown \"gene\" is to mutate it \\& study the effect on the cell functioning. It has been successful for prokaryotes. In one such study in eukaryotes, in order to determine the function of a mouse gene ' $X$ ', a plasmid was constructed which contained the gene ' $X$ '. Then the gene neo ' ('promoter less' gene that assigns neomycin resistance to the cells) was introduced within the gene ' $X$ '. Mouse stem cells transfected with this plasmid were cultured in vitro in a neomycin containing medium and only a few cells survived. Note that stem cells are sensitive to neomycin.\n\nWhich of the following is correct?\nA: The expression of the neo ${ }^{r}$ gene from the plasmid conferred resistance and allowed the cells to grow in media containing neomycin.\nB: The cells that grew in the culture media are likely to have mutant ' $X$ ' gene in the genome.\nC: It can be deduced that gene ' $X$ ' is crucial for cell growth as most of the stem cells died in culture media.\nD: In vivo multiplication of plasmid led to its lateral transfer to other cells, making them antibiotic resistant.\nE: The gene for antibiotic resistance got transferred by homologous recombination in stem cells.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nOne of the classical ways to study the function of any unknown \"gene\" is to mutate it \\& study the effect on the cell functioning. It has been successful for prokaryotes. In one such study in eukaryotes, in order to determine the function of a mouse gene ' $X$ ', a plasmid was constructed which contained the gene ' $X$ '. Then the gene neo ' ('promoter less' gene that assigns neomycin resistance to the cells) was introduced within the gene ' $X$ '. Mouse stem cells transfected with this plasmid were cultured in vitro in a neomycin containing medium and only a few cells survived. Note that stem cells are sensitive to neomycin.\n\nWhich of the following is correct?\n\nA: The expression of the neo ${ }^{r}$ gene from the plasmid conferred resistance and allowed the cells to grow in media containing neomycin.\nB: The cells that grew in the culture media are likely to have mutant ' $X$ ' gene in the genome.\nC: It can be deduced that gene ' $X$ ' is crucial for cell growth as most of the stem cells died in culture media.\nD: In vivo multiplication of plasmid led to its lateral transfer to other cells, making them antibiotic resistant.\nE: The gene for antibiotic resistance got transferred by homologous recombination in stem cells.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_428",
        "problem": "蜂毒素是工蜂毒腺分泌的多肽, 具有抗菌、抗病毒及抗肿瘤等广泛的生物学效应。下图表示胃癌细胞在不同浓度的蜂毒素培养液中培养一定时间后, 胃癌细胞的凋亡率和调亡基因 Bax、Bel-2 的表达率。推测下列说法不正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 工蜂毒腺细胞合成蜂毒素需要核糖体, 不需要线粒体\nB: 癌细胞无限增殖时其细胞周期较正常细胞明显缩短\nC: 一定浓度的蜂毒素能诱导胃癌细胞凋亡, 并随浓度增大诱导效应增强\nD: 蜂毒素通过促进 Bax 蛋白产生和抑制 Bel-2 蛋白产生诱导胃癌细胞调亡\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n蜂毒素是工蜂毒腺分泌的多肽, 具有抗菌、抗病毒及抗肿瘤等广泛的生物学效应。下图表示胃癌细胞在不同浓度的蜂毒素培养液中培养一定时间后, 胃癌细胞的凋亡率和调亡基因 Bax、Bel-2 的表达率。推测下列说法不正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 工蜂毒腺细胞合成蜂毒素需要核糖体, 不需要线粒体\nB: 癌细胞无限增殖时其细胞周期较正常细胞明显缩短\nC: 一定浓度的蜂毒素能诱导胃癌细胞凋亡, 并随浓度增大诱导效应增强\nD: 蜂毒素通过促进 Bax 蛋白产生和抑制 Bel-2 蛋白产生诱导胃癌细胞调亡\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-027.jpg?height=403&width=625&top_left_y=435&top_left_x=404",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-027.jpg?height=400&width=651&top_left_y=440&top_left_x=1042"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_651",
        "problem": "某二倍体昆虫, 以正常翅 (AA) 为父本, 残翅 (aa) 为母本进行杂交, 由于某亲本在减数分裂中发生染色体变异, 子代出现了一只基因型为 AAa 的可育正常翅雄性个体。请判断下列叙述不正确的是（ ）\nA: 该异常个体产生的原因是父本在形成配子时发生异常形成基因组成为 AA 的配子\nB: 该亲本发生的变异类型可能是个别染色体数目增加、染色体片段重复和易位\nC: 可利用该变异个体精巢中的细胞观察有丝分裂中期染色体确定发生的变异类型\nD: 将该变异个体与残翅 (aa) 个体杂交, 若后代出现正常翅与残翅比例为 $3: 1$, 则变异个体产生的原因是染色体易位\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二倍体昆虫, 以正常翅 (AA) 为父本, 残翅 (aa) 为母本进行杂交, 由于某亲本在减数分裂中发生染色体变异, 子代出现了一只基因型为 AAa 的可育正常翅雄性个体。请判断下列叙述不正确的是（ ）\n\nA: 该异常个体产生的原因是父本在形成配子时发生异常形成基因组成为 AA 的配子\nB: 该亲本发生的变异类型可能是个别染色体数目增加、染色体片段重复和易位\nC: 可利用该变异个体精巢中的细胞观察有丝分裂中期染色体确定发生的变异类型\nD: 将该变异个体与残翅 (aa) 个体杂交, 若后代出现正常翅与残翅比例为 $3: 1$, 则变异个体产生的原因是染色体易位\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-75.jpg?height=508&width=1434&top_left_y=337&top_left_x=354"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_592",
        "problem": "某繁殖力超强鼠的自然种群中, 体色有黄色、黑色、灰色三种, 体色色素的转化关系如图所示。已知控制色素合成的两对基因位于两对常染色体上，基因 B 能完全抑制基因 $\\mathrm{b}$ 的表达，不含基因 $\\mathrm{A}$ 的个体会由于黄色素在体内过多积累而导致 50\\%的个体死亡。下列叙述不正确的是（ ）\n\n[图1]\n\n酶 1\n\n## 基因 B\n\n酶 2\n\n## 黄色素 $\\longrightarrow$ 中间产物 $\\longrightarrow$ 黑色素\n\n酶 3 匹基因 $b$\n\n## 灰色素\nA: 黄色鼠个体可有三种基因型\nB: 若让一只黄色雌鼠与一只灰色雄鼠交配, $F_{1}$ 全为黑色鼠, 则双亲的基因型为 aaBB 和 AAbb\nC: 两只黑色鼠交配, 子代只有黄色和黑色, 且比例接近 1: 6, 则双亲中一定有一只基因型是 $\\mathrm{AaBb}$\nD: 基因型为 $\\mathrm{AaBb}$ 的雌、雄鼠自由交配, 子代个体表型及比例为黄: 黑: 灰 $=2$ : 9: 3\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某繁殖力超强鼠的自然种群中, 体色有黄色、黑色、灰色三种, 体色色素的转化关系如图所示。已知控制色素合成的两对基因位于两对常染色体上，基因 B 能完全抑制基因 $\\mathrm{b}$ 的表达，不含基因 $\\mathrm{A}$ 的个体会由于黄色素在体内过多积累而导致 50\\%的个体死亡。下列叙述不正确的是（ ）\n\n[图1]\n\n酶 1\n\n## 基因 B\n\n酶 2\n\n## 黄色素 $\\longrightarrow$ 中间产物 $\\longrightarrow$ 黑色素\n\n酶 3 匹基因 $b$\n\n## 灰色素\n\nA: 黄色鼠个体可有三种基因型\nB: 若让一只黄色雌鼠与一只灰色雄鼠交配, $F_{1}$ 全为黑色鼠, 则双亲的基因型为 aaBB 和 AAbb\nC: 两只黑色鼠交配, 子代只有黄色和黑色, 且比例接近 1: 6, 则双亲中一定有一只基因型是 $\\mathrm{AaBb}$\nD: 基因型为 $\\mathrm{AaBb}$ 的雌、雄鼠自由交配, 子代个体表型及比例为黄: 黑: 灰 $=2$ : 9: 3\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-100.jpg?height=174&width=171&top_left_y=1626&top_left_x=517"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_22",
        "problem": "A gene-regulatory protein $\\mathrm{X}$ controls cell proliferation. Protein $\\mathrm{X}$ is found in the cytosol and has no typical nuclear localization signal (NLS). When cells are treated with a specific growth hormone, protein $\\mathrm{X}$ re-localizes from the cytoplasm into the nucleus where it activates the transcription factors involved in cell proliferation.\n\nRecently, a protein $(\\mathrm{Y})$ that interacts with protein $\\mathrm{X}$ has been identified in unstimulated cells. To investigate the function of protein $\\mathrm{Y}$, a mutant lacking the gene encoding protein $\\mathrm{Y}$ was generated. Fractionation of cells from the wild type and mutant produced membrane (M), cytoplasmic (C), and nuclear (N) fractions for each cell type. Proteins extracted from each fraction were separated by SDS-PAGE and analyzed by Western blotting for the presence of proteins X and Y.\n\n[figure1]\n\nOn the basis of the results shown above, which of the following statements is the most plausible characterization of protein $\\mathrm{Y}$ ?\nA: In the absence of growth hormone, protein $\\mathrm{Y}$ associates with protein $\\mathrm{X}$, and the $\\mathrm{X} / \\mathrm{Y}$ complex is subjected to a degradation pathway.\nB: In the presence of growth hormone, protein $\\mathrm{Y}$ interacts with protein $\\mathrm{X}$, and the complex remains in the cytoplasm.\nC: Protein $\\mathrm{X}$ interacts with protein $\\mathrm{Y}$ in the absence of growth hormone. Upon growth hormone treatment, protein $\\mathrm{X}$ is released from protein $\\mathrm{Y}$ and re-localizes to the nucleus.\nD: Protein $\\mathrm{Y}$ is a membrane-associated protein and re-localizes with protein $\\mathrm{X}$ to the nucleus upon the growth hormone treatment.\nE: Protein $\\mathrm{Y}$ is one of the nuclear import proteins and the growth hormone does not induceprotein $\\mathrm{Y}$ to translocate protein $\\mathrm{X}$ to the nucleus.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA gene-regulatory protein $\\mathrm{X}$ controls cell proliferation. Protein $\\mathrm{X}$ is found in the cytosol and has no typical nuclear localization signal (NLS). When cells are treated with a specific growth hormone, protein $\\mathrm{X}$ re-localizes from the cytoplasm into the nucleus where it activates the transcription factors involved in cell proliferation.\n\nRecently, a protein $(\\mathrm{Y})$ that interacts with protein $\\mathrm{X}$ has been identified in unstimulated cells. To investigate the function of protein $\\mathrm{Y}$, a mutant lacking the gene encoding protein $\\mathrm{Y}$ was generated. Fractionation of cells from the wild type and mutant produced membrane (M), cytoplasmic (C), and nuclear (N) fractions for each cell type. Proteins extracted from each fraction were separated by SDS-PAGE and analyzed by Western blotting for the presence of proteins X and Y.\n\n[figure1]\n\nOn the basis of the results shown above, which of the following statements is the most plausible characterization of protein $\\mathrm{Y}$ ?\n\nA: In the absence of growth hormone, protein $\\mathrm{Y}$ associates with protein $\\mathrm{X}$, and the $\\mathrm{X} / \\mathrm{Y}$ complex is subjected to a degradation pathway.\nB: In the presence of growth hormone, protein $\\mathrm{Y}$ interacts with protein $\\mathrm{X}$, and the complex remains in the cytoplasm.\nC: Protein $\\mathrm{X}$ interacts with protein $\\mathrm{Y}$ in the absence of growth hormone. Upon growth hormone treatment, protein $\\mathrm{X}$ is released from protein $\\mathrm{Y}$ and re-localizes to the nucleus.\nD: Protein $\\mathrm{Y}$ is a membrane-associated protein and re-localizes with protein $\\mathrm{X}$ to the nucleus upon the growth hormone treatment.\nE: Protein $\\mathrm{Y}$ is one of the nuclear import proteins and the growth hormone does not induceprotein $\\mathrm{Y}$ to translocate protein $\\mathrm{X}$ to the nucleus.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-07.jpg?height=331&width=985&top_left_y=1108&top_left_x=593"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_5",
        "problem": "Figure Q. 20 shows the models of four types of common human congenital heart defects.\n\n[figure1]\n\nNormal\n\n[figure2]\n\n[figure3]\n\n[figure4]\n\n[figure5]\n\nFigure Q.20.\nA: In type 1 , the blood volume going to the lungs is lower than normal.\nB: In type 2 , the stroke volume of the left ventricle is increased.\nC: In type 3, the systolic blood pressure at the arms is higher than that in the normal type.\nD: In type 4, the pulmonary blood pressure is increased.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFigure Q. 20 shows the models of four types of common human congenital heart defects.\n\n[figure1]\n\nNormal\n\n[figure2]\n\n[figure3]\n\n[figure4]\n\n[figure5]\n\nFigure Q.20.\n\nA: In type 1 , the blood volume going to the lungs is lower than normal.\nB: In type 2 , the stroke volume of the left ventricle is increased.\nC: In type 3, the systolic blood pressure at the arms is higher than that in the normal type.\nD: In type 4, the pulmonary blood pressure is increased.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-046.jpg?height=502&width=391&top_left_y=481&top_left_x=360",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-046.jpg?height=538&width=374&top_left_y=1067&top_left_x=377",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-046.jpg?height=544&width=368&top_left_y=497&top_left_x=813",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-046.jpg?height=532&width=351&top_left_y=1070&top_left_x=830",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-046.jpg?height=544&width=374&top_left_y=1064&top_left_x=1246"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_191",
        "problem": "Neural tube in the vertebrate embryos, lies the anterior-posterior (AP) axis and creates a variety of structures. There is a large body of evidence showing that AP patterning of the vertebrate embryos is controlled by $H O X$ genes. Spatial and temporal expression patterns of $H O X$ genes are controlled by factors usually present as gradients in the AP axis of embryos. One of these gradients is generated by retinoic acid (RA), a derivative of vitamin A, with a maximum concentration in the posterior region of the embryos (Figure below).\n\ngene II gene IV\n\ngene I gene III\n\n[figure1]\n\nHead\n\nTail\n\n[figure2]\n\nRA concentration\nA: Overuse of vitamin A by pregnant women may cause abnormalities in the embryos.\nB: In an embryo that has been affected by excessive amounts of RA in the early stages of embryogenesis, the forebrain may not form.\nC: Loss of function of a $H O X$ gene in the embryo may have the same effect as an excessive amount of RA.\nD: In an embryo whose expression of RA nuclear receptors has been downregulated, the midbrain may develop similar to the forebrain.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nNeural tube in the vertebrate embryos, lies the anterior-posterior (AP) axis and creates a variety of structures. There is a large body of evidence showing that AP patterning of the vertebrate embryos is controlled by $H O X$ genes. Spatial and temporal expression patterns of $H O X$ genes are controlled by factors usually present as gradients in the AP axis of embryos. One of these gradients is generated by retinoic acid (RA), a derivative of vitamin A, with a maximum concentration in the posterior region of the embryos (Figure below).\n\ngene II gene IV\n\ngene I gene III\n\n[figure1]\n\nHead\n\nTail\n\n[figure2]\n\nRA concentration\n\nA: Overuse of vitamin A by pregnant women may cause abnormalities in the embryos.\nB: In an embryo that has been affected by excessive amounts of RA in the early stages of embryogenesis, the forebrain may not form.\nC: Loss of function of a $H O X$ gene in the embryo may have the same effect as an excessive amount of RA.\nD: In an embryo whose expression of RA nuclear receptors has been downregulated, the midbrain may develop similar to the forebrain.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-32.jpg?height=431&width=788&top_left_y=935&top_left_x=631",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-32.jpg?height=143&width=622&top_left_y=1485&top_left_x=637"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1025",
        "problem": "In the picture below - taken from the work of $\\mathbf{N}$. Tinbergen - a cardinal feeds minnows, which rose to the surface looking for food. The bird fed the fish for several weeks, probably because its nest had been destroyed.\n\n[figure1]\n\n\nThe cardinal's behavior is best understood as:\nA: Habituation\nB: Imprinting\nC: Fixed action pattern\nD: Associative learning\nE: Operant conditioning\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the picture below - taken from the work of $\\mathbf{N}$. Tinbergen - a cardinal feeds minnows, which rose to the surface looking for food. The bird fed the fish for several weeks, probably because its nest had been destroyed.\n\n[figure1]\n\n\nThe cardinal's behavior is best understood as:\n\nA: Habituation\nB: Imprinting\nC: Fixed action pattern\nD: Associative learning\nE: Operant conditioning\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_162746b98e552b6160afg-08.jpg?height=688&width=808&top_left_y=762&top_left_x=363"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_528",
        "problem": "下图表示某二倍体生物不同细胞分裂过程中同源染色体对数的变化情况。下列叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 图 1 中 $\\mathrm{BC}$ 段和图 2 中 $\\mathrm{NH}$ 段都与染色体的着丝粒分裂有关\nB: 图 1 中 $\\mathrm{EF}$ 段和图 2 中 $\\mathrm{HI}$ 段, 细胞中都不存在姐妹染色单体\nC: 若细胞质正发生不均等分裂, 则其可能处于图 2 中的 MN 段\nD: 同源染色体的分离发生在 DE 段, 此时细胞发生基因重组\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图表示某二倍体生物不同细胞分裂过程中同源染色体对数的变化情况。下列叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 图 1 中 $\\mathrm{BC}$ 段和图 2 中 $\\mathrm{NH}$ 段都与染色体的着丝粒分裂有关\nB: 图 1 中 $\\mathrm{EF}$ 段和图 2 中 $\\mathrm{HI}$ 段, 细胞中都不存在姐妹染色单体\nC: 若细胞质正发生不均等分裂, 则其可能处于图 2 中的 MN 段\nD: 同源染色体的分离发生在 DE 段, 此时细胞发生基因重组\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-51.jpg?height=474&width=714&top_left_y=1054&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-51.jpg?height=471&width=708&top_left_y=1055&top_left_x=1048"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_473",
        "problem": "人的某条染色体上 $\\mathrm{B} 、 \\mathrm{C}$ 两个基因紧密排列，这两个基因各有上百个等位基因（例如: $\\mathrm{B}_{1} \\sim \\mathrm{B}_{\\mathrm{n}}$ 均为 $\\mathrm{B}$ 的等位基因)。父母及孩子的基因组成如下表, 不考虑 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段。下列叙述正确的是（）\n\n| 项目 | 父亲 | 母亲 | 儿子 | 女儿 |\n| :---: | :---: | :---: | :---: | :---: |\n| 基因组成 | $\\mathrm{B}_{7} \\mathrm{~B}_{35} \\mathrm{C}_{2} \\mathrm{C}_{4}$ | $\\mathrm{~B}_{8} \\mathrm{~B}_{44} \\mathrm{C}_{5} \\mathrm{C}_{9}$ | $\\mathrm{~B}_{7} \\mathrm{~B}_{8} \\mathrm{C}_{4} \\mathrm{C}_{5}$ | $\\mathrm{~B}_{35} \\mathrm{~B}_{44} \\mathrm{C}_{2} \\mathrm{C}_{9}$ |\nA: $\\mathrm{B}_{1} \\sim \\mathrm{B}_{\\mathrm{n}}$ 基因的出现说明基因突变具有随机性\nB: B 基因与人类红绿色盲基因间的遗传遵循自由组合定律\nC: 若不发生互换, 父亲的其中一条染色体上的基因组成是 $\\mathrm{B}_{7} \\mathrm{C}_{2}$\nD: 上述实例说明基因在染色体上呈非线性排列\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人的某条染色体上 $\\mathrm{B} 、 \\mathrm{C}$ 两个基因紧密排列，这两个基因各有上百个等位基因（例如: $\\mathrm{B}_{1} \\sim \\mathrm{B}_{\\mathrm{n}}$ 均为 $\\mathrm{B}$ 的等位基因)。父母及孩子的基因组成如下表, 不考虑 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段。下列叙述正确的是（）\n\n| 项目 | 父亲 | 母亲 | 儿子 | 女儿 |\n| :---: | :---: | :---: | :---: | :---: |\n| 基因组成 | $\\mathrm{B}_{7} \\mathrm{~B}_{35} \\mathrm{C}_{2} \\mathrm{C}_{4}$ | $\\mathrm{~B}_{8} \\mathrm{~B}_{44} \\mathrm{C}_{5} \\mathrm{C}_{9}$ | $\\mathrm{~B}_{7} \\mathrm{~B}_{8} \\mathrm{C}_{4} \\mathrm{C}_{5}$ | $\\mathrm{~B}_{35} \\mathrm{~B}_{44} \\mathrm{C}_{2} \\mathrm{C}_{9}$ |\n\nA: $\\mathrm{B}_{1} \\sim \\mathrm{B}_{\\mathrm{n}}$ 基因的出现说明基因突变具有随机性\nB: B 基因与人类红绿色盲基因间的遗传遵循自由组合定律\nC: 若不发生互换, 父亲的其中一条染色体上的基因组成是 $\\mathrm{B}_{7} \\mathrm{C}_{2}$\nD: 上述实例说明基因在染色体上呈非线性排列\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_383",
        "problem": "下列关于细胞生命历程的叙述错误的是（）\nA: 细胞周期检验点基因发生突变可能导致细胞癌变生长\nB: 细胞内外部环境诱导细胞分化而发生基因的选择性表达\nC: 过氧化氢酶可以及时清除过氧化氢，防止氧化性损伤导致的细胞衰老\nD: 细胞调亡和细胞坏死往往是基因与内外环境共同作用的结果\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列关于细胞生命历程的叙述错误的是（）\n\nA: 细胞周期检验点基因发生突变可能导致细胞癌变生长\nB: 细胞内外部环境诱导细胞分化而发生基因的选择性表达\nC: 过氧化氢酶可以及时清除过氧化氢，防止氧化性损伤导致的细胞衰老\nD: 细胞调亡和细胞坏死往往是基因与内外环境共同作用的结果\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_958",
        "problem": "Which component of Fick's Law of Diffusion is MOST optimized in the ventilation of fish gills compared to mammalian lungs?\nA: Diffusion coefficient\nB: Gradient for diffusion\nC: Surface area\nD: Path length\nE: Temperature\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich component of Fick's Law of Diffusion is MOST optimized in the ventilation of fish gills compared to mammalian lungs?\n\nA: Diffusion coefficient\nB: Gradient for diffusion\nC: Surface area\nD: Path length\nE: Temperature\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_834",
        "problem": "某家系甲病和乙病的系谱图如图所示。已知两病独立遗传, 各由一对等位基因控制,且基因不位于 Y 染色体。甲病在人群中的发病率为 $1 / 2500$ 。下列有关说法错误的是\n\n[图1]\nA: 甲病为常染色体隐性病\nB: 乙病为常染色体显性遗传病或 X 染色体显性遗传病\nC: 若乙病为常染色体显性遗传病, $\\mathrm{IIII}_{3}$ 与表型正常男子结婚, 所生子女同时患两种病的概率为 $2 / 459$\nD: 若需判断乙病遗传方式, 可选择用乙病的正常基因和致病基因设计的探针与 $\\mathrm{I}_{2}$ 、 $\\mathrm{II}_{3}$ 个体进行核酸杂交\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某家系甲病和乙病的系谱图如图所示。已知两病独立遗传, 各由一对等位基因控制,且基因不位于 Y 染色体。甲病在人群中的发病率为 $1 / 2500$ 。下列有关说法错误的是\n\n[图1]\n\nA: 甲病为常染色体隐性病\nB: 乙病为常染色体显性遗传病或 X 染色体显性遗传病\nC: 若乙病为常染色体显性遗传病, $\\mathrm{IIII}_{3}$ 与表型正常男子结婚, 所生子女同时患两种病的概率为 $2 / 459$\nD: 若需判断乙病遗传方式, 可选择用乙病的正常基因和致病基因设计的探针与 $\\mathrm{I}_{2}$ 、 $\\mathrm{II}_{3}$ 个体进行核酸杂交\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-92.jpg?height=462&width=1437&top_left_y=1785&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_505",
        "problem": "脊髓性肌萎缩症（SMA）是一种单基因遗传病, 主要由运动神经元存活基因 1\n\n（SMN1）突变引起, SMA 在胎儿发育期常常不表现出结构畸形, 出生后表现为肌无力和肌萎缩, 具有严重的致死性。现有一对正常夫妻, 生出一个 SMA 男孩 (染色体组成为 $\\mathrm{XXY}$ ）后他们进行了基因检测，电泳结果如图（正常基因一条带，致病基因为另一不同的条带，不考虑 XY 同源区段)。下列叙述正确的是（）\n\n[图1]\nA: 患病男孩同时患有常染色体隐性遗传病和染色体异常遗传病\nB: SMA 的致病原因是 SMN1 基因碱基的缺失引起的基因突变\nC: 患病男孩的染色体异常是由于母亲或父亲减数分裂I性染色体未分离导致的\nD: SMA 遗传病可以通过 B 超检查等产前诊断进行治疗\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n脊髓性肌萎缩症（SMA）是一种单基因遗传病, 主要由运动神经元存活基因 1\n\n（SMN1）突变引起, SMA 在胎儿发育期常常不表现出结构畸形, 出生后表现为肌无力和肌萎缩, 具有严重的致死性。现有一对正常夫妻, 生出一个 SMA 男孩 (染色体组成为 $\\mathrm{XXY}$ ）后他们进行了基因检测，电泳结果如图（正常基因一条带，致病基因为另一不同的条带，不考虑 XY 同源区段)。下列叙述正确的是（）\n\n[图1]\n\nA: 患病男孩同时患有常染色体隐性遗传病和染色体异常遗传病\nB: SMA 的致病原因是 SMN1 基因碱基的缺失引起的基因突变\nC: 患病男孩的染色体异常是由于母亲或父亲减数分裂I性染色体未分离导致的\nD: SMA 遗传病可以通过 B 超检查等产前诊断进行治疗\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-10.jpg?height=288&width=653&top_left_y=2043&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1393",
        "problem": "The three recognised colours of Labradors are black, chocolate and golden. These coat colours result from differences in two genetic loci which have alleles B/b and E/e respectively.\n\nSupplied is a simplified flowchart representing pigment deposition in labradors. $\\mathrm{E}$ is the allele for functional MRC1 protein, while e is a mutated allele resulting in no MRC1 protein. B is the allele for functional TYRP1 enzyme, while $b$ is a mutated allele resulting in no TYRP1 enzyme activity. If there is no pigment deposition, the labrador is considered golden.\n\n[figure1]\n\nIn a cross of a chocolate female with a chocolate male, which phenotypic ratio is IMPOSSIBLE?\nA: All chocolate\nB: 3 chocolate to 1 golden\nC: 1 black to 1 chocolate\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe three recognised colours of Labradors are black, chocolate and golden. These coat colours result from differences in two genetic loci which have alleles B/b and E/e respectively.\n\nSupplied is a simplified flowchart representing pigment deposition in labradors. $\\mathrm{E}$ is the allele for functional MRC1 protein, while e is a mutated allele resulting in no MRC1 protein. B is the allele for functional TYRP1 enzyme, while $b$ is a mutated allele resulting in no TYRP1 enzyme activity. If there is no pigment deposition, the labrador is considered golden.\n\n[figure1]\n\nIn a cross of a chocolate female with a chocolate male, which phenotypic ratio is IMPOSSIBLE?\n\nA: All chocolate\nB: 3 chocolate to 1 golden\nC: 1 black to 1 chocolate\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-19.jpg?height=780&width=926&top_left_y=1312&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_37",
        "problem": "Pundamilia pundalilia and $P$. nyererei are a closely related sister species pair of cichlids in Lake Victoria. These two species are distinct in male nuptial body colors, in that the former and latter are blue and red, respectively. By contrast, the females of the two species are not distinct, both possessing cryptic body coloration. $P$. pundamilia and $P$. nyererei inhabit shallow and deep environments, respectively. The light component in Lake Victoria is oriented to be blue (short wavelength) in shallow and red (long wavelength) in deep environments. The opsin protein of the two species are shifted to the same wavelength of their habitat light components. In addition, inter-species hybridization occurs under the specific light condition, where red and blue lights cannot be distinguished.\nA: The speciation of the two species is considered to have been caused by mating preference of males to females.\nB: During evolution, each of the two species is considered to have adapted their visual cues to their habitat light environment.\nC: The consistency between the male nuptial colorations and the light components of their habitats are explained by natural selection for camouflage\nD: The sequences of the opsin gene differ between males and females in each species.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPundamilia pundalilia and $P$. nyererei are a closely related sister species pair of cichlids in Lake Victoria. These two species are distinct in male nuptial body colors, in that the former and latter are blue and red, respectively. By contrast, the females of the two species are not distinct, both possessing cryptic body coloration. $P$. pundamilia and $P$. nyererei inhabit shallow and deep environments, respectively. The light component in Lake Victoria is oriented to be blue (short wavelength) in shallow and red (long wavelength) in deep environments. The opsin protein of the two species are shifted to the same wavelength of their habitat light components. In addition, inter-species hybridization occurs under the specific light condition, where red and blue lights cannot be distinguished.\n\nA: The speciation of the two species is considered to have been caused by mating preference of males to females.\nB: During evolution, each of the two species is considered to have adapted their visual cues to their habitat light environment.\nC: The consistency between the male nuptial colorations and the light components of their habitats are explained by natural selection for camouflage\nD: The sequences of the opsin gene differ between males and females in each species.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_289",
        "problem": "Goats are fed with alfalfa and corn stubble. At time 0 , they were also fed with Mimosa seeds. The presence of viable Mimosa seeds in goat faeces was recorded with a germination experiment with egested and control seeds.\n\n[figure1]\n\nFigure Q.47.1. Percent of seeds of Mimosa found in goat faeces as a function of time since seeds were ingested. Goat pellets were collected every 8 hours over a period of 80 hours after ingestion.\n\n[figure2]\n\nFigure Q.47.2. Probability of not germinating of egested (stippled line) and control (solid line) Mimosa seeds.\nA: Mimosa seed can survive up to 3 days in the goat digestive system.\nB: The passage through the goat digestive system decreases seed germination.\nC: Number of seeds egested after ingestion is highest after 24 hours.\nD: Goats could act as legitimate disperser of Mimosa seeds.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nGoats are fed with alfalfa and corn stubble. At time 0 , they were also fed with Mimosa seeds. The presence of viable Mimosa seeds in goat faeces was recorded with a germination experiment with egested and control seeds.\n\n[figure1]\n\nFigure Q.47.1. Percent of seeds of Mimosa found in goat faeces as a function of time since seeds were ingested. Goat pellets were collected every 8 hours over a period of 80 hours after ingestion.\n\n[figure2]\n\nFigure Q.47.2. Probability of not germinating of egested (stippled line) and control (solid line) Mimosa seeds.\n\nA: Mimosa seed can survive up to 3 days in the goat digestive system.\nB: The passage through the goat digestive system decreases seed germination.\nC: Number of seeds egested after ingestion is highest after 24 hours.\nD: Goats could act as legitimate disperser of Mimosa seeds.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-102.jpg?height=796&width=1375&top_left_y=498&top_left_x=405",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-102.jpg?height=939&width=1144&top_left_y=1613&top_left_x=523"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1016",
        "problem": "A certain plant you are studying has alleles $R$ and $r$ for pointed and round leaves, and alleles $B$ and $b$ for blue and white petals. You perform a cross between homozygous dominant (RRBB) and homozygous recessive (rrbb) plants. The offspring are then crossed with a homozygous recessive (rrbb) plant. You observe the results. 1600 offspring are produced. Which of the following data is CLOSEST to what you would expect given a recombination frequency of $25 \\%$ ?\nA: $100 \\mathrm{rrbb}, 300 \\mathrm{rrBb}, 300 \\mathrm{Rrbb}, 900 \\mathrm{RrBb}$\nB: $600 \\mathrm{rrbb}, 600 \\mathrm{rrBb}, 200 \\mathrm{Rrbb}, 200 \\mathrm{RrBb}$\nC: $200 \\mathrm{rrbb}, 200 \\mathrm{rrBb}, 600 \\mathrm{Rrbb}, 600 \\mathrm{RrBb}$\nD: $900 \\mathrm{rrbb}, 300 \\mathrm{rrBb}, 300 \\mathrm{Rrbb}, 100 \\mathrm{RrBb}$\nE: $600 \\mathrm{rrbb}, 200 \\mathrm{rrBb}, 200 \\mathrm{Rrbb}, 600 \\mathrm{RrBb}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA certain plant you are studying has alleles $R$ and $r$ for pointed and round leaves, and alleles $B$ and $b$ for blue and white petals. You perform a cross between homozygous dominant (RRBB) and homozygous recessive (rrbb) plants. The offspring are then crossed with a homozygous recessive (rrbb) plant. You observe the results. 1600 offspring are produced. Which of the following data is CLOSEST to what you would expect given a recombination frequency of $25 \\%$ ?\n\nA: $100 \\mathrm{rrbb}, 300 \\mathrm{rrBb}, 300 \\mathrm{Rrbb}, 900 \\mathrm{RrBb}$\nB: $600 \\mathrm{rrbb}, 600 \\mathrm{rrBb}, 200 \\mathrm{Rrbb}, 200 \\mathrm{RrBb}$\nC: $200 \\mathrm{rrbb}, 200 \\mathrm{rrBb}, 600 \\mathrm{Rrbb}, 600 \\mathrm{RrBb}$\nD: $900 \\mathrm{rrbb}, 300 \\mathrm{rrBb}, 300 \\mathrm{Rrbb}, 100 \\mathrm{RrBb}$\nE: $600 \\mathrm{rrbb}, 200 \\mathrm{rrBb}, 200 \\mathrm{Rrbb}, 600 \\mathrm{RrBb}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_707",
        "problem": "某动物的一个精原细胞中含有 $2 \\mathrm{n}$ 条染色体, 其 DNA 双链均只含 ${ }^{31} \\mathrm{P}$ 。现将该精原细胞置于含 ${ }^{32} \\mathrm{P}$ 的培养基中培养，细胞能正常进行分裂，下列相关叙述正确的是（）\nA: 若该精原细胞产生的 4 个子细胞的每条染色体均含 ${ }^{32} \\mathrm{P}$, 则进行的一定是减数分裂\nB: 若该精原细胞只进行减数分裂, 则在减数分裂II后期含 ${ }^{31} \\mathrm{P}$ 的染色体数目为 $2 \\mathrm{n}$\nC: 若该精原细胞进行了两次有丝分裂, 第二次有丝分裂中期细胞中每条染色单体均含 ${ }^{31} \\mathrm{P}$\nD: 若该精原细胞先进行一次有丝分裂再进行减数分裂, 则精细胞中含 ${ }^{31} \\mathrm{P}$ 的染色体数目为 $\\mathrm{n}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某动物的一个精原细胞中含有 $2 \\mathrm{n}$ 条染色体, 其 DNA 双链均只含 ${ }^{31} \\mathrm{P}$ 。现将该精原细胞置于含 ${ }^{32} \\mathrm{P}$ 的培养基中培养，细胞能正常进行分裂，下列相关叙述正确的是（）\n\nA: 若该精原细胞产生的 4 个子细胞的每条染色体均含 ${ }^{32} \\mathrm{P}$, 则进行的一定是减数分裂\nB: 若该精原细胞只进行减数分裂, 则在减数分裂II后期含 ${ }^{31} \\mathrm{P}$ 的染色体数目为 $2 \\mathrm{n}$\nC: 若该精原细胞进行了两次有丝分裂, 第二次有丝分裂中期细胞中每条染色单体均含 ${ }^{31} \\mathrm{P}$\nD: 若该精原细胞先进行一次有丝分裂再进行减数分裂, 则精细胞中含 ${ }^{31} \\mathrm{P}$ 的染色体数目为 $\\mathrm{n}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_489",
        "problem": "某二倍体动物的某细胞内含有 10 条染色体、 10 个 DNA 分子, 且细胞膜开始缢缩, 则该细胞（）\nA: 处理有丝分裂中期\nB: 正在发生基因自由组合\nC: 将形成配子\nD: 正在发生 DNA 复制\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二倍体动物的某细胞内含有 10 条染色体、 10 个 DNA 分子, 且细胞膜开始缢缩, 则该细胞（）\n\nA: 处理有丝分裂中期\nB: 正在发生基因自由组合\nC: 将形成配子\nD: 正在发生 DNA 复制\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_67",
        "problem": "Some restriction enzymes have different recognition sequences but create the same sticky ends, as shown below for SalI and XhoI.\n\n[figure1]\n\nThe following gel electrophoresis image shows linearized DNAs obtained after complete restriction enzyme digestion of non-recombinant (empty vector) and recombinant (containing gene $\\mathrm{X}$ ) expression plasmid vectors. The expression vector includes a strong promoter near its cloning site. The insert of the recombinant plasmid was a product of SalI digestion. The insert was ligated into the vector which had been cut with XhoI.\n\n[figure2]\n\n* All recombinant vectors gave the same pattern in the gel\nA: The data indicate that two copies of the insert were cloned in the expression vector.\nB: Half of the clones are expected to be able to transcribe the mRNA of gene X.\nC: An XhoI site exists outside of the insert in the recombinant vector of the experiment.\nD: The $2 \\mathrm{~kb}$ fragment seen in the gel can be used as probe to screen for the recombinant vectors\nE: If the insert produced by SalI digestion was ligated into a vector that had been cut with both XhoI and SalI enzymes and the recombinant plasmid subsequently cut with $X h o I$, a $4 \\mathrm{~kb}$ product would be obtained. True False\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSome restriction enzymes have different recognition sequences but create the same sticky ends, as shown below for SalI and XhoI.\n\n[figure1]\n\nThe following gel electrophoresis image shows linearized DNAs obtained after complete restriction enzyme digestion of non-recombinant (empty vector) and recombinant (containing gene $\\mathrm{X}$ ) expression plasmid vectors. The expression vector includes a strong promoter near its cloning site. The insert of the recombinant plasmid was a product of SalI digestion. The insert was ligated into the vector which had been cut with XhoI.\n\n[figure2]\n\n* All recombinant vectors gave the same pattern in the gel\n\nA: The data indicate that two copies of the insert were cloned in the expression vector.\nB: Half of the clones are expected to be able to transcribe the mRNA of gene X.\nC: An XhoI site exists outside of the insert in the recombinant vector of the experiment.\nD: The $2 \\mathrm{~kb}$ fragment seen in the gel can be used as probe to screen for the recombinant vectors\nE: If the insert produced by SalI digestion was ligated into a vector that had been cut with both XhoI and SalI enzymes and the recombinant plasmid subsequently cut with $X h o I$, a $4 \\mathrm{~kb}$ product would be obtained. True False\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-12.jpg?height=205&width=922&top_left_y=497&top_left_x=567",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-12.jpg?height=574&width=1254&top_left_y=958&top_left_x=401"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_637",
        "problem": "下列有关生物体遗传物质的叙述, 正确的是 ( )\nA: 显宛豆的遗传物质主要是 DNA\nB: 酵母菌的遗传物质主要分布在染色体上\nC: $\\mathrm{T}_{2}$ 噬菌体的遗传物质含有硫元素\nD: HIV 的遗传物质水解产生 4 种脱氧核苷酸\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列有关生物体遗传物质的叙述, 正确的是 ( )\n\nA: 显宛豆的遗传物质主要是 DNA\nB: 酵母菌的遗传物质主要分布在染色体上\nC: $\\mathrm{T}_{2}$ 噬菌体的遗传物质含有硫元素\nD: HIV 的遗传物质水解产生 4 种脱氧核苷酸\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1536",
        "problem": "The RECOVERY trial run by the NHS is the most successful trial to identify proven treatments for COVID-19, and has disproven several popular candidates.\n\nMore data from the Oxford/AstraZeneca COVID-19 vaccine trial was released last month looking at people given two full doses of vaccine 12 -weeks apart.\n\n- They found 15 out of 2038 people given the COVID-19 vaccine caught COVID-19.\n- They found 76 out of 2093 people given a meningitis vaccine caught COVID-19.\n\nVaccine efficiency is calculated as the percentage decrease in cases caused by vaccination.\n\nCalculate the vaccine efficiency. Pick the nearest number below.\nA: $60 \\%$\nB: $70 \\%$\nC: $80 \\%$\nD: $90 \\%$\nE: $100 \\%$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe RECOVERY trial run by the NHS is the most successful trial to identify proven treatments for COVID-19, and has disproven several popular candidates.\n\nMore data from the Oxford/AstraZeneca COVID-19 vaccine trial was released last month looking at people given two full doses of vaccine 12 -weeks apart.\n\n- They found 15 out of 2038 people given the COVID-19 vaccine caught COVID-19.\n- They found 76 out of 2093 people given a meningitis vaccine caught COVID-19.\n\nVaccine efficiency is calculated as the percentage decrease in cases caused by vaccination.\n\nCalculate the vaccine efficiency. Pick the nearest number below.\n\nA: $60 \\%$\nB: $70 \\%$\nC: $80 \\%$\nD: $90 \\%$\nE: $100 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_321",
        "problem": "果蝇的灰身与黑身、长翅与短翅各由一对等位基因控制（两对等位基因均不位于 Y 染色体上)。一只纯合黑身长翅雌果蝇与一只纯合灰身短翅雄果蝇杂交得到 $F_{1}, F_{1}$ 雌雄相互交配, $F_{2}$ 表型及比例为灰身长翅:灰身短翅: 黑身长翅: 黑身短翅 $=9: 3: 3: 1$ 。 $F_{2}$ 表型中不可能出现的现象是（ ）\nA: 长翅果蝇中雌雄比 2:1\nB: 短翅果蝇中雌雄比 $2: 1$\nC: 长翅果蝇中雌雄比 $1: 1$\nD: 黑身果蝇中雌雄比 $1: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的灰身与黑身、长翅与短翅各由一对等位基因控制（两对等位基因均不位于 Y 染色体上)。一只纯合黑身长翅雌果蝇与一只纯合灰身短翅雄果蝇杂交得到 $F_{1}, F_{1}$ 雌雄相互交配, $F_{2}$ 表型及比例为灰身长翅:灰身短翅: 黑身长翅: 黑身短翅 $=9: 3: 3: 1$ 。 $F_{2}$ 表型中不可能出现的现象是（ ）\n\nA: 长翅果蝇中雌雄比 2:1\nB: 短翅果蝇中雌雄比 $2: 1$\nC: 长翅果蝇中雌雄比 $1: 1$\nD: 黑身果蝇中雌雄比 $1: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_634",
        "problem": "细胞质基质中的 PDHEl $\\alpha$ 通过促进磷脂酶 PPMIB 对 IKK $\\beta S 177 / 181$ 的去磷酸化可以削弱 NF-kB 信号通路的激活, 增强炎症因子及细胞毒性 $\\mathrm{T}$ 细胞诱导的肿瘤细胞死亡。而某些致癌信号激活会导致 PDHEI $\\alpha$ 上某位点的磷酸化并转位到线粒体, 细胞质基质中的 PDHEla。水平下降活化了 NF-kB 信号通路; 同时线粒体中的 PDHEla 增多间接促进了炎症因子刺激下肿瘤细胞的 ROS（活性氧）解毒从而降低其对肿瘤细胞的毒害。\n\nNF- $\\kappa B$ 信号通路活化和 ROS 的清除共同促进了肿瘤细胞在炎症因子刺激下的存活, 增\n强了肿瘤细胞对细胞毒性 $\\mathrm{T}$ 细胞的耐受性, 最终促进了肿瘤的免疫逃逸。下列说法正确的是 ( )\nA: 炎症因子及细胞毒性 $\\mathrm{T}$ 细胞诱导的肿瘤细胞死亡属于细胞坏死 B磷脂酶 PPM1B 磷酸化可以抑制肿瘤细胞的生长\nB: 细胞质基质中的 PDHEl $\\alpha$ 能抑制 NF-kB 信号通路激活\nC: 抑制 PDHEla 的磷酸化可以阻断肿瘤的免疫逃逸并提高肿瘤免疫治疗的疗效\nD: 磷脂酶 PPM1B 磷酸化可以抑制肿瘤细胞的生长\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n细胞质基质中的 PDHEl $\\alpha$ 通过促进磷脂酶 PPMIB 对 IKK $\\beta S 177 / 181$ 的去磷酸化可以削弱 NF-kB 信号通路的激活, 增强炎症因子及细胞毒性 $\\mathrm{T}$ 细胞诱导的肿瘤细胞死亡。而某些致癌信号激活会导致 PDHEI $\\alpha$ 上某位点的磷酸化并转位到线粒体, 细胞质基质中的 PDHEla。水平下降活化了 NF-kB 信号通路; 同时线粒体中的 PDHEla 增多间接促进了炎症因子刺激下肿瘤细胞的 ROS（活性氧）解毒从而降低其对肿瘤细胞的毒害。\n\nNF- $\\kappa B$ 信号通路活化和 ROS 的清除共同促进了肿瘤细胞在炎症因子刺激下的存活, 增\n强了肿瘤细胞对细胞毒性 $\\mathrm{T}$ 细胞的耐受性, 最终促进了肿瘤的免疫逃逸。下列说法正确的是 ( )\n\nA: 炎症因子及细胞毒性 $\\mathrm{T}$ 细胞诱导的肿瘤细胞死亡属于细胞坏死 B磷脂酶 PPM1B 磷酸化可以抑制肿瘤细胞的生长\nB: 细胞质基质中的 PDHEl $\\alpha$ 能抑制 NF-kB 信号通路激活\nC: 抑制 PDHEla 的磷酸化可以阻断肿瘤的免疫逃逸并提高肿瘤免疫治疗的疗效\nD: 磷脂酶 PPM1B 磷酸化可以抑制肿瘤细胞的生长\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_612",
        "problem": "下列关于探索 DNA 是遗传物质实验的相关叙述, 正确的是(\nA: 格里菲思实验中肺炎双球菌 $\\mathrm{R}$ 型转化为 $\\mathrm{S}$ 型是基因突变的结果\nB: 格里菲思实验证明了 DNA 是肺炎双球菌的遗传物质\nC: 赫尔希和蔡斯实验中 $\\mathrm{T}_{2}$ 噬菌体的 DNA 是用 ${ }^{32} \\mathrm{P}$ 直接标记的\nD: 赫尔希和蔡斯实验证明了 DNA 是 $T_{2}$ 噬菌体的遗传物质\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列关于探索 DNA 是遗传物质实验的相关叙述, 正确的是(\n\nA: 格里菲思实验中肺炎双球菌 $\\mathrm{R}$ 型转化为 $\\mathrm{S}$ 型是基因突变的结果\nB: 格里菲思实验证明了 DNA 是肺炎双球菌的遗传物质\nC: 赫尔希和蔡斯实验中 $\\mathrm{T}_{2}$ 噬菌体的 DNA 是用 ${ }^{32} \\mathrm{P}$ 直接标记的\nD: 赫尔希和蔡斯实验证明了 DNA 是 $T_{2}$ 噬菌体的遗传物质\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_120",
        "problem": "Animals living in deserts like kangaroo rats achieve the ability to sustain themselves on a limited supply of water through incredibly well adapted kidney. To remove waste without losing water, species have developed mechanisms to concentrate their urine. There are two types of nephrons that concentrate urine, a type with a short Henle loop located in the renal cortex (cortex: C) and a type with a long Henle loop located near the renal medulla (juxtamedullary: JM). The ratio of these two types of nephrons differs depending on the animal. The table shows the habitat of each animal species and the urea concentration in urine. The graph plots the juxtamedullary-cortex ratio (the number of JM type loop/the number of C type loop) in each animal species.\n\n| Species | Habitat | Urine concentration <br> $(\\mathrm{mOsm} / \\mathrm{L})$ |\n| :---: | :---: | :---: |\n| Rat | moderate | 2900 |\n| Domestic cat | moderate | 3100 |\n| Kangaroo rat | dry | 5500 |\n| Beaver | freshwater/land | 520 |\n| Human | moderate | 1400 |\n| Porpoise | marine | 1800 |\n| Eland | dry | 1880 |\n| Camel | dry | 2800 |\n\n[figure1]\nA: Beavers seem not to possess the cortex type nephron.\nB: The $\\mathrm{JM} / \\mathrm{C}$ ratio of the kangaroo rat is estimated at 1.5 or more.\nC: Longer Henle loops can efficiently reabsorb salts, resulting in urine concentration.\nD: Animals living in dry regions have a higher proportion of cortex type nephrons than those living in freshwater.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAnimals living in deserts like kangaroo rats achieve the ability to sustain themselves on a limited supply of water through incredibly well adapted kidney. To remove waste without losing water, species have developed mechanisms to concentrate their urine. There are two types of nephrons that concentrate urine, a type with a short Henle loop located in the renal cortex (cortex: C) and a type with a long Henle loop located near the renal medulla (juxtamedullary: JM). The ratio of these two types of nephrons differs depending on the animal. The table shows the habitat of each animal species and the urea concentration in urine. The graph plots the juxtamedullary-cortex ratio (the number of JM type loop/the number of C type loop) in each animal species.\n\n| Species | Habitat | Urine concentration <br> $(\\mathrm{mOsm} / \\mathrm{L})$ |\n| :---: | :---: | :---: |\n| Rat | moderate | 2900 |\n| Domestic cat | moderate | 3100 |\n| Kangaroo rat | dry | 5500 |\n| Beaver | freshwater/land | 520 |\n| Human | moderate | 1400 |\n| Porpoise | marine | 1800 |\n| Eland | dry | 1880 |\n| Camel | dry | 2800 |\n\n[figure1]\n\nA: Beavers seem not to possess the cortex type nephron.\nB: The $\\mathrm{JM} / \\mathrm{C}$ ratio of the kangaroo rat is estimated at 1.5 or more.\nC: Longer Henle loops can efficiently reabsorb salts, resulting in urine concentration.\nD: Animals living in dry regions have a higher proportion of cortex type nephrons than those living in freshwater.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-31.jpg?height=537&width=765&top_left_y=1545&top_left_x=651"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_172",
        "problem": "The HapMap project was designed to estimate the amount of variation among the genomes of different individuals. One of the outcomes of the project was identification of many SNPs (single nucleotide polymorphisms) in the human genome. It was observed that the vast majority (here, assume all) of SNPs exist as only one of two (not four) nucleotides.\n\nTherefore, for a region of the genome consisting of $n$ SNPs, $2^{n}$ combinations of the SNPs are conceivable. In fact, sequencing of the SNPs in hundreds of individuals has revealed that generally a much lower number of combinations exist. The combinations of SNPs in fact observed in a region of the genome are named the SNP haplotypes of that region. The figure below is a representation of this finding. It shows the genotype of 26 neighbouring SNPs in a region of human chromosome 5 from 20 individuals of different populations throughout the world. Only one copy of chromosome 5 was isolated from each individual. The chromosomes are grouped on basis of having the same combination of genotypes for all the SNPs (i.e. having the same haplotype).\n[figure1]\nA: Recombination events are not expected to affect the number of haplotypes observed in a population.\nB: If in fact all conceivable haplotypes of the 26 SNPs analysed were present in the human population, one would not expect any two of the 20 chromosomes analysed to have the same haplotype.\nC: Based on data presented in the table, knowledge of the nucleotide at only SNP 19 and SNP 23 of chromosome 5 of newly investigated individuals from the same cohort would allow prediction of the most likely nucleotides present at the remaining 24 SNP positions of chromosome 5 of those individuals.\nD: The data on the 20 chromosomes suggest that among the four major haplotypes, haplotype IV has the minimum number of sequence variations\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe HapMap project was designed to estimate the amount of variation among the genomes of different individuals. One of the outcomes of the project was identification of many SNPs (single nucleotide polymorphisms) in the human genome. It was observed that the vast majority (here, assume all) of SNPs exist as only one of two (not four) nucleotides.\n\nTherefore, for a region of the genome consisting of $n$ SNPs, $2^{n}$ combinations of the SNPs are conceivable. In fact, sequencing of the SNPs in hundreds of individuals has revealed that generally a much lower number of combinations exist. The combinations of SNPs in fact observed in a region of the genome are named the SNP haplotypes of that region. The figure below is a representation of this finding. It shows the genotype of 26 neighbouring SNPs in a region of human chromosome 5 from 20 individuals of different populations throughout the world. Only one copy of chromosome 5 was isolated from each individual. The chromosomes are grouped on basis of having the same combination of genotypes for all the SNPs (i.e. having the same haplotype).\n[figure1]\n\nA: Recombination events are not expected to affect the number of haplotypes observed in a population.\nB: If in fact all conceivable haplotypes of the 26 SNPs analysed were present in the human population, one would not expect any two of the 20 chromosomes analysed to have the same haplotype.\nC: Based on data presented in the table, knowledge of the nucleotide at only SNP 19 and SNP 23 of chromosome 5 of newly investigated individuals from the same cohort would allow prediction of the most likely nucleotides present at the remaining 24 SNP positions of chromosome 5 of those individuals.\nD: The data on the 20 chromosomes suggest that among the four major haplotypes, haplotype IV has the minimum number of sequence variations\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://i.postimg.cc/zvvSKB15/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_66",
        "problem": "Photosynthesis of submerged aquatic plants is severely impeded by many environment factors. In seawater and freshwater, light density and its spectrum is changed with depth in the water column and thus influence photosynthesis. Other factors affecting photosynthesis include level of carbon dioxide $\\left(\\mathrm{CO}_{2}\\right)$ and oxygen $\\left(\\mathrm{O}_{2}\\right)$.\n\nSwamp Raspwort (Meionectes brownie) is a wetland plant species but can grow as a submerged aquatic plant in freshwater. An experiment was conducted to study the photosynthesis of the aquatic vegetation. Diurnal fluctuations in surface irradiance, partial pressure of $\\mathrm{O}_{2}, \\mathrm{CO}_{2}$ concentration and $\\mathrm{pH}$ of the water in Swamp Raspwort-rich ponds are shown in Fig.Q64.\n\n[figure1]\n\nB\n\n[figure2]\n\nC\n\n[figure3]\n\nFig.Q64\nA: In the underwater of ponds, light limitation appears early in the moming and colimitation of both light and $\\mathrm{CO}_{2}$ takes place early in the afternoon.\nB: The decrease in level of $\\mathrm{O}_{2}$ in water column during the night is caused by the Swamp Raspwort respiration.\nC: In water column of ponds, $\\mathrm{CO}_{2}$ molecules are directly produced by respiration of Swamp Raspwort and by conversion from $\\mathrm{HCO}_{3}^{-}$at $\\mathrm{pH}$ neutral results in increasing $\\mathrm{CO}_{2}$ level.\nD: As indicated in the figure, temperature variation in ponds rich Swamp Raspwort is from 13 to $20^{\\circ} \\mathrm{C}$. The alteration in temperature is mainly maintained by high density of this plant species.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPhotosynthesis of submerged aquatic plants is severely impeded by many environment factors. In seawater and freshwater, light density and its spectrum is changed with depth in the water column and thus influence photosynthesis. Other factors affecting photosynthesis include level of carbon dioxide $\\left(\\mathrm{CO}_{2}\\right)$ and oxygen $\\left(\\mathrm{O}_{2}\\right)$.\n\nSwamp Raspwort (Meionectes brownie) is a wetland plant species but can grow as a submerged aquatic plant in freshwater. An experiment was conducted to study the photosynthesis of the aquatic vegetation. Diurnal fluctuations in surface irradiance, partial pressure of $\\mathrm{O}_{2}, \\mathrm{CO}_{2}$ concentration and $\\mathrm{pH}$ of the water in Swamp Raspwort-rich ponds are shown in Fig.Q64.\n\n[figure1]\n\nB\n\n[figure2]\n\nC\n\n[figure3]\n\nFig.Q64\n\nA: In the underwater of ponds, light limitation appears early in the moming and colimitation of both light and $\\mathrm{CO}_{2}$ takes place early in the afternoon.\nB: The decrease in level of $\\mathrm{O}_{2}$ in water column during the night is caused by the Swamp Raspwort respiration.\nC: In water column of ponds, $\\mathrm{CO}_{2}$ molecules are directly produced by respiration of Swamp Raspwort and by conversion from $\\mathrm{HCO}_{3}^{-}$at $\\mathrm{pH}$ neutral results in increasing $\\mathrm{CO}_{2}$ level.\nD: As indicated in the figure, temperature variation in ponds rich Swamp Raspwort is from 13 to $20^{\\circ} \\mathrm{C}$. The alteration in temperature is mainly maintained by high density of this plant species.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-035.jpg?height=487&width=1282&top_left_y=953&top_left_x=433",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-035.jpg?height=462&width=1006&top_left_y=1478&top_left_x=568",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-035.jpg?height=587&width=1148&top_left_y=1977&top_left_x=548"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_107",
        "problem": "Uljanin in 1870, discovered a group of marine, soft-bodied, unsegmented hermaphroditic worms without hindgut and coelom. The mouth in these worms opens to a central digestive parenchyma. They moved with their multiciliated epidermis, although, many were 'surprisingly muscular'. Most of the species were free-living and some were ectocommensals. It has now been showed that several species form obligate symbioses with green algae, making them functional photoautotroph organisms. Later, phylogenetic position of these multicellular organisms was studied using the complete mitochondrial genome of a member of this group (Symsagittifera roscoffensis) which has been given in the figure below.\n\n[figure1]\nA: Based on this tree, Acoelomorpha is a sister group of Deuterostomia.\nB: Based on the above given description, the identified worm is a triploblastic acoelomate.\nC: Based on the above given text, the identified worm has incomplete digestive system.\nD: The provided data is consistent with the hypothesis that the identified worm belongs to the earliest diverging lineage of Bilateria.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nUljanin in 1870, discovered a group of marine, soft-bodied, unsegmented hermaphroditic worms without hindgut and coelom. The mouth in these worms opens to a central digestive parenchyma. They moved with their multiciliated epidermis, although, many were 'surprisingly muscular'. Most of the species were free-living and some were ectocommensals. It has now been showed that several species form obligate symbioses with green algae, making them functional photoautotroph organisms. Later, phylogenetic position of these multicellular organisms was studied using the complete mitochondrial genome of a member of this group (Symsagittifera roscoffensis) which has been given in the figure below.\n\n[figure1]\n\nA: Based on this tree, Acoelomorpha is a sister group of Deuterostomia.\nB: Based on the above given description, the identified worm is a triploblastic acoelomate.\nC: Based on the above given text, the identified worm has incomplete digestive system.\nD: The provided data is consistent with the hypothesis that the identified worm belongs to the earliest diverging lineage of Bilateria.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-27.jpg?height=1716&width=1307&top_left_y=758&top_left_x=366"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_833",
        "problem": "下图是某种单基因遗传病的家系图谱。已知 $\\mathrm{II}_{16}$ 和 $\\mathrm{II}_{17}$ 婚后生育了一个正常小孩, 该小孩携带致病基因的概率是（ ）\n\n[图1]\nA: $2 / 5$\nB: $16 / 37$\nC: $4 / 9$\nD: $21 / 64$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图是某种单基因遗传病的家系图谱。已知 $\\mathrm{II}_{16}$ 和 $\\mathrm{II}_{17}$ 婚后生育了一个正常小孩, 该小孩携带致病基因的概率是（ ）\n\n[图1]\n\nA: $2 / 5$\nB: $16 / 37$\nC: $4 / 9$\nD: $21 / 64$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-02.jpg?height=399&width=962&top_left_y=837&top_left_x=364"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_44",
        "problem": "The following photograph shows filamentous growth of a kind of cyanobacteria, Nostoc sp. The bacteria form heterocysts (thick-walled cells), when nitrogen sources such as ammonia or nitrates are deficient in the environment.\n\n[figure1]\n\nWhich of the following statements describing these heterocysts is/are true?\n\nI. Nitrogen is fixed in the heterocyst.\n\nII. Photosystem I does not function in the heterocyst.\n\nIII. Photosystem II does not function in the heterocyst.\nA: Only I\nB: Only II\nC: Only I and II\nD: Only I and III\nE: Only II and III\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following photograph shows filamentous growth of a kind of cyanobacteria, Nostoc sp. The bacteria form heterocysts (thick-walled cells), when nitrogen sources such as ammonia or nitrates are deficient in the environment.\n\n[figure1]\n\nWhich of the following statements describing these heterocysts is/are true?\n\nI. Nitrogen is fixed in the heterocyst.\n\nII. Photosystem I does not function in the heterocyst.\n\nIII. Photosystem II does not function in the heterocyst.\n\nA: Only I\nB: Only II\nC: Only I and II\nD: Only I and III\nE: Only II and III\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-06.jpg?height=400&width=597&top_left_y=565&top_left_x=752"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_793",
        "problem": "动物细胞线粒体内环状双链 ( $\\mathrm{H}$ 链和 L 链) DNA 分子的复制过程为: $\\mathrm{H}$ 链上的复制起始区 $\\mathrm{O}$ 先启动, 以 $\\mathrm{L}$ 链为模板, 先合成一段 RNA 引物, 引物引导 $\\mathrm{H}^{\\prime}$ 链合成, $\\mathrm{H}^{\\prime}$ 链取代原来老 $\\mathrm{H}$ 链的位置, 而被取代的老 $\\mathrm{H}$ 链以环的形式游离出来。当 $\\mathrm{H}^{\\prime}$ 链合成约 $2 / 3$ 时, $\\mathrm{L}$ 链上的复制起始区 $\\mathrm{O}_{\\mathrm{L}}$ 启动, 以老 $\\mathrm{H}$ 链为模板, 经过与 $\\mathrm{H}^{\\prime}$ 链相似的过程合成 $\\mathrm{L}^{\\prime}$ 链。具体过程如图所示。下列说法错误的是（）\n\n[图1]\nA: 该 DNA 分子两条链的复制不同步进行\nB: 复制过程中子链和模板链之间按照碱基互补配对原则通过氢键连接\nC: 该 DNA 分子连续复制 2 次共需要 8 个 RNA 引物\nD: 该 DNA 分子连续复制 $\\mathrm{n}$ 次新合成的 L'链有 2 \"-1 条\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n动物细胞线粒体内环状双链 ( $\\mathrm{H}$ 链和 L 链) DNA 分子的复制过程为: $\\mathrm{H}$ 链上的复制起始区 $\\mathrm{O}$ 先启动, 以 $\\mathrm{L}$ 链为模板, 先合成一段 RNA 引物, 引物引导 $\\mathrm{H}^{\\prime}$ 链合成, $\\mathrm{H}^{\\prime}$ 链取代原来老 $\\mathrm{H}$ 链的位置, 而被取代的老 $\\mathrm{H}$ 链以环的形式游离出来。当 $\\mathrm{H}^{\\prime}$ 链合成约 $2 / 3$ 时, $\\mathrm{L}$ 链上的复制起始区 $\\mathrm{O}_{\\mathrm{L}}$ 启动, 以老 $\\mathrm{H}$ 链为模板, 经过与 $\\mathrm{H}^{\\prime}$ 链相似的过程合成 $\\mathrm{L}^{\\prime}$ 链。具体过程如图所示。下列说法错误的是（）\n\n[图1]\n\nA: 该 DNA 分子两条链的复制不同步进行\nB: 复制过程中子链和模板链之间按照碱基互补配对原则通过氢键连接\nC: 该 DNA 分子连续复制 2 次共需要 8 个 RNA 引物\nD: 该 DNA 分子连续复制 $\\mathrm{n}$ 次新合成的 L'链有 2 \"-1 条\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-07.jpg?height=300&width=1148&top_left_y=201&top_left_x=320"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1540",
        "problem": "The Linnaean classification system allows scientists to understand the ancestry and characteristics of organisms at a glance.\n\n[figure1]\n\nSort these taxonomic ranks into order, starting with the oldest/biggest at the top.\nA: Animalia (animals), Eukarya (eukaryotes), Chordata (inc.vertebrates/backbones), Mammalia (mammals), Carnivora (true carnivores), Canines (canids), Canis (dog-like), Canis lupus (Grey wolf), Canis lupus familiaris (tame dogs)\nB: Mammalia (mammals), Chordata (inc.vertebrates/backbones), Eukarya (eukaryotes), Animalia (animals), Carnivora (true carnivores), Canines (canids), Canis (dog-like), Canis lupus (Grey wolf), Canis lupus familiaris (tame dogs)\nC: Animalia (animals), Chordata (inc.vertebrates/backbones), Eukarya (eukaryotes), Carnivora (true carnivores), Canines (canids), Mammalia (mammals), Canis (dog-like), Canis lupus (Grey wolf), Canis lupus familiaris (tame dogs)\nD: Eukarya (eukaryotes), Animalia (animals), Chordata (inc.vertebrates/backbones), Carnivora (true carnivores), Canines (canids), Mammalia (mammals), Canis (dog-like), Canis lupus (Grey wolf), Canis lupus familiaris (tame dogs)\nE: Eukarya (eukaryotes), Animalia (animals), Chordata (inc.vertebrates/backbones), Mammalia (mammals), Carnivora (true carnivores), Canines (canids), Canis (dog-like), Canis lupus (Grey wolf), Canis lupus familiaris (tame dogs)\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe Linnaean classification system allows scientists to understand the ancestry and characteristics of organisms at a glance.\n\n[figure1]\n\nSort these taxonomic ranks into order, starting with the oldest/biggest at the top.\n\nA: Animalia (animals), Eukarya (eukaryotes), Chordata (inc.vertebrates/backbones), Mammalia (mammals), Carnivora (true carnivores), Canines (canids), Canis (dog-like), Canis lupus (Grey wolf), Canis lupus familiaris (tame dogs)\nB: Mammalia (mammals), Chordata (inc.vertebrates/backbones), Eukarya (eukaryotes), Animalia (animals), Carnivora (true carnivores), Canines (canids), Canis (dog-like), Canis lupus (Grey wolf), Canis lupus familiaris (tame dogs)\nC: Animalia (animals), Chordata (inc.vertebrates/backbones), Eukarya (eukaryotes), Carnivora (true carnivores), Canines (canids), Mammalia (mammals), Canis (dog-like), Canis lupus (Grey wolf), Canis lupus familiaris (tame dogs)\nD: Eukarya (eukaryotes), Animalia (animals), Chordata (inc.vertebrates/backbones), Carnivora (true carnivores), Canines (canids), Mammalia (mammals), Canis (dog-like), Canis lupus (Grey wolf), Canis lupus familiaris (tame dogs)\nE: Eukarya (eukaryotes), Animalia (animals), Chordata (inc.vertebrates/backbones), Mammalia (mammals), Carnivora (true carnivores), Canines (canids), Canis (dog-like), Canis lupus (Grey wolf), Canis lupus familiaris (tame dogs)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-26.jpg?height=782&width=603&top_left_y=480&top_left_x=241"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_315",
        "problem": "将果蝇 $(2 \\mathrm{~N}=8)$ 的精原细胞 $(2 \\mathrm{~N}=8)$ 的所有染色体 DNA 链都用放射性同位素 ${ }^{32} \\mathrm{P}$标记, 再置于含 ${ }^{31} \\mathrm{P}$ 的培养液中培养。实验期间收集细胞甲、乙、丙、丁，统计样本放射性标记的染色体数和核 DNA 数, 不考虑突变及互换, 根据表格情况分析, 下列叙述正确的是（）\n\n| 细胞 | ${ }^{31} \\mathrm{P}$ 标记染色体 | ${ }^{32} \\mathrm{P}$ 标记 DNA |\n| :--- | :--- | :--- |\n\n\n|  | 数 | 数 |\n| :---: | :---: | :---: |\n| 甲 | 8 | 16 |\n| 乙 | 8 | 8 |\n| 丙 | 4 | 8 |\n| 丁 | 6 | 6 |\nA: 细胞甲可能处于第二次有丝分裂中期\nB: 细胞乙可能处于减数分裂 II 中期\nC: 细胞丙可能处于有丝分裂后的减数分裂 I 后期\nD: 细胞丁可能处于第三次有丝分裂后期\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将果蝇 $(2 \\mathrm{~N}=8)$ 的精原细胞 $(2 \\mathrm{~N}=8)$ 的所有染色体 DNA 链都用放射性同位素 ${ }^{32} \\mathrm{P}$标记, 再置于含 ${ }^{31} \\mathrm{P}$ 的培养液中培养。实验期间收集细胞甲、乙、丙、丁，统计样本放射性标记的染色体数和核 DNA 数, 不考虑突变及互换, 根据表格情况分析, 下列叙述正确的是（）\n\n| 细胞 | ${ }^{31} \\mathrm{P}$ 标记染色体 | ${ }^{32} \\mathrm{P}$ 标记 DNA |\n| :--- | :--- | :--- |\n\n\n|  | 数 | 数 |\n| :---: | :---: | :---: |\n| 甲 | 8 | 16 |\n| 乙 | 8 | 8 |\n| 丙 | 4 | 8 |\n| 丁 | 6 | 6 |\n\nA: 细胞甲可能处于第二次有丝分裂中期\nB: 细胞乙可能处于减数分裂 II 中期\nC: 细胞丙可能处于有丝分裂后的减数分裂 I 后期\nD: 细胞丁可能处于第三次有丝分裂后期\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_202",
        "problem": "Steroid hormones affect expression of primary response genes and secondary response genes in cells as schematically shown in the figure below.\n\n[figure1]\n\nOne primary-response protein shuts off primary-response genes\n\nAnother primary-response protein turns on secondary-response genes\n\nPrimary and secondary response genes can be distinguished by ...\nA: inhibition of DNA replication simultaneously with hormone administration\nB: inhibition of transcription simultaneously with hormone administration\nC: inhibition of translation simultaneously with hormone administration\nD: inhibition of transcription and translation simultaneously with hormone administration\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSteroid hormones affect expression of primary response genes and secondary response genes in cells as schematically shown in the figure below.\n\n[figure1]\n\nOne primary-response protein shuts off primary-response genes\n\nAnother primary-response protein turns on secondary-response genes\n\nPrimary and secondary response genes can be distinguished by ...\n\nA: inhibition of DNA replication simultaneously with hormone administration\nB: inhibition of transcription simultaneously with hormone administration\nC: inhibition of translation simultaneously with hormone administration\nD: inhibition of transcription and translation simultaneously with hormone administration\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-10.jpg?height=462&width=910&top_left_y=497&top_left_x=493"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1189",
        "problem": "Geological analysis suggests the initial separation of New Zealand from Gondwana took place approximately 80 million years ago. However, it is hypothesised that full separation was not complete until 60 million years ago. Some scientists propose that a common ancestor of the Moa and the Kiwi (a proto-Moa) floated away from Gondwana during this time. Others propose that Moa and Kiwi arrived in New Zealand at two different times. Four possible phylogenetic trees, describing different possible evolutionary relationships within the ratites, are shown below. Use these trees to answer the following questions.\n[figure1]Which tree incorporates mitochondrial DNA evidence showing that the Cassowary shared a common ancestor with Kiwi 60 mya and with Moa 70 mya.\nA: Tree 1\nB: Tree 2\nC: Tree 3\nD: Tree 4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nGeological analysis suggests the initial separation of New Zealand from Gondwana took place approximately 80 million years ago. However, it is hypothesised that full separation was not complete until 60 million years ago. Some scientists propose that a common ancestor of the Moa and the Kiwi (a proto-Moa) floated away from Gondwana during this time. Others propose that Moa and Kiwi arrived in New Zealand at two different times. Four possible phylogenetic trees, describing different possible evolutionary relationships within the ratites, are shown below. Use these trees to answer the following questions.\n[figure1]\n\nproblem:\nWhich tree incorporates mitochondrial DNA evidence showing that the Cassowary shared a common ancestor with Kiwi 60 mya and with Moa 70 mya.\n\nA: Tree 1\nB: Tree 2\nC: Tree 3\nD: Tree 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-26.jpg?height=1120&width=1812&top_left_y=505&top_left_x=150"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1061",
        "problem": "Environmental variables that limit respiratory gas exchange in animals can be explained by Fick's Law:\n\n$$\nQ=D \\cdot A \\cdot \\frac{P 1-P 2}{L}\n$$\n\nWhere, $\\mathrm{Q}=$ rate of diffusion. It is affected by\n\ni. $\\quad D=$ diffusion coefficient (depends on the type of diffusing substance, medium and temperature)\n\nii. $\\quad A=$ cross sectional area\n\niii. P1and P2 = partial pressures across two locations\n\niv. $L=$ path length\n\nIn fish, countercurrent exchange of gases across the gill lamellae results in very efficient gas exchange. Which of the following factors has the maximum influence on it?\nA: D\nB: A\nC: P1-P2\nD: L\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEnvironmental variables that limit respiratory gas exchange in animals can be explained by Fick's Law:\n\n$$\nQ=D \\cdot A \\cdot \\frac{P 1-P 2}{L}\n$$\n\nWhere, $\\mathrm{Q}=$ rate of diffusion. It is affected by\n\ni. $\\quad D=$ diffusion coefficient (depends on the type of diffusing substance, medium and temperature)\n\nii. $\\quad A=$ cross sectional area\n\niii. P1and P2 = partial pressures across two locations\n\niv. $L=$ path length\n\nIn fish, countercurrent exchange of gases across the gill lamellae results in very efficient gas exchange. Which of the following factors has the maximum influence on it?\n\nA: D\nB: A\nC: P1-P2\nD: L\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1018",
        "problem": "Which two major climatic factors dictate what kind of vegetation is found on the landscape?\nA: Temperature and latitude\nB: Temperature and precipitation\nC: Temperature and soil type\nD: Precipitation and soil type\nE: Precipitation and solar angle\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich two major climatic factors dictate what kind of vegetation is found on the landscape?\n\nA: Temperature and latitude\nB: Temperature and precipitation\nC: Temperature and soil type\nD: Precipitation and soil type\nE: Precipitation and solar angle\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1201",
        "problem": "A vacuolated plant cell is equilibrated in pure water, and then transferred to a sucrose solution with a water potential of $-800 \\mathrm{kPa}$. Which one of the rows $A$ - $\\mathbf{D}$ shows what would happen next?\nA: Direction of net <br> water movement: into cell, Pressure potential <br> $\\left(\\psi_{p}\\right)$: increases  \nB: Direction of net <br> water movement: out of cell, Pressure potential <br> $\\left(\\psi_{p}\\right)$: decreases\nC: Direction of net <br> water movement: into cell, Pressure potential <br> $\\left(\\psi_{p}\\right)$: decreases\nD: Direction of net <br> water movement: out of cell, Pressure potential <br> $\\left(\\psi_{p}\\right)$: increases\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA vacuolated plant cell is equilibrated in pure water, and then transferred to a sucrose solution with a water potential of $-800 \\mathrm{kPa}$. Which one of the rows $A$ - $\\mathbf{D}$ shows what would happen next?\n\nA: Direction of net <br> water movement: into cell, Pressure potential <br> $\\left(\\psi_{p}\\right)$: increases  \nB: Direction of net <br> water movement: out of cell, Pressure potential <br> $\\left(\\psi_{p}\\right)$: decreases\nC: Direction of net <br> water movement: into cell, Pressure potential <br> $\\left(\\psi_{p}\\right)$: decreases\nD: Direction of net <br> water movement: out of cell, Pressure potential <br> $\\left(\\psi_{p}\\right)$: increases\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_75",
        "problem": "Förster resonance energy transfer (FRET) is a distance-dependent process whereby energy from an excited fluorescent molecule (donor) is transferred to a second, non-excited molecule (acceptor) in its vicinity. The FRET efficiency depends on the spectral overlap of the donor emission spectrum and the acceptor absorption spectrum. Acceptors in various forms of FRET may or may not emit fluorescence after receipt of energy from donor.\n\nQuantum dots (QDs) are fluorescent nanoparticles that can act as FRET donors and can be used in biosensor systems. The binding of some molecules to QDs enhances the emission properties of the QDs. In this study, $560 \\mathrm{~nm}$ emitting QDs (QD-560) were used. Maltose Binding Protein (MBP)bounded QD-560 has higher emission than unbound QD-560.\n\nThe effect of native MBP or penta-histidine tagged MBP (MBP-5His) binding to QD-560 on fluorescence intensity was tested over a range of protein-to-QD ratios (Figure 1). Histidine residues can bind zinc ions present on the surface of the QD particles.\n\n[figure1]\n\nFigure 1: Fluorimetric measurements of QDs bound to MBPs or to MBP-5His at $560 \\mathrm{~nm}$.\n\nFor construction of a Maltose detecting biosensor, the saccharide binding pocket of each QDcoordinated MBP-5His was preloaded with a maltose analogue named $\\beta$-cyclodextrin bound to QSY9 ( $\\beta$-CD-QSY9); the QSY9 component absorbs light. Figure 2 shows the biosensor.\n\n[figure2]\n\nFigure 2: Schematic representation of maltose detecting biosensor.\nA\n\n[figure3]\n\nB\n\n[figure4]\n\nFigure 3: Absorption (A) and emission (B) spectra of various molecules.\nA: Displacement of $\\beta-\\mathrm{CD}-\\mathrm{QSY} 9$ by maltose on the biosensor will result in reduced fluorescence intensity.\nB: In the absence of maltose, FRET cannot occur in the biosensor.\nC: It is expected that relative fluorescence intensity levels under conditions below are $\\mathrm{C}>\\mathrm{B}>\\mathrm{A}$  A: QD with MBP-5His and $\\beta-\\mathrm{CD}-\\mathrm{QSY} 9$ dye  B: QD with MBP-5His and free QSY9 dye  $\\mathrm{C}$ : QD with MBP-5His\nD:  QD incubated with MBP that was preloaded with $\\beta-\\mathrm{CD}-\\mathrm{QSY} 9$ will not show FRET.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFörster resonance energy transfer (FRET) is a distance-dependent process whereby energy from an excited fluorescent molecule (donor) is transferred to a second, non-excited molecule (acceptor) in its vicinity. The FRET efficiency depends on the spectral overlap of the donor emission spectrum and the acceptor absorption spectrum. Acceptors in various forms of FRET may or may not emit fluorescence after receipt of energy from donor.\n\nQuantum dots (QDs) are fluorescent nanoparticles that can act as FRET donors and can be used in biosensor systems. The binding of some molecules to QDs enhances the emission properties of the QDs. In this study, $560 \\mathrm{~nm}$ emitting QDs (QD-560) were used. Maltose Binding Protein (MBP)bounded QD-560 has higher emission than unbound QD-560.\n\nThe effect of native MBP or penta-histidine tagged MBP (MBP-5His) binding to QD-560 on fluorescence intensity was tested over a range of protein-to-QD ratios (Figure 1). Histidine residues can bind zinc ions present on the surface of the QD particles.\n\n[figure1]\n\nFigure 1: Fluorimetric measurements of QDs bound to MBPs or to MBP-5His at $560 \\mathrm{~nm}$.\n\nFor construction of a Maltose detecting biosensor, the saccharide binding pocket of each QDcoordinated MBP-5His was preloaded with a maltose analogue named $\\beta$-cyclodextrin bound to QSY9 ( $\\beta$-CD-QSY9); the QSY9 component absorbs light. Figure 2 shows the biosensor.\n\n[figure2]\n\nFigure 2: Schematic representation of maltose detecting biosensor.\nA\n\n[figure3]\n\nB\n\n[figure4]\n\nFigure 3: Absorption (A) and emission (B) spectra of various molecules.\n\nA: Displacement of $\\beta-\\mathrm{CD}-\\mathrm{QSY} 9$ by maltose on the biosensor will result in reduced fluorescence intensity.\nB: In the absence of maltose, FRET cannot occur in the biosensor.\nC: It is expected that relative fluorescence intensity levels under conditions below are $\\mathrm{C}>\\mathrm{B}>\\mathrm{A}$  A: QD with MBP-5His and $\\beta-\\mathrm{CD}-\\mathrm{QSY} 9$ dye  B: QD with MBP-5His and free QSY9 dye  $\\mathrm{C}$ : QD with MBP-5His\nD:  QD incubated with MBP that was preloaded with $\\beta-\\mathrm{CD}-\\mathrm{QSY} 9$ will not show FRET.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-19.jpg?height=543&width=688&top_left_y=979&top_left_x=684",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-19.jpg?height=353&width=874&top_left_y=1774&top_left_x=591",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-20.jpg?height=500&width=671&top_left_y=298&top_left_x=378",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-20.jpg?height=499&width=602&top_left_y=296&top_left_x=1061"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1067",
        "problem": "In an experiment, a researcher incubated plant tissue with a bacterial strain for different time durations followed by heat treatment to specifically kill the bacteria. The plant material was analyzed ten days after each experiment for appearance of disease symptoms. Results of this study, which was done in three replicates and with appropriate controls, are tabulated below.\n\n| Duration of bacterial incubation | Heat treatment | Disease symptom |\n| :--- | :---: | :--- |\n| 0 (minus infection) | + | No |\n| 0 (minus infection) | - | No |\n| 12 hours | + | No |\n| 12 hours | - | Yes |\n| 24 hours | + | No |\n| 24 hours | - | Yes |\n| 48 hours | + | No |\n| 48 hours | - | Yes |\n| 72 hours | + | Yes |\n| 72 hours | - | Yes |\n\nWhich of the following statement/s is/are likely explanation/s of the above observations?\n\n(i) The bacterium is incapable of causing infection in the absence of heat treatment.\n\n(ii) Continuous contact with live bacteria is not essential to cause disease symptoms.\n\n(iii) The bacterium undergoes spontaneous mutations to become heat-resistant.\n\n(iv) Contact duration of 72 hours with live bacteria is essential to cause disease symptoms.\nA: ii and iv only\nB: i, iii and iv only\nC: ii only\nD: i and iii only\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn an experiment, a researcher incubated plant tissue with a bacterial strain for different time durations followed by heat treatment to specifically kill the bacteria. The plant material was analyzed ten days after each experiment for appearance of disease symptoms. Results of this study, which was done in three replicates and with appropriate controls, are tabulated below.\n\n| Duration of bacterial incubation | Heat treatment | Disease symptom |\n| :--- | :---: | :--- |\n| 0 (minus infection) | + | No |\n| 0 (minus infection) | - | No |\n| 12 hours | + | No |\n| 12 hours | - | Yes |\n| 24 hours | + | No |\n| 24 hours | - | Yes |\n| 48 hours | + | No |\n| 48 hours | - | Yes |\n| 72 hours | + | Yes |\n| 72 hours | - | Yes |\n\nWhich of the following statement/s is/are likely explanation/s of the above observations?\n\n(i) The bacterium is incapable of causing infection in the absence of heat treatment.\n\n(ii) Continuous contact with live bacteria is not essential to cause disease symptoms.\n\n(iii) The bacterium undergoes spontaneous mutations to become heat-resistant.\n\n(iv) Contact duration of 72 hours with live bacteria is essential to cause disease symptoms.\n\nA: ii and iv only\nB: i, iii and iv only\nC: ii only\nD: i and iii only\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_365",
        "problem": "某科研团队开发出了无细胞蛋白质合成体系, 发现其中存在一些小 RNA, 为研究其功能进行了实验。实验一: 将大量的 ${ }^{14} \\mathrm{C}$ 亮氨酸加入到一种无细胞体系中，一段时间后分离出小 RNA 并进行检测, 结果如图甲所示; 实验二: 将 ${ }^{14} \\mathrm{C}$ 亮氨酸一小 RNA 复合物与附着有核糖体的内质网提取物混合，在不同时间检测分离出的新合成的蛋白质和小 RNA 的放射性，结果如图乙所示。下列有关说法正确的是（ ）\n[图1]\nA: 实验一中使用的无细胞体系中应含有核糖体和 mRNA\nB: 两个实验中均存在亮氨酸与小 RNA 结合和分离的过程\nC: 与亮氨酸结合的小 RNA 中含有亮氨酸的密码子\nD: 根据实验结果推测, 无细胞蛋白质合成体系的小 RNA 应为 tRNA\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某科研团队开发出了无细胞蛋白质合成体系, 发现其中存在一些小 RNA, 为研究其功能进行了实验。实验一: 将大量的 ${ }^{14} \\mathrm{C}$ 亮氨酸加入到一种无细胞体系中，一段时间后分离出小 RNA 并进行检测, 结果如图甲所示; 实验二: 将 ${ }^{14} \\mathrm{C}$ 亮氨酸一小 RNA 复合物与附着有核糖体的内质网提取物混合，在不同时间检测分离出的新合成的蛋白质和小 RNA 的放射性，结果如图乙所示。下列有关说法正确的是（ ）\n[图1]\n\nA: 实验一中使用的无细胞体系中应含有核糖体和 mRNA\nB: 两个实验中均存在亮氨酸与小 RNA 结合和分离的过程\nC: 与亮氨酸结合的小 RNA 中含有亮氨酸的密码子\nD: 根据实验结果推测, 无细胞蛋白质合成体系的小 RNA 应为 tRNA\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-19.jpg?height=630&width=1352&top_left_y=160&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_767",
        "problem": "某精原细胞同源染色体中的一条发生倒位, 如图甲。减数分裂过程中, 由于染色体倒位, 同源染色体联会时会形成倒位环, 此时经常伴随同源染色体的交叉互换, 如图乙。完成分裂后, 若配子中出现染色体片段缺失, 染色体上增加某个相同片段, 则不能存活,而出现倒位的配子能存活。下列叙述正确的是（ ）\n\n[图1]\n\n(1)、(2)、(3)、(4)代表断裂位点\n\n图甲\n\n[图2]\n\nI 、II 、III IV 代表不同染色单体\n\n图乙\nA: 图甲发生了 (1)至(3)区段的倒位\nB: 图乙细胞中 II 和 III 发生交叉互换\nC: 该精原细胞减数分裂时染色体有片段缺失\nD: 该精原细胞共产生了 3 种类型的可育雄配子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某精原细胞同源染色体中的一条发生倒位, 如图甲。减数分裂过程中, 由于染色体倒位, 同源染色体联会时会形成倒位环, 此时经常伴随同源染色体的交叉互换, 如图乙。完成分裂后, 若配子中出现染色体片段缺失, 染色体上增加某个相同片段, 则不能存活,而出现倒位的配子能存活。下列叙述正确的是（ ）\n\n[图1]\n\n(1)、(2)、(3)、(4)代表断裂位点\n\n图甲\n\n[图2]\n\nI 、II 、III IV 代表不同染色单体\n\n图乙\n\nA: 图甲发生了 (1)至(3)区段的倒位\nB: 图乙细胞中 II 和 III 发生交叉互换\nC: 该精原细胞减数分裂时染色体有片段缺失\nD: 该精原细胞共产生了 3 种类型的可育雄配子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-72.jpg?height=432&width=645&top_left_y=1483&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-72.jpg?height=451&width=671&top_left_y=1465&top_left_x=978"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_32",
        "problem": "DNA helicase, a key enzyme for DNA replication, separates double-stranded DNA into single-stranded DNA. The following describes an experiment to find out the characteristics of this enzyme.\n\nA linear $6 \\mathrm{~kb}$ ssDNA was annealed with a short (300 bp) complementary ssDNA that is labeled with radioactive nucleotides (a). The annealed DNA was then treated in one of three ways: with DNA helicase, boiling without helicase, or boiled helicase. Treated DNA samples were electrophoresed on an agarose gel. The gel in $\\boldsymbol{b}$ shows the DNA bands that could be detected in the gel by autoradiography. (It is assumed that the ATP energy needed for this enzyme reaction was provided during the treatment of DNA helicase).\n\n[figure1]\n\nWhich of the following explanation about this experiment is correct?\nA: The band appearing in the top part of the gel is the $6.3 \\mathrm{~kb}$ ssDNA only.\nB: The band appearing in the lower part of the gel is the labelled 300 bp DNA.\nC: If the annealed DNA is treated only with DNA helicase and the reaction is complete, the band pattern looks like the lane 3 in $\\boldsymbol{b}$.\nD: If the annealed DNA is treated only with the boiling without helicase treatment, the band pattern will look like lane 2 in $\\boldsymbol{b}$.\nE: If the annealed DNA is treated only with boiled helicase, the band pattern will look like lane 1 in $\\boldsymbol{b}$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDNA helicase, a key enzyme for DNA replication, separates double-stranded DNA into single-stranded DNA. The following describes an experiment to find out the characteristics of this enzyme.\n\nA linear $6 \\mathrm{~kb}$ ssDNA was annealed with a short (300 bp) complementary ssDNA that is labeled with radioactive nucleotides (a). The annealed DNA was then treated in one of three ways: with DNA helicase, boiling without helicase, or boiled helicase. Treated DNA samples were electrophoresed on an agarose gel. The gel in $\\boldsymbol{b}$ shows the DNA bands that could be detected in the gel by autoradiography. (It is assumed that the ATP energy needed for this enzyme reaction was provided during the treatment of DNA helicase).\n\n[figure1]\n\nWhich of the following explanation about this experiment is correct?\n\nA: The band appearing in the top part of the gel is the $6.3 \\mathrm{~kb}$ ssDNA only.\nB: The band appearing in the lower part of the gel is the labelled 300 bp DNA.\nC: If the annealed DNA is treated only with DNA helicase and the reaction is complete, the band pattern looks like the lane 3 in $\\boldsymbol{b}$.\nD: If the annealed DNA is treated only with the boiling without helicase treatment, the band pattern will look like lane 2 in $\\boldsymbol{b}$.\nE: If the annealed DNA is treated only with boiled helicase, the band pattern will look like lane 1 in $\\boldsymbol{b}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-12.jpg?height=462&width=1168&top_left_y=1188&top_left_x=524"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_531",
        "problem": "豌豆的花色和花的位置分别由基因 $A 、 a$ 和 $B 、 b$ 控制, 基因型为 $A a B b$ 的豌豆植株\n自交获得的子代表型及比例是红花顶生：白花顶生：红花腋生：白花腋生=9：3：3：1。下列说法正确的是（ ）\nA: 若子代中红花顶生个体自交, 则产生纯合红花顶生个体的概率为 $1 / 3$\nB: 若子代中的白花顶生个体经处理可进行自由交配, 则产生顶生花的概率为 $1 / 9$\nC: 让子代白花顶生与红花腋生杂交, 后代中红花顶生的概率为 $4 / 9$\nD: 若 $A 、 a$ 和 $B 、 b$ 位于同一对同源染色体上, 则后代不会出现四种表现型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n豌豆的花色和花的位置分别由基因 $A 、 a$ 和 $B 、 b$ 控制, 基因型为 $A a B b$ 的豌豆植株\n自交获得的子代表型及比例是红花顶生：白花顶生：红花腋生：白花腋生=9：3：3：1。下列说法正确的是（ ）\n\nA: 若子代中红花顶生个体自交, 则产生纯合红花顶生个体的概率为 $1 / 3$\nB: 若子代中的白花顶生个体经处理可进行自由交配, 则产生顶生花的概率为 $1 / 9$\nC: 让子代白花顶生与红花腋生杂交, 后代中红花顶生的概率为 $4 / 9$\nD: 若 $A 、 a$ 和 $B 、 b$ 位于同一对同源染色体上, 则后代不会出现四种表现型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1349",
        "problem": "Increasing the supply of food from a limited area of land could best be achieved by increasing the\nA: gross production of the producers\nB: net production of the producers\nC: rate of energy turnover of the producers\nD: net production of the primary consumers\nE: density of the primary consumers\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIncreasing the supply of food from a limited area of land could best be achieved by increasing the\n\nA: gross production of the producers\nB: net production of the producers\nC: rate of energy turnover of the producers\nD: net production of the primary consumers\nE: density of the primary consumers\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1052",
        "problem": "The following pedigree depicts the transmission of myopia (near sightedness) in a family.\n\n[figure1]\n\nWhat is the probability that individual III 2 will have myopia? Indicate the answer as fraction.",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe following pedigree depicts the transmission of myopia (near sightedness) in a family.\n\n[figure1]\n\nWhat is the probability that individual III 2 will have myopia? Indicate the answer as fraction.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-26.jpg?height=455&width=1390&top_left_y=778&top_left_x=384"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_287",
        "problem": "Self-incompatibility (SI) in flowering plants is the most common mechanism preventing self-pollination, which is mediated by a single $S$ locus with multiple alleles. In gametophytic self-incompatibility (GSI), the incompatibility of pollen is determined by the haploid pollen genotype at the $\\boldsymbol{S}$ locus. In sporophytic self-incompatibility (SSI), the incompatibility is determined by the diploid $S$ genotype of the parent pollen wall. The table below shows the SI type and pollen/style $\\boldsymbol{S}$-gene genotypes of two plants crossed for fertilization. $\\boldsymbol{S}_{1}$ and $\\boldsymbol{S}_{2}$ alleles are codominant in pollen wall.\n\n|  | SI type | Expressed genotype |  |\n| :---: | :---: | :---: | :---: |\n|  |  | Pollen of plant 1 | Style of plant 2 |\n| I | GSI | $S_{1}$ or $S_{2}$ | $S_{2} S_{3}$ |\n| II | GSI | $S_{2}$ or $S_{3}$ | $S_{2} S_{3}$ |\n| III | SSI | $S_{1}$ or $S_{2}$ | $S_{1} S_{3}$ |\n| IV | SSI | $S_{1}$ or $S_{2}$ | $S_{3} S_{4}$ |\n\nWhich of these crosses (I, II, III, and IV) result in successful fertilizations?\nA: I and II\nB: I and III\nC: I and IV\nD: II and III\nE: II and IV\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSelf-incompatibility (SI) in flowering plants is the most common mechanism preventing self-pollination, which is mediated by a single $S$ locus with multiple alleles. In gametophytic self-incompatibility (GSI), the incompatibility of pollen is determined by the haploid pollen genotype at the $\\boldsymbol{S}$ locus. In sporophytic self-incompatibility (SSI), the incompatibility is determined by the diploid $S$ genotype of the parent pollen wall. The table below shows the SI type and pollen/style $\\boldsymbol{S}$-gene genotypes of two plants crossed for fertilization. $\\boldsymbol{S}_{1}$ and $\\boldsymbol{S}_{2}$ alleles are codominant in pollen wall.\n\n|  | SI type | Expressed genotype |  |\n| :---: | :---: | :---: | :---: |\n|  |  | Pollen of plant 1 | Style of plant 2 |\n| I | GSI | $S_{1}$ or $S_{2}$ | $S_{2} S_{3}$ |\n| II | GSI | $S_{2}$ or $S_{3}$ | $S_{2} S_{3}$ |\n| III | SSI | $S_{1}$ or $S_{2}$ | $S_{1} S_{3}$ |\n| IV | SSI | $S_{1}$ or $S_{2}$ | $S_{3} S_{4}$ |\n\nWhich of these crosses (I, II, III, and IV) result in successful fertilizations?\n\nA: I and II\nB: I and III\nC: I and IV\nD: II and III\nE: II and IV\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1287",
        "problem": "The two main factors taken into account when measuring biological diversity are richness and evenness. Richness is a measure of the number of different kinds of organisms present in a particular area. Evenness is a measure of the relative abundance of the different species making up the richness of an area. As species richness and evenness increase, so diversity increases. Simpson's Diversity Index (D) is a measure of diversity which takes into account both richness and evenness. Its formula is:\n\n[figure1]\n\nNote: $\\sum$ means to sum.\n\nThe value of $D$ ranges between 0 and 1 with 0 representing infinite diversity and 1 , no diversity. That is, the larger the value of $D$, the lower the diversity. This is neither intuitive nor logical, so to get over this problem, $D$ is often subtracted from 1 to give Simpson's Index of Diversity 1 - D. The value of this index also ranges between 0 and 1 , but now, the greater the value, the greater the sample diversity.\n\nCalculate Simpson's Index of Diversity for a single transect sample of canopy trees in native bush near Rotorua.\n\n| Species | Number (n) | n(n-1) |\n| :---: | :---: | :---: |\n| Rimu | 2 |  |\n| Rewarewa | 8 |  |\n| Rata | 1 |  |\n| Tawa | 1 | $\\mathbf{6 4}$ |\n| Tree fern | 3 |  |\n| Total $(\\mathbf{N})$ | $\\mathbf{1 5}$ |  |\nA: 0.3\nB: 3.0\nC: 0.7\nD: 7.0\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe two main factors taken into account when measuring biological diversity are richness and evenness. Richness is a measure of the number of different kinds of organisms present in a particular area. Evenness is a measure of the relative abundance of the different species making up the richness of an area. As species richness and evenness increase, so diversity increases. Simpson's Diversity Index (D) is a measure of diversity which takes into account both richness and evenness. Its formula is:\n\n[figure1]\n\nNote: $\\sum$ means to sum.\n\nThe value of $D$ ranges between 0 and 1 with 0 representing infinite diversity and 1 , no diversity. That is, the larger the value of $D$, the lower the diversity. This is neither intuitive nor logical, so to get over this problem, $D$ is often subtracted from 1 to give Simpson's Index of Diversity 1 - D. The value of this index also ranges between 0 and 1 , but now, the greater the value, the greater the sample diversity.\n\nCalculate Simpson's Index of Diversity for a single transect sample of canopy trees in native bush near Rotorua.\n\n| Species | Number (n) | n(n-1) |\n| :---: | :---: | :---: |\n| Rimu | 2 |  |\n| Rewarewa | 8 |  |\n| Rata | 1 |  |\n| Tawa | 1 | $\\mathbf{6 4}$ |\n| Tree fern | 3 |  |\n| Total $(\\mathbf{N})$ | $\\mathbf{1 5}$ |  |\n\nA: 0.3\nB: 3.0\nC: 0.7\nD: 7.0\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-08.jpg?height=180&width=425&top_left_y=487&top_left_x=170"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1468",
        "problem": "Auxin (IAA) is a hormone produced primarily in shoot apical meristems and young leaves. Root apical meristems also produce some auxin, although the root depends on the shoot for much of its auxin. Amongst its many functions in plant growth, auxin plays an important role in gravitropism.\n\nFigure A shows the anatomy of a root. Figure B shows the relationship between auxin concentration and growth of roots and shoots.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\nWhich of the statement/s is/are true?\n\nI. At low concentrations, auxin stimulates the root cap to elongate.\n\nII. The effects of auxin are confined to the regions in which it is synthesised.\n\nIII. Auxin stimulates gravitropism in root hairs.\n\nIV. Auxin is transported through the plant in an active, cell-to-cell manner.\nA: I only\nB: I and II\nC: II and III\nD: I, III and IV\nE: IV only\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAuxin (IAA) is a hormone produced primarily in shoot apical meristems and young leaves. Root apical meristems also produce some auxin, although the root depends on the shoot for much of its auxin. Amongst its many functions in plant growth, auxin plays an important role in gravitropism.\n\nFigure A shows the anatomy of a root. Figure B shows the relationship between auxin concentration and growth of roots and shoots.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\nWhich of the statement/s is/are true?\n\nI. At low concentrations, auxin stimulates the root cap to elongate.\n\nII. The effects of auxin are confined to the regions in which it is synthesised.\n\nIII. Auxin stimulates gravitropism in root hairs.\n\nIV. Auxin is transported through the plant in an active, cell-to-cell manner.\n\nA: I only\nB: I and II\nC: II and III\nD: I, III and IV\nE: IV only\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-17.jpg?height=731&width=599&top_left_y=700&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-17.jpg?height=671&width=683&top_left_y=767&top_left_x=1052"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1452",
        "problem": "The $K_{M}$ of an enzymatic reaction is defined as the substrate concentration where there is enzymatic activity is half that of $\\mathrm{Vmax}$ (maximum enzyme activity). Which of the following statements is correct?\nA: A high $K_{M}$ is indicative of high enzyme efficiency.\nB: A low $K_{M}$ is indicative of a high enzyme efficiency.\nC: $K_{M}$ is calculated by doubling Vmax.\nD: $K_{M}$ is calculated by halving Vmax.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe $K_{M}$ of an enzymatic reaction is defined as the substrate concentration where there is enzymatic activity is half that of $\\mathrm{Vmax}$ (maximum enzyme activity). Which of the following statements is correct?\n\nA: A high $K_{M}$ is indicative of high enzyme efficiency.\nB: A low $K_{M}$ is indicative of a high enzyme efficiency.\nC: $K_{M}$ is calculated by doubling Vmax.\nD: $K_{M}$ is calculated by halving Vmax.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1083",
        "problem": "Apoptosis is programmed cell death characterized by the formation of apoptotic bodies whereas necrosis is characterized by bursting of cells. Which of the following condition/s lead/s to necrosis?\n\ni. Gaps and nicks formed in DNA during replication cannot be repaired.\n\nii. Developing neurons fail to make synapses with neighbouring cells.\n\niii. Heart muscle cells damaged by oxygen depletion followed by cardiac infarction.\nA: i and ii\nB: ii and iii\nC: i only\nD: iii only\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nApoptosis is programmed cell death characterized by the formation of apoptotic bodies whereas necrosis is characterized by bursting of cells. Which of the following condition/s lead/s to necrosis?\n\ni. Gaps and nicks formed in DNA during replication cannot be repaired.\n\nii. Developing neurons fail to make synapses with neighbouring cells.\n\niii. Heart muscle cells damaged by oxygen depletion followed by cardiac infarction.\n\nA: i and ii\nB: ii and iii\nC: i only\nD: iii only\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1505",
        "problem": "The way groups of organisms live depends on fundamental aspects of their biology. For each of these characteristics, state whether they belong to plants, animals, fungi and/or bacteria.\n\nIncludes many autotrophs (producers).\nA: Plants, animals, fungi\nB: Plants, bacteria, fungi\nC: Animals, fungi, bacteria\nD: Plants, animals, bacteria\nE: Plants, animals, bacteria, fungi\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe way groups of organisms live depends on fundamental aspects of their biology. For each of these characteristics, state whether they belong to plants, animals, fungi and/or bacteria.\n\nIncludes many autotrophs (producers).\n\nA: Plants, animals, fungi\nB: Plants, bacteria, fungi\nC: Animals, fungi, bacteria\nD: Plants, animals, bacteria\nE: Plants, animals, bacteria, fungi\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1245",
        "problem": "Evolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).Considering these data what is the best conclusion/s?\nA: Vicariant speciation provides a poor explanation of the close relationship between elephant birds and kiwi.\nB: Madagascar and New Zealand have been directly connected in the geological past.\nC: Kiwi and elephant birds diverged after the breakup of Gondwana.\nD: A and B only.\nE: A and C only.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nEvolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).\n\nproblem:\nConsidering these data what is the best conclusion/s?\n\nA: Vicariant speciation provides a poor explanation of the close relationship between elephant birds and kiwi.\nB: Madagascar and New Zealand have been directly connected in the geological past.\nC: Kiwi and elephant birds diverged after the breakup of Gondwana.\nD: A and B only.\nE: A and C only.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-28.jpg?height=1868&width=1627&top_left_y=748&top_left_x=206"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_29",
        "problem": "Chloroplast symbiosis or kleptoplasty refers to a naturally occurring process that results in maintenance of chloroplasts of one species in host cells of another species. This process has occurred in sacoglossan sea slugs (Figure 1: Pierce and Curtis, 2012). Cells that line the digestive diverticula of the slugs maintain chloroplasts that originated from algal food. Sea slug species with these chloroplasts are able to tolerate starvation for periods up to one year. It has been shown that some sea slugs are able to retain functional plastids for a year or more, or even to the end of their life.\n\nSeveral hypotheses were tested to find the mechanisms behind this kind of symbiosis. All earlier attempts to locate algal nuclei in such sea slugs failed. However, recent studies have demonstrated presence of Vaucheria litorea algal nuclear genes in genomic DNA of the larvae of the slug Elysia chlorotica which have never fed on $V$. litorea. Note that use of animal proteins by chloroplasts seems implausible.\n\n[figure1]\n\nA: E. chlorotica\n\n[figure2]\n\nB: E. clarki\nA: Horizontal gene transfer might have resulted in transfer of certain genes of the harvested chloroplast into the genome of some slugs, particularly in slugs that\nB: Activity of lysosomal enzymes in host cells must be minimal in order to prevent chloroplast ingestion.\nC: The sea slugs with algal chloroplasts also need algal mitochondria to ensure a source of energy during periods of starvation.\nD: Nuclear algal RNA present in ingested chloroplasts could somehow persist for as long as one year in the sea slug.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nChloroplast symbiosis or kleptoplasty refers to a naturally occurring process that results in maintenance of chloroplasts of one species in host cells of another species. This process has occurred in sacoglossan sea slugs (Figure 1: Pierce and Curtis, 2012). Cells that line the digestive diverticula of the slugs maintain chloroplasts that originated from algal food. Sea slug species with these chloroplasts are able to tolerate starvation for periods up to one year. It has been shown that some sea slugs are able to retain functional plastids for a year or more, or even to the end of their life.\n\nSeveral hypotheses were tested to find the mechanisms behind this kind of symbiosis. All earlier attempts to locate algal nuclei in such sea slugs failed. However, recent studies have demonstrated presence of Vaucheria litorea algal nuclear genes in genomic DNA of the larvae of the slug Elysia chlorotica which have never fed on $V$. litorea. Note that use of animal proteins by chloroplasts seems implausible.\n\n[figure1]\n\nA: E. chlorotica\n\n[figure2]\n\nB: E. clarki\n\nA: Horizontal gene transfer might have resulted in transfer of certain genes of the harvested chloroplast into the genome of some slugs, particularly in slugs that\nB: Activity of lysosomal enzymes in host cells must be minimal in order to prevent chloroplast ingestion.\nC: The sea slugs with algal chloroplasts also need algal mitochondria to ensure a source of energy during periods of starvation.\nD: Nuclear algal RNA present in ingested chloroplasts could somehow persist for as long as one year in the sea slug.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-39.jpg?height=314&width=371&top_left_y=914&top_left_x=640",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-39.jpg?height=314&width=373&top_left_y=914&top_left_x=1047"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_290",
        "problem": "The action potential of cardiac muscle cells differs from that of other cells such as skeletal muscle cells and neurons. Figure Q. 70 displays the different phases of action potential in cardiac muscle cells.\n\n[figure1]\n\nFigure Q. 71\nA: A substance that inhibits the reuptake of $\\mathrm{Ca}^{2+}$ into the sarcoplasmic reticulum increases the time interval from 3 to 4 .\nB: The concentration of $\\mathrm{K}^{+}$in the sarcoplasm at position 2 is higher than that at position 3.\nC: Injection of adrenaline decreases the time interval from 1 to 5 .\nD: The height of membrane potential (from 1 to 2 ) is decreased when the sarcoplasmic level of $\\mathrm{Na}^{+}$is higher than the normal level.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe action potential of cardiac muscle cells differs from that of other cells such as skeletal muscle cells and neurons. Figure Q. 70 displays the different phases of action potential in cardiac muscle cells.\n\n[figure1]\n\nFigure Q. 71\n\nA: A substance that inhibits the reuptake of $\\mathrm{Ca}^{2+}$ into the sarcoplasmic reticulum increases the time interval from 3 to 4 .\nB: The concentration of $\\mathrm{K}^{+}$in the sarcoplasm at position 2 is higher than that at position 3.\nC: Injection of adrenaline decreases the time interval from 1 to 5 .\nD: The height of membrane potential (from 1 to 2 ) is decreased when the sarcoplasmic level of $\\mathrm{Na}^{+}$is higher than the normal level.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-052.jpg?height=829&width=1259&top_left_y=634&top_left_x=407"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1543",
        "problem": "At a temple in Bali, macaques steal things from tourists and barter them for food. The macaques steal high-end items like smartphones more often than easily-accessible items like hats. The macaques then wait until the tourist offers them fruit. The macaques only return the most valuable items when a large amount of fruit has been offered, whereas cheaper items are returned for less fruit.\n\n[figure1]\n\nHow do the macaques know when to accept an offer of fruit instead of holding out in the hope of more?\nA: Reasoning\nB: Instinct\nC: Association/conditioning\nD: Habituation\nE: Imprinting\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAt a temple in Bali, macaques steal things from tourists and barter them for food. The macaques steal high-end items like smartphones more often than easily-accessible items like hats. The macaques then wait until the tourist offers them fruit. The macaques only return the most valuable items when a large amount of fruit has been offered, whereas cheaper items are returned for less fruit.\n\n[figure1]\n\nHow do the macaques know when to accept an offer of fruit instead of holding out in the hope of more?\n\nA: Reasoning\nB: Instinct\nC: Association/conditioning\nD: Habituation\nE: Imprinting\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-15.jpg?height=1397&width=945&top_left_y=555&top_left_x=227"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1251",
        "problem": "Conservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.What proportion of the Maui's dolphin population sampled in 2010 and 2011 is female?\nA: $45.5 \\%$\nB: $41.8 \\%$\nC: $59.5 \\%$\nD: $54.7 \\%$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nConservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.\n\nproblem:\nWhat proportion of the Maui's dolphin population sampled in 2010 and 2011 is female?\n\nA: $45.5 \\%$\nB: $41.8 \\%$\nC: $59.5 \\%$\nD: $54.7 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_686",
        "problem": "一个基因型为 AaXBY 的精原细胞, 用苂光分子标记基因 $\\mathrm{A} 、 \\mathrm{a}$ 和 $\\mathrm{B}$, 分裂过程中不同时期的 4 个细胞中染色体组数和核 DNA 数如图所示。不考虑基因突变、交叉互换和染色体畸变，下列叙述正确的是（ ）\n\n[图1]\nA: 甲细胞中一定无同源染色体, 细胞内至少有 4 个苂光标记\nB: 乙细胞中一定有同源染色体, 染色体数: 核 DNA 分子数 $=1: 1$\nC: 丙细胞中一定有姐妹染色单体, 可能存在四分体\nD: 丁细胞一定处于有丝分裂, 细胞内有 4 条 Y 染色体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一个基因型为 AaXBY 的精原细胞, 用苂光分子标记基因 $\\mathrm{A} 、 \\mathrm{a}$ 和 $\\mathrm{B}$, 分裂过程中不同时期的 4 个细胞中染色体组数和核 DNA 数如图所示。不考虑基因突变、交叉互换和染色体畸变，下列叙述正确的是（ ）\n\n[图1]\n\nA: 甲细胞中一定无同源染色体, 细胞内至少有 4 个苂光标记\nB: 乙细胞中一定有同源染色体, 染色体数: 核 DNA 分子数 $=1: 1$\nC: 丙细胞中一定有姐妹染色单体, 可能存在四分体\nD: 丁细胞一定处于有丝分裂, 细胞内有 4 条 Y 染色体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-71.jpg?height=643&width=756&top_left_y=158&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_862",
        "problem": "有一果蝇品系, 其一种突变体的 $\\mathrm{X}$ 染色体上存在 CIB 区段 (用 $\\mathrm{X}^{\\mathrm{CIB}}$ 表示)。B 基因表现显性棒眼性状; I 基因的纯合子在胚胎期死亡 ( $\\mathrm{X}^{\\mathrm{CIB}} \\mathrm{X}^{\\mathrm{CIB}}$ 与 $\\mathrm{X}^{\\mathrm{CIB}} \\mathrm{Y}$ 不能存活); CIB 存在时, X 染色体间非姐妹染色单体不发生交换, 正常果蝇 X 染色体无 CIB 区段 (用 $X^{+}$表示)。果蝇的长翅 ( $\\mathrm{Vg})$ 对残翅 ( $\\mathrm{vg}$ ) 为显性, 基因位于常染色体上。下图是一对果蝇的杂交实验示意图。下列相关叙述正确的是（）\n\n[图1]\n\n棒眼长翅\n\n[图2]\n\n○\n\n[图3]\n\nor\n\n[图4]\n\n正常眼长翅\n\n## $F_{1}$ 棒眼长翅 棒眼残翅 正常眼长翅 正常眼残翅\nA: $F_{1}$ 中残翅雄果蝇:长翅雄果蝇 $=1: 2$\nB: $\\mathrm{F}_{1}$ 中棒眼雌果蝇:正常眼雌果蝇 $=3: 1$\nC: $F_{1}$ 中长翅果蝇自由交配, 产生残翅雄果蝇的概率是 $1 / 9$\nD: $F_{1}$ 中果蝇自由交配, 产生棒眼雌果蝇的概率是 $1 / 7$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n有一果蝇品系, 其一种突变体的 $\\mathrm{X}$ 染色体上存在 CIB 区段 (用 $\\mathrm{X}^{\\mathrm{CIB}}$ 表示)。B 基因表现显性棒眼性状; I 基因的纯合子在胚胎期死亡 ( $\\mathrm{X}^{\\mathrm{CIB}} \\mathrm{X}^{\\mathrm{CIB}}$ 与 $\\mathrm{X}^{\\mathrm{CIB}} \\mathrm{Y}$ 不能存活); CIB 存在时, X 染色体间非姐妹染色单体不发生交换, 正常果蝇 X 染色体无 CIB 区段 (用 $X^{+}$表示)。果蝇的长翅 ( $\\mathrm{Vg})$ 对残翅 ( $\\mathrm{vg}$ ) 为显性, 基因位于常染色体上。下图是一对果蝇的杂交实验示意图。下列相关叙述正确的是（）\n\n[图1]\n\n棒眼长翅\n\n[图2]\n\n○\n\n[图3]\n\nor\n\n[图4]\n\n正常眼长翅\n\n## $F_{1}$ 棒眼长翅 棒眼残翅 正常眼长翅 正常眼残翅\n\nA: $F_{1}$ 中残翅雄果蝇:长翅雄果蝇 $=1: 2$\nB: $\\mathrm{F}_{1}$ 中棒眼雌果蝇:正常眼雌果蝇 $=3: 1$\nC: $F_{1}$ 中长翅果蝇自由交配, 产生残翅雄果蝇的概率是 $1 / 9$\nD: $F_{1}$ 中果蝇自由交配, 产生棒眼雌果蝇的概率是 $1 / 7$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-72.jpg?height=211&width=323&top_left_y=1145&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-72.jpg?height=305&width=329&top_left_y=1144&top_left_x=652",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-72.jpg?height=272&width=214&top_left_y=1166&top_left_x=978",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-72.jpg?height=208&width=215&top_left_y=1141&top_left_x=1206",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-72.jpg?height=54&width=1393&top_left_y=2309&top_left_x=346"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_342",
        "problem": "下列关于病毒的叙述, 错误的是( )\nA: 从烟草花叶病毒中可以提取到 RNA\nB: $\\mathrm{T}_{2}$ 噬菌体可感染肺炎双球菌导致其裂解\nC: HIV 可引起人的获得性免疫缺陷综合征\nD: 阻断病毒的传播可降低其所致疾病的发病率\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列关于病毒的叙述, 错误的是( )\n\nA: 从烟草花叶病毒中可以提取到 RNA\nB: $\\mathrm{T}_{2}$ 噬菌体可感染肺炎双球菌导致其裂解\nC: HIV 可引起人的获得性免疫缺陷综合征\nD: 阻断病毒的传播可降低其所致疾病的发病率\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_166",
        "problem": "The diagram on the left depicts the different areas of the spinal cord from the cervical region to the coccyx. The statements on the right provide descriptions about the spinal cord.\n\n[figure1]\n\nSuppose that an athlete injures the left half of the spinal cord T4 during a football game.\n\nChoose the correct statement concerning this patient's sensory or motor function.\nA: Abnormal touch sensation in the right foot.\nB: Disability in the movement of the right leg.\nC: Normal pain perception in the left leg.\nD: No cutaneous sensation in the left hand.\nE: Normal pain perception in the right leg.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram on the left depicts the different areas of the spinal cord from the cervical region to the coccyx. The statements on the right provide descriptions about the spinal cord.\n\n[figure1]\n\nSuppose that an athlete injures the left half of the spinal cord T4 during a football game.\n\nChoose the correct statement concerning this patient's sensory or motor function.\n\nA: Abnormal touch sensation in the right foot.\nB: Disability in the movement of the right leg.\nC: Normal pain perception in the left leg.\nD: No cutaneous sensation in the left hand.\nE: Normal pain perception in the right leg.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-31.jpg?height=1362&width=1584&top_left_y=470&top_left_x=293"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_679",
        "problem": "马蛔虫 $(2 n=4)$ 常用于研究细胞的增殖, 如图是基因型为 $A a X^{B} Y$ 的马蛔虫精巢中两个细胞的染色体组成和基因分布情况，其中一个细胞处于着丝粒整齐排列在赤道板上的时期。下列叙述错误的是（ ）\n\n[图1]\nA: 不考虑其他变异情况, 形成乙细胞的精原细胞完成分裂过程后, 可产生一半的正常配子\nB: 该个体的性腺适合作观察减数分裂的材料, 在精巢观察到的细胞为 1 个或 2 个染色体组\nC: 形成乙细胞的过程中至少发生了两种可遗传变异\nD: 甲乙细胞正常分裂都能形成两种基因型的子细胞\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n马蛔虫 $(2 n=4)$ 常用于研究细胞的增殖, 如图是基因型为 $A a X^{B} Y$ 的马蛔虫精巢中两个细胞的染色体组成和基因分布情况，其中一个细胞处于着丝粒整齐排列在赤道板上的时期。下列叙述错误的是（ ）\n\n[图1]\n\nA: 不考虑其他变异情况, 形成乙细胞的精原细胞完成分裂过程后, 可产生一半的正常配子\nB: 该个体的性腺适合作观察减数分裂的材料, 在精巢观察到的细胞为 1 个或 2 个染色体组\nC: 形成乙细胞的过程中至少发生了两种可遗传变异\nD: 甲乙细胞正常分裂都能形成两种基因型的子细胞\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-34.jpg?height=386&width=622&top_left_y=478&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_832",
        "problem": "细胞内部与外部通过细胞膜进行营养物质和代谢产物的交换, 以保障细胞内部环境相对稳定。当细胞内部环境恶化到一定程度时会触发细胞分裂、调亡过程, 其关系如图所示 (横坐标表示细胞内部环境, 范围 $0 \\sim 1,0$ 为最差, 1 为最好)。下列相关叙述错误的是 ( )\n\n[图1]\nA: 随细胞分裂的进行, 细胞内部环境的恶化程度增大\nB: 细胞分裂、凋亡是内部环境恶化程度决定的被动过程\nC: 若某细胞发生突变, 使 B 点右移、A 点左移则该细胞可能变成癌细胞\nD: 若细胞内部环境经常处于略低于 B 的状态, 则细胞更容易发生癌变\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n细胞内部与外部通过细胞膜进行营养物质和代谢产物的交换, 以保障细胞内部环境相对稳定。当细胞内部环境恶化到一定程度时会触发细胞分裂、调亡过程, 其关系如图所示 (横坐标表示细胞内部环境, 范围 $0 \\sim 1,0$ 为最差, 1 为最好)。下列相关叙述错误的是 ( )\n\n[图1]\n\nA: 随细胞分裂的进行, 细胞内部环境的恶化程度增大\nB: 细胞分裂、凋亡是内部环境恶化程度决定的被动过程\nC: 若某细胞发生突变, 使 B 点右移、A 点左移则该细胞可能变成癌细胞\nD: 若细胞内部环境经常处于略低于 B 的状态, 则细胞更容易发生癌变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-001.jpg?height=434&width=740&top_left_y=800&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1005",
        "problem": "The food source of Rhizopus stolonifer (black bread mold, a zygomycete) has been depleted. Which of these is most likely to happen?\nA: Zygosporangia form.\nB: Mycelia increase the rate of asexual reproduction.\nC: Mycelia produce conidia (pigmented haploid spores).\nD: The Rhiropus forms mycorrhizal interactions with root plants.\nE: The Rhizopus forms flagellated zoospores to spread via aquatic transportation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe food source of Rhizopus stolonifer (black bread mold, a zygomycete) has been depleted. Which of these is most likely to happen?\n\nA: Zygosporangia form.\nB: Mycelia increase the rate of asexual reproduction.\nC: Mycelia produce conidia (pigmented haploid spores).\nD: The Rhiropus forms mycorrhizal interactions with root plants.\nE: The Rhizopus forms flagellated zoospores to spread via aquatic transportation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_300",
        "problem": "The seed germination of plants is mainly controlled by the action of two plant hormones called gibberellin and abscisic acid. Gibberellin promotes germination and abscisic acid suppresses germination. Through the actions of these two plant hormones, plant seeds are regulated so that germination is induced in an appropriate environment.\n\nIn plants, gibberellin is biosynthesized from a molecule called geranylgeranyl diphosphate. Geranylgeranyl diphosphate is converted into ent-kaurene through the action of ent-kaurene synthase. ent-Kaurene is then converted into gibberellin through the action of several enzymes such as GA20 oxidase. Biosynthetic intermediates such as ent-kaurene do not have germination-inducing activity (Figure 1).\n\nOn the other hand, abscisic acid is biosynthesized from carotenoid pigments. Abscisic acid is converted into 8'hydroxyabscisic acid by an oxidase called CYP707A. Seeds of Arabidopsis mutants lacking the gene encoding CYP707A were observed to have significantly delayed germination as compared to seeds of wild-type plants. In addition, the germination of the seeds of plants in which the CYP707A gene was overexpressed were promoted more than the wild-type seeds. In this experiment, the administered compounds play a similar function of endogenous hormones.\n\n[figure1]\nA: In the mutant lacking the ent-kaurene synthase gene, germination is delayed compared to the wild-type plants.\nB: When a mutant lacking the ent-kaurene synthase gene is treated with ent-kaurene, germination is promoted\nC: ent-Kaurene treatment to a mutant lacking the gene encoding GA20-oxidase promotes germination.\nD: 8'-Hydroxyabscisic acid has a stronger germination-inhibiting activity than abscisic acid.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe seed germination of plants is mainly controlled by the action of two plant hormones called gibberellin and abscisic acid. Gibberellin promotes germination and abscisic acid suppresses germination. Through the actions of these two plant hormones, plant seeds are regulated so that germination is induced in an appropriate environment.\n\nIn plants, gibberellin is biosynthesized from a molecule called geranylgeranyl diphosphate. Geranylgeranyl diphosphate is converted into ent-kaurene through the action of ent-kaurene synthase. ent-Kaurene is then converted into gibberellin through the action of several enzymes such as GA20 oxidase. Biosynthetic intermediates such as ent-kaurene do not have germination-inducing activity (Figure 1).\n\nOn the other hand, abscisic acid is biosynthesized from carotenoid pigments. Abscisic acid is converted into 8'hydroxyabscisic acid by an oxidase called CYP707A. Seeds of Arabidopsis mutants lacking the gene encoding CYP707A were observed to have significantly delayed germination as compared to seeds of wild-type plants. In addition, the germination of the seeds of plants in which the CYP707A gene was overexpressed were promoted more than the wild-type seeds. In this experiment, the administered compounds play a similar function of endogenous hormones.\n\n[figure1]\n\nA: In the mutant lacking the ent-kaurene synthase gene, germination is delayed compared to the wild-type plants.\nB: When a mutant lacking the ent-kaurene synthase gene is treated with ent-kaurene, germination is promoted\nC: ent-Kaurene treatment to a mutant lacking the gene encoding GA20-oxidase promotes germination.\nD: 8'-Hydroxyabscisic acid has a stronger germination-inhibiting activity than abscisic acid.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-19.jpg?height=822&width=1496&top_left_y=1442&top_left_x=197"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_928",
        "problem": "已知果蝇的眼色由位于常染色体上的等位基因 $\\mathrm{A} / \\mathrm{a}$ 和位于 $\\mathrm{X}$ 染色体上的等位基因 $\\mathrm{B} / \\mathrm{b}$ 控制, $\\mathrm{A}$ 和 $\\mathrm{B}$ 基因同时存在时表现为红眼, 含 $\\mathrm{B}$ 基因但不含 $\\mathrm{A}$ 基因时表现为粉眼,无 B 基因时表现为白眼。让白眼雄果蝇与纯合粉眼雌果蝇杂交, $F_{1}$ 中既有红眼, 也有粉眼。不考虑突变, 下列相关叙述错误的是（）\nA: 亲本中白眼雄果蝇的基因型为 $\\mathrm{AaX}^{\\mathrm{b}} \\mathrm{Y}$\nB: $\\mathrm{F}_{1}$ 中红眼果蝇的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$ 和 $\\mathrm{AaX} \\mathrm{X}^{\\mathrm{B}} \\mathrm{Y}$\nC: $F_{1}$ 中的雌雄粉眼果蝇杂交, 其后代中不可能出现白眼果蝇\nD: $F_{1}$ 中的雌雄果蝇随机交配, $F_{2}$ 雌果蝇中粉眼:红眼 $=9: 7$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知果蝇的眼色由位于常染色体上的等位基因 $\\mathrm{A} / \\mathrm{a}$ 和位于 $\\mathrm{X}$ 染色体上的等位基因 $\\mathrm{B} / \\mathrm{b}$ 控制, $\\mathrm{A}$ 和 $\\mathrm{B}$ 基因同时存在时表现为红眼, 含 $\\mathrm{B}$ 基因但不含 $\\mathrm{A}$ 基因时表现为粉眼,无 B 基因时表现为白眼。让白眼雄果蝇与纯合粉眼雌果蝇杂交, $F_{1}$ 中既有红眼, 也有粉眼。不考虑突变, 下列相关叙述错误的是（）\n\nA: 亲本中白眼雄果蝇的基因型为 $\\mathrm{AaX}^{\\mathrm{b}} \\mathrm{Y}$\nB: $\\mathrm{F}_{1}$ 中红眼果蝇的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$ 和 $\\mathrm{AaX} \\mathrm{X}^{\\mathrm{B}} \\mathrm{Y}$\nC: $F_{1}$ 中的雌雄粉眼果蝇杂交, 其后代中不可能出现白眼果蝇\nD: $F_{1}$ 中的雌雄果蝇随机交配, $F_{2}$ 雌果蝇中粉眼:红眼 $=9: 7$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_749",
        "problem": "少数蜜蜂幼虫因取食蜂王浆发育成蜂王, 大多数蜜蜂幼虫因取食花粉或者花蜜发育成工蜂。已知 DNMT3 蛋白是 DNMT3 基因表达的一种 DNA 甲基化转移酶, 能使 DNA 某些区域添加甲基基团, 进而影响基因的表达。敲除 DNMT3 基因后, 蜜蜂幼虫则发\n育成蜂王, 这与取食蜂王浆有相同的效果。下列相关叙述正确的是（ ）\nA: 蜜蜂的 DNA 甲基化会改变 DNA 的碱基序列\nB: 表观遗传中的分子修饰只发生在 DNA 上, 不能发生在其他物质上\nC: 甲基化可能通过影响 RNA 聚合酶与调控序列的结合抑制转录过程\nD: DNA 甲基化造成的表观遗传现象仅存在于生物体生长发育的特定时期\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n少数蜜蜂幼虫因取食蜂王浆发育成蜂王, 大多数蜜蜂幼虫因取食花粉或者花蜜发育成工蜂。已知 DNMT3 蛋白是 DNMT3 基因表达的一种 DNA 甲基化转移酶, 能使 DNA 某些区域添加甲基基团, 进而影响基因的表达。敲除 DNMT3 基因后, 蜜蜂幼虫则发\n育成蜂王, 这与取食蜂王浆有相同的效果。下列相关叙述正确的是（ ）\n\nA: 蜜蜂的 DNA 甲基化会改变 DNA 的碱基序列\nB: 表观遗传中的分子修饰只发生在 DNA 上, 不能发生在其他物质上\nC: 甲基化可能通过影响 RNA 聚合酶与调控序列的结合抑制转录过程\nD: DNA 甲基化造成的表观遗传现象仅存在于生物体生长发育的特定时期\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_606",
        "problem": "基因型为 $\\mathrm{Mm}$ 的某植株产生的含 $\\mathrm{m}$ 基因的花粉中, 有 $2 / 3$ 的比例是致死的。则该植株自花传粉产生的子代中，基因型为 $\\mathrm{MM} 、 \\mathrm{Mm} 、 \\mathrm{~mm}$ 的个体的数量比为（）\nA: 3: 4: 1\nB: 9: $6: 1$\nC: 3: 5:2\nD: 1:2: 1\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n基因型为 $\\mathrm{Mm}$ 的某植株产生的含 $\\mathrm{m}$ 基因的花粉中, 有 $2 / 3$ 的比例是致死的。则该植株自花传粉产生的子代中，基因型为 $\\mathrm{MM} 、 \\mathrm{Mm} 、 \\mathrm{~mm}$ 的个体的数量比为（）\n\nA: 3: 4: 1\nB: 9: $6: 1$\nC: 3: 5:2\nD: 1:2: 1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1528",
        "problem": "Norwegian lemmings are more brightly coloured than normal lemmings. They are also very violent, and often attack predators.\n\n[figure1]\n\nWhich animal has colouration performing a similar function?\nA: Tiger ![]([figure2])\nB: Anglerfish ![]([figure3])\nC: Wasp ![]([figure4])\nD: Peacock (peafowl) ![]([figure5])\nE: Cuttlefish ![]([figure6])\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nNorwegian lemmings are more brightly coloured than normal lemmings. They are also very violent, and often attack predators.\n\n[figure1]\n\nWhich animal has colouration performing a similar function?\n\nA: Tiger ![]([figure2])\nB: Anglerfish ![]([figure3])\nC: Wasp ![]([figure4])\nD: Peacock (peafowl) ![]([figure5])\nE: Cuttlefish ![]([figure6])\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-06.jpg?height=674&width=917&top_left_y=474&top_left_x=227",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-06.jpg?height=428&width=628&top_left_y=1462&top_left_x=226",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-06.jpg?height=406&width=694&top_left_y=2050&top_left_x=224",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-07.jpg?height=645&width=494&top_left_y=317&top_left_x=238",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-07.jpg?height=437&width=711&top_left_y=1049&top_left_x=227",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-07.jpg?height=423&width=734&top_left_y=1573&top_left_x=227"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_349",
        "problem": "蝴蝶的性别决定方式为 $\\mathrm{ZW}$ 型, 某种蝴蝶的眼色受 $\\mathrm{A} 、 \\mathrm{a}$ 基因控制, 口器长短受 B、 $\\mathrm{b}$ 基因控制。研究发现纯合白眼与绿眼个体正反交, 后代只有绿眼一种表型; 长口器与短口器个体正反交的结果不同, 反交得到的后代均为短口器, 不考虑基因位于 $\\mathrm{Z} 、 \\mathrm{~W}$ 同源区段的情况。下列分析错误的是（ ）\nA: 因细胞质中不存在等位基因可确定控制眼色的基因不在细胞质中\nB: 控制眼色的基因与控制口器长短的基因在遗传上遵循自由组合定律\nC: 如果正交的子代中只有雄性个体是短口器, 则亲本中母本为短口器\nD: 选择反交后代为亲本随机交配，子代雌性中基因型为 $\\mathrm{aaZ}^{\\mathrm{b}} \\mathrm{W}$ 的概率为 $1 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n蝴蝶的性别决定方式为 $\\mathrm{ZW}$ 型, 某种蝴蝶的眼色受 $\\mathrm{A} 、 \\mathrm{a}$ 基因控制, 口器长短受 B、 $\\mathrm{b}$ 基因控制。研究发现纯合白眼与绿眼个体正反交, 后代只有绿眼一种表型; 长口器与短口器个体正反交的结果不同, 反交得到的后代均为短口器, 不考虑基因位于 $\\mathrm{Z} 、 \\mathrm{~W}$ 同源区段的情况。下列分析错误的是（ ）\n\nA: 因细胞质中不存在等位基因可确定控制眼色的基因不在细胞质中\nB: 控制眼色的基因与控制口器长短的基因在遗传上遵循自由组合定律\nC: 如果正交的子代中只有雄性个体是短口器, 则亲本中母本为短口器\nD: 选择反交后代为亲本随机交配，子代雌性中基因型为 $\\mathrm{aaZ}^{\\mathrm{b}} \\mathrm{W}$ 的概率为 $1 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1402",
        "problem": "The three recognised colours of Labradors are black, chocolate and golden. These coat colours result from differences in two genetic loci which have alleles B/b and E/e respectively.\n\nSupplied is a simplified flowchart representing pigment deposition in labradors. $\\mathrm{E}$ is the allele for functional MRC1 protein, while e is a mutated allele resulting in no MRC1 protein. B is the allele for functional TYRP1 enzyme, while $b$ is a mutated allele resulting in no TYRP1 enzyme activity. If there is no pigment deposition, the labrador is considered golden.\n\n[figure1]\n\nIf a double heterozygote female black Labrador is mated with a male Labrador of the same genotype, what is the phenotypic ratio of their offspring expected to be?\nA: 9 black: 3 chocolate: 4 golden\nB: 9 black: 6 chocolate: 1 golden\nC: 12 black: 3 chocolate: 1 golden\nD: 1 black: 2 chocolate: 1 golden\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe three recognised colours of Labradors are black, chocolate and golden. These coat colours result from differences in two genetic loci which have alleles B/b and E/e respectively.\n\nSupplied is a simplified flowchart representing pigment deposition in labradors. $\\mathrm{E}$ is the allele for functional MRC1 protein, while e is a mutated allele resulting in no MRC1 protein. B is the allele for functional TYRP1 enzyme, while $b$ is a mutated allele resulting in no TYRP1 enzyme activity. If there is no pigment deposition, the labrador is considered golden.\n\n[figure1]\n\nIf a double heterozygote female black Labrador is mated with a male Labrador of the same genotype, what is the phenotypic ratio of their offspring expected to be?\n\nA: 9 black: 3 chocolate: 4 golden\nB: 9 black: 6 chocolate: 1 golden\nC: 12 black: 3 chocolate: 1 golden\nD: 1 black: 2 chocolate: 1 golden\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-19.jpg?height=780&width=926&top_left_y=1312&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_221",
        "problem": "Following is a description regarding a population of a diploid organism, species A, with a special focus on the locus $C$ that is involved in body color.\nA: Information: Species A consists of two color morphs, black and yellow, controlled by a single locus $\\mathrm{C}$ : the allele $C^{\\mathrm{B}}$ for the black type and the allele $C^{\\mathrm{Y}}$ for the yellow type. Statement: If the allele $C^{\\mathrm{B}}$ is completely dominant to the allele $C^{\\mathrm{Y}}$ and the frequency of the yellow type individuals is $9 \\%$, the genotype frequency of $C^{\\mathrm{B}} C^{\\mathrm{B}}$ is about $70 \\%$. Note that the population is assumed to be under Hardy-Weinberg equilibrium.\nB: Information: When the body colors of ten species belonging to the same genus with species A were examined, they were all yellow. Statement: In this case, the body color of the ancestral species A just after splitting from these closely related species must have been yellow under a parsimony principle.\nC: Information: A small portion of individuals in the population of species A was isolated due to diastrophism (large-scale deformation of Earth's crust) and formed a new population A'. Statement: The drastic inter-generation changes of allele frequency of locus $C$ in population $\\mathrm{A}^{\\prime}$ can be best explained by natural selection.\nD: Information: A slightly deleterious mutation with exactly the same effect on the fitness of an individual independently occurred in both the small population $\\mathrm{A}^{\\prime}$ and the larger parental population A. Statement: The fixation probability of this mutation is the same in both populations.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFollowing is a description regarding a population of a diploid organism, species A, with a special focus on the locus $C$ that is involved in body color.\n\nA: Information: Species A consists of two color morphs, black and yellow, controlled by a single locus $\\mathrm{C}$ : the allele $C^{\\mathrm{B}}$ for the black type and the allele $C^{\\mathrm{Y}}$ for the yellow type. Statement: If the allele $C^{\\mathrm{B}}$ is completely dominant to the allele $C^{\\mathrm{Y}}$ and the frequency of the yellow type individuals is $9 \\%$, the genotype frequency of $C^{\\mathrm{B}} C^{\\mathrm{B}}$ is about $70 \\%$. Note that the population is assumed to be under Hardy-Weinberg equilibrium.\nB: Information: When the body colors of ten species belonging to the same genus with species A were examined, they were all yellow. Statement: In this case, the body color of the ancestral species A just after splitting from these closely related species must have been yellow under a parsimony principle.\nC: Information: A small portion of individuals in the population of species A was isolated due to diastrophism (large-scale deformation of Earth's crust) and formed a new population A'. Statement: The drastic inter-generation changes of allele frequency of locus $C$ in population $\\mathrm{A}^{\\prime}$ can be best explained by natural selection.\nD: Information: A slightly deleterious mutation with exactly the same effect on the fitness of an individual independently occurred in both the small population $\\mathrm{A}^{\\prime}$ and the larger parental population A. Statement: The fixation probability of this mutation is the same in both populations.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_504",
        "problem": "铂类药物可抑制细胞增殖, 顺铂是临床上治疗晚期宫颈癌的首选药物。为研究大蒜素与顺铂联合给药对癌细胞增殖的影响及机制, 科研人员检测大蒜素、顺铂和联合给药对癌细胞周期的影响, 结果如图甲和乙所示。已知 P27、CDK2、CyclinD1 等多个基因参与 $\\mathrm{G}_{2}$ 期到 $\\mathrm{S}$ 期转换, 科研人员也检测了相应蛋白的表达量, 结果如图丙。下列叙述错误的是（）\n[图1]\n\n注： $\\mathrm{G}_{0}$ : 细胞暂不分裂 $\\mathrm{G}_{1}$ : DNA 合成前 $\\mathrm{S}$ :DNA 合成期 $\\mathrm{G}_{2}$ :DNA 合成后 $\\mathrm{M}$ 期：分裂\n\n期\nA: 细胞癌变存在基因的选择性表达, 癌变后一般不会正常调亡 B．大蒜素和顺铂在抑制癌细胞增殖方面存在协同作用\nB: CDK2 和 CyclinD1 基因参与抑制 $\\mathrm{G}_{1}$ 期到 $\\mathrm{S}$ 期的转换\nC: 联合给药通过调控相关基因表达使更多细胞停留在（在 $\\mathrm{G}_{0}+\\mathrm{G}_{1}$ 期）\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n铂类药物可抑制细胞增殖, 顺铂是临床上治疗晚期宫颈癌的首选药物。为研究大蒜素与顺铂联合给药对癌细胞增殖的影响及机制, 科研人员检测大蒜素、顺铂和联合给药对癌细胞周期的影响, 结果如图甲和乙所示。已知 P27、CDK2、CyclinD1 等多个基因参与 $\\mathrm{G}_{2}$ 期到 $\\mathrm{S}$ 期转换, 科研人员也检测了相应蛋白的表达量, 结果如图丙。下列叙述错误的是（）\n[图1]\n\n注： $\\mathrm{G}_{0}$ : 细胞暂不分裂 $\\mathrm{G}_{1}$ : DNA 合成前 $\\mathrm{S}$ :DNA 合成期 $\\mathrm{G}_{2}$ :DNA 合成后 $\\mathrm{M}$ 期：分裂\n\n期\n\nA: 细胞癌变存在基因的选择性表达, 癌变后一般不会正常调亡 B．大蒜素和顺铂在抑制癌细胞增殖方面存在协同作用\nB: CDK2 和 CyclinD1 基因参与抑制 $\\mathrm{G}_{1}$ 期到 $\\mathrm{S}$ 期的转换\nC: 联合给药通过调控相关基因表达使更多细胞停留在（在 $\\mathrm{G}_{0}+\\mathrm{G}_{1}$ 期）\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-43.jpg?height=336&width=1444&top_left_y=1228&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_465",
        "problem": "在人群中有两种独立遗传的单基因遗传病甲病（基因 $\\mathrm{A} / \\mathrm{a}$ 控制）和乙病（基因 $\\mathrm{B} / \\mathrm{b}$控制), 通过调查数据计算, 甲、乙两种遗传病的发病率分别为 $1 / 250$ 和 $1 / 100$ 。下图为调查过程中记录的两个家庭的家系图 (无两病均患个体)。以下说法正确的是（）\n[图1]\nA: 甲病的遗传方式为常染色体隐性遗传, 乙病的遗传方式为常染色体显性遗传\nB: 系谱中的I2、I4 和II4 的基因型相同, I1 和II1 的基因型相同的概率是 2/3\nC: II2 与人群中表型正常的男性婚配, 生出患病的孩子的概率为 $1 / 33$\nD: II1 与II6 结婚, 生出只患一种病的孩子的概率为 $11 / 18$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在人群中有两种独立遗传的单基因遗传病甲病（基因 $\\mathrm{A} / \\mathrm{a}$ 控制）和乙病（基因 $\\mathrm{B} / \\mathrm{b}$控制), 通过调查数据计算, 甲、乙两种遗传病的发病率分别为 $1 / 250$ 和 $1 / 100$ 。下图为调查过程中记录的两个家庭的家系图 (无两病均患个体)。以下说法正确的是（）\n[图1]\n\nA: 甲病的遗传方式为常染色体隐性遗传, 乙病的遗传方式为常染色体显性遗传\nB: 系谱中的I2、I4 和II4 的基因型相同, I1 和II1 的基因型相同的概率是 2/3\nC: II2 与人群中表型正常的男性婚配, 生出患病的孩子的概率为 $1 / 33$\nD: II1 与II6 结婚, 生出只患一种病的孩子的概率为 $11 / 18$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-09.jpg?height=238&width=946&top_left_y=1068&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_633",
        "problem": "黄跖鱼 ( $2 \\mathrm{~N}=52$ ) 是我国重要的经济鱼类。雄性比雌性生长速度快 $2 \\sim 3$ 倍, 我国科学家培育获得子代全部为雄性个体的黄欼鱼, 让养殖户获得了较高经济效益。培育过程如下:\n\n(1)通过相关激素诱导, 将雄鱼性逆转为生理雌鱼;\n\n(2)使用低温休克法干预生理雌鱼产生配子的相关过程, 获得无需受精即可发育成个体的卵细胞, 从而获得 YY 超雄鱼（该鱼能正常发育且具有繁殖能力）; 但此方法获得超雄鱼的成功率较低，难以规模化生产。\n\n(3)让超雄鱼与正常雌鱼交配, 后代全为雄鱼。下列分析正确的是（ )\nA: (1)中雄鱼和生理雌鱼体细胞中遗传信息的执行情况几乎完全相同\nB: (2)中低温休克法作用的时期是 MII期，超雄鱼体细胞含 52 条染色体\nC: 生理雌性黄颡鱼经 (2)过程可以得到与其遗传物质组成相同的雌鱼\nD: 将超雄鱼性逆转为生理雌鱼与超雄鱼交配, 可实现超雄鱼的规模化生产\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n黄跖鱼 ( $2 \\mathrm{~N}=52$ ) 是我国重要的经济鱼类。雄性比雌性生长速度快 $2 \\sim 3$ 倍, 我国科学家培育获得子代全部为雄性个体的黄欼鱼, 让养殖户获得了较高经济效益。培育过程如下:\n\n(1)通过相关激素诱导, 将雄鱼性逆转为生理雌鱼;\n\n(2)使用低温休克法干预生理雌鱼产生配子的相关过程, 获得无需受精即可发育成个体的卵细胞, 从而获得 YY 超雄鱼（该鱼能正常发育且具有繁殖能力）; 但此方法获得超雄鱼的成功率较低，难以规模化生产。\n\n(3)让超雄鱼与正常雌鱼交配, 后代全为雄鱼。下列分析正确的是（ )\n\nA: (1)中雄鱼和生理雌鱼体细胞中遗传信息的执行情况几乎完全相同\nB: (2)中低温休克法作用的时期是 MII期，超雄鱼体细胞含 52 条染色体\nC: 生理雌性黄颡鱼经 (2)过程可以得到与其遗传物质组成相同的雌鱼\nD: 将超雄鱼性逆转为生理雌鱼与超雄鱼交配, 可实现超雄鱼的规模化生产\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_314",
        "problem": "鹤鹑的羽色由三对等位基因共同控制 (见下表), 其中 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 均位于 $\\mathrm{Z}$ 染色体上,\n\n$\\mathrm{H} / \\mathrm{h}$ 位于常染色体上。科研人员取纯系黑羽雄性鹤鸴和纯系白羽雌性鹤鹑进行杂交实验, $F_{1}$ 表型均为不完全黑羽, $F_{1}$ 随机交配得到 $F_{2}$ 。下列叙述正确的是 $(\\quad)$\n\n| 基因组成 | $\\mathrm{A} 、 \\mathrm{~B}$ 同时存在, <br> 且 $\\mathrm{H}$ 基因纯合 | $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{H} 、 \\mathrm{~h}$ 同时 <br> 存在 | $\\mathrm{A} 、 \\mathrm{~B}$ 同时存在,且 $\\mathrm{h}$ 基 <br> 因纯合 | $\\mathrm{b}$ 基因纯 <br> 合 |\n| :--- | :--- | :--- | :--- | :--- |\n| 表型 | 栗羽 | 不完全黑羽 | 黑羽 | 白羽 |\nA: $\\mathrm{A} / \\mathrm{a}$ 与 $\\mathrm{B} / \\mathrm{b}$ 两对基因的遗传遵循自由组合定律\nB: 亲代黑羽、白羽基因型依次为 $h h Z^{A B} W 、 H H Z^{A b} Z^{A b}$\nC: $F_{2}$ 不完全黑羽的雌雄比例为 $1: 1$\nD: $F_{2}$ 中栗羽: 不完全黑羽：黑羽：白羽的比例为 3:6:3:4\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n鹤鹑的羽色由三对等位基因共同控制 (见下表), 其中 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 均位于 $\\mathrm{Z}$ 染色体上,\n\n$\\mathrm{H} / \\mathrm{h}$ 位于常染色体上。科研人员取纯系黑羽雄性鹤鸴和纯系白羽雌性鹤鹑进行杂交实验, $F_{1}$ 表型均为不完全黑羽, $F_{1}$ 随机交配得到 $F_{2}$ 。下列叙述正确的是 $(\\quad)$\n\n| 基因组成 | $\\mathrm{A} 、 \\mathrm{~B}$ 同时存在, <br> 且 $\\mathrm{H}$ 基因纯合 | $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{H} 、 \\mathrm{~h}$ 同时 <br> 存在 | $\\mathrm{A} 、 \\mathrm{~B}$ 同时存在,且 $\\mathrm{h}$ 基 <br> 因纯合 | $\\mathrm{b}$ 基因纯 <br> 合 |\n| :--- | :--- | :--- | :--- | :--- |\n| 表型 | 栗羽 | 不完全黑羽 | 黑羽 | 白羽 |\n\nA: $\\mathrm{A} / \\mathrm{a}$ 与 $\\mathrm{B} / \\mathrm{b}$ 两对基因的遗传遵循自由组合定律\nB: 亲代黑羽、白羽基因型依次为 $h h Z^{A B} W 、 H H Z^{A b} Z^{A b}$\nC: $F_{2}$ 不完全黑羽的雌雄比例为 $1: 1$\nD: $F_{2}$ 中栗羽: 不完全黑羽：黑羽：白羽的比例为 3:6:3:4\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1490",
        "problem": "In bacteria, one transcription factor often controls multiple genes. The expression of genes consumes energy so groups of genes controlled by the same factor are finely adapted to help the survival strategy of the bacterium. Floating individuals move vigorously in search of nutrients, but bacteria in biofilms rarely move.\n\nSort whether these genes are more likely to be expressed with the transcription factors controlling biofilm formation or free-floating individual formation.\n\nWhich genes turn on for biofilm formation?\nA: Genes for secreted defensive compounds, Genes for plasmid sharing\nB: Genes for secreted defensive compounds, Flagella forming genes, Nutrient sensors\nC: Flagella forming genes, Genes for surviving starvation, Nutrient sensors\nD: Flagella forming genes, Genes for surviving starvation, Genes for plasmid sharing\nE: Genes for secreted defensive compounds, Genes for plasmid sharing, Flagella forming genes\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn bacteria, one transcription factor often controls multiple genes. The expression of genes consumes energy so groups of genes controlled by the same factor are finely adapted to help the survival strategy of the bacterium. Floating individuals move vigorously in search of nutrients, but bacteria in biofilms rarely move.\n\nSort whether these genes are more likely to be expressed with the transcription factors controlling biofilm formation or free-floating individual formation.\n\nWhich genes turn on for biofilm formation?\n\nA: Genes for secreted defensive compounds, Genes for plasmid sharing\nB: Genes for secreted defensive compounds, Flagella forming genes, Nutrient sensors\nC: Flagella forming genes, Genes for surviving starvation, Nutrient sensors\nD: Flagella forming genes, Genes for surviving starvation, Genes for plasmid sharing\nE: Genes for secreted defensive compounds, Genes for plasmid sharing, Flagella forming genes\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1176",
        "problem": "Blue jays are omnivorous birds who supplement their vegetarian diet with insects. When a blue jay eats a monarch butterfly the foul taste of the butterfly causes the blue jay to vomit up the food. Blue jays who have previously consumed a monarch butterfly will avoid them in future with the result that wild monarch butterflies are less susceptible to natural predation by blue jays than other species of butterfly. The blue jays avoidance of monarch butterflies as a food is a result of:\nA: Habituation\nB: Classical conditioning\nC: Operant conditioning\nD: Observational learning\nE: Insight learning\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBlue jays are omnivorous birds who supplement their vegetarian diet with insects. When a blue jay eats a monarch butterfly the foul taste of the butterfly causes the blue jay to vomit up the food. Blue jays who have previously consumed a monarch butterfly will avoid them in future with the result that wild monarch butterflies are less susceptible to natural predation by blue jays than other species of butterfly. The blue jays avoidance of monarch butterflies as a food is a result of:\n\nA: Habituation\nB: Classical conditioning\nC: Operant conditioning\nD: Observational learning\nE: Insight learning\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_117",
        "problem": "The following figure shows a mushroom belonging to the Basidiomycetes.\n\n[figure1]\n\nWhich of the following combination is correct for the nuclear ploidy states of structures $a \\sim c$ ?\n\n|  | Ploidy state of structure |  |  |\n| :---: | :---: | :---: | :---: |\n|  | $b$ |  |  |\n| A | $\\mathrm{n}$ | $2 \\mathrm{n}$ | $\\mathrm{n}+\\mathrm{n}$ |\n| B | $2 \\mathrm{n}$ | $\\mathrm{n}$ | $\\mathrm{n}+\\mathrm{n}$ |\n| C | $\\mathrm{n}$ | $\\mathrm{n}$ | $\\mathrm{n}$ |\n| D | $\\mathrm{n}+\\mathrm{n}$ | $2 \\mathrm{n}$ | $\\mathrm{n}$ |\n| E | $\\mathrm{n}+\\mathrm{n}$ | $\\mathrm{n}$ | $\\mathrm{n}+\\mathrm{n}$ |\nA: A\nB: B\nC: C\nD: D\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following figure shows a mushroom belonging to the Basidiomycetes.\n\n[figure1]\n\nWhich of the following combination is correct for the nuclear ploidy states of structures $a \\sim c$ ?\n\n|  | Ploidy state of structure |  |  |\n| :---: | :---: | :---: | :---: |\n|  | $b$ |  |  |\n| A | $\\mathrm{n}$ | $2 \\mathrm{n}$ | $\\mathrm{n}+\\mathrm{n}$ |\n| B | $2 \\mathrm{n}$ | $\\mathrm{n}$ | $\\mathrm{n}+\\mathrm{n}$ |\n| C | $\\mathrm{n}$ | $\\mathrm{n}$ | $\\mathrm{n}$ |\n| D | $\\mathrm{n}+\\mathrm{n}$ | $2 \\mathrm{n}$ | $\\mathrm{n}$ |\n| E | $\\mathrm{n}+\\mathrm{n}$ | $\\mathrm{n}$ | $\\mathrm{n}+\\mathrm{n}$ |\n\nA: A\nB: B\nC: C\nD: D\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-62.jpg?height=605&width=691&top_left_y=380&top_left_x=794"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_661",
        "problem": "NPM1 是一种核仁蛋白，可在核质之间穿梭，参与核糖体蛋白的组装和运输，与中心粒结合后可抑制中心粒的复制。NPM1 与抑癌基因 P53 结合后会增强 P53 的转录。下列分析错误的是（ ）\nA: NPM1 可能通过核孔进出细胞核\nB: 代谢旺盛的细胞中 NPM1 基因表达增强\nC: 细胞质中的 NPM1 增多可能会抑制动物细胞分裂\nD: NPM1 基因突变后可能会抑制细胞发生癌变\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nNPM1 是一种核仁蛋白，可在核质之间穿梭，参与核糖体蛋白的组装和运输，与中心粒结合后可抑制中心粒的复制。NPM1 与抑癌基因 P53 结合后会增强 P53 的转录。下列分析错误的是（ ）\n\nA: NPM1 可能通过核孔进出细胞核\nB: 代谢旺盛的细胞中 NPM1 基因表达增强\nC: 细胞质中的 NPM1 增多可能会抑制动物细胞分裂\nD: NPM1 基因突变后可能会抑制细胞发生癌变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_145",
        "problem": "A researcher used a patch clamp technique to record the current of only a cation channel (single channel recording). The pipette contains $150 \\mathrm{mM}$ of $\\mathrm{KCl}$ surrounded by a bath solution containing $150 \\mathrm{mM}$ of $\\mathrm{NaCl}$, and measures the channel current by clamping the voltage in different values. Current-Voltage curve (I-V curve) is shown in the following figure. $\\Delta V=V_{\\text {interior }}-V_{\\text {exterior }}$.\n\nA negative current value (i.e., inward current) can reflect either the movement of positive ions (cations) into the cell or negative ions (anions) out of the cell.\n\nA positive current value (i.e., outward current) can reflect either the movement of positive ions (cations) out of the cell or negative ions (anions) into the cell.\n\n[figure1]\nA: This channel is not selective, and both $\\mathrm{K}^{+}$and $\\mathrm{Na}^{+}$ions pass through this channel.\nB: The $\\mathrm{Na}^{+}$current decreases with increasing voltage.\nC: The $\\mathrm{K}^{+}$current increases with increasing voltage.\nD: The resistance of the channel to negative voltages is less than positive voltages.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA researcher used a patch clamp technique to record the current of only a cation channel (single channel recording). The pipette contains $150 \\mathrm{mM}$ of $\\mathrm{KCl}$ surrounded by a bath solution containing $150 \\mathrm{mM}$ of $\\mathrm{NaCl}$, and measures the channel current by clamping the voltage in different values. Current-Voltage curve (I-V curve) is shown in the following figure. $\\Delta V=V_{\\text {interior }}-V_{\\text {exterior }}$.\n\nA negative current value (i.e., inward current) can reflect either the movement of positive ions (cations) into the cell or negative ions (anions) out of the cell.\n\nA positive current value (i.e., outward current) can reflect either the movement of positive ions (cations) out of the cell or negative ions (anions) into the cell.\n\n[figure1]\n\nA: This channel is not selective, and both $\\mathrm{K}^{+}$and $\\mathrm{Na}^{+}$ions pass through this channel.\nB: The $\\mathrm{Na}^{+}$current decreases with increasing voltage.\nC: The $\\mathrm{K}^{+}$current increases with increasing voltage.\nD: The resistance of the channel to negative voltages is less than positive voltages.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-26.jpg?height=666&width=1156&top_left_y=795&top_left_x=450"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1142",
        "problem": "Which one of the following pieces of information provides the MOST direct evidence that individuals who are heterozygous for the sickle cell condition have a selective advantage in areas where malaria is endemic?\nA: The geographical distribution of malaria and high sickle cell allele frequency are closely correlated\nB: Mortality from malaria is lower in heterozygous individuals than in those with normal blood\nC: Heterozygous individuals are less susceptible to experimentally induced malaria than are homozygous individuals\nD: Counts of malarial parasites in the blood of heterozygous individuals are lower than in those with normal blood\nE: Malarial parasites grow more slowly in heterozygous individuals than in those with normal blood\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following pieces of information provides the MOST direct evidence that individuals who are heterozygous for the sickle cell condition have a selective advantage in areas where malaria is endemic?\n\nA: The geographical distribution of malaria and high sickle cell allele frequency are closely correlated\nB: Mortality from malaria is lower in heterozygous individuals than in those with normal blood\nC: Heterozygous individuals are less susceptible to experimentally induced malaria than are homozygous individuals\nD: Counts of malarial parasites in the blood of heterozygous individuals are lower than in those with normal blood\nE: Malarial parasites grow more slowly in heterozygous individuals than in those with normal blood\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1267",
        "problem": "If light were a limiting factor, the rate at which bubbles of gas would be given off by Elodea (a pondweed), when exposed to a small but bright light source, would be increased four-fold if\nA: the light were placed at half of the original distance from the plant\nB: the light were placed at a quarter of the original distance from the plant\nC: the temperature were raised $10{ }^{\\circ} \\mathrm{C}$\nD: the carbon dioxide supply were increased four-fold\nE: the carbon dioxide supply were doubled\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf light were a limiting factor, the rate at which bubbles of gas would be given off by Elodea (a pondweed), when exposed to a small but bright light source, would be increased four-fold if\n\nA: the light were placed at half of the original distance from the plant\nB: the light were placed at a quarter of the original distance from the plant\nC: the temperature were raised $10{ }^{\\circ} \\mathrm{C}$\nD: the carbon dioxide supply were increased four-fold\nE: the carbon dioxide supply were doubled\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_238",
        "problem": "In Adélie penguins when both parent leave the nest to find food, three-weeks-old chicks group together for protection. In this species, social defence behaviour can occur in unusual ways. When adult Adélie penguins reach the coast, they gather in groups before diving into the ocean, where usually their predator (leopard seal) could be lurking. The penguins have two choices, first to dive together, or wait until one of penguins dives so they could make sure if there is a predator or not.\nA: This is an example of altruism behaviour as the penguin that jump first could be hunted by possible predators while transferring its genes by helping its\nB: According to above example we can say one individual have influence on what other penguins are doing.\nC: If a new strategy arose which would make a penguin trip another one into the ocean without retaliation from others, everyone in the population would adopt this new strategy in the long run.\nD: A mutation that results in postponing to jump as long as possible would eventually go to fixation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn Adélie penguins when both parent leave the nest to find food, three-weeks-old chicks group together for protection. In this species, social defence behaviour can occur in unusual ways. When adult Adélie penguins reach the coast, they gather in groups before diving into the ocean, where usually their predator (leopard seal) could be lurking. The penguins have two choices, first to dive together, or wait until one of penguins dives so they could make sure if there is a predator or not.\n\nA: This is an example of altruism behaviour as the penguin that jump first could be hunted by possible predators while transferring its genes by helping its\nB: According to above example we can say one individual have influence on what other penguins are doing.\nC: If a new strategy arose which would make a penguin trip another one into the ocean without retaliation from others, everyone in the population would adopt this new strategy in the long run.\nD: A mutation that results in postponing to jump as long as possible would eventually go to fixation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1133",
        "problem": "omparison of related sequences from different species can give clues to evolutionary relationships among proteins. A phylogenetic tree representing relationship between tubulin sequences is depicted below.\n\n[figure1]\n\nFilled circles indicate common ancestors. Mark the correct statement about the likely evolutionary mechanism.\nA: Tubulin gene most likely diverged into $\\alpha$ and $\\beta$ forms after speciation.\nB: A gene duplication event led to two identical copies of the gene within each species and then the sequences diverged further.\nC: A gene duplication event occurred first followed by sequence divergence. Each sequence diverged further during speciation.\nD: Earliest eukaryotic cell most likely contained two tubulin genes which further diverged from each other during the course of speciation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nomparison of related sequences from different species can give clues to evolutionary relationships among proteins. A phylogenetic tree representing relationship between tubulin sequences is depicted below.\n\n[figure1]\n\nFilled circles indicate common ancestors. Mark the correct statement about the likely evolutionary mechanism.\n\nA: Tubulin gene most likely diverged into $\\alpha$ and $\\beta$ forms after speciation.\nB: A gene duplication event led to two identical copies of the gene within each species and then the sequences diverged further.\nC: A gene duplication event occurred first followed by sequence divergence. Each sequence diverged further during speciation.\nD: Earliest eukaryotic cell most likely contained two tubulin genes which further diverged from each other during the course of speciation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-10.jpg?height=441&width=1304&top_left_y=2116&top_left_x=492"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_520",
        "problem": "如图 1、图 2、图 3 分别表示雄果蝇 $(2 n=8)$ 体内细胞分裂过程中有关物质的变化。下列相关叙述错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\nA: 图 1 可用于表示精原细胞增殖或产生精子的过程中 DNA 与染色体的数量关系\nB: 图 2 中 $\\mathrm{GH}$ 段产生的原因是着丝粒分裂, 其 $\\mathrm{HJ}$ 段与图 3 中的II代表同一时期\nC: 图 3 所示的细胞分裂可发生在该果蝇的精巢中, a、c 可分别代表染色体和核 DNA\nD: 图 3 中，I时期的细胞中不含姐妹染色单体，其可表示减数分裂II后期\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图 1、图 2、图 3 分别表示雄果蝇 $(2 n=8)$ 体内细胞分裂过程中有关物质的变化。下列相关叙述错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\n\nA: 图 1 可用于表示精原细胞增殖或产生精子的过程中 DNA 与染色体的数量关系\nB: 图 2 中 $\\mathrm{GH}$ 段产生的原因是着丝粒分裂, 其 $\\mathrm{HJ}$ 段与图 3 中的II代表同一时期\nC: 图 3 所示的细胞分裂可发生在该果蝇的精巢中, a、c 可分别代表染色体和核 DNA\nD: 图 3 中，I时期的细胞中不含姐妹染色单体，其可表示减数分裂II后期\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-16.jpg?height=325&width=397&top_left_y=200&top_left_x=344",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-16.jpg?height=328&width=417&top_left_y=196&top_left_x=768",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-16.jpg?height=343&width=600&top_left_y=180&top_left_x=1185"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1450",
        "problem": "Cellular respiration is a complex process involving many steps. Generally, these steps are grouped into 3 main phases: glycolysis, the citric acid cycle, and the electron transport chain. The process of glycolysis can be summarised as follows:\n[figure1]\n\nGlycolysis utilises a total of 10 enzymes, including those mentioned in the above diagram. How might the cell benefit from using many different enzymes in cellular respiration?\nA: Activation energy of a reaction is decreased, allowing it to occur more quickly.\nB: The cell needs fewer points at which it can promote function.\nC: Using specialised enzymes allows each step of respiration to occur more slowly.\nD: The cell needs fewer points at which it can limit enzyme function to control overall energy output.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCellular respiration is a complex process involving many steps. Generally, these steps are grouped into 3 main phases: glycolysis, the citric acid cycle, and the electron transport chain. The process of glycolysis can be summarised as follows:\n[figure1]\n\nGlycolysis utilises a total of 10 enzymes, including those mentioned in the above diagram. How might the cell benefit from using many different enzymes in cellular respiration?\n\nA: Activation energy of a reaction is decreased, allowing it to occur more quickly.\nB: The cell needs fewer points at which it can promote function.\nC: Using specialised enzymes allows each step of respiration to occur more slowly.\nD: The cell needs fewer points at which it can limit enzyme function to control overall energy output.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-34.jpg?height=1204&width=832&top_left_y=532&top_left_x=286"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_822",
        "problem": "已知一小鼠的基因型为 $X^{B} X^{b}$ 。该小鼠次级卵母细胞减数分裂的结果是( )\nA: 第二极体中同时含有 $\\mathrm{B}$ 基因和 $\\mathrm{b}$ 基因\nB: 卵细胞中可能含有 $\\mathrm{B}$ 基因或含有 $\\mathrm{b}$ 基因\nC: 若卵细胞中含有 $\\mathrm{B}$ 基因，则极体中必不含有 $\\mathrm{B}$ 基因\nD: 卵细胞中必然含有 $\\mathrm{B}$ 基因，极体中可能含有 $\\mathrm{b}$ 基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知一小鼠的基因型为 $X^{B} X^{b}$ 。该小鼠次级卵母细胞减数分裂的结果是( )\n\nA: 第二极体中同时含有 $\\mathrm{B}$ 基因和 $\\mathrm{b}$ 基因\nB: 卵细胞中可能含有 $\\mathrm{B}$ 基因或含有 $\\mathrm{b}$ 基因\nC: 若卵细胞中含有 $\\mathrm{B}$ 基因，则极体中必不含有 $\\mathrm{B}$ 基因\nD: 卵细胞中必然含有 $\\mathrm{B}$ 基因，极体中可能含有 $\\mathrm{b}$ 基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1456",
        "problem": "Myasthenia gravis is a chronic autoimmune condition that affects the junction between neurons and muscles, causing serious muscle weakness. In most cases, it is caused by the production of antibodies against acetylcholine neurotransmitter receptors on the muscle.\n\nWhich of the following options is least likely to be used as a treatment for people with myasthenia gravis?\nA: Anti-acetylcholinesterase agents to prevent the breakdown of acetylcholine in the neuromuscular junction.\nB: Immunosuppressants to reduce the activity of the immune system.\nC: Surgical removal of the thymus to prevent antibody production.\nD: A high sodium diet to increase the strength of action potentials.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMyasthenia gravis is a chronic autoimmune condition that affects the junction between neurons and muscles, causing serious muscle weakness. In most cases, it is caused by the production of antibodies against acetylcholine neurotransmitter receptors on the muscle.\n\nWhich of the following options is least likely to be used as a treatment for people with myasthenia gravis?\n\nA: Anti-acetylcholinesterase agents to prevent the breakdown of acetylcholine in the neuromuscular junction.\nB: Immunosuppressants to reduce the activity of the immune system.\nC: Surgical removal of the thymus to prevent antibody production.\nD: A high sodium diet to increase the strength of action potentials.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1195",
        "problem": "Understanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.The percentage of plant material consumed by orangefronted parakeets on Maud Island was?\nA: $96 \\%$\nB: $90 \\%$\nC: $76 \\%$\nD: $71 \\%$\nE: $15 \\%$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nUnderstanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.\n\nproblem:\nThe percentage of plant material consumed by orangefronted parakeets on Maud Island was?\n\nA: $96 \\%$\nB: $90 \\%$\nC: $76 \\%$\nD: $71 \\%$\nE: $15 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_352",
        "problem": "缺乏某蛋白的小鼠会表现为侏儒鼠。小鼠体内的 $\\mathrm{H}$ 基因能控制该蛋白的合成, $\\mathrm{h}$ 基因则不能。H 基因的表达受 $\\mathrm{a}$ 序列（一段 DNA 序列）的调控，如图所示。a 序列在精子中是非甲基化, 传给子代后能正常表达; 在卵细胞中是甲基化（甲基化需要甲基化酶\n[图1]\nA: 基因型为 $\\mathrm{Hh}$ 的侏儒鼠, $\\mathrm{H}$ 基因只能来自母本\nB: 降低发育中的侏儒鼠甲基化酶的活性, 侏儒症状均能一定程度上缓解\nC: 两只侏儒鼠交配, 子代小鼠不一定是侏儒鼠\nD: 两只正常鼠交配，子代小鼠不一定是正常鼠\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n缺乏某蛋白的小鼠会表现为侏儒鼠。小鼠体内的 $\\mathrm{H}$ 基因能控制该蛋白的合成, $\\mathrm{h}$ 基因则不能。H 基因的表达受 $\\mathrm{a}$ 序列（一段 DNA 序列）的调控，如图所示。a 序列在精子中是非甲基化, 传给子代后能正常表达; 在卵细胞中是甲基化（甲基化需要甲基化酶\n[图1]\n\nA: 基因型为 $\\mathrm{Hh}$ 的侏儒鼠, $\\mathrm{H}$ 基因只能来自母本\nB: 降低发育中的侏儒鼠甲基化酶的活性, 侏儒症状均能一定程度上缓解\nC: 两只侏儒鼠交配, 子代小鼠不一定是侏儒鼠\nD: 两只正常鼠交配，子代小鼠不一定是正常鼠\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-82.jpg?height=408&width=1398&top_left_y=247&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_3",
        "problem": "Phytopathogens use their secreted proteins \"secretomes\" to attack plant hosts. The yeast secretion trap (YST) functional screen is a method used for isolation and identification of these secreted proteins. This method involves generating a vector library that includes cDNAs (containing 5' non-coding sequences) synthesized using phytopathogen RNAs fused to a mutated form of the Saccharomyces cerevisiae suc2 reporter gene.\n\nSuc2 encodes invertase which is the only protein used by yeast for sucrose degradation. Degradation occurs in the extra-cellular medium. The mutated form of the gene suc2-SP encodes an invertase that lacks the signal peptide for secretion at the amino terminal. The fusion library is used to transfect an invertase-deficient yeast strain, and the cells are subsequently plated on a sucrose selection medium. Each rescued cell is expected to contain a recombinant vector whose cDNA that encodes a signal peptide for the chimeric protein. Vectors are isolated and their cDNAs are sequenced to identify the secreted proteins.\n\nFigure 1 is a schematic presentation of the vector, and Figure 2 summarizes the protocol described above.\n\n[figure1]\n\nFigure 1. GAATTC: EcoR1 recognition site; GCGGCCGC: Not1 recognition site\n\nmRNA isolation and synthesis of cDNAs bound\n\nby EcoRI and NoTI sites\n\n[figure2]\n\nFigure 2.\nA: For the purpose of this screen, oligo-dT primers can be used for synthesis of cDNAs\nB: Variable number of Cytidine nucleotide(s) after the Not1 site in the vector ensures achieving a correct reading frame.\nC: Presence of signal peptide in the fusion protein will not necessarily ensure the growth of yeast on the selection medium.\nD: For introduction of EcoRI and NotI sites to ends of cDNA molecules, linkers (which are short double stranded oligonucleotides; example shown below) are better than adapters (which have single stranded ends; example shown below). Example of EcoRI linker: -GAATTC- - CTTAAG- Example of EcoRI adapters: -G- and -AATTC- -CTTAA- -G-\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPhytopathogens use their secreted proteins \"secretomes\" to attack plant hosts. The yeast secretion trap (YST) functional screen is a method used for isolation and identification of these secreted proteins. This method involves generating a vector library that includes cDNAs (containing 5' non-coding sequences) synthesized using phytopathogen RNAs fused to a mutated form of the Saccharomyces cerevisiae suc2 reporter gene.\n\nSuc2 encodes invertase which is the only protein used by yeast for sucrose degradation. Degradation occurs in the extra-cellular medium. The mutated form of the gene suc2-SP encodes an invertase that lacks the signal peptide for secretion at the amino terminal. The fusion library is used to transfect an invertase-deficient yeast strain, and the cells are subsequently plated on a sucrose selection medium. Each rescued cell is expected to contain a recombinant vector whose cDNA that encodes a signal peptide for the chimeric protein. Vectors are isolated and their cDNAs are sequenced to identify the secreted proteins.\n\nFigure 1 is a schematic presentation of the vector, and Figure 2 summarizes the protocol described above.\n\n[figure1]\n\nFigure 1. GAATTC: EcoR1 recognition site; GCGGCCGC: Not1 recognition site\n\nmRNA isolation and synthesis of cDNAs bound\n\nby EcoRI and NoTI sites\n\n[figure2]\n\nFigure 2.\n\nA: For the purpose of this screen, oligo-dT primers can be used for synthesis of cDNAs\nB: Variable number of Cytidine nucleotide(s) after the Not1 site in the vector ensures achieving a correct reading frame.\nC: Presence of signal peptide in the fusion protein will not necessarily ensure the growth of yeast on the selection medium.\nD: For introduction of EcoRI and NotI sites to ends of cDNA molecules, linkers (which are short double stranded oligonucleotides; example shown below) are better than adapters (which have single stranded ends; example shown below). Example of EcoRI linker: -GAATTC- - CTTAAG- Example of EcoRI adapters: -G- and -AATTC- -CTTAA- -G-\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-16.jpg?height=508&width=1202&top_left_y=1065&top_left_x=427",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-16.jpg?height=734&width=1496&top_left_y=1832&top_left_x=286"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_635",
        "problem": "山羊的毛色有白色和灰色, 角形有盘状和直角状。山羊甲、乙交配. 所得 $F_{1}$ 均表现为灰毛盘状角, $F_{1}$ 雌雄交配所得 $F_{2}$ 的表型及比例为灰毛盘状角：白毛盘状角：灰毛直角: 白毛直角 $=27: 21: 9: 7$ 。上述所涉及的基因位于非同源染色体上, 下列推断错误的是 ( )\nA: 山羊的毛色和角形分别受一对等位基因控制\nB: 盘状角对直角为显性，若仅考虑角形， $F_{1}$ 的基因型相同\nC: $F_{2}$ 中白毛直角杂合子雌雄交配, $F_{3}$ 中灰毛直角个体占 $1 / 8$\nD: $F_{2}$ 的灰毛个体中, 约有 $1 / 9$ 的个体测交子代均为灰毛\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n山羊的毛色有白色和灰色, 角形有盘状和直角状。山羊甲、乙交配. 所得 $F_{1}$ 均表现为灰毛盘状角, $F_{1}$ 雌雄交配所得 $F_{2}$ 的表型及比例为灰毛盘状角：白毛盘状角：灰毛直角: 白毛直角 $=27: 21: 9: 7$ 。上述所涉及的基因位于非同源染色体上, 下列推断错误的是 ( )\n\nA: 山羊的毛色和角形分别受一对等位基因控制\nB: 盘状角对直角为显性，若仅考虑角形， $F_{1}$ 的基因型相同\nC: $F_{2}$ 中白毛直角杂合子雌雄交配, $F_{3}$ 中灰毛直角个体占 $1 / 8$\nD: $F_{2}$ 的灰毛个体中, 约有 $1 / 9$ 的个体测交子代均为灰毛\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_278",
        "problem": "A mutated male mouse that is phenotypically normal shows reproductive anomalies when compared with a normal male as shown in table Q. 38 .\n\nTable Q. 38\n\n|  | Embryo (mean No.) |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n| Mating | Implanted in <br> uterine wall | Degenerating after <br> implantation | Normal | Degeneration <br> $(\\%)$ |\n| Mutated ठ X normal क | 8.7 | 5.0 | 3.7 | 57.5 |\n| Normal ठ X normal के | 9.5 | 0.6 | 8.9 | 6.5 |\nA: The mutated male mouse could have a chromosome deletion.\nB: The mutated male mouse can be a chromosomal translocation heterozygote.\nC: The mutated male mouse can be a chromosomal inversion heterozygote.\nD: The genetic defect in mutated male could be verified by cytological observation of meiotic cells in the mouse.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA mutated male mouse that is phenotypically normal shows reproductive anomalies when compared with a normal male as shown in table Q. 38 .\n\nTable Q. 38\n\n|  | Embryo (mean No.) |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n| Mating | Implanted in <br> uterine wall | Degenerating after <br> implantation | Normal | Degeneration <br> $(\\%)$ |\n| Mutated ठ X normal क | 8.7 | 5.0 | 3.7 | 57.5 |\n| Normal ठ X normal के | 9.5 | 0.6 | 8.9 | 6.5 |\n\nA: The mutated male mouse could have a chromosome deletion.\nB: The mutated male mouse can be a chromosomal translocation heterozygote.\nC: The mutated male mouse can be a chromosomal inversion heterozygote.\nD: The genetic defect in mutated male could be verified by cytological observation of meiotic cells in the mouse.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_199",
        "problem": "A 55 -year-old man has a resting cardiac output of $7000 \\mathrm{~mL} /$ minute. His arterial pressure is $125 / 85 \\mathrm{mmHg}$. His heart rate is 100 beats/ $\\mathrm{min}$ and his body temperature is normal. Figure Q. 18 represents the changes in left ventricular pressure and blood volume during a cardiac cycle.\n\n[figure1]\n\nFigure Q. 18\nA: At $Q$, the left atrioventricular valve is opened.\nB: Ventricular ejection ends at $\\mathrm{S}$.\nC: The distance from $\\mathrm{P}$ to $\\mathrm{S}$ should be longer if there is aortic valve stenosis.\nD: In period R-S, blood does not flow into both atria and ventricles.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA 55 -year-old man has a resting cardiac output of $7000 \\mathrm{~mL} /$ minute. His arterial pressure is $125 / 85 \\mathrm{mmHg}$. His heart rate is 100 beats/ $\\mathrm{min}$ and his body temperature is normal. Figure Q. 18 represents the changes in left ventricular pressure and blood volume during a cardiac cycle.\n\n[figure1]\n\nFigure Q. 18\n\nA: At $Q$, the left atrioventricular valve is opened.\nB: Ventricular ejection ends at $\\mathrm{S}$.\nC: The distance from $\\mathrm{P}$ to $\\mathrm{S}$ should be longer if there is aortic valve stenosis.\nD: In period R-S, blood does not flow into both atria and ventricles.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-042.jpg?height=1057&width=1558&top_left_y=785&top_left_x=286"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_901",
        "problem": "在 DNA 复制时, 5 -溴尿嘧啶脱氧核苷 (BrdU) 可作为原料, 与腺嘌呤配对, 掺入新合成的子链。用 Giemsa 染料对复制后的染色体进行染色, DNA 分子的双链都含有 $\\mathrm{BrdU}$ 的染色单体呈浅蓝色, 只有一条链含有 BrdU 的染色单体呈深蓝色。现将某植物器官放在含有 $\\mathrm{BrdU}$ 的培养液中培养使其进行一次有丝分裂再进行减数分裂, 用 Giemsa 染料染色后，观察细胞分裂中期染色体的着色情况。下列推测错误的是（ ）\nA: 减数第一次分裂中期每条染色体的两条染色单体都呈深蓝色\nB: 减数第二次分裂中期每条染色体的两条染色单体着色都不同\nC: 有丝分裂中期每条染色体的两条染色单体都呈深蓝色\nD: 在分裂过程中不会出现某条染色体的两条单体都呈浅蓝色\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在 DNA 复制时, 5 -溴尿嘧啶脱氧核苷 (BrdU) 可作为原料, 与腺嘌呤配对, 掺入新合成的子链。用 Giemsa 染料对复制后的染色体进行染色, DNA 分子的双链都含有 $\\mathrm{BrdU}$ 的染色单体呈浅蓝色, 只有一条链含有 BrdU 的染色单体呈深蓝色。现将某植物器官放在含有 $\\mathrm{BrdU}$ 的培养液中培养使其进行一次有丝分裂再进行减数分裂, 用 Giemsa 染料染色后，观察细胞分裂中期染色体的着色情况。下列推测错误的是（ ）\n\nA: 减数第一次分裂中期每条染色体的两条染色单体都呈深蓝色\nB: 减数第二次分裂中期每条染色体的两条染色单体着色都不同\nC: 有丝分裂中期每条染色体的两条染色单体都呈深蓝色\nD: 在分裂过程中不会出现某条染色体的两条单体都呈浅蓝色\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1237",
        "problem": "In humans, identical twins occur in about one in 300 births. Assuming that red hair is recessive to dark hair, what are the chances of a dark-haired couple, each of whom had a red-haired parent, having red-haired identical twin boys?\nA: $1 / 600$\nB: $1 / 1200$\nC: $1 / 2400$\nD: $1 / 4800$\nE: $1 / 9600$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn humans, identical twins occur in about one in 300 births. Assuming that red hair is recessive to dark hair, what are the chances of a dark-haired couple, each of whom had a red-haired parent, having red-haired identical twin boys?\n\nA: $1 / 600$\nB: $1 / 1200$\nC: $1 / 2400$\nD: $1 / 4800$\nE: $1 / 9600$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1091",
        "problem": "Which of the following properties of plasmids are central to the evolution of bacteria?\n\ni. Plasmids are genetically dispensable for host cells in the absence of a selection pressure.\n\nii. Some plasmids have the ability to get transferred horizontally from one cell to another.\n\niii. Some plasmids permit interaction between bacterial cells of different species.\n\niv. Plasmids are naked DNA molecules incapable of survival outside a living cell.\nA: i \\& ii only\nB: ii \\& iii only\nC: i\\& iii only\nD: i, ii \\& iv\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following properties of plasmids are central to the evolution of bacteria?\n\ni. Plasmids are genetically dispensable for host cells in the absence of a selection pressure.\n\nii. Some plasmids have the ability to get transferred horizontally from one cell to another.\n\niii. Some plasmids permit interaction between bacterial cells of different species.\n\niv. Plasmids are naked DNA molecules incapable of survival outside a living cell.\n\nA: i \\& ii only\nB: ii \\& iii only\nC: i\\& iii only\nD: i, ii \\& iv\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_213",
        "problem": "mRNA in cytoplasm of eukaryote cells sometimes forms closed loop by circularization.\nA: Circularization is due to a phosphodiester bond between the 5 'end and the 3'end of mRNA.\nB: Circularization increases stability of mRNAs.\nC: Circularization enhances translocation speed of the ribosomes.\nD: Controlling circularization is a mechanism of post-transcriptional regulation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nmRNA in cytoplasm of eukaryote cells sometimes forms closed loop by circularization.\n\nA: Circularization is due to a phosphodiester bond between the 5 'end and the 3'end of mRNA.\nB: Circularization increases stability of mRNAs.\nC: Circularization enhances translocation speed of the ribosomes.\nD: Controlling circularization is a mechanism of post-transcriptional regulation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-088.jpg?height=661&width=658&top_left_y=1393&top_left_x=833"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_405",
        "problem": "大肠杆菌的蛋白质翻译起始需要核糖体、特异性起始 tRNA、mRNA 以及三个翻译起始因子（IF-1、IF-2 和 IF-3)。首先，IF-3、mRNA 和 30 S 核糖体亚基形成复合物。随后, $1 \\mathrm{~F}-2$ 和有特异性识别作用的甲酰甲硫氨酰-tRNA 形成复合物。接着, 甲酰甲硫氨酰 -tRNA 与 mRNA 结合, 两个复合物连同 IF-1 以及 GTP 分子共同构成 $30 \\mathrm{~S}$ 起始复合物。最后， GTP 水解， $50 \\mathrm{~S}$ 核糖体亚基结合到复合物中，起始因子被释放，形成完整的 $70 \\mathrm{~S}$核糖体-mRNA 复合物并进行翻译。如图为蛋白质翻译延伸示意图。下列说法正确的是\n\n[图1]\nA: GTP 水解, 可以为翻译过程提供能量\nB: 翻译过程中如果 I 可以和多种碱基配对, 会引起基因突变\nC: 翻译时, mRNA 先与 $50 \\mathrm{~S}$ 核糖体亚基结合, 形成复合物\nD: 大肠杆菌特异性的起始 tRNA 上含有特定的起始密码子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n大肠杆菌的蛋白质翻译起始需要核糖体、特异性起始 tRNA、mRNA 以及三个翻译起始因子（IF-1、IF-2 和 IF-3)。首先，IF-3、mRNA 和 30 S 核糖体亚基形成复合物。随后, $1 \\mathrm{~F}-2$ 和有特异性识别作用的甲酰甲硫氨酰-tRNA 形成复合物。接着, 甲酰甲硫氨酰 -tRNA 与 mRNA 结合, 两个复合物连同 IF-1 以及 GTP 分子共同构成 $30 \\mathrm{~S}$ 起始复合物。最后， GTP 水解， $50 \\mathrm{~S}$ 核糖体亚基结合到复合物中，起始因子被释放，形成完整的 $70 \\mathrm{~S}$核糖体-mRNA 复合物并进行翻译。如图为蛋白质翻译延伸示意图。下列说法正确的是\n\n[图1]\n\nA: GTP 水解, 可以为翻译过程提供能量\nB: 翻译过程中如果 I 可以和多种碱基配对, 会引起基因突变\nC: 翻译时, mRNA 先与 $50 \\mathrm{~S}$ 核糖体亚基结合, 形成复合物\nD: 大肠杆菌特异性的起始 tRNA 上含有特定的起始密码子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-29.jpg?height=445&width=826&top_left_y=180&top_left_x=318"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1425",
        "problem": "This diagram summarises the phases of the cell cycle.\n\n[figure1]\n\nA sample of somatic cells in various stages of the cell cycle are labelled with a fluorescent dye that exclusively binds to DNA. A technique called flow cytometry is then used to:\n\n1. assess the intensity of the fluorescence in each single cell,\n2. sort cells with different fluorescent intensities into collection tubes, and\n3. count the number of cells in each collection tube.\n\nThe following graph is generated.\n\n[figure2]\n\nWhat is the LEAST plausible reason as to why some cells may display a relative amount of DNA less than 1 arbitrary unit?\nA: Some cells may be dead, and their DNA may have degraded over time.\nB: The measurement of fluorescence may not be completely exact, leading to random fluctuations.\nC: The fluorescent dye may not bind to DNA perfectly.\nD: They are haploid cells lacking multiple chromosomes due to non-disjunction.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThis diagram summarises the phases of the cell cycle.\n\n[figure1]\n\nA sample of somatic cells in various stages of the cell cycle are labelled with a fluorescent dye that exclusively binds to DNA. A technique called flow cytometry is then used to:\n\n1. assess the intensity of the fluorescence in each single cell,\n2. sort cells with different fluorescent intensities into collection tubes, and\n3. count the number of cells in each collection tube.\n\nThe following graph is generated.\n\n[figure2]\n\nWhat is the LEAST plausible reason as to why some cells may display a relative amount of DNA less than 1 arbitrary unit?\n\nA: Some cells may be dead, and their DNA may have degraded over time.\nB: The measurement of fluorescence may not be completely exact, leading to random fluctuations.\nC: The fluorescent dye may not bind to DNA perfectly.\nD: They are haploid cells lacking multiple chromosomes due to non-disjunction.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-21.jpg?height=1085&width=1253&top_left_y=391&top_left_x=367",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-22.jpg?height=990&width=631&top_left_y=236&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1364",
        "problem": "Androgens and oestrogens are steroid (lipid) hormones that affect specific populations of cells in the human body. Which one of the following statements about this action is correct?\nA: Oestrogen receptors and androgen receptors are found in identical regions of the body\nB: The steroid hormones act by binding to receptors concentrated on the plasma membrane\nC: Androgen and oestrogen signalling control only one pathway in the body\nD: The lipid solubility of steroid hormones means that all cells in the body are exposed to circulating steroid hormones\nE: Androgen and oestrogen can dissolve in water and thus travel in blood plasma without a reliance on protein carriers\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAndrogens and oestrogens are steroid (lipid) hormones that affect specific populations of cells in the human body. Which one of the following statements about this action is correct?\n\nA: Oestrogen receptors and androgen receptors are found in identical regions of the body\nB: The steroid hormones act by binding to receptors concentrated on the plasma membrane\nC: Androgen and oestrogen signalling control only one pathway in the body\nD: The lipid solubility of steroid hormones means that all cells in the body are exposed to circulating steroid hormones\nE: Androgen and oestrogen can dissolve in water and thus travel in blood plasma without a reliance on protein carriers\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_597",
        "problem": "植物--丛枝菌根真菌 (AM, 脂质缺陷型) 共生是自然界中广泛存在的共生形式, AM 菌丝进入根皮层细胞内经过连续的双叉分枝形成从枝。AM 可以促进植物根系吸收磷酸盐, 植物为 $\\mathrm{AM}$ 提供脂质促进从枝形成。科学家研究了苜宿是如何调节脂质合成及输出,从而维持互惠共生稳定的分子机制, 结果如图, 下列说法正确的是（ ）\n[图1]\n\n细胞核\n\n$\\longrightarrow$ 激活转录, 线条越粗, 激活效果越强\n\n## 円抑制转录\n\n细胞核 STR: 脂质转运蛋白基因, 介导脂质向AM转运 PT4: 磷酸盐转运蛋白基因 $F a t M$ : 脂质合成基因 $E R M:$ 转录因子, 促进转录\nA: 从共生初期到后期, STR 基因表达量逐渐升高, 促进从枝形成\nB: ERF12 与 ERM 相互作用并在从枝形成过程中起协同作用\nC: ERF12 自我转录抑制的负反馈调节机制确保了共生的稳定性\nD: 敲除 STR、PT4 和 FatM 基因后萛落因减少脂质输出而长势更好\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n植物--丛枝菌根真菌 (AM, 脂质缺陷型) 共生是自然界中广泛存在的共生形式, AM 菌丝进入根皮层细胞内经过连续的双叉分枝形成从枝。AM 可以促进植物根系吸收磷酸盐, 植物为 $\\mathrm{AM}$ 提供脂质促进从枝形成。科学家研究了苜宿是如何调节脂质合成及输出,从而维持互惠共生稳定的分子机制, 结果如图, 下列说法正确的是（ ）\n[图1]\n\n细胞核\n\n$\\longrightarrow$ 激活转录, 线条越粗, 激活效果越强\n\n## 円抑制转录\n\n细胞核 STR: 脂质转运蛋白基因, 介导脂质向AM转运 PT4: 磷酸盐转运蛋白基因 $F a t M$ : 脂质合成基因 $E R M:$ 转录因子, 促进转录\n\nA: 从共生初期到后期, STR 基因表达量逐渐升高, 促进从枝形成\nB: ERF12 与 ERM 相互作用并在从枝形成过程中起协同作用\nC: ERF12 自我转录抑制的负反馈调节机制确保了共生的稳定性\nD: 敲除 STR、PT4 和 FatM 基因后萛落因减少脂质输出而长势更好\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-011.jpg?height=760&width=1302&top_left_y=161&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1085",
        "problem": "The rate of evolution varies in different lineages. For example: It is higher in rat lineage than in human lineage. Which of the following statements is correct?\nA: Rate of evolution would be the same for the coding and non-coding regions for a given species.\nB: Errors during DNA replication of somatic cells are the major source of mutations that leads to evolution.\nC: Rats have shorter generation time as compared to humans. Thus, more rounds of germ cell divisions would lead to more DNA replication errors. This can hasten rate of evolution.\nD: Humans show lower rate of metabolism than rats. This would lead to fewer errors during DNA replication thereby reducing the rate of evolution.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe rate of evolution varies in different lineages. For example: It is higher in rat lineage than in human lineage. Which of the following statements is correct?\n\nA: Rate of evolution would be the same for the coding and non-coding regions for a given species.\nB: Errors during DNA replication of somatic cells are the major source of mutations that leads to evolution.\nC: Rats have shorter generation time as compared to humans. Thus, more rounds of germ cell divisions would lead to more DNA replication errors. This can hasten rate of evolution.\nD: Humans show lower rate of metabolism than rats. This would lead to fewer errors during DNA replication thereby reducing the rate of evolution.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1262",
        "problem": "Vampire bats feed solely on blood and can only survive about two days without a meal. When a bat fails to find food it will \"beg\" from other members of the colony. The other bat will then regurgitate a small amount of blood, even if the \"begging\" bat is unrelated. This reciprocal altruism is\nA: largely limited to species living in stable social groups\nB: adaptive only if the aided individual returns the favour at a later date\nC: seen when there are likely to be negative consequences associated with not returning favours\nD: B and C only\nE: A, B and C\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nVampire bats feed solely on blood and can only survive about two days without a meal. When a bat fails to find food it will \"beg\" from other members of the colony. The other bat will then regurgitate a small amount of blood, even if the \"begging\" bat is unrelated. This reciprocal altruism is\n\nA: largely limited to species living in stable social groups\nB: adaptive only if the aided individual returns the favour at a later date\nC: seen when there are likely to be negative consequences associated with not returning favours\nD: B and C only\nE: A, B and C\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1377",
        "problem": "Cellular respiration is a complex process involving many steps. Generally, these steps are grouped into 3 main phases: glycolysis, the citric acid cycle, and the electron transport chain. The process of glycolysis can be summarised as follows:\n\n[figure1]\n\nGiven this information, how many carbon atoms does a pyruvate molecule contain?\nA: 1\nB: 2\nC: 3\nD: 4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCellular respiration is a complex process involving many steps. Generally, these steps are grouped into 3 main phases: glycolysis, the citric acid cycle, and the electron transport chain. The process of glycolysis can be summarised as follows:\n\n[figure1]\n\nGiven this information, how many carbon atoms does a pyruvate molecule contain?\n\nA: 1\nB: 2\nC: 3\nD: 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-34.jpg?height=1204&width=832&top_left_y=532&top_left_x=286"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1193",
        "problem": "The graph shows changes in mass of three individual sheep ticks when they were removed and kept for two days at low humidity and then four days at high humidity.\n\n[figure1]\n\nThe best deduction from these data is that\nA: At high humidity ticks absorb water from the atmosphere.\nB: At day 6 the ticks have recovered their body water.\nC: The loss of mass during the first two days is due to starvation.\nD: Individual ticks contain different amounts of water.\nE: Individual ticks lose water at the same rate when held in low humidity conditions.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows changes in mass of three individual sheep ticks when they were removed and kept for two days at low humidity and then four days at high humidity.\n\n[figure1]\n\nThe best deduction from these data is that\n\nA: At high humidity ticks absorb water from the atmosphere.\nB: At day 6 the ticks have recovered their body water.\nC: The loss of mass during the first two days is due to starvation.\nD: Individual ticks contain different amounts of water.\nE: Individual ticks lose water at the same rate when held in low humidity conditions.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-19.jpg?height=509&width=805&top_left_y=428&top_left_x=114"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1321",
        "problem": "Bacterial contamination of meat was studied in Karachi, Pakistan in 2010. Of the 342 bacterial pathogens isolated from meat samples, 120 (35\\%) were identified as Escherichia coli and 51 (15\\%) of these E. coli isolates were characterized as serotype $0157 ; \\mathrm{H} 7$, which is known to cause hemorrhagic colitis. Other potentially pathogenic isolates were Listeria species 14 (4\\%), Klebsiella 27 (8\\%), Enterobacter species 51 (15\\%), and Staphylococcus aureus 24 (7\\%). Antibiotic susceptibility was tested on the total microbial population and the results are shown below.\n\n[figure1]\n\nAli, et al. 2010. Microbial contamination of raw meat and its environment in retail shops in Karachi, Pakistan. J Infect Dev Ctries; $4(6): 382-388$.\n\nWhat is the probability that a pathogenic bacterium\nchosen at random would be E col 0157;H7 that is\nresistant to Novobiocin?\nA: 0%\nB: 10.50%\nC: 12.60%\nD: 70.15%\nE: 24.5%\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBacterial contamination of meat was studied in Karachi, Pakistan in 2010. Of the 342 bacterial pathogens isolated from meat samples, 120 (35\\%) were identified as Escherichia coli and 51 (15\\%) of these E. coli isolates were characterized as serotype $0157 ; \\mathrm{H} 7$, which is known to cause hemorrhagic colitis. Other potentially pathogenic isolates were Listeria species 14 (4\\%), Klebsiella 27 (8\\%), Enterobacter species 51 (15\\%), and Staphylococcus aureus 24 (7\\%). Antibiotic susceptibility was tested on the total microbial population and the results are shown below.\n\n[figure1]\n\nAli, et al. 2010. Microbial contamination of raw meat and its environment in retail shops in Karachi, Pakistan. J Infect Dev Ctries; $4(6): 382-388$.\n\nWhat is the probability that a pathogenic bacterium\nchosen at random would be E col 0157;H7 that is\nresistant to Novobiocin?\n\nA: 0%\nB: 10.50%\nC: 12.60%\nD: 70.15%\nE: 24.5%\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-21.jpg?height=726&width=1228&top_left_y=448&top_left_x=128"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_249",
        "problem": "DNA polymerase III holoenzyme consists of the pol III* and the $\\beta$-clamp which hold pol III* and DNA template together. You are investigating whether during lagging strand synthesis, the pol III* synthesizing the lagging strand dissociates from the $\\beta$-clamp as it finishes one Okazaki fragment and re-associates with another $\\beta$-clamp to begin making the next Okazaki fragment. You prepared a primed M13phage (a single stranded DNA and a primer) template (M13mp18) as a donor and loaded a $\\beta$ clamp and pol III* onto it.\n\nAs acceptors, you then added two more primed phage DNA templates, one (M13Gori) preloaded with a $\\beta$-clamp and the other ( $\\Phi$ X174) lacking a $\\beta$-clamp (Figure A). You incubated the templates together under replication conditions for $90 \\mathrm{~min}$ - long enough for the donor and acceptors to be replicated then performed gel electrophoresis (Figure B, lane 1-4). You then repeated the experiment but now loading a $\\beta$-clamp on X174 instead of M13Gori (Figure B, lane 5-8).\nA.\n\n[figure1]\nB.\n\n[figure2]\nA: After replication has finished, Pol III* dissociates from its original $\\beta$-clamp.\nB: The data are consistent with the notion that the $\\beta$-clamp is not absolutely required for replication. \nC: Pol III* prefers an acceptor template preloaded with a $\\beta$-clamp.\nD: Based on the results, one $\\beta$-clamp is enough to synthesize all Okazaki fragments on the lagging strand of each replication fork.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nDNA polymerase III holoenzyme consists of the pol III* and the $\\beta$-clamp which hold pol III* and DNA template together. You are investigating whether during lagging strand synthesis, the pol III* synthesizing the lagging strand dissociates from the $\\beta$-clamp as it finishes one Okazaki fragment and re-associates with another $\\beta$-clamp to begin making the next Okazaki fragment. You prepared a primed M13phage (a single stranded DNA and a primer) template (M13mp18) as a donor and loaded a $\\beta$ clamp and pol III* onto it.\n\nAs acceptors, you then added two more primed phage DNA templates, one (M13Gori) preloaded with a $\\beta$-clamp and the other ( $\\Phi$ X174) lacking a $\\beta$-clamp (Figure A). You incubated the templates together under replication conditions for $90 \\mathrm{~min}$ - long enough for the donor and acceptors to be replicated then performed gel electrophoresis (Figure B, lane 1-4). You then repeated the experiment but now loading a $\\beta$-clamp on X174 instead of M13Gori (Figure B, lane 5-8).\nA.\n\n[figure1]\nB.\n\n[figure2]\n\nA: After replication has finished, Pol III* dissociates from its original $\\beta$-clamp.\nB: The data are consistent with the notion that the $\\beta$-clamp is not absolutely required for replication. \nC: Pol III* prefers an acceptor template preloaded with a $\\beta$-clamp.\nD: Based on the results, one $\\beta$-clamp is enough to synthesize all Okazaki fragments on the lagging strand of each replication fork.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-16.jpg?height=539&width=1100&top_left_y=947&top_left_x=478",
            "https://i.postimg.cc/4y1b8L6h/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1088",
        "problem": "A cladogram that classifies five species ( $M, N, O, P$ and $Q$ ) is shown below.\n\n[figure1]\n\nWhich of the six dots in the cladogram corresponds to the most recent common ancestor of $N$ and $P$ ?\nA: 2\nB: 3\nC: 5\nD: 6\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA cladogram that classifies five species ( $M, N, O, P$ and $Q$ ) is shown below.\n\n[figure1]\n\nWhich of the six dots in the cladogram corresponds to the most recent common ancestor of $N$ and $P$ ?\n\nA: 2\nB: 3\nC: 5\nD: 6\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-13.jpg?height=674&width=1090&top_left_y=1517&top_left_x=515"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_573",
        "problem": "某家系甲、乙两种单基因遗传病的系谱图如图 1 所示。控制两病的基因均不在 Y 染色体上, 已死亡个体无法知道其性状。图 2 是对控制甲病的基因通过 PCR 技术后用某限制酶处理并电泳的结果图。外周血相对骨髓而言，指被造血器官释放到血管里参与血液循环的血。下列相关叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 乙病是常染色体隐性遗传病的依据是 $\\mathrm{II}_{3} 、 \\mathrm{II}_{4}$ 正常, 生出了患乙病的 $\\mathrm{III}_{10}$\nB: 判定甲病是伴 X 显性遗传病, $\\mathrm{II}_{7}$ 和 II 8 再生育一位健康女孩的概率是 $1 / 6$\nC: 从电泳的结果可知致病基因的限制酶的切割位点移往原基因长的片段\nD: $\\mathrm{II}_{6}$ 和一位家族无相关病史的女性结婚, 抽取母体外周血进行基因检测可以判断胎儿是否患病\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某家系甲、乙两种单基因遗传病的系谱图如图 1 所示。控制两病的基因均不在 Y 染色体上, 已死亡个体无法知道其性状。图 2 是对控制甲病的基因通过 PCR 技术后用某限制酶处理并电泳的结果图。外周血相对骨髓而言，指被造血器官释放到血管里参与血液循环的血。下列相关叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 乙病是常染色体隐性遗传病的依据是 $\\mathrm{II}_{3} 、 \\mathrm{II}_{4}$ 正常, 生出了患乙病的 $\\mathrm{III}_{10}$\nB: 判定甲病是伴 X 显性遗传病, $\\mathrm{II}_{7}$ 和 II 8 再生育一位健康女孩的概率是 $1 / 6$\nC: 从电泳的结果可知致病基因的限制酶的切割位点移往原基因长的片段\nD: $\\mathrm{II}_{6}$ 和一位家族无相关病史的女性结婚, 抽取母体外周血进行基因检测可以判断胎儿是否患病\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-86.jpg?height=320&width=577&top_left_y=183&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-86.jpg?height=360&width=851&top_left_y=160&top_left_x=934"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_983",
        "problem": "[figure1]\n\nThe diagram above depicts:\nA: Developing human sperm cells\nB: Human blood smear\nC: Human lung section\nD: Frog eggs\nE: Frog blood smear\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nThe diagram above depicts:\n\nA: Developing human sperm cells\nB: Human blood smear\nC: Human lung section\nD: Frog eggs\nE: Frog blood smear\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_162746b98e552b6160afg-07.jpg?height=819&width=1030&top_left_y=290&top_left_x=320"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_211",
        "problem": "The glycoprotein 4-1BB is a receptor that is highly expressed on the surface of active Tcells. The 4-1BB ligand (4-1BBL) is a molecule that binds to and activates 4-1BB. It is found strongly expressed on antigen-presenting cells. Bidirectional signals of 4-1BB and 4-1BBL interaction increase the activity of white blood cells and increase the production and secretion of cytokines, such as MCP-1 that is an important factor promoting the infiltration of leukocytes (Figure Q.70). Currently, many studies have shown a relationship between the signaling pathways via $4-1 \\mathrm{BB} / 4-1 \\mathrm{BBL}$ interaction and several human diseases, including those related to metabolism.\n\n[figure1]\n\nFigure Q. 70.\nA: Inhibition of 4-1BB expression diminishes the development of atherosclerosis.\nB: Activation of 4-1BB limits the effect of autoimmune diseases on the body.\nC: All three kinds of cells, macrophages, dendritic cells and natural killer cells strongly express 4-1BBL.\nD: Blockade of the 4-1BB and 4-1BBL interaction increases graft tolerance.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe glycoprotein 4-1BB is a receptor that is highly expressed on the surface of active Tcells. The 4-1BB ligand (4-1BBL) is a molecule that binds to and activates 4-1BB. It is found strongly expressed on antigen-presenting cells. Bidirectional signals of 4-1BB and 4-1BBL interaction increase the activity of white blood cells and increase the production and secretion of cytokines, such as MCP-1 that is an important factor promoting the infiltration of leukocytes (Figure Q.70). Currently, many studies have shown a relationship between the signaling pathways via $4-1 \\mathrm{BB} / 4-1 \\mathrm{BBL}$ interaction and several human diseases, including those related to metabolism.\n\n[figure1]\n\nFigure Q. 70.\n\nA: Inhibition of 4-1BB expression diminishes the development of atherosclerosis.\nB: Activation of 4-1BB limits the effect of autoimmune diseases on the body.\nC: All three kinds of cells, macrophages, dendritic cells and natural killer cells strongly express 4-1BBL.\nD: Blockade of the 4-1BB and 4-1BBL interaction increases graft tolerance.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-050.jpg?height=869&width=1128&top_left_y=1024&top_left_x=422"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_704",
        "problem": "某家系中有甲（设基因为 $A 、 a$ ）、乙（设基因为 $B 、 b$ ）两种单基因遗传病（如下图)，其中一种是伴性遗传病。不考虑基因突变，下列分析错误的是\n\n[图1]\nA: $\\mathrm{II}_{3}$ 的致病基因来自 $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$\nB: $\\mathrm{II}_{2}$ 的基因型为 $\\mathrm{aaX} \\mathrm{X}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}, \\mathrm{III}_{8}$ 基因型有 4 种可能\nC: 若 $\\mathrm{III}_{6}$ 的性染色体组成为 $\\mathrm{XXY}$, 原因是 $\\mathrm{II}_{4}$ 形成卵细胞时减数第二次分裂出现异常\nD: 若 $\\mathrm{III}_{4}$ 与 $\\mathrm{III}_{5}$ 结婚, 生育一患两种病孩子的概率是 $1 / 12$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某家系中有甲（设基因为 $A 、 a$ ）、乙（设基因为 $B 、 b$ ）两种单基因遗传病（如下图)，其中一种是伴性遗传病。不考虑基因突变，下列分析错误的是\n\n[图1]\n\nA: $\\mathrm{II}_{3}$ 的致病基因来自 $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$\nB: $\\mathrm{II}_{2}$ 的基因型为 $\\mathrm{aaX} \\mathrm{X}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}, \\mathrm{III}_{8}$ 基因型有 4 种可能\nC: 若 $\\mathrm{III}_{6}$ 的性染色体组成为 $\\mathrm{XXY}$, 原因是 $\\mathrm{II}_{4}$ 形成卵细胞时减数第二次分裂出现异常\nD: 若 $\\mathrm{III}_{4}$ 与 $\\mathrm{III}_{5}$ 结婚, 生育一患两种病孩子的概率是 $1 / 12$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-049.jpg?height=374&width=1246&top_left_y=190&top_left_x=362"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1257",
        "problem": "## MARINE BIODIVERSITY - THE CENSUS OF MARINE LIFE\n\nThe Census of Marine Life (2000-2010) is the largest global research programme on marine biodiversity. Its findings have recently been integrated in an article authored by Mark Costello who is based at Leigh Marine Laboratory, University of Auckland (PLoS ONE 5(8): e12110). The study found that many habitats were poorly sampled and that there are major gaps in our knowledge of marine organisms worldwide that limit our ability to understand species of economic and ecological importance.\n\nThe table below gives the number of endemic plants, invertebrates, and vertebrates reported for specific geographic regions. Endemic species are those found only in one specific geographical area.\n\n|  |  |  |  |  |  |  |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| NRIC region | Plants | Invertebrates | Fish | Other vertebrates | Total | Number of species | $\\%$ endemics |\n| Antarctica | - | - | - | - | 3,700 | 8,200 | 45 |\n| Australia | - | 7987 | 1298 | - | 9,286 | 32,889 | 28 |\n| Baltic | 1 | 0 | 0 | 0 | 1 | 5,865 | 2 |\n| Caribbean | - | 868 | 704 | 1 | 1,573 | 12,046 | 13 |\n| China | 142 | 1387 | 70 | 2 | 1,601 | 22,365 | 7 |\n| Japan | - | 1508 | 364 | 0 | 1,872 | 32,777 | 6 |\n| Mediterranean | 171 | 844 | 80 | 3 | 1,098 | 16,845 | 7 |\n| New Zealand | 225 | 6014 | 278 | 43 | 6,560 | 12,780 | 51 |\n| South Africa | - | 3269 | 280 | - | 3,549 | 12,715 | 28 |\n| Total | 538 | 21,639 | 3,074 | 49 | 25,300 | 150,617 | 17 |\n\ndoi:10.1371/journal.pone.0012110.t004\n\nNote: - means no specific data is available.Which area has the greatest proportion of endemic species?\nA: Antarctica\nB: Australia\nC: Japan\nD: New Zealand\nE: South Africa\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## MARINE BIODIVERSITY - THE CENSUS OF MARINE LIFE\n\nThe Census of Marine Life (2000-2010) is the largest global research programme on marine biodiversity. Its findings have recently been integrated in an article authored by Mark Costello who is based at Leigh Marine Laboratory, University of Auckland (PLoS ONE 5(8): e12110). The study found that many habitats were poorly sampled and that there are major gaps in our knowledge of marine organisms worldwide that limit our ability to understand species of economic and ecological importance.\n\nThe table below gives the number of endemic plants, invertebrates, and vertebrates reported for specific geographic regions. Endemic species are those found only in one specific geographical area.\n\n|  |  |  |  |  |  |  |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| NRIC region | Plants | Invertebrates | Fish | Other vertebrates | Total | Number of species | $\\%$ endemics |\n| Antarctica | - | - | - | - | 3,700 | 8,200 | 45 |\n| Australia | - | 7987 | 1298 | - | 9,286 | 32,889 | 28 |\n| Baltic | 1 | 0 | 0 | 0 | 1 | 5,865 | 2 |\n| Caribbean | - | 868 | 704 | 1 | 1,573 | 12,046 | 13 |\n| China | 142 | 1387 | 70 | 2 | 1,601 | 22,365 | 7 |\n| Japan | - | 1508 | 364 | 0 | 1,872 | 32,777 | 6 |\n| Mediterranean | 171 | 844 | 80 | 3 | 1,098 | 16,845 | 7 |\n| New Zealand | 225 | 6014 | 278 | 43 | 6,560 | 12,780 | 51 |\n| South Africa | - | 3269 | 280 | - | 3,549 | 12,715 | 28 |\n| Total | 538 | 21,639 | 3,074 | 49 | 25,300 | 150,617 | 17 |\n\ndoi:10.1371/journal.pone.0012110.t004\n\nNote: - means no specific data is available.\n\nproblem:\nWhich area has the greatest proportion of endemic species?\n\nA: Antarctica\nB: Australia\nC: Japan\nD: New Zealand\nE: South Africa\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_532",
        "problem": "现有 DNA 分子的两条单链均只含有 ${ }^{14} \\mathrm{~N}$ (表示为 ${ }^{14} \\mathrm{~N}^{14} \\mathrm{~N}$ ) 的大肠杆菌, 若将该大肠杆菌在含有 ${ }^{15} \\mathrm{~N}$ 的培养基中繁殖两代，再转到含有 ${ }^{14} \\mathrm{~N}$ 的培养基中繁殖一代，则理论上 DNA 分子的组成类型和比例分别是\nA: 有 ${ }^{15} \\mathrm{~N}^{14} \\mathrm{~N}$ 和 ${ }^{14} \\mathrm{~N}{ }^{14} \\mathrm{~N}$ 两种，其比例为 1：3\nB: 有 ${ }^{15} \\mathrm{~N}^{15} \\mathrm{~N}$ 和 ${ }^{14} \\mathrm{~N}^{14} \\mathrm{~N}$ 两种, 其比例为 1：1\nC: 有 ${ }^{15} \\mathrm{~N}^{15} \\mathrm{~N}$ 和 ${ }^{14} \\mathrm{~N} \\mathrm{~N}^{14} \\mathrm{~N}$ 两种，其比例为 3: 1\nD: 有 ${ }^{15} \\mathrm{~N}{ }^{14} \\mathrm{~N}$ 和 ${ }^{14} \\mathrm{~N} 1{ }^{4} \\mathrm{~N}$ 两种, 其比例为 3: 1\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n现有 DNA 分子的两条单链均只含有 ${ }^{14} \\mathrm{~N}$ (表示为 ${ }^{14} \\mathrm{~N}^{14} \\mathrm{~N}$ ) 的大肠杆菌, 若将该大肠杆菌在含有 ${ }^{15} \\mathrm{~N}$ 的培养基中繁殖两代，再转到含有 ${ }^{14} \\mathrm{~N}$ 的培养基中繁殖一代，则理论上 DNA 分子的组成类型和比例分别是\n\nA: 有 ${ }^{15} \\mathrm{~N}^{14} \\mathrm{~N}$ 和 ${ }^{14} \\mathrm{~N}{ }^{14} \\mathrm{~N}$ 两种，其比例为 1：3\nB: 有 ${ }^{15} \\mathrm{~N}^{15} \\mathrm{~N}$ 和 ${ }^{14} \\mathrm{~N}^{14} \\mathrm{~N}$ 两种, 其比例为 1：1\nC: 有 ${ }^{15} \\mathrm{~N}^{15} \\mathrm{~N}$ 和 ${ }^{14} \\mathrm{~N} \\mathrm{~N}^{14} \\mathrm{~N}$ 两种，其比例为 3: 1\nD: 有 ${ }^{15} \\mathrm{~N}{ }^{14} \\mathrm{~N}$ 和 ${ }^{14} \\mathrm{~N} 1{ }^{4} \\mathrm{~N}$ 两种, 其比例为 3: 1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1481",
        "problem": "The figure shows oxygen consumption (respiration) in mitochondria when ADP, or drugs (DNP or DCCD) are added. The solution already contains respiratory substrates, oxygen and phosphate.\n\n## $\\mathrm{O}_{2}$ concentration\n\n[figure1]\n\n\nWhich is false?\nA: The mitochondria are able to use the added ADP.\nB: In normal conditions, the mitochondria respire only when ATP can be produced.\nC: DNP stimulates ATP synthesis.\nD: DCCD inhibits ATP synthesis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe figure shows oxygen consumption (respiration) in mitochondria when ADP, or drugs (DNP or DCCD) are added. The solution already contains respiratory substrates, oxygen and phosphate.\n\n## $\\mathrm{O}_{2}$ concentration\n\n[figure1]\n\n\nWhich is false?\n\nA: The mitochondria are able to use the added ADP.\nB: In normal conditions, the mitochondria respire only when ATP can be produced.\nC: DNP stimulates ATP synthesis.\nD: DCCD inhibits ATP synthesis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-30.jpg?height=706&width=1354&top_left_y=532&top_left_x=498"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_740",
        "problem": "转铁蛋白受体参与细胞对 $\\mathrm{Fe}^{3+}$ 的吸收, $\\mathrm{Fe}^{3+}$ 大量积累对细胞有毒害。下图是细胞中 $\\mathrm{Fe}^{3+}$含量对转铁蛋白受体 mRNA 稳定性的调节过程 (图中铁反应元件是转铁蛋白受体 mRNA 上一段富含碱基 $\\mathrm{A} 、 \\mathrm{U}$ 的序列）。结合图文分析，描述错误的是（）\n\n[图1]\nA: 铁反应元件形成的茎环结构能影响转铁蛋白受体的氨基酸序列, 从而影响该受体的结构\nB: 铁反应元件能形成茎环结构的原因是该片段存在能自身互补配对的碱基序列\nC: 合成转铁蛋白受体的模板是 mRNA, 该模板被彻底水解的产物是磷酸、核糖和含氮碱基\nD: 这种调节机制既可以避免 $\\mathrm{Fe}^{3+}$ 对细胞的毒性影响, 又可以减少细胞内物质和能量的浪费\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n转铁蛋白受体参与细胞对 $\\mathrm{Fe}^{3+}$ 的吸收, $\\mathrm{Fe}^{3+}$ 大量积累对细胞有毒害。下图是细胞中 $\\mathrm{Fe}^{3+}$含量对转铁蛋白受体 mRNA 稳定性的调节过程 (图中铁反应元件是转铁蛋白受体 mRNA 上一段富含碱基 $\\mathrm{A} 、 \\mathrm{U}$ 的序列）。结合图文分析，描述错误的是（）\n\n[图1]\n\nA: 铁反应元件形成的茎环结构能影响转铁蛋白受体的氨基酸序列, 从而影响该受体的结构\nB: 铁反应元件能形成茎环结构的原因是该片段存在能自身互补配对的碱基序列\nC: 合成转铁蛋白受体的模板是 mRNA, 该模板被彻底水解的产物是磷酸、核糖和含氮碱基\nD: 这种调节机制既可以避免 $\\mathrm{Fe}^{3+}$ 对细胞的毒性影响, 又可以减少细胞内物质和能量的浪费\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-006.jpg?height=500&width=1016&top_left_y=178&top_left_x=380"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1479",
        "problem": "Dynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\nThe start of two different alignments was given in the example.\n\nHow should line 1 be extended? i.e. the missing characters: $G C A-T X X X X X X X X X X X$\nA: GCT\nB: CGT\nC: CCT\nD: G-T\nE: $T C G$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\nThe start of two different alignments was given in the example.\n\nHow should line 1 be extended? i.e. the missing characters: $G C A-T X X X X X X X X X X X$\n\nA: GCT\nB: CGT\nC: CCT\nD: G-T\nE: $T C G$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=925&width=928&top_left_y=354&top_left_x=238",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=922&width=928&top_left_y=1458&top_left_x=238"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1478",
        "problem": "Blue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nWhich method is better for estimating the change in whale population?\nA: Mark-recapture.\nB: Counting from images.\nC: They are equal.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBlue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nWhich method is better for estimating the change in whale population?\n\nA: Mark-recapture.\nB: Counting from images.\nC: They are equal.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-36.jpg?height=948&width=1673&top_left_y=657&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_725",
        "problem": "某种动物的眼色由两对独立遗传的等位基因 (A、a 和 $B 、 b$ )控制, 具体控制关系如图。相关叙述正确的是( )\n\n[图1]\nA: A 基因正常表达时, 以任一链为模板转录和翻译产生酶 A\nB: B 基因上可结合多个核糖体，以提高酶 B 的合成效率\nC: 该动物群体中无色眼的基因型只有 1 种，猩红色眼对应的基因型有 4 种\nD: 若一对无色眼亲本所形成的受精卵中基因 $\\mathrm{a}$ 或 $\\mathrm{b}$ 发生突变, 发育成的子代为深红色眼\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种动物的眼色由两对独立遗传的等位基因 (A、a 和 $B 、 b$ )控制, 具体控制关系如图。相关叙述正确的是( )\n\n[图1]\n\nA: A 基因正常表达时, 以任一链为模板转录和翻译产生酶 A\nB: B 基因上可结合多个核糖体，以提高酶 B 的合成效率\nC: 该动物群体中无色眼的基因型只有 1 种，猩红色眼对应的基因型有 4 种\nD: 若一对无色眼亲本所形成的受精卵中基因 $\\mathrm{a}$ 或 $\\mathrm{b}$ 发生突变, 发育成的子代为深红色眼\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-003.jpg?height=311&width=1131&top_left_y=730&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1295",
        "problem": "## MARINE BIODIVERSITY - THE CENSUS OF MARINE LIFE\n\nThe Census of Marine Life (2000-2010) is the largest global research programme on marine biodiversity. Its findings have recently been integrated in an article authored by Mark Costello who is based at Leigh Marine Laboratory, University of Auckland (PLoS ONE 5(8): e12110). The study found that many habitats were poorly sampled and that there are major gaps in our knowledge of marine organisms worldwide that limit our ability to understand species of economic and ecological importance.\n\nThe table below gives the number of endemic plants, invertebrates, and vertebrates reported for specific geographic regions. Endemic species are those found only in one specific geographical area.\n\n|  |  |  |  |  |  |  |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| NRIC region | Plants | Invertebrates | Fish | Other vertebrates | Total | Number of species | $\\%$ endemics |\n| Antarctica | - | - | - | - | 3,700 | 8,200 | 45 |\n| Australia | - | 7987 | 1298 | - | 9,286 | 32,889 | 28 |\n| Baltic | 1 | 0 | 0 | 0 | 1 | 5,865 | 2 |\n| Caribbean | - | 868 | 704 | 1 | 1,573 | 12,046 | 13 |\n| China | 142 | 1387 | 70 | 2 | 1,601 | 22,365 | 7 |\n| Japan | - | 1508 | 364 | 0 | 1,872 | 32,777 | 6 |\n| Mediterranean | 171 | 844 | 80 | 3 | 1,098 | 16,845 | 7 |\n| New Zealand | 225 | 6014 | 278 | 43 | 6,560 | 12,780 | 51 |\n| South Africa | - | 3269 | 280 | - | 3,549 | 12,715 | 28 |\n| Total | 538 | 21,639 | 3,074 | 49 | 25,300 | 150,617 | 17 |\n\ndoi:10.1371/journal.pone.0012110.t004\n\nNote: - means no specific data is available.Which of the following statements is INCORRECT?\nA: Antarctica has no endemic fish.\nB: Australia has the greatest number of reported marine species.\nC: The Baltic has low endemism.\nD: Japan has no endemic other vertebrates reported.\nE: South Africa has a high proportion of marine invertebrates that are endemic.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## MARINE BIODIVERSITY - THE CENSUS OF MARINE LIFE\n\nThe Census of Marine Life (2000-2010) is the largest global research programme on marine biodiversity. Its findings have recently been integrated in an article authored by Mark Costello who is based at Leigh Marine Laboratory, University of Auckland (PLoS ONE 5(8): e12110). The study found that many habitats were poorly sampled and that there are major gaps in our knowledge of marine organisms worldwide that limit our ability to understand species of economic and ecological importance.\n\nThe table below gives the number of endemic plants, invertebrates, and vertebrates reported for specific geographic regions. Endemic species are those found only in one specific geographical area.\n\n|  |  |  |  |  |  |  |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| NRIC region | Plants | Invertebrates | Fish | Other vertebrates | Total | Number of species | $\\%$ endemics |\n| Antarctica | - | - | - | - | 3,700 | 8,200 | 45 |\n| Australia | - | 7987 | 1298 | - | 9,286 | 32,889 | 28 |\n| Baltic | 1 | 0 | 0 | 0 | 1 | 5,865 | 2 |\n| Caribbean | - | 868 | 704 | 1 | 1,573 | 12,046 | 13 |\n| China | 142 | 1387 | 70 | 2 | 1,601 | 22,365 | 7 |\n| Japan | - | 1508 | 364 | 0 | 1,872 | 32,777 | 6 |\n| Mediterranean | 171 | 844 | 80 | 3 | 1,098 | 16,845 | 7 |\n| New Zealand | 225 | 6014 | 278 | 43 | 6,560 | 12,780 | 51 |\n| South Africa | - | 3269 | 280 | - | 3,549 | 12,715 | 28 |\n| Total | 538 | 21,639 | 3,074 | 49 | 25,300 | 150,617 | 17 |\n\ndoi:10.1371/journal.pone.0012110.t004\n\nNote: - means no specific data is available.\n\nproblem:\nWhich of the following statements is INCORRECT?\n\nA: Antarctica has no endemic fish.\nB: Australia has the greatest number of reported marine species.\nC: The Baltic has low endemism.\nD: Japan has no endemic other vertebrates reported.\nE: South Africa has a high proportion of marine invertebrates that are endemic.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_292",
        "problem": "In figure a, the pain and temperature sensory pathway is depicted. The first-order neuron enters the spinal cord and forms a synapse with the second-order neuron, which its axon ends in the thalamus. The third-order neuron transmits the information to the brain cortex. Figure $b$ shows the vibration and proprioception (joint position) sensory pathway. The axon terminals of the first-order neurons are located in the brain stem and form synapses with second-order neuron, which crosses the midline and their axons end in the thalamus. finally, the sensory information is conducted to the cortex via the third-order neurons.\n\nFollowing a spinal injury, the right ventral and left dorsal sides of the white matter in thoracolumbar spinal cord are damaged.\n\n[figure1]\nA: The patient has impaired temperature sensation in right hand.\nB: The patient has impaired vibration sensation in left leg.\nC: The patient has impaired joint position sensation in left leg and temperature sensation in right leg.\nD: The patient has impaired pain sensation in left leg.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn figure a, the pain and temperature sensory pathway is depicted. The first-order neuron enters the spinal cord and forms a synapse with the second-order neuron, which its axon ends in the thalamus. The third-order neuron transmits the information to the brain cortex. Figure $b$ shows the vibration and proprioception (joint position) sensory pathway. The axon terminals of the first-order neurons are located in the brain stem and form synapses with second-order neuron, which crosses the midline and their axons end in the thalamus. finally, the sensory information is conducted to the cortex via the third-order neurons.\n\nFollowing a spinal injury, the right ventral and left dorsal sides of the white matter in thoracolumbar spinal cord are damaged.\n\n[figure1]\n\nA: The patient has impaired temperature sensation in right hand.\nB: The patient has impaired vibration sensation in left leg.\nC: The patient has impaired joint position sensation in left leg and temperature sensation in right leg.\nD: The patient has impaired pain sensation in left leg.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-19.jpg?height=694&width=1564&top_left_y=818&top_left_x=246"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_779",
        "problem": "有氧运动能改变骨骼肌细胞中的 DNA 甲基化状态, 引发骨骼肌的结构和代谢变化,改善肥胖、延缓衰老。下列相关叙述正确的是（ ）\nA: DNA 甲基化能改变骨骼肌细胞中基因的碱基序列\nB: 骨骼肌细胞中的 DNA 甲基化状态可以遗传给后代\nC: DNA 甲基化程度可能影响代谢相关酶基因的转录\nD: 甲基化若发生在染色体的组蛋白上则不影响基因表达\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n有氧运动能改变骨骼肌细胞中的 DNA 甲基化状态, 引发骨骼肌的结构和代谢变化,改善肥胖、延缓衰老。下列相关叙述正确的是（ ）\n\nA: DNA 甲基化能改变骨骼肌细胞中基因的碱基序列\nB: 骨骼肌细胞中的 DNA 甲基化状态可以遗传给后代\nC: DNA 甲基化程度可能影响代谢相关酶基因的转录\nD: 甲基化若发生在染色体的组蛋白上则不影响基因表达\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_994",
        "problem": "Which one of these hormones is NOT secreted by the anterior pituitary gland?\nA: Growth hormone\nB: Prolactin\nC: Antidiuretic hormone\nD: Luteinizing hormone\nE: Adrenocorticotropic hormone\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of these hormones is NOT secreted by the anterior pituitary gland?\n\nA: Growth hormone\nB: Prolactin\nC: Antidiuretic hormone\nD: Luteinizing hormone\nE: Adrenocorticotropic hormone\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1038",
        "problem": "From which geological period over 500 million years ago did fossils first show all of the major body plans we see today?\nA: Burgess period\nB: Cambrian period\nC: Carboniferous period\nD: Cretaceous period\nE: Ediacaran period\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFrom which geological period over 500 million years ago did fossils first show all of the major body plans we see today?\n\nA: Burgess period\nB: Cambrian period\nC: Carboniferous period\nD: Cretaceous period\nE: Ediacaran period\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_403",
        "problem": "某自花传粉机物紫茎（A）对绿茎（a）为显性，抗病（B）对感病（b）为显性,这两对基因分别位于两对同源染色体上、且当在杨舍 $\\mathrm{AB}$ 基因时不能萌发长出花粉管,因而不能参与受精作用。下列叙述不正确的是（）\nA: 这两对基因的遗传遵循基因的自由组合定律\nB: 基因型为 $\\mathrm{AaBb}$ 和 $a a b b$ 的植株正反交子代性状及比例不同\nC: 两紫茎抗病性状植株杂交, 后代可能不会出现性状分离\nD: 用单倍体育种的方法不能得到基因型为 AABB 的植株\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某自花传粉机物紫茎（A）对绿茎（a）为显性，抗病（B）对感病（b）为显性,这两对基因分别位于两对同源染色体上、且当在杨舍 $\\mathrm{AB}$ 基因时不能萌发长出花粉管,因而不能参与受精作用。下列叙述不正确的是（）\n\nA: 这两对基因的遗传遵循基因的自由组合定律\nB: 基因型为 $\\mathrm{AaBb}$ 和 $a a b b$ 的植株正反交子代性状及比例不同\nC: 两紫茎抗病性状植株杂交, 后代可能不会出现性状分离\nD: 用单倍体育种的方法不能得到基因型为 AABB 的植株\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_224",
        "problem": "There are four types of bases used for RNA - A, C, G and U - while for DNA there are four types of bases, A, C, G, and T. I wondered why thymine T could only be used for DNA and looked closely at the base-pairing pattern (Figure 1).\n\nIt is reported that the mutant strain of the certain gene in Escherichia coli sometimes incorporates dUTP in place of thymine to include bases in the DNA strand. This frequently results in a new mutation. In a chemistry lecture, I learned that compounds such as bases could undergo chemical changes (mainly hydrolytic deamination) by reacting with certain water molecules even under in vivo conditions.\n[figure1]\n\nFigure 1\nA: Chemical changes made to RNA bases are not repaired.\nB: Chemical changes that occur in cytosine bases are the main reason that thymine bases are used only in DNA.\nC: E. coli mutant strains that incorporate uracil instead of thymine are more likely to mutate the A-T base pair.\nD: E. coli mutant cells containing uracil bases in the DNA chain are susceptible to chemical changes in uracil bases, so that new mutations occur frequently.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThere are four types of bases used for RNA - A, C, G and U - while for DNA there are four types of bases, A, C, G, and T. I wondered why thymine T could only be used for DNA and looked closely at the base-pairing pattern (Figure 1).\n\nIt is reported that the mutant strain of the certain gene in Escherichia coli sometimes incorporates dUTP in place of thymine to include bases in the DNA strand. This frequently results in a new mutation. In a chemistry lecture, I learned that compounds such as bases could undergo chemical changes (mainly hydrolytic deamination) by reacting with certain water molecules even under in vivo conditions.\n[figure1]\n\nFigure 1\n\nA: Chemical changes made to RNA bases are not repaired.\nB: Chemical changes that occur in cytosine bases are the main reason that thymine bases are used only in DNA.\nC: E. coli mutant strains that incorporate uracil instead of thymine are more likely to mutate the A-T base pair.\nD: E. coli mutant cells containing uracil bases in the DNA chain are susceptible to chemical changes in uracil bases, so that new mutations occur frequently.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-14.jpg?height=688&width=966&top_left_y=1018&top_left_x=242"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1516",
        "problem": "At a temple in Bali, macaques steal things from tourists and barter them for food. The macaques steal high-end items like smartphones more often than easily-accessible items like hats. The macaques then wait until the tourist offers them fruit. The macaques only return the most valuable items when a large amount of fruit has been offered, whereas cheaper items are returned for less fruit.\n\n[figure1]\n\nWhy do the macaques want fruit?\nA: Reasoning\nB: Instinct\nC: Association/conditioning\nD: Habituation\nE: Imprinting\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAt a temple in Bali, macaques steal things from tourists and barter them for food. The macaques steal high-end items like smartphones more often than easily-accessible items like hats. The macaques then wait until the tourist offers them fruit. The macaques only return the most valuable items when a large amount of fruit has been offered, whereas cheaper items are returned for less fruit.\n\n[figure1]\n\nWhy do the macaques want fruit?\n\nA: Reasoning\nB: Instinct\nC: Association/conditioning\nD: Habituation\nE: Imprinting\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-15.jpg?height=1397&width=945&top_left_y=555&top_left_x=227"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1241",
        "problem": "The following question relates to organisms from the planet Orti.\n\nDogids, ratids, catids and humanids have red spots.\n\nCatids and humanids both have blue skin.\n\nRatids and goatids have yellow ears.\n\nHumanids, dogids and catids have green ears.\n\nWhich of the following groups represents a possible evolution of the above characteristics, provided that each characteristic only evolved once?\nA: ![]([figure1])\nB: ![]([figure2])\nC: ![]([figure3])\nD: ![]([figure4])\nE: ![]([figure5])\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following question relates to organisms from the planet Orti.\n\nDogids, ratids, catids and humanids have red spots.\n\nCatids and humanids both have blue skin.\n\nRatids and goatids have yellow ears.\n\nHumanids, dogids and catids have green ears.\n\nWhich of the following groups represents a possible evolution of the above characteristics, provided that each characteristic only evolved once?\n\nA: ![]([figure1])\nB: ![]([figure2])\nC: ![]([figure3])\nD: ![]([figure4])\nE: ![]([figure5])\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-04.jpg?height=377&width=508&top_left_y=1665&top_left_x=197",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-04.jpg?height=342&width=508&top_left_y=1665&top_left_x=1162",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-04.jpg?height=334&width=511&top_left_y=2015&top_left_x=196",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-04.jpg?height=339&width=505&top_left_y=2052&top_left_x=1164",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-04.jpg?height=340&width=511&top_left_y=2354&top_left_x=193"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_279",
        "problem": "Five distinctive characteristics that, taken together, set chordates apart from all other phyla are notochord, dorsal tubular nerve cord, pharyngeal pouches or slits, an iodine secreting organ (endostyle), and post-anal tail. These characteristics are always found at some embryonic stage, although they may be altered or may disappear in later stages of the life cycle. Tunicates belong to Deuterostomia group of Metazoa and are found in all seas from near shoreline to great depths. Most are sessile as adults, although some are free-living. The name \"tunicate\" refers to the usually tough and non-living tunic that surrounds the animal and contains cellulose-like components. As adults, tunicates are highly specialized chordates, for in most species only the larval form, which resembles a microscopic tadpole, bears all the chordate hallmarks. Tadpole undergoes metamorphosis and changes its morphology and anatomy.\n\n[figure1]\nA: Adult of tunicates shows two shared derived characters, which are present in all chordates.\nB: Main pathway for seawater is: incurrent siphon, gut, anus, atrium and excurrent siphon.\nC: In Ciona, the notochord is homologous to backbone of vertebrates.\nD: Ciona belongs to the coelomate vertebrate group and coelom forms from fusion of enterocoelous pouches.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFive distinctive characteristics that, taken together, set chordates apart from all other phyla are notochord, dorsal tubular nerve cord, pharyngeal pouches or slits, an iodine secreting organ (endostyle), and post-anal tail. These characteristics are always found at some embryonic stage, although they may be altered or may disappear in later stages of the life cycle. Tunicates belong to Deuterostomia group of Metazoa and are found in all seas from near shoreline to great depths. Most are sessile as adults, although some are free-living. The name \"tunicate\" refers to the usually tough and non-living tunic that surrounds the animal and contains cellulose-like components. As adults, tunicates are highly specialized chordates, for in most species only the larval form, which resembles a microscopic tadpole, bears all the chordate hallmarks. Tadpole undergoes metamorphosis and changes its morphology and anatomy.\n\n[figure1]\n\nA: Adult of tunicates shows two shared derived characters, which are present in all chordates.\nB: Main pathway for seawater is: incurrent siphon, gut, anus, atrium and excurrent siphon.\nC: In Ciona, the notochord is homologous to backbone of vertebrates.\nD: Ciona belongs to the coelomate vertebrate group and coelom forms from fusion of enterocoelous pouches.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-36.jpg?height=1013&width=714&top_left_y=844&top_left_x=677"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_864",
        "problem": "某果蝇的一个精原细胞的基因型为 $\\mathrm{AaBb}$, 这两对等位基因独立遗传。在不考虑其他基因的情况下，下列关于该精原细胞的叙述错误的是（ ）\nA: 若该细胞经过一次分裂产生的子细胞中含有两个染色体组, 则其进行的是有丝分裂\nB: 若该细胞经过一次分裂产生的子细胞中含有姐妹染色单体, 则其进行的是有丝分裂\nC: 若该细胞减数分裂产生了一个基因型为 $\\mathrm{AaB}$ 的精子, 原因可能是同源染色体未分离\nD: 若该细胞减数分裂产生了基因型不同的四种精子，则该过程可能发生了染色体互换\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某果蝇的一个精原细胞的基因型为 $\\mathrm{AaBb}$, 这两对等位基因独立遗传。在不考虑其他基因的情况下，下列关于该精原细胞的叙述错误的是（ ）\n\nA: 若该细胞经过一次分裂产生的子细胞中含有两个染色体组, 则其进行的是有丝分裂\nB: 若该细胞经过一次分裂产生的子细胞中含有姐妹染色单体, 则其进行的是有丝分裂\nC: 若该细胞减数分裂产生了一个基因型为 $\\mathrm{AaB}$ 的精子, 原因可能是同源染色体未分离\nD: 若该细胞减数分裂产生了基因型不同的四种精子，则该过程可能发生了染色体互换\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1353",
        "problem": "Four important steps in meiosis are\n\nI pairing of homologous chromosomes\n\nII chromatids moving apart\n\nIII division of the centromeres\n\nIV replication of DNA\n\nThe correct order of these steps is\nA: IV $\\rightarrow \\mathrm{I} \\rightarrow \\mathrm{II} \\rightarrow \\mathrm{II}$\nB: IV $\\rightarrow \\mathrm{III} \\rightarrow \\mathrm{I} \\rightarrow \\mathrm{II}$\nC: IV $\\rightarrow \\mathrm{II} \\rightarrow \\mathrm{I} \\rightarrow \\mathrm{III}$\nD: IV $\\rightarrow \\mathrm{I} \\rightarrow \\mathrm{III} \\rightarrow \\mathrm{II}$\nE: I $\\rightarrow \\mathrm{II} \\rightarrow \\mathrm{IV} \\rightarrow \\mathrm{III}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFour important steps in meiosis are\n\nI pairing of homologous chromosomes\n\nII chromatids moving apart\n\nIII division of the centromeres\n\nIV replication of DNA\n\nThe correct order of these steps is\n\nA: IV $\\rightarrow \\mathrm{I} \\rightarrow \\mathrm{II} \\rightarrow \\mathrm{II}$\nB: IV $\\rightarrow \\mathrm{III} \\rightarrow \\mathrm{I} \\rightarrow \\mathrm{II}$\nC: IV $\\rightarrow \\mathrm{II} \\rightarrow \\mathrm{I} \\rightarrow \\mathrm{III}$\nD: IV $\\rightarrow \\mathrm{I} \\rightarrow \\mathrm{III} \\rightarrow \\mathrm{II}$\nE: I $\\rightarrow \\mathrm{II} \\rightarrow \\mathrm{IV} \\rightarrow \\mathrm{III}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1356",
        "problem": "THE RENA OIL SPILL - BIOLOGY WHEN DISASTER STRIKES\n\n[figure1]\n\n\nThe grounding of the CV Rena on October 5, 2011 was New Zealand's worst maritime environmental disaster when the container ship spilled approximately 350 tonnes of heavy fuel oil into the ocean. The container ship ran aground on Otaiti (Astrolabe Reef) just off Mōtitīi Island in the Bay of Plenty, releasing oil into a relatively pristine marine environment. Te Mauri Moana, a group of scientists led by the University of Waikato as part of the Government's \\$ 4.2 million Rena Long-term Environmental Recovery Plan, undertook one of the most comprehensive, multi-disciplinary studies ever done following a marine pollution event. Their findings are summarised in the report; \"Rena Environmental Recovery Monitoring Programme 2011-2013.\"\n\nBy international standards the oil spill was relatively minor but it occurred in an otherwise uncontaminated coastline in an area renowned for its beauty and highly valued for tourism, recreation and fisheries. Iwi, government, commercial stakeholders and the public were rightly concerned there would be long-lasting negative impacts on beaches, reefs and fisheries. Te Mauri Moana reports on fifteen monitoring and research programmes focused on the range of ecosystems and habitats within the Bay of Plenty.\n\nChemical contamination of the sediments and fauna of Otaiti Reef and Môtitī Island and the surrounding soft sediments was examined to determine if the discharge of fuel oil (and later release of container debris) from the Rena resulted in significant chemical contamination of the marine environment and marine organisms. Amongst other contaminants the presence of polycyclic aromatic hydrocarbons (PAHs) from fuel oil was examined in sediments and from selected organisms.\n\nThe maps below shows the PAH levels in sediments collected from A. on Otaiti Reef and B. off-reef.\n\n\nA.\n\n[figure2]\n\nB.\n\n[figure3]\n\n[figure4]\n\nThe graph at left shows the PAH levels in sea urchins collected from Otaiti Reef and nearby islands. The numbers in brackets indicate the number of samples collected at each site. All sea urchins on Otaiti reef showed elevated PAH levels in the gonad with an average level of $0.057 \\mathrm{mg} \\mathrm{kg}^{-1}$. Gut levels averaged $1.58 \\mathrm{mg} \\mathrm{kg}^{-1}$ but at Astro 2 and 3 were orders of magnitude higher than at other sites on Otaiti Reef. PAH levels in the gonad and gut from urchins from Mōtîti Island averaged $0.026 \\mathrm{mg} \\mathrm{kg}^{-1}$ and $0.03 \\mathrm{mg} \\mathrm{kg}^{-1}$ respectively and were similar to levels found elsewhere, including East Cape.The shallowest on-reef sampling site was?\nA: Astro $2 / 3$\nB: Astro 4\nC: Astro 5\nD: Astro 6\nE: Astro 7\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nTHE RENA OIL SPILL - BIOLOGY WHEN DISASTER STRIKES\n\n[figure1]\n\n\nThe grounding of the CV Rena on October 5, 2011 was New Zealand's worst maritime environmental disaster when the container ship spilled approximately 350 tonnes of heavy fuel oil into the ocean. The container ship ran aground on Otaiti (Astrolabe Reef) just off Mōtitīi Island in the Bay of Plenty, releasing oil into a relatively pristine marine environment. Te Mauri Moana, a group of scientists led by the University of Waikato as part of the Government's \\$ 4.2 million Rena Long-term Environmental Recovery Plan, undertook one of the most comprehensive, multi-disciplinary studies ever done following a marine pollution event. Their findings are summarised in the report; \"Rena Environmental Recovery Monitoring Programme 2011-2013.\"\n\nBy international standards the oil spill was relatively minor but it occurred in an otherwise uncontaminated coastline in an area renowned for its beauty and highly valued for tourism, recreation and fisheries. Iwi, government, commercial stakeholders and the public were rightly concerned there would be long-lasting negative impacts on beaches, reefs and fisheries. Te Mauri Moana reports on fifteen monitoring and research programmes focused on the range of ecosystems and habitats within the Bay of Plenty.\n\nChemical contamination of the sediments and fauna of Otaiti Reef and Môtitī Island and the surrounding soft sediments was examined to determine if the discharge of fuel oil (and later release of container debris) from the Rena resulted in significant chemical contamination of the marine environment and marine organisms. Amongst other contaminants the presence of polycyclic aromatic hydrocarbons (PAHs) from fuel oil was examined in sediments and from selected organisms.\n\nThe maps below shows the PAH levels in sediments collected from A. on Otaiti Reef and B. off-reef.\n\n\nA.\n\n[figure2]\n\nB.\n\n[figure3]\n\n[figure4]\n\nThe graph at left shows the PAH levels in sea urchins collected from Otaiti Reef and nearby islands. The numbers in brackets indicate the number of samples collected at each site. All sea urchins on Otaiti reef showed elevated PAH levels in the gonad with an average level of $0.057 \\mathrm{mg} \\mathrm{kg}^{-1}$. Gut levels averaged $1.58 \\mathrm{mg} \\mathrm{kg}^{-1}$ but at Astro 2 and 3 were orders of magnitude higher than at other sites on Otaiti Reef. PAH levels in the gonad and gut from urchins from Mōtîti Island averaged $0.026 \\mathrm{mg} \\mathrm{kg}^{-1}$ and $0.03 \\mathrm{mg} \\mathrm{kg}^{-1}$ respectively and were similar to levels found elsewhere, including East Cape.\n\nproblem:\nThe shallowest on-reef sampling site was?\n\nA: Astro $2 / 3$\nB: Astro 4\nC: Astro 5\nD: Astro 6\nE: Astro 7\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=556&width=1834&top_left_y=276&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=657&width=894&top_left_y=1876&top_left_x=113",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=548&width=854&top_left_y=1942&top_left_x=1092",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-25.jpg?height=591&width=802&top_left_y=201&top_left_x=136"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_153",
        "problem": "In recent years, a genome editing technology called the CRISPR-Cas9 method has been widely used for biology research. In the CRISPR-Cas9 method, an enzyme called Cas9 is guided to the target gene by forming a complex with a guide RNA with a sequence complementary to a part of the target gene. Then, Cas9 cleaves the doublestranded DNA of the target gene specifically with its activity of cleaving double-stranded DNA. Cas9 recognizes a 3-base sequence (NGG) called PAM sequence and cuts the DNA strand 3 to 4 bases upstream of PAM. The cleaved DNA chain is repaired by the DNA repair system, but at that time, a few bases are frequently deleted or inserted.\n\nThe CRISPR-Cas9 method was applied by targeting the region close to the translation start codon of the most upstream exon of a gene encoding enzyme A of a certain animal. The base sequence of the target region was determined for each of the four mutants obtained (Figure 1).\n\nOriginal sequence TA TCT TAC $\\underline{\\text { ATG ATC CTA CAA GTA CCT TAC GCT CGG CAG GAA G }}$\n\n[figure1]\n\n$\\square$ : Pam sequence recognized by Cas9<br>Start codon : ATG (underlined)<br>Stop codon : TAA, TAG, TGA\n\n## Figure 1\nA: It is highly likely that the activity of enzyme $\\mathrm{A}$ is retained in mutant 1. .\nB: It is highly likely that the activity of enzyme A is retained in mutant 2.\nC: It is possible that the activity of enzyme A is retained in mutant 3 .\nD: It is highly likely that the activity of enzyme $\\mathrm{A}$ is lost in mutant 4 .\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn recent years, a genome editing technology called the CRISPR-Cas9 method has been widely used for biology research. In the CRISPR-Cas9 method, an enzyme called Cas9 is guided to the target gene by forming a complex with a guide RNA with a sequence complementary to a part of the target gene. Then, Cas9 cleaves the doublestranded DNA of the target gene specifically with its activity of cleaving double-stranded DNA. Cas9 recognizes a 3-base sequence (NGG) called PAM sequence and cuts the DNA strand 3 to 4 bases upstream of PAM. The cleaved DNA chain is repaired by the DNA repair system, but at that time, a few bases are frequently deleted or inserted.\n\nThe CRISPR-Cas9 method was applied by targeting the region close to the translation start codon of the most upstream exon of a gene encoding enzyme A of a certain animal. The base sequence of the target region was determined for each of the four mutants obtained (Figure 1).\n\nOriginal sequence TA TCT TAC $\\underline{\\text { ATG ATC CTA CAA GTA CCT TAC GCT CGG CAG GAA G }}$\n\n[figure1]\n\n$\\square$ : Pam sequence recognized by Cas9<br>Start codon : ATG (underlined)<br>Stop codon : TAA, TAG, TGA\n\n## Figure 1\n\nA: It is highly likely that the activity of enzyme $\\mathrm{A}$ is retained in mutant 1. .\nB: It is highly likely that the activity of enzyme A is retained in mutant 2.\nC: It is possible that the activity of enzyme A is retained in mutant 3 .\nD: It is highly likely that the activity of enzyme $\\mathrm{A}$ is lost in mutant 4 .\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-23.jpg?height=282&width=1642&top_left_y=1321&top_left_x=227"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_395",
        "problem": "某二倍体高等动物（2 $\\mathrm{n}=24 ）$ 的某个精原细胞（全部 DNA 被 ${ }^{32} \\mathrm{P}$ 标记）在 ${ }^{31} \\mathrm{P}$ 培养液中培养一段时间, 分裂过程中形成的其中一个细胞如图所示 (1-8 表示基因), 图中有两条染色体含有 ${ }^{32} \\mathrm{P}$ ，下列叙述正确的是（）\n\n[图1]\nA: 图示细胞正在经历第 2 次胞质分裂\nB: 该精原细胞分裂结束后可形成 4 个精细胞\nC: 图示细胞形成的子细胞可能不带有 ${ }^{32} \\mathrm{P}$\nD: 1 与 2 都可突变体现了基因突变的不定向性\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二倍体高等动物（2 $\\mathrm{n}=24 ）$ 的某个精原细胞（全部 DNA 被 ${ }^{32} \\mathrm{P}$ 标记）在 ${ }^{31} \\mathrm{P}$ 培养液中培养一段时间, 分裂过程中形成的其中一个细胞如图所示 (1-8 表示基因), 图中有两条染色体含有 ${ }^{32} \\mathrm{P}$ ，下列叙述正确的是（）\n\n[图1]\n\nA: 图示细胞正在经历第 2 次胞质分裂\nB: 该精原细胞分裂结束后可形成 4 个精细胞\nC: 图示细胞形成的子细胞可能不带有 ${ }^{32} \\mathrm{P}$\nD: 1 与 2 都可突变体现了基因突变的不定向性\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-068.jpg?height=557&width=356&top_left_y=173&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_930",
        "problem": "下图为蛋白质工程操作的基本思路, 下列叙述正确的是（）\n\n[图1]\nA: 代表蛋白质工程操作思路的过程是 (1) (4)\nB: 可以运用重组 DNA 技术来实现的环节是(1)(2)\nC: (1)代表 DNA 复制, (2)(3)代表转录和翻译, (4)代表分子设计, (5)代表 DNA 合成\nD: 蛋白质工程是对基因的组合类型进行定向设计, 并通过实验室条件下的基因重组来实现\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为蛋白质工程操作的基本思路, 下列叙述正确的是（）\n\n[图1]\n\nA: 代表蛋白质工程操作思路的过程是 (1) (4)\nB: 可以运用重组 DNA 技术来实现的环节是(1)(2)\nC: (1)代表 DNA 复制, (2)(3)代表转录和翻译, (4)代表分子设计, (5)代表 DNA 合成\nD: 蛋白质工程是对基因的组合类型进行定向设计, 并通过实验室条件下的基因重组来实现\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-06.jpg?height=205&width=1210&top_left_y=523&top_left_x=366"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_378",
        "problem": "在番茄中, 圆形果 $(R)$ 对卵圆形果 $(r)$ 为显性, 单一花序 (E)对复状花序 (e)是显性。对某单一花序圆形果植株进行测交，测交后代表型及其株数为: 单一花序圆形果 22 株、单一花序卵圆形果 83 株、复状花序圆形果 85 株、复状花序卵圆形 20 株。据此判断，下列四图中，能正确表示该单一花序圆形果植株基因与染色体关系的是（）\nA: [图1]\nB: [图2]\nC: [图3]\nD: [图4]\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在番茄中, 圆形果 $(R)$ 对卵圆形果 $(r)$ 为显性, 单一花序 (E)对复状花序 (e)是显性。对某单一花序圆形果植株进行测交，测交后代表型及其株数为: 单一花序圆形果 22 株、单一花序卵圆形果 83 株、复状花序圆形果 85 株、复状花序卵圆形 20 株。据此判断，下列四图中，能正确表示该单一花序圆形果植株基因与染色体关系的是（）\n\nA: [图1]\nB: [图2]\nC: [图3]\nD: [图4]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-11.jpg?height=251&width=231&top_left_y=1091&top_left_x=244",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-11.jpg?height=254&width=249&top_left_y=1089&top_left_x=618",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-11.jpg?height=254&width=239&top_left_y=1089&top_left_x=977",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-11.jpg?height=254&width=251&top_left_y=1087&top_left_x=1368"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_508",
        "problem": "某二倍体雌雄同株的植物, 紫花对白花为显性（分别由基因 A 和 a 控制), 现用纯合紫花和白花杂交，子代中出现了甲、乙两株基因型为 AAa 的可育紫花植株。研究人员让甲与白花植株杂交, 让乙自交, 后代紫花与白花的分离比均为 3: 1。甲、乙两植株产生过程中所发生的变异类型分别是（ ）\nA: 非整倍体变异，染色体片段易位\nB: 非整倍体变异, 染色体片段重复\nC: 染色体片段易位，染色体片段重复\nD: 染色体片段易位, 个别染色体数量变异\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二倍体雌雄同株的植物, 紫花对白花为显性（分别由基因 A 和 a 控制), 现用纯合紫花和白花杂交，子代中出现了甲、乙两株基因型为 AAa 的可育紫花植株。研究人员让甲与白花植株杂交, 让乙自交, 后代紫花与白花的分离比均为 3: 1。甲、乙两植株产生过程中所发生的变异类型分别是（ ）\n\nA: 非整倍体变异，染色体片段易位\nB: 非整倍体变异, 染色体片段重复\nC: 染色体片段易位，染色体片段重复\nD: 染色体片段易位, 个别染色体数量变异\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1492",
        "problem": "Viruses which can deliver genes into human cells are essential for research, gene-therapy and vaccines.\n\nSomehow, scientists must make virus particles which are infectious, but only carry the useful gene and will not produce new virus within the person. To do this, scientists infect factory cells in large vats with different genes. The factory cells then release viral particles which can be harvested and used.\n\nA schematic of a generic virus genome is shown below.\n\n[figure1]\n\n- 1 is the binding site for the virus transcription factor and replicating-polymerase.\n- 2 is the gene encoding the transcription factor.\n- 3 is the gene encoding the replicating-polymerase.\n- 4 is the gene encoding the envelope proteins which make up the outside of the virus particle. They bind site 1 to package the genome inside the viral particle.\n\nIn the course of infection by a generic virus, put these steps in order starting with entry into the cell at the top.\nA: Envelope proteins, transcription factor and polymerase is made, Polymerase and transcription factor bind the viral genome, Viral genes are transcribed and replicated, Envelope proteins form viral particles with genome and proteins inside\nB: Envelope proteins, transcription factor and polymerase is made, Viral genes are transcribed and replicated, Polymerase and transcription factor bind the viral genome, Envelope proteins form viral particles with genome and proteins inside\nC: Viral genes are transcribed and replicated, Envelope proteins, transcription factor and polymerase is made, Polymerase and transcription factor bind the viral genome, Envelope proteins form viral particles with genome and proteins inside\nD: Polymerase and transcription factor bind the viral genome, Viral genes are transcribed and replicated, Envelope proteins, transcription factor and polymerase is made, Envelope proteins form viral particles with genome and proteins inside.\nE: Envelope proteins form viral particles with genome and proteins inside, Polymerase and transcription factor bind the viral genome, Viral genes are transcribed and replicated, Envelope proteins, transcription factor and polymerase is made.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nViruses which can deliver genes into human cells are essential for research, gene-therapy and vaccines.\n\nSomehow, scientists must make virus particles which are infectious, but only carry the useful gene and will not produce new virus within the person. To do this, scientists infect factory cells in large vats with different genes. The factory cells then release viral particles which can be harvested and used.\n\nA schematic of a generic virus genome is shown below.\n\n[figure1]\n\n- 1 is the binding site for the virus transcription factor and replicating-polymerase.\n- 2 is the gene encoding the transcription factor.\n- 3 is the gene encoding the replicating-polymerase.\n- 4 is the gene encoding the envelope proteins which make up the outside of the virus particle. They bind site 1 to package the genome inside the viral particle.\n\nIn the course of infection by a generic virus, put these steps in order starting with entry into the cell at the top.\n\nA: Envelope proteins, transcription factor and polymerase is made, Polymerase and transcription factor bind the viral genome, Viral genes are transcribed and replicated, Envelope proteins form viral particles with genome and proteins inside\nB: Envelope proteins, transcription factor and polymerase is made, Viral genes are transcribed and replicated, Polymerase and transcription factor bind the viral genome, Envelope proteins form viral particles with genome and proteins inside\nC: Viral genes are transcribed and replicated, Envelope proteins, transcription factor and polymerase is made, Polymerase and transcription factor bind the viral genome, Envelope proteins form viral particles with genome and proteins inside\nD: Polymerase and transcription factor bind the viral genome, Viral genes are transcribed and replicated, Envelope proteins, transcription factor and polymerase is made, Envelope proteins form viral particles with genome and proteins inside.\nE: Envelope proteins form viral particles with genome and proteins inside, Polymerase and transcription factor bind the viral genome, Viral genes are transcribed and replicated, Envelope proteins, transcription factor and polymerase is made.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-55.jpg?height=128&width=688&top_left_y=741&top_left_x=227"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_544",
        "problem": "将某一经 ${ }^{3} \\mathrm{H}$ 充分标记 DNA 的雄性动物细胞（染色体数为 $2 \\mathrm{~N}$ ) 置于不含 ${ }^{3} \\mathrm{H}$ 的培养基中培养，该细胞经过两次连续分裂后形成 4 个大小相等的子细胞。下列有关的说法正确的是（）\nA: 若子细胞中染色体数为 $2 \\mathrm{~N}$, 则其中含 ${ }^{3} \\mathrm{H}$ 的染色体数一定为 $\\mathrm{N}$\nB: 若子细胞中染色体数为 $\\mathrm{N}$, 则其中含 ${ }^{3} \\mathrm{H}$ 的 DNA 分子数为 1\nC: 若子细胞中染色体都含 ${ }^{3} \\mathrm{H}$, 则细胞分裂过程中会发生基因重组 \nD: 若子细胞中有的染色体不含 ${ }^{3} \\mathrm{H}$ ，则原因是同源染色体分离\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将某一经 ${ }^{3} \\mathrm{H}$ 充分标记 DNA 的雄性动物细胞（染色体数为 $2 \\mathrm{~N}$ ) 置于不含 ${ }^{3} \\mathrm{H}$ 的培养基中培养，该细胞经过两次连续分裂后形成 4 个大小相等的子细胞。下列有关的说法正确的是（）\n\nA: 若子细胞中染色体数为 $2 \\mathrm{~N}$, 则其中含 ${ }^{3} \\mathrm{H}$ 的染色体数一定为 $\\mathrm{N}$\nB: 若子细胞中染色体数为 $\\mathrm{N}$, 则其中含 ${ }^{3} \\mathrm{H}$ 的 DNA 分子数为 1\nC: 若子细胞中染色体都含 ${ }^{3} \\mathrm{H}$, 则细胞分裂过程中会发生基因重组 \nD: 若子细胞中有的染色体不含 ${ }^{3} \\mathrm{H}$ ，则原因是同源染色体分离\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_630",
        "problem": "水稻的杂交是育种的重要手段, 但不同品种的水稻之间的杂交种常有育性下降的现\n\n象, 为育种带来麻烦。研究人员发现水稻 8 号染色体上有一对等位基因 $\\mathrm{T} / \\mathrm{t}, 4$ 号染色体上有一对等位基因 $\\mathrm{G} / \\mathrm{g}$ 。在花粉发育过程中至少需要含有 $\\mathrm{T}$ 蛋白或 $\\mathrm{G}$ 蛋白, $\\mathrm{t}$ 和 $\\mathrm{g}$ 基因无法表达有正常功能的蛋白质。将基因型为 TTgg 的栽培稻和基因型 $\\mathrm{ttGG}$ 的野生稻杂交得到 $F_{1}$, 下列选项中说法错误的是（）\nA: $F_{1}$ 仅有 $3 / 4$ 的花粉发育正常\nB: $F_{1}$ 自交其后代的基因型有 8 种\nC: $F_{1}$ 作父本与亲本回交, 子代中与 $F_{1}$ 基因型相同的个体所占比例为 $1 / 3$\nD: $F_{1}$ 自交后代中, 花粉发育全部正常的个体所占比例是 $7 / 15$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n水稻的杂交是育种的重要手段, 但不同品种的水稻之间的杂交种常有育性下降的现\n\n象, 为育种带来麻烦。研究人员发现水稻 8 号染色体上有一对等位基因 $\\mathrm{T} / \\mathrm{t}, 4$ 号染色体上有一对等位基因 $\\mathrm{G} / \\mathrm{g}$ 。在花粉发育过程中至少需要含有 $\\mathrm{T}$ 蛋白或 $\\mathrm{G}$ 蛋白, $\\mathrm{t}$ 和 $\\mathrm{g}$ 基因无法表达有正常功能的蛋白质。将基因型为 TTgg 的栽培稻和基因型 $\\mathrm{ttGG}$ 的野生稻杂交得到 $F_{1}$, 下列选项中说法错误的是（）\n\nA: $F_{1}$ 仅有 $3 / 4$ 的花粉发育正常\nB: $F_{1}$ 自交其后代的基因型有 8 种\nC: $F_{1}$ 作父本与亲本回交, 子代中与 $F_{1}$ 基因型相同的个体所占比例为 $1 / 3$\nD: $F_{1}$ 自交后代中, 花粉发育全部正常的个体所占比例是 $7 / 15$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_179",
        "problem": "Fragaria chiloensis is a stolon*-bearing perennial herb that grows on coastal sand dunes. In coastal sand dunes, nitrogen-fixing shrubs often create small patches of lower photon flux density (PFD) but higher soil nitrogen availability. The presence of such patches frequently makes a difference in the resource availability between stolon-connected ramets**. To examine effects of stolon connection, researchers compared the growth of connected ramets and severed ramets; one ramet in each pair was provided with high PFD but a low level of soil nitrogen, and the other ramet was provided with low PFD but a high level of soil nitrogen (Figure 1). As a result, combined dry biomass of connected ramets was $54 \\%$ higher than that of severed ramets.\n\n*Stolon: a stem that grows along the soil surface and forms buds and roots at the nodes for clonal propagation.\n\n**Ramet: an individual unit of a clonal colony.\n\n[figure1]\n\nFigure 1 Schematic cartoon of experimental setup\nA: Shoot/root ratio of the connected ramet provided with high PFD and low nitrogen was higher than that of the severed ramet provided with high PFD and low nitrogen.\nB: In the severed ramet provided with low PFD and high nitrogen, PFD was not a limiting factor for plant growth.\nC: Severing of stolons does not affect the combined dry mass of ramets when resources (i.e. PFD and nitrogen) are distributed uniformly.\nD: Assimilation products and nitrogen can be translocated via stolons in Fragaria chiloensis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFragaria chiloensis is a stolon*-bearing perennial herb that grows on coastal sand dunes. In coastal sand dunes, nitrogen-fixing shrubs often create small patches of lower photon flux density (PFD) but higher soil nitrogen availability. The presence of such patches frequently makes a difference in the resource availability between stolon-connected ramets**. To examine effects of stolon connection, researchers compared the growth of connected ramets and severed ramets; one ramet in each pair was provided with high PFD but a low level of soil nitrogen, and the other ramet was provided with low PFD but a high level of soil nitrogen (Figure 1). As a result, combined dry biomass of connected ramets was $54 \\%$ higher than that of severed ramets.\n\n*Stolon: a stem that grows along the soil surface and forms buds and roots at the nodes for clonal propagation.\n\n**Ramet: an individual unit of a clonal colony.\n\n[figure1]\n\nFigure 1 Schematic cartoon of experimental setup\n\nA: Shoot/root ratio of the connected ramet provided with high PFD and low nitrogen was higher than that of the severed ramet provided with high PFD and low nitrogen.\nB: In the severed ramet provided with low PFD and high nitrogen, PFD was not a limiting factor for plant growth.\nC: Severing of stolons does not affect the combined dry mass of ramets when resources (i.e. PFD and nitrogen) are distributed uniformly.\nD: Assimilation products and nitrogen can be translocated via stolons in Fragaria chiloensis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-47.jpg?height=546&width=1334&top_left_y=1186&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_547",
        "problem": "螺美波纹蛾（性别决定类型为 ZW 型）的体色有棕色、灰色和绿色，由常染色体上的 $\\mathrm{M} / \\mathrm{m}$ 基因和 $\\mathrm{Z}$ 染色体上的 $\\mathrm{N} / \\mathrm{n}$ 基因共同控制。已知含有 $\\mathrm{N}$ 基因的个体均为棕色, 含 $\\mathrm{M}$ 基因但不含 $\\mathrm{N}$ 基因的个体均为灰色，其余情况为绿色。现有一只棕色个体与一只棕色个体交配，产生足够多的子代，子代中绿色雌性个体占 $1 / 16$, 下列相关叙述错误的是\nA: 雄性亲本表型为棕色, 基因型为 $M m Z^{N} Z^{n}$\nB: 雌性亲本产生基因型为 $m Z^{n}$ 配子的概率为 $1 / 4$\nC: 子代中体色为棕色的个体所占比例约为 $1 / 2$\nD: 子代中灰色雌性个体所占的比例为 $3 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n螺美波纹蛾（性别决定类型为 ZW 型）的体色有棕色、灰色和绿色，由常染色体上的 $\\mathrm{M} / \\mathrm{m}$ 基因和 $\\mathrm{Z}$ 染色体上的 $\\mathrm{N} / \\mathrm{n}$ 基因共同控制。已知含有 $\\mathrm{N}$ 基因的个体均为棕色, 含 $\\mathrm{M}$ 基因但不含 $\\mathrm{N}$ 基因的个体均为灰色，其余情况为绿色。现有一只棕色个体与一只棕色个体交配，产生足够多的子代，子代中绿色雌性个体占 $1 / 16$, 下列相关叙述错误的是\n\nA: 雄性亲本表型为棕色, 基因型为 $M m Z^{N} Z^{n}$\nB: 雌性亲本产生基因型为 $m Z^{n}$ 配子的概率为 $1 / 4$\nC: 子代中体色为棕色的个体所占比例约为 $1 / 2$\nD: 子代中灰色雌性个体所占的比例为 $3 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1",
        "problem": "You are given four figures referring to typical representatives of four major phyla of animals.\n[figure1]\nA: Organism $\\mathbf{A}$ belongs to a taxon characterized by unique aquiferous system, radial or no symmetry and lack of tissue and organ systems.\nB: Organism B belongs to a taxon characterized by bilateral symmetry, lack of coelom, protonephridia and ladder-like nervous system.\nC: Organism $\\mathbf{C}$ belongs to a taxon characterized by entrocoelic development, pentaradial symmetry and metamorphosis.\nD: Organism $\\mathbf{D}$ belongs to a taxon characterized by radial symmetry, hydrostatic skeleton and schizocoelic development.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nYou are given four figures referring to typical representatives of four major phyla of animals.\n[figure1]\n\nA: Organism $\\mathbf{A}$ belongs to a taxon characterized by unique aquiferous system, radial or no symmetry and lack of tissue and organ systems.\nB: Organism B belongs to a taxon characterized by bilateral symmetry, lack of coelom, protonephridia and ladder-like nervous system.\nC: Organism $\\mathbf{C}$ belongs to a taxon characterized by entrocoelic development, pentaradial symmetry and metamorphosis.\nD: Organism $\\mathbf{D}$ belongs to a taxon characterized by radial symmetry, hydrostatic skeleton and schizocoelic development.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-37.jpg?height=620&width=1240&top_left_y=478&top_left_x=408"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1357",
        "problem": "## DEAR ENEMIES \\& NASTY NEIGHBORS\n\n\"Dear Enemy, I curse you, and hope that something slightly unpleasant happens to you, like an onion falling on your head.\"\n\n## Blackadder I, The Archbishop\n\nMany species show territorial behaviour (actively defending an area containing resources) allowing the species to monopolise resources such as food or mates. Territoriality is costly in terms of energy and time so some species reduce these costs by being less aggressive towards their neighbours than towards unfamiliar strangers, the so called \"dear enemy\" hypothesis. The converse are \"nasty neighbours\" in which species are more, not less aggressive towards their neighbours.\n\nDr Brunton's research group, at the Institute of Natural Sciences at Massey University, has been studying the behaviour of the New Zealand bellbird,\n\n[figure1]\n(Anthornis melanura) on Tiritiri Matangi Island. Both sexes sing prolifically and are known to use vocalisations to recognise individuals. Male and female bellbirds counter-sing in response to the vocalisations of their neighbours in a territorial behaviour that may lead to chasing of an individual bird intruding on a territory.\n\nThe researchers used speakers to play the song of neighbouring females or stranger females at different breeding stages. They recorded the responses of male and female bellbirds, including counter-singing and whether the birds approached the speaker. An asterisk indicates significant differences between the response to neighbours and strangers. Due to small sample sizes, the responses to the neighbours versus strangers were not tested during the incubation stage. Control 1 was silence, control 2 played back the\n\n[figure2]\nsong of a different species.\n[figure3]\n\nb) Counter-singing\n[figure4]\n\nUsing the information and the graphs above answer Questions 54 - 58 on the following page.Female bellbird counter-singing occurs at a much lower frequency during incubation. Which of the following is the MOST plausible explanation for this change in behaviour over the breeding season?\nA: Male bellbirds are likely to eat eggs that they discover so females remain quiet when incubating.\nB: Noise from the incubating female bellbird may attract predators to the nest resulting in a loss of eggs.\nC: Female bellbirds do not require any resources whilst incubating so they have no need to exhibit territorial behaviours.\nD: During incubation male bellbirds leave the area so females do not need to counter-sing.\nE: During chick rearing the female bellbirds need to communicate with their young so counter-singing increases.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## DEAR ENEMIES \\& NASTY NEIGHBORS\n\n\"Dear Enemy, I curse you, and hope that something slightly unpleasant happens to you, like an onion falling on your head.\"\n\n## Blackadder I, The Archbishop\n\nMany species show territorial behaviour (actively defending an area containing resources) allowing the species to monopolise resources such as food or mates. Territoriality is costly in terms of energy and time so some species reduce these costs by being less aggressive towards their neighbours than towards unfamiliar strangers, the so called \"dear enemy\" hypothesis. The converse are \"nasty neighbours\" in which species are more, not less aggressive towards their neighbours.\n\nDr Brunton's research group, at the Institute of Natural Sciences at Massey University, has been studying the behaviour of the New Zealand bellbird,\n\n[figure1]\n(Anthornis melanura) on Tiritiri Matangi Island. Both sexes sing prolifically and are known to use vocalisations to recognise individuals. Male and female bellbirds counter-sing in response to the vocalisations of their neighbours in a territorial behaviour that may lead to chasing of an individual bird intruding on a territory.\n\nThe researchers used speakers to play the song of neighbouring females or stranger females at different breeding stages. They recorded the responses of male and female bellbirds, including counter-singing and whether the birds approached the speaker. An asterisk indicates significant differences between the response to neighbours and strangers. Due to small sample sizes, the responses to the neighbours versus strangers were not tested during the incubation stage. Control 1 was silence, control 2 played back the\n\n[figure2]\nsong of a different species.\n[figure3]\n\nb) Counter-singing\n[figure4]\n\nUsing the information and the graphs above answer Questions 54 - 58 on the following page.\n\nproblem:\nFemale bellbird counter-singing occurs at a much lower frequency during incubation. Which of the following is the MOST plausible explanation for this change in behaviour over the breeding season?\n\nA: Male bellbirds are likely to eat eggs that they discover so females remain quiet when incubating.\nB: Noise from the incubating female bellbird may attract predators to the nest resulting in a loss of eggs.\nC: Female bellbirds do not require any resources whilst incubating so they have no need to exhibit territorial behaviours.\nD: During incubation male bellbirds leave the area so females do not need to counter-sing.\nE: During chick rearing the female bellbirds need to communicate with their young so counter-singing increases.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=357&width=571&top_left_y=524&top_left_x=1368",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=485&width=311&top_left_y=931&top_left_x=1495",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=486&width=1734&top_left_y=1493&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=534&width=1744&top_left_y=2042&top_left_x=108"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_876",
        "problem": "关于真核细胞中生命活动与能量关系的叙述, 错误的是( )\nA: DNA 复制需要消耗能量\nB: 光合作用的暗反应阶段需要消耗能量\nC: 物质通过协助扩散进出细胞时需要消耗 ATP\nD: 细胞代谢所需的 ATP 可在细胞质基质中产生\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n关于真核细胞中生命活动与能量关系的叙述, 错误的是( )\n\nA: DNA 复制需要消耗能量\nB: 光合作用的暗反应阶段需要消耗能量\nC: 物质通过协助扩散进出细胞时需要消耗 ATP\nD: 细胞代谢所需的 ATP 可在细胞质基质中产生\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_148",
        "problem": "Polymorphic DNA sequences are widely used for molecular identification. Short tandem repeat (STR) is composed of multiple repeats of 2-8 nucleotides flanking by two conserved sequences. Each STR locus normally has more than two alleles. Single nucleotide polymorphism (SNP) is a variation at a single position in a DNA sequence among individuals. Each SNP usually has only two alleles. Seven individuals were genotyping for two autosomal and two mitochondrial (mtDNA) SNPs, two autosomal and two Y-linked (NRY) STRs (Table Q.91).\n\nTable Q. 91\n\n|  | Autosomes |  |  |  | NRY |  | mtDNA |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Individuals | SNP1 | SNP2 | STR1 | STR2 | STR1 | STR2 | SNP1 | SNP2 |\n| Ind_1 | A/A | A/A | $13 / 15$ | $18 / 20$ | 13 | 12 | C | A |\n| Ind_2 | A/C | A/G | $12 / 14$ | $18 / 21$ | 13 | 15 | T | A |\n| Ind_3 | C/C | A/G | $14 / 15$ | $18 / 21$ | 13 | 15 | C | G |\n| Ind_4 | A/C | G/G | $13 / 15$ | $19 / 19$ | 11 | 14 | T | G |\n| Ind_5 | C/C | A/G | $14 / 15$ | $18 / 19$ | - | - | C | G |\n| Ind_6 | A/C | G/G | $14 / 14$ | $18 / 19$ | - | - | T | G |\n| Ind_7 | C/C | G/G | $14 / 16$ | $19 / 21$ | - | - | C | A |\nA: If the same number of SNPs or STRs is used, SNPs are better marker to distinguish individuals than STRs.\nB: Ind_ 6 is more likely a child of Ind_ 2 and Ind_ 5 than Ind_ 3 is.\nC: Ind_ 4 is possibly a brother of Ind_6.\nD: It is possible that Ind_ 7 is a granddaughter of Ind_1 and Ind_6.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPolymorphic DNA sequences are widely used for molecular identification. Short tandem repeat (STR) is composed of multiple repeats of 2-8 nucleotides flanking by two conserved sequences. Each STR locus normally has more than two alleles. Single nucleotide polymorphism (SNP) is a variation at a single position in a DNA sequence among individuals. Each SNP usually has only two alleles. Seven individuals were genotyping for two autosomal and two mitochondrial (mtDNA) SNPs, two autosomal and two Y-linked (NRY) STRs (Table Q.91).\n\nTable Q. 91\n\n|  | Autosomes |  |  |  | NRY |  | mtDNA |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Individuals | SNP1 | SNP2 | STR1 | STR2 | STR1 | STR2 | SNP1 | SNP2 |\n| Ind_1 | A/A | A/A | $13 / 15$ | $18 / 20$ | 13 | 12 | C | A |\n| Ind_2 | A/C | A/G | $12 / 14$ | $18 / 21$ | 13 | 15 | T | A |\n| Ind_3 | C/C | A/G | $14 / 15$ | $18 / 21$ | 13 | 15 | C | G |\n| Ind_4 | A/C | G/G | $13 / 15$ | $19 / 19$ | 11 | 14 | T | G |\n| Ind_5 | C/C | A/G | $14 / 15$ | $18 / 19$ | - | - | C | G |\n| Ind_6 | A/C | G/G | $14 / 14$ | $18 / 19$ | - | - | T | G |\n| Ind_7 | C/C | G/G | $14 / 16$ | $19 / 21$ | - | - | C | A |\n\nA: If the same number of SNPs or STRs is used, SNPs are better marker to distinguish individuals than STRs.\nB: Ind_ 6 is more likely a child of Ind_ 2 and Ind_ 5 than Ind_ 3 is.\nC: Ind_ 4 is possibly a brother of Ind_6.\nD: It is possible that Ind_ 7 is a granddaughter of Ind_1 and Ind_6.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_758",
        "problem": "某二倍体高等动物 $(2 n=4)$ 的卵原细胞的 DNA 被 ${ }^{32} \\mathrm{P}$ 完全标记, 在 ${ }^{31} \\mathrm{P}$ 的培养液中进行减数分裂产生卵细胞。该卵细胞与 DNA 被 ${ }^{32} \\mathrm{P}$ 完全标记的精子形成的受精卯在 ${ }^{31} \\mathrm{P}$ 的培养液中进行一次有丝分裂, 分裂过程中形成的某时期的细胞如下图所示, 其中(1)2染色体上的 DNA 无 ${ }^{32} \\mathrm{P}$ 标记。下列叙述错误的是（ ）\n\n[图1]\nA: 受精卵有丝分裂分裂过程中发生了基因突变\nB: 图示细胞中有 4 对同源染色体, 4 套遗传信息\nC: 图示细胞中含 ${ }^{31} \\mathrm{P}$ 的核 DNA 为 8 个, 含 ${ }^{32} \\mathrm{P}$ 的 核 DNA 为 4 个\nD: 若产生精子的精原细胞为纯合子，则精原细胞的基因型可能有两种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二倍体高等动物 $(2 n=4)$ 的卵原细胞的 DNA 被 ${ }^{32} \\mathrm{P}$ 完全标记, 在 ${ }^{31} \\mathrm{P}$ 的培养液中进行减数分裂产生卵细胞。该卵细胞与 DNA 被 ${ }^{32} \\mathrm{P}$ 完全标记的精子形成的受精卯在 ${ }^{31} \\mathrm{P}$ 的培养液中进行一次有丝分裂, 分裂过程中形成的某时期的细胞如下图所示, 其中(1)2染色体上的 DNA 无 ${ }^{32} \\mathrm{P}$ 标记。下列叙述错误的是（ ）\n\n[图1]\n\nA: 受精卵有丝分裂分裂过程中发生了基因突变\nB: 图示细胞中有 4 对同源染色体, 4 套遗传信息\nC: 图示细胞中含 ${ }^{31} \\mathrm{P}$ 的核 DNA 为 8 个, 含 ${ }^{32} \\mathrm{P}$ 的 核 DNA 为 4 个\nD: 若产生精子的精原细胞为纯合子，则精原细胞的基因型可能有两种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-092.jpg?height=548&width=508&top_left_y=154&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1497",
        "problem": "The complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nWhich pathway causes animal red blood cells to burst if they are transfused into humans?\nA: Pathway A\nB: Pathway B\nC: Pathway C\nD: Pathway D\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nWhich pathway causes animal red blood cells to burst if they are transfused into humans?\n\nA: Pathway A\nB: Pathway B\nC: Pathway C\nD: Pathway D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=790&width=1714&top_left_y=486&top_left_x=228",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=930&width=1806&top_left_y=1335&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_884",
        "problem": "果蝇的某条染色体上 $\\mathrm{D} 、 \\mathrm{E} 、 \\mathrm{~F}$ 三个基因不发生互换。这三个基因各有多个等位基\n\n因（例如： $\\mathrm{D}$ 的等位基因包括 $\\mathrm{D}_{1} \\sim \\mathrm{D}_{\\mathrm{n}}$ )。亲代果蝇与两只子代果蝇的基因组成如下表。下列叙述正确的是（）\n\n| 个体 | $\\mathrm{P}$ 雄蝇 | $\\mathrm{P}$ 雌蝇 | $\\mathrm{F}_{1}$ 雄蝇 | $\\mathrm{F}_{1}$ 雌蝇 |\n| :---: | :---: | :---: | :---: | :---: |\n| 基因组成 | $\\mathrm{D}_{2} \\mathrm{D}_{5} \\mathrm{E}_{7} \\mathrm{E}_{5} \\mathrm{~F}_{2} \\mathrm{~F}_{4}$ | $\\mathrm{D}_{3} \\mathrm{D}_{4} \\mathrm{E}_{8} \\mathrm{E}_{4} \\mathrm{~F}_{3} \\mathrm{~F}_{5}$ | $\\mathrm{D}_{4} \\mathrm{D}_{5} \\mathrm{E}_{7} \\mathrm{E}_{8} \\mathrm{~F}_{4} \\mathrm{~F}_{5}$ | $\\mathrm{D}_{3} \\mathrm{D}_{2} \\mathrm{E}_{5} \\mathrm{E}_{4} \\mathrm{~F}_{2} \\mathrm{~F}_{3}$ |\nA: 基因 $D 、 E 、 F$ 的遗传方式肯定是伴 X 染色体遗传\nB: 亲代雌蝇的其中一条染色体上基因组成是 $\\mathrm{D}_{3} \\mathrm{E}_{4} \\mathrm{~F}_{3}$\nC: 基因 $\\mathrm{D}$ 与基因 $\\mathrm{E}$ 的遗传符合基因的自由组合定律\nD: 若再产一只 $\\mathrm{F}_{1}$ 果蝇, 该果蝇 $\\mathrm{D}$ 基因组成为 $\\mathrm{D}_{2} \\mathrm{D}_{4}$, 则其 $\\mathrm{F}$ 基因组成为 $\\mathrm{F}_{4} \\mathrm{~F}_{5}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的某条染色体上 $\\mathrm{D} 、 \\mathrm{E} 、 \\mathrm{~F}$ 三个基因不发生互换。这三个基因各有多个等位基\n\n因（例如： $\\mathrm{D}$ 的等位基因包括 $\\mathrm{D}_{1} \\sim \\mathrm{D}_{\\mathrm{n}}$ )。亲代果蝇与两只子代果蝇的基因组成如下表。下列叙述正确的是（）\n\n| 个体 | $\\mathrm{P}$ 雄蝇 | $\\mathrm{P}$ 雌蝇 | $\\mathrm{F}_{1}$ 雄蝇 | $\\mathrm{F}_{1}$ 雌蝇 |\n| :---: | :---: | :---: | :---: | :---: |\n| 基因组成 | $\\mathrm{D}_{2} \\mathrm{D}_{5} \\mathrm{E}_{7} \\mathrm{E}_{5} \\mathrm{~F}_{2} \\mathrm{~F}_{4}$ | $\\mathrm{D}_{3} \\mathrm{D}_{4} \\mathrm{E}_{8} \\mathrm{E}_{4} \\mathrm{~F}_{3} \\mathrm{~F}_{5}$ | $\\mathrm{D}_{4} \\mathrm{D}_{5} \\mathrm{E}_{7} \\mathrm{E}_{8} \\mathrm{~F}_{4} \\mathrm{~F}_{5}$ | $\\mathrm{D}_{3} \\mathrm{D}_{2} \\mathrm{E}_{5} \\mathrm{E}_{4} \\mathrm{~F}_{2} \\mathrm{~F}_{3}$ |\n\nA: 基因 $D 、 E 、 F$ 的遗传方式肯定是伴 X 染色体遗传\nB: 亲代雌蝇的其中一条染色体上基因组成是 $\\mathrm{D}_{3} \\mathrm{E}_{4} \\mathrm{~F}_{3}$\nC: 基因 $\\mathrm{D}$ 与基因 $\\mathrm{E}$ 的遗传符合基因的自由组合定律\nD: 若再产一只 $\\mathrm{F}_{1}$ 果蝇, 该果蝇 $\\mathrm{D}$ 基因组成为 $\\mathrm{D}_{2} \\mathrm{D}_{4}$, 则其 $\\mathrm{F}$ 基因组成为 $\\mathrm{F}_{4} \\mathrm{~F}_{5}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_728",
        "problem": "某种鼠中, 黄鼠基因 $\\mathrm{A}$ 对灰鼠基因 $\\mathrm{a}$ 显性, 短尾基因 $\\mathrm{B}$ 对长尾基因 $\\mathrm{b}$ 显性, 且基因 $\\mathrm{A}$ 和基因 $\\mathrm{B}$ 均纯合时使胚胎致死，两对基因独立遗传。现有两只黄色短尾鼠交配，理论上子代表现型比例不可能为\nA: $2: 1$\nB: $5: 2$\nC: $4: 2: 2: 1$\nD: $8: 3: 3: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种鼠中, 黄鼠基因 $\\mathrm{A}$ 对灰鼠基因 $\\mathrm{a}$ 显性, 短尾基因 $\\mathrm{B}$ 对长尾基因 $\\mathrm{b}$ 显性, 且基因 $\\mathrm{A}$ 和基因 $\\mathrm{B}$ 均纯合时使胚胎致死，两对基因独立遗传。现有两只黄色短尾鼠交配，理论上子代表现型比例不可能为\n\nA: $2: 1$\nB: $5: 2$\nC: $4: 2: 2: 1$\nD: $8: 3: 3: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_938",
        "problem": "某严格自花传粉植物开红花, 研究人员偶尔发现白花甲、白花乙两株隐性突变植株。已知基因 A 控制合成的酶能将白色底物 1 合成白色底物 2 , 基因 B 控制合成的酶能将白色底物 2 合成红色色素。为研究白花甲和白花乙的突变基因是否相同, 研究人员利用两白花花瓣的研磨液（简称“甲液”“乙液”）进行了如下实验。实验结果不能表明（）\n\n| 实验 | 过程 | 结果 |\n| :---: | :---: | :---: |\n| 一 | 甲液和乙液混合 | 红色 |\n| 二 | 将甲液加热后冷却至室温, 与乙液混合 | 红色 |\n| 三 | 将乙液加热后冷却至室温, 与甲液混合 | 白色 |\n\n说明: 加热后白色底物 1 、白色底物 2 的化学性质没有改变\nA: 甲、乙两白花的突变基因不同\nB: 甲液中含有白色底物 2, 乙液中含有白色底物 1\nC: 白花甲、白花乙的基因型分别为 AAbb、aaBB\nD: A（a）和 B（b）分别位于非同源染色体上\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某严格自花传粉植物开红花, 研究人员偶尔发现白花甲、白花乙两株隐性突变植株。已知基因 A 控制合成的酶能将白色底物 1 合成白色底物 2 , 基因 B 控制合成的酶能将白色底物 2 合成红色色素。为研究白花甲和白花乙的突变基因是否相同, 研究人员利用两白花花瓣的研磨液（简称“甲液”“乙液”）进行了如下实验。实验结果不能表明（）\n\n| 实验 | 过程 | 结果 |\n| :---: | :---: | :---: |\n| 一 | 甲液和乙液混合 | 红色 |\n| 二 | 将甲液加热后冷却至室温, 与乙液混合 | 红色 |\n| 三 | 将乙液加热后冷却至室温, 与甲液混合 | 白色 |\n\n说明: 加热后白色底物 1 、白色底物 2 的化学性质没有改变\n\nA: 甲、乙两白花的突变基因不同\nB: 甲液中含有白色底物 2, 乙液中含有白色底物 1\nC: 白花甲、白花乙的基因型分别为 AAbb、aaBB\nD: A（a）和 B（b）分别位于非同源染色体上\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1467",
        "problem": "A Poikilotherm is an organism whose internal temperature varies considerably with their environment. In comparison, a homeotherm is able to maintain a stable internal body temperature regardless of external influence.\n\n[figure1]\n\nBiochemical reaction rates vary with temperature. To function, poikilotherms may have four to ten enzyme systems that operate at different temperatures for an important biochemical reaction. Compared to homeotherms in similar ecological niches, poikilotherms are likely to have:\nA: more complex organ systems\nB: larger, more complex genomes\nC: enzymes that are significantly more resistant to denaturation\nD: enzymes that may perform very different functions depending on the temperature\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA Poikilotherm is an organism whose internal temperature varies considerably with their environment. In comparison, a homeotherm is able to maintain a stable internal body temperature regardless of external influence.\n\n[figure1]\n\nBiochemical reaction rates vary with temperature. To function, poikilotherms may have four to ten enzyme systems that operate at different temperatures for an important biochemical reaction. Compared to homeotherms in similar ecological niches, poikilotherms are likely to have:\n\nA: more complex organ systems\nB: larger, more complex genomes\nC: enzymes that are significantly more resistant to denaturation\nD: enzymes that may perform very different functions depending on the temperature\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-06.jpg?height=683&width=763&top_left_y=1466&top_left_x=638"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_665",
        "problem": "某种小鼠的毛色受 $A^{Y}$ (黄色)、 $\\mathrm{A}$ (鼠色)、a（黑色） 3 个基因控制, 三者互为等位基因, $\\mathrm{A}^{\\mathrm{Y}}$ 对 $\\mathrm{A} 、 \\mathrm{a}$ 为完全显性, $\\mathrm{A}$ 对 $\\mathrm{a}$ 为完全显性, 并且基因型 $\\mathrm{A}^{\\mathrm{Y}} \\mathrm{A}^{\\mathrm{Y}}$ 胚胎致死 (不计人个体数)。下列叙述错误的是 ( )\nA: 若 $\\mathrm{A}^{\\mathrm{Y}} \\mathrm{a}$ 个体与 $\\mathrm{A}^{\\mathrm{Y}} \\mathrm{A}$ 个体杂交, 则 $\\mathrm{F}_{1}$ 有 3 种基因型\nB: 若 $\\mathrm{A}^{\\mathrm{Y}} \\mathrm{a}$ 个体与 $\\mathrm{Aa}$ 个体杂交, 则 $\\mathrm{F}_{1}$ 有 3 种表现型\nC: 若 1 只黄色雄鼠与若干只黑色雌鼠杂交, 则 $F_{1}$ 可同时出现鼠色个体与黑色个体\nD: 若 1 只黄色雄鼠与若干只纯合鼠色雌鼠杂交, 则 $\\mathrm{F}_{1}$ 可同时出现黄色个体与鼠色个体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种小鼠的毛色受 $A^{Y}$ (黄色)、 $\\mathrm{A}$ (鼠色)、a（黑色） 3 个基因控制, 三者互为等位基因, $\\mathrm{A}^{\\mathrm{Y}}$ 对 $\\mathrm{A} 、 \\mathrm{a}$ 为完全显性, $\\mathrm{A}$ 对 $\\mathrm{a}$ 为完全显性, 并且基因型 $\\mathrm{A}^{\\mathrm{Y}} \\mathrm{A}^{\\mathrm{Y}}$ 胚胎致死 (不计人个体数)。下列叙述错误的是 ( )\n\nA: 若 $\\mathrm{A}^{\\mathrm{Y}} \\mathrm{a}$ 个体与 $\\mathrm{A}^{\\mathrm{Y}} \\mathrm{A}$ 个体杂交, 则 $\\mathrm{F}_{1}$ 有 3 种基因型\nB: 若 $\\mathrm{A}^{\\mathrm{Y}} \\mathrm{a}$ 个体与 $\\mathrm{Aa}$ 个体杂交, 则 $\\mathrm{F}_{1}$ 有 3 种表现型\nC: 若 1 只黄色雄鼠与若干只黑色雌鼠杂交, 则 $F_{1}$ 可同时出现鼠色个体与黑色个体\nD: 若 1 只黄色雄鼠与若干只纯合鼠色雌鼠杂交, 则 $\\mathrm{F}_{1}$ 可同时出现黄色个体与鼠色个体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_875",
        "problem": "水稻的高秆对矮秆为完全显性, 由一对等位基因 A、a 控制, 抗病对易感病为完全显性, 由另一对等位基因 $\\mathrm{B} 、 \\mathrm{~b}$ 控制, 现有纯合高秆抗病和纯合矮秆易感病的两种亲本杂交, 所得 $F_{1}$ 自交, 多次重复实验, 统计 $F_{2}$ 的表现型及比例都近似有如下结果：高秆抗病：高秆易感病：矮秆抗病：矮秆易感病=66：9：9：16。下列叙述错误的是（）\nA: 控制抗病和易感病的等位基因遵循基因的分离定律\nB: 上述两对等位基因的遗传遵循基因的自由组合定律\nC: $F_{2}$ 中出现新表型是因为减 I 前期非等位基因重新组合\nD: $\\mathrm{F}_{1}$ 产生的雌雄配子的比例都是 $\\mathrm{AB}: \\mathrm{Ab}: \\mathrm{aB}: \\mathrm{ab}=4: 1: 1: 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n水稻的高秆对矮秆为完全显性, 由一对等位基因 A、a 控制, 抗病对易感病为完全显性, 由另一对等位基因 $\\mathrm{B} 、 \\mathrm{~b}$ 控制, 现有纯合高秆抗病和纯合矮秆易感病的两种亲本杂交, 所得 $F_{1}$ 自交, 多次重复实验, 统计 $F_{2}$ 的表现型及比例都近似有如下结果：高秆抗病：高秆易感病：矮秆抗病：矮秆易感病=66：9：9：16。下列叙述错误的是（）\n\nA: 控制抗病和易感病的等位基因遵循基因的分离定律\nB: 上述两对等位基因的遗传遵循基因的自由组合定律\nC: $F_{2}$ 中出现新表型是因为减 I 前期非等位基因重新组合\nD: $\\mathrm{F}_{1}$ 产生的雌雄配子的比例都是 $\\mathrm{AB}: \\mathrm{Ab}: \\mathrm{aB}: \\mathrm{ab}=4: 1: 1: 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_627",
        "problem": "杂交水稻的无融合生殖是指不发生雌、雄配子的细胞核融合而产生种子的一种无性繁殖方式。无融合生殖过程主要由 2 个基因控制: 含基因 A 的植株形成雌配子时, 减数分裂 $\\mathrm{I}$ 常, 导致雌配子染色体数目加倍; 含基因 $\\mathrm{P}$ 的植株产生的雌配子不经过受精作用，直接发育成个体。雄配子的发育不受基因 $\\mathrm{A} 、 \\mathrm{P}$ 的影响。下列有关叙述，错误的是 ( )\nA: 基因型为 $\\mathrm{AaPp}$ 的水稻自交，子代基因型与亲代相同\nB: 基因型为 Aapp 的水稻自交, 子代染色体数与亲代相同\nC: 利用无融合生殖技术可以获得母本的单倍体子代植株\nD: 利用无融合生殖技术可以保持作物的优良性状\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n杂交水稻的无融合生殖是指不发生雌、雄配子的细胞核融合而产生种子的一种无性繁殖方式。无融合生殖过程主要由 2 个基因控制: 含基因 A 的植株形成雌配子时, 减数分裂 $\\mathrm{I}$ 常, 导致雌配子染色体数目加倍; 含基因 $\\mathrm{P}$ 的植株产生的雌配子不经过受精作用，直接发育成个体。雄配子的发育不受基因 $\\mathrm{A} 、 \\mathrm{P}$ 的影响。下列有关叙述，错误的是 ( )\n\nA: 基因型为 $\\mathrm{AaPp}$ 的水稻自交，子代基因型与亲代相同\nB: 基因型为 Aapp 的水稻自交, 子代染色体数与亲代相同\nC: 利用无融合生殖技术可以获得母本的单倍体子代植株\nD: 利用无融合生殖技术可以保持作物的优良性状\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_957",
        "problem": "In birds, the keel is an extension of the sternum where the powerful pectoral flight muscles attach. Which of the following orders of birds is most likely to have a reduced keel?\nA: Piciformes\nB: Anseriformes\nC: Sphenisciformes\nD: Struthioniformes\nE: Galliformes\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn birds, the keel is an extension of the sternum where the powerful pectoral flight muscles attach. Which of the following orders of birds is most likely to have a reduced keel?\n\nA: Piciformes\nB: Anseriformes\nC: Sphenisciformes\nD: Struthioniformes\nE: Galliformes\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1247",
        "problem": "The diagram shows the pedigree of a human family in which the individuals marked with shaded symbols have no sweat glands. This is a rare condition.\n\n[figure1]\n\nWhich of the following would be the best inference from the data?\nA: A mutation occurred in individual $P$\nB: The mutant allele is on the $X$ chromosome\nC: The normal allele is incompletely dominant\nD: Individual $\\mathrm{Q}$ is carrying the mutant allele\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram shows the pedigree of a human family in which the individuals marked with shaded symbols have no sweat glands. This is a rare condition.\n\n[figure1]\n\nWhich of the following would be the best inference from the data?\n\nA: A mutation occurred in individual $P$\nB: The mutant allele is on the $X$ chromosome\nC: The normal allele is incompletely dominant\nD: Individual $\\mathrm{Q}$ is carrying the mutant allele\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-17.jpg?height=428&width=1257&top_left_y=236&top_left_x=434"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1532",
        "problem": "A section of DNA is shown.\n\n[figure1]\n\nThe bond labelled 3 links the carbon atom to...\nA: another sugar molecule.\nB: an amino acid.\nC: a phosphate group.\nD: a lipid/fat.\nE: a hydroxyl/alcohol group\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA section of DNA is shown.\n\n[figure1]\n\nThe bond labelled 3 links the carbon atom to...\n\nA: another sugar molecule.\nB: an amino acid.\nC: a phosphate group.\nD: a lipid/fat.\nE: a hydroxyl/alcohol group\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-08.jpg?height=845&width=1382&top_left_y=431&top_left_x=226"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_298",
        "problem": "ATP is an important energy source for maintaining normal membrane potential in nerve cells. Figure 1 shows the result of an experiment demonstrating $\\mathrm{Na}^{+}$efflux from an isolated squid giant nerve axon after injecting a buffer solution (artificial cytoplasm) that contains radioactive ${ }^{24} \\mathrm{Na}^{+}$.\n\n[figure1]\n\nFigure 1 Investigation of the efflux rate of radioisotope ${ }^{24} \\mathrm{Na}$ from a squid giant axon to the external solution (seawater). At 0 min, a buffer solution containing ${ }^{24} \\mathrm{Na}^{+}$was injected into the giant axon. For 100-190 min, the external seawater was replaced with a solution (seawater) containing $0.2 \\mathrm{mM}$ DNP (dinitrophenol), an uncoupler of oxidative phosphorylation.\nA: This experiment should have been carried out under the condition with sufficient oxygen to maintain the activity of ATP production by mitochondria. .....\nB: The efflux of ${ }^{24} \\mathrm{Na}^{+}$observed in seawater without DNP indicates the leaking of $\\mathrm{Na}$ ions out of the cell by nonspecific transportation.\nC: Delayed decrease of ${ }^{24} \\mathrm{Na}^{+}$efflux after using DNP reflects some amount of ATP storage inside the axon, including that being produced by glycolysis. .....\nD: Active transport of sodium ions was estimated to increase internal sodium concentration by $10 \\%$ in $50 \\mathrm{~min}$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nATP is an important energy source for maintaining normal membrane potential in nerve cells. Figure 1 shows the result of an experiment demonstrating $\\mathrm{Na}^{+}$efflux from an isolated squid giant nerve axon after injecting a buffer solution (artificial cytoplasm) that contains radioactive ${ }^{24} \\mathrm{Na}^{+}$.\n\n[figure1]\n\nFigure 1 Investigation of the efflux rate of radioisotope ${ }^{24} \\mathrm{Na}$ from a squid giant axon to the external solution (seawater). At 0 min, a buffer solution containing ${ }^{24} \\mathrm{Na}^{+}$was injected into the giant axon. For 100-190 min, the external seawater was replaced with a solution (seawater) containing $0.2 \\mathrm{mM}$ DNP (dinitrophenol), an uncoupler of oxidative phosphorylation.\n\nA: This experiment should have been carried out under the condition with sufficient oxygen to maintain the activity of ATP production by mitochondria. .....\nB: The efflux of ${ }^{24} \\mathrm{Na}^{+}$observed in seawater without DNP indicates the leaking of $\\mathrm{Na}$ ions out of the cell by nonspecific transportation.\nC: Delayed decrease of ${ }^{24} \\mathrm{Na}^{+}$efflux after using DNP reflects some amount of ATP storage inside the axon, including that being produced by glycolysis. .....\nD: Active transport of sodium ions was estimated to increase internal sodium concentration by $10 \\%$ in $50 \\mathrm{~min}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-09.jpg?height=619&width=876&top_left_y=770&top_left_x=453"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1254",
        "problem": "The diagram represents the life cycle of a fern, with the stages drawn to different scales.\n\n[figure1]\n\nWhich one of the rows $A$ to $E$ correctly represents the nuclear events occurring at stages $\\mathrm{K}, \\mathrm{L}$ and $\\mathrm{M}$ ?\nA: $\\mathrm{K}$: mitosis, $\\mathrm{L}$: meiosis, $\\mathrm{M}$:  mitosis\nB: $\\mathrm{K}$: mitosis, $\\mathrm{L}$: mitosis, $\\mathrm{M}$:  meiosis\nC: $\\mathrm{K}$: mitosis, $\\mathrm{L}$: meiosis, $\\mathrm{M}$:  meiosis\nD: $\\mathrm{K}$: meiosis, $\\mathrm{L}$: meiosis, $\\mathrm{M}$:  mitosis\nE: $\\mathrm{K}$: meiosis, $\\mathrm{L}$: mitosis, $\\mathrm{M}$:  mitosis\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram represents the life cycle of a fern, with the stages drawn to different scales.\n\n[figure1]\n\nWhich one of the rows $A$ to $E$ correctly represents the nuclear events occurring at stages $\\mathrm{K}, \\mathrm{L}$ and $\\mathrm{M}$ ?\n\nA: $\\mathrm{K}$: mitosis, $\\mathrm{L}$: meiosis, $\\mathrm{M}$:  mitosis\nB: $\\mathrm{K}$: mitosis, $\\mathrm{L}$: mitosis, $\\mathrm{M}$:  meiosis\nC: $\\mathrm{K}$: mitosis, $\\mathrm{L}$: meiosis, $\\mathrm{M}$:  meiosis\nD: $\\mathrm{K}$: meiosis, $\\mathrm{L}$: meiosis, $\\mathrm{M}$:  mitosis\nE: $\\mathrm{K}$: meiosis, $\\mathrm{L}$: mitosis, $\\mathrm{M}$:  mitosis\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-10.jpg?height=493&width=764&top_left_y=301&top_left_x=686"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1320",
        "problem": "The diagram below shows the anatomical planes and directions of a fish.\n\n[figure1]\n\nhttp://en.wikipedia.org/wiki/Anatomical terms of location\\#mediaviewer/File:Anatomical Directions and Axes.JPG\n\n[figure2]\nImage from Purves et al., Life: The Science of Biology, 4th Edition\n\n\nFully describe the anatomical position of the gonad relative to the gut in a fish.\nA: The gonad is a paired organ and each gonad lies dorsal to the gut.\nB: The gonad is a paired organ and each gonad lies dorsal to the gut just lateral to the midline.\nC: The gonad lies dorsal to the gut and extends laterally.\nD: The gonad is a paired organ and each gonad lies lateral to the gut.\nE: The paired gonad is lateral to the gut.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram below shows the anatomical planes and directions of a fish.\n\n[figure1]\n\nhttp://en.wikipedia.org/wiki/Anatomical terms of location\\#mediaviewer/File:Anatomical Directions and Axes.JPG\n\n[figure2]\nImage from Purves et al., Life: The Science of Biology, 4th Edition\n\n\nFully describe the anatomical position of the gonad relative to the gut in a fish.\n\nA: The gonad is a paired organ and each gonad lies dorsal to the gut.\nB: The gonad is a paired organ and each gonad lies dorsal to the gut just lateral to the midline.\nC: The gonad lies dorsal to the gut and extends laterally.\nD: The gonad is a paired organ and each gonad lies lateral to the gut.\nE: The paired gonad is lateral to the gut.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-12.jpg?height=856&width=1834&top_left_y=326&top_left_x=111",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-12.jpg?height=668&width=543&top_left_y=1245&top_left_x=1085"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_529",
        "problem": "遗传调查中发现一种罕见的单基因遗传病, 是某个关键基因中有一个 G-C 碱基对突变成 T-A 碱基对所致。图甲是一个该遗传病家系，对 5-10 号成员与该病相关等位基因进行测序, 发现 6、8、9 号个体只有一种碱基序列, 5、7、10 号个体均有两种。图乙是该突变基因表达过程图，（沙）表示突变位点。相关密码子：(1)AGU（丝氨酸)\n\n(2) CGU (精氨酸) (3) GAG (谷氨酸) (4) GUG (倾氨酸) (5)UAA (终止密码) (6)UAG (终止密码)。下列叙述不正确的是（）\n\n[图1]\n\n图甲突变基因\n\n[图2]\n\n## mRNA\n\n## 翻译 』\n\n0000000 肽链缩短\n\n图乙\nA: 这种遗传病的遗传方式是 X 染色体隐性遗传\nB: 图中所有女性个体的该基因相关的基因型均相同\nC: 该基因突变后转录形成的 mRNA 上相应密码子可能由(3)变成了 (6)\nD: 3 号与 4 号再生一个表现型正常的男孩的概率是 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n遗传调查中发现一种罕见的单基因遗传病, 是某个关键基因中有一个 G-C 碱基对突变成 T-A 碱基对所致。图甲是一个该遗传病家系，对 5-10 号成员与该病相关等位基因进行测序, 发现 6、8、9 号个体只有一种碱基序列, 5、7、10 号个体均有两种。图乙是该突变基因表达过程图，（沙）表示突变位点。相关密码子：(1)AGU（丝氨酸)\n\n(2) CGU (精氨酸) (3) GAG (谷氨酸) (4) GUG (倾氨酸) (5)UAA (终止密码) (6)UAG (终止密码)。下列叙述不正确的是（）\n\n[图1]\n\n图甲突变基因\n\n[图2]\n\n## mRNA\n\n## 翻译 』\n\n0000000 肽链缩短\n\n图乙\n\nA: 这种遗传病的遗传方式是 X 染色体隐性遗传\nB: 图中所有女性个体的该基因相关的基因型均相同\nC: 该基因突变后转录形成的 mRNA 上相应密码子可能由(3)变成了 (6)\nD: 3 号与 4 号再生一个表现型正常的男孩的概率是 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-021.jpg?height=394&width=371&top_left_y=1573&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-021.jpg?height=160&width=394&top_left_y=1616&top_left_x=745"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_31",
        "problem": "Trung cloned a coding sequence (CDS) of a gene into a vector, named pVN2016. This fragment was inserted at the SacII recognition site which is located in the multi cloning site (MCS) region within the lacZ gene of the vector (Fig.Q.93A). The inserted fragment has a $P s t I$ restriction site located $0.8 \\mathrm{~kb}$ upstream of its stop codon. To identify the size and direction of the inserted CDS (3'-end to $5^{\\prime}$ 'end of template strand), Trung digested this plasmid by different restriction enzymes, and the results of the digestions are shown in Fig.Q.93B.\n\nA\n\n[figure1]\n\nFig.Q.93. (A) A schematic map of the vector, numbers indicate positions of restriction enzyme recognition sites located in the vector (B) A schematic electrophoresis of digestive products using different restriction enzymes, M: 1 kb DNA ladder.\nA: The CDS is $2.6 \\mathrm{~kb}$ in length and has an EcoRI recognition site at about $0.5 \\mathrm{~kb}$ from one of its ends.\nB: Spel can be used to determine the orientation of the CDS.\nC: The CDS is oriented in reverse direction compared to lacZ gene.\nD: If plasmid pVN2016 is digested by both enzymes SpeI and EcoRI in Tango buffer (EcoRI and SpeI cut at $100 \\%$ and $20 \\%$ efficiency, respectively), five fragments of $0.5,0.8,1.3,2.1$ and $3.0 \\mathrm{~kb}$ could be detected by gel electrophoresis assuming that fragments smaller than $50 \\mathrm{bp}$ are not visible.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nTrung cloned a coding sequence (CDS) of a gene into a vector, named pVN2016. This fragment was inserted at the SacII recognition site which is located in the multi cloning site (MCS) region within the lacZ gene of the vector (Fig.Q.93A). The inserted fragment has a $P s t I$ restriction site located $0.8 \\mathrm{~kb}$ upstream of its stop codon. To identify the size and direction of the inserted CDS (3'-end to $5^{\\prime}$ 'end of template strand), Trung digested this plasmid by different restriction enzymes, and the results of the digestions are shown in Fig.Q.93B.\n\nA\n\n[figure1]\n\nFig.Q.93. (A) A schematic map of the vector, numbers indicate positions of restriction enzyme recognition sites located in the vector (B) A schematic electrophoresis of digestive products using different restriction enzymes, M: 1 kb DNA ladder.\n\nA: The CDS is $2.6 \\mathrm{~kb}$ in length and has an EcoRI recognition site at about $0.5 \\mathrm{~kb}$ from one of its ends.\nB: Spel can be used to determine the orientation of the CDS.\nC: The CDS is oriented in reverse direction compared to lacZ gene.\nD: If plasmid pVN2016 is digested by both enzymes SpeI and EcoRI in Tango buffer (EcoRI and SpeI cut at $100 \\%$ and $20 \\%$ efficiency, respectively), five fragments of $0.5,0.8,1.3,2.1$ and $3.0 \\mathrm{~kb}$ could be detected by gel electrophoresis assuming that fragments smaller than $50 \\mathrm{bp}$ are not visible.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-094.jpg?height=607&width=643&top_left_y=1138&top_left_x=251"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_917",
        "problem": "下图为甲、乙两种单基因遗传病的遗传家系图, 其中一种遗传病为伴性遗传。人群中纯合子甲病患者出现的频率约 $1 / 625$ 。下列叙述正确的是（ ）\n\n[图1]\nA: 甲病是伴 X 染色体显性遗传病\nB: $\\mathrm{I}_{3}$ 和 $\\mathrm{III}_{5}$ 的基因型相同的概率为 $5 / 9$\nC: 若 $\\mathrm{III}_{3}$ 与人群中只患甲病男性结婚, 所生正常孩子的概率约为 $18 / 49$\nD: 若 $\\mathrm{III}_{3}$ 性染色体组成为 XXX, 应是 $\\mathrm{II}_{3} 、 \\mathrm{II}_{4}$ 染色单体在减数第二次分裂时未分离\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为甲、乙两种单基因遗传病的遗传家系图, 其中一种遗传病为伴性遗传。人群中纯合子甲病患者出现的频率约 $1 / 625$ 。下列叙述正确的是（ ）\n\n[图1]\n\nA: 甲病是伴 X 染色体显性遗传病\nB: $\\mathrm{I}_{3}$ 和 $\\mathrm{III}_{5}$ 的基因型相同的概率为 $5 / 9$\nC: 若 $\\mathrm{III}_{3}$ 与人群中只患甲病男性结婚, 所生正常孩子的概率约为 $18 / 49$\nD: 若 $\\mathrm{III}_{3}$ 性染色体组成为 XXX, 应是 $\\mathrm{II}_{3} 、 \\mathrm{II}_{4}$ 染色单体在减数第二次分裂时未分离\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-52.jpg?height=439&width=1402&top_left_y=1845&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1192",
        "problem": "The graph shows the relationship between the volume, water potential $(\\Psi)$ and solute potential $\\left(\\Psi_{s}\\right)$ of plant cells immersed in a series of sucrose solutions of decreasing concentration. The cells were allowed to reach equilibrium with the bathing solutions, so that water was being neither lost nor gained, before the measurements were made.\n\nPressure potential $\\Psi_{p}$ is mechanical pressure. It increases as water enters a cell because the water present inside the cell exerts an outward pressure that is opposed by the structural rigidity of the cell wall. By creating this pressure, the plant can maintain turgor.\n\nWater potential $(\\Psi)=$ pressure potential $\\left(\\Psi_{\\mathrm{p}}\\right)+$ solute potential $\\left(\\Psi_{\\mathrm{s}}\\right)$\n\nWhen the volume of the cell is 103 units, the pressure potential of the cell $\\left(\\Psi_{\\mathrm{p}}\\right)$ is\n\n[figure1]\nA: $-320 \\mathrm{kPa}$\nB: $+320 \\mathrm{kPa}$\nC: $-590 \\mathrm{kPa}$\nD: $-880 \\mathrm{kPa}$\nE: $+880 \\mathrm{kPa}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows the relationship between the volume, water potential $(\\Psi)$ and solute potential $\\left(\\Psi_{s}\\right)$ of plant cells immersed in a series of sucrose solutions of decreasing concentration. The cells were allowed to reach equilibrium with the bathing solutions, so that water was being neither lost nor gained, before the measurements were made.\n\nPressure potential $\\Psi_{p}$ is mechanical pressure. It increases as water enters a cell because the water present inside the cell exerts an outward pressure that is opposed by the structural rigidity of the cell wall. By creating this pressure, the plant can maintain turgor.\n\nWater potential $(\\Psi)=$ pressure potential $\\left(\\Psi_{\\mathrm{p}}\\right)+$ solute potential $\\left(\\Psi_{\\mathrm{s}}\\right)$\n\nWhen the volume of the cell is 103 units, the pressure potential of the cell $\\left(\\Psi_{\\mathrm{p}}\\right)$ is\n\n[figure1]\n\nA: $-320 \\mathrm{kPa}$\nB: $+320 \\mathrm{kPa}$\nC: $-590 \\mathrm{kPa}$\nD: $-880 \\mathrm{kPa}$\nE: $+880 \\mathrm{kPa}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-13.jpg?height=1082&width=988&top_left_y=304&top_left_x=997"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_701",
        "problem": "紫薯是重要的粮食作物, 含 $\\mathrm{A}$ 基因时普通紫薯蔗糖含量低, 基本无甜味。科研工作者偶然发现一个单基因突变纯合子 $\\mathrm{aaBBCC}$, 微甜。继续培育高甜度紫薯品种过程中,得到了两个超甜紫薯品种甲（aabbCC）和乙（aaBBcc），其相关基因位置及基因控制相关物质合成途径如图所示。为验证甲、乙的基因型，分别与普通紫薯（AABBCC）杂交得 $F_{1}$ 再让 $F_{1}$ 自交得 $F_{2}$ (不考虑互换)。下列叙述正确的是（）\n\n[图1]\nA: 基因 $\\mathrm{A} / \\mathrm{a}$ 和 $\\mathrm{C} / \\mathrm{c}$ 的遗传遵循分离定律, 具有甜味的紫薯基因型最多有 16 种\nB: 若 $F_{2}$ 出现普通紫薯:超甜紫薯 $=3: 1$, 则超甜紫薯的基因型为 aaBBcc\nC: 基因型 aaBBCC 微甜的原因是酶甲催化合成的部分蔗糖被用于合成淀粉\nD: 若 $F_{2}$ 出现普通紫薯:超甜紫薯=3:1, 则超甜紫薯的基因型为 aabbCC\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n紫薯是重要的粮食作物, 含 $\\mathrm{A}$ 基因时普通紫薯蔗糖含量低, 基本无甜味。科研工作者偶然发现一个单基因突变纯合子 $\\mathrm{aaBBCC}$, 微甜。继续培育高甜度紫薯品种过程中,得到了两个超甜紫薯品种甲（aabbCC）和乙（aaBBcc），其相关基因位置及基因控制相关物质合成途径如图所示。为验证甲、乙的基因型，分别与普通紫薯（AABBCC）杂交得 $F_{1}$ 再让 $F_{1}$ 自交得 $F_{2}$ (不考虑互换)。下列叙述正确的是（）\n\n[图1]\n\nA: 基因 $\\mathrm{A} / \\mathrm{a}$ 和 $\\mathrm{C} / \\mathrm{c}$ 的遗传遵循分离定律, 具有甜味的紫薯基因型最多有 16 种\nB: 若 $F_{2}$ 出现普通紫薯:超甜紫薯 $=3: 1$, 则超甜紫薯的基因型为 aaBBcc\nC: 基因型 aaBBCC 微甜的原因是酶甲催化合成的部分蔗糖被用于合成淀粉\nD: 若 $F_{2}$ 出现普通紫薯:超甜紫薯=3:1, 则超甜紫薯的基因型为 aabbCC\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-47.jpg?height=363&width=1382&top_left_y=752&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_543",
        "problem": "某单基因突变引起的遗传病在人群中的发病率为 $1 / 100$, 由位于染色体上的基因 D、 $\\mathrm{d}$ 控制 (不考虑 XY 的同源区段)。图甲为某生物兴趣小组对该病三代之家调查绘制出的遗传系谱图 (图中各成员正常和患病的情况忘记标注), 图乙为图甲中部分成员关于该病基因酶切并电泳后的条带情况。下列叙述错误的是（ ）\n[图1]\nA: 由图乙电泳条带可知该病是常染色体隐性遗传病\nB: 对 3 号个体相关基因酶切后进行电泳, 预测电泳条带与 1 号个体完全相同的概率是 $1 / 2$\nC: 8 号个体与人群中一个正常的女性婚配, 生育患该病孩子的概率为 9/200\nD: 若 7 号个体正常，则 7 号个体的基因型一定为 Dd\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某单基因突变引起的遗传病在人群中的发病率为 $1 / 100$, 由位于染色体上的基因 D、 $\\mathrm{d}$ 控制 (不考虑 XY 的同源区段)。图甲为某生物兴趣小组对该病三代之家调查绘制出的遗传系谱图 (图中各成员正常和患病的情况忘记标注), 图乙为图甲中部分成员关于该病基因酶切并电泳后的条带情况。下列叙述错误的是（ ）\n[图1]\n\nA: 由图乙电泳条带可知该病是常染色体隐性遗传病\nB: 对 3 号个体相关基因酶切后进行电泳, 预测电泳条带与 1 号个体完全相同的概率是 $1 / 2$\nC: 8 号个体与人群中一个正常的女性婚配, 生育患该病孩子的概率为 9/200\nD: 若 7 号个体正常，则 7 号个体的基因型一定为 Dd\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-63.jpg?height=438&width=1414&top_left_y=518&top_left_x=364"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_619",
        "problem": "鸟面综合征是一种先天性遗传疾病, 又称领面骨发育不全及耳䶮综合征。某生物兴趣小组同学查阅相关资料得知, 该病由单基因控制。该小组同学又进一步做了相关调查,绘制了如图所示的遗传图谱并做了有关分析。下列分析正确的是 ( )\n\n[图1]\nA: 由图可知, 该致病基因位于 X 染色体上\nB: 若该病为显性遗传病, 则 I-2、III-2 一定为杂合子\nC: 若该病为隐性遗传病, 则 I-4、II-3 一定为杂合子\nD: 若该病为常染色体显性遗传病, 则II-2 的所有细胞中均含 1 个致病基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n鸟面综合征是一种先天性遗传疾病, 又称领面骨发育不全及耳䶮综合征。某生物兴趣小组同学查阅相关资料得知, 该病由单基因控制。该小组同学又进一步做了相关调查,绘制了如图所示的遗传图谱并做了有关分析。下列分析正确的是 ( )\n\n[图1]\n\nA: 由图可知, 该致病基因位于 X 染色体上\nB: 若该病为显性遗传病, 则 I-2、III-2 一定为杂合子\nC: 若该病为隐性遗传病, 则 I-4、II-3 一定为杂合子\nD: 若该病为常染色体显性遗传病, 则II-2 的所有细胞中均含 1 个致病基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-58.jpg?height=300&width=900&top_left_y=2117&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_975",
        "problem": "Which of the following classes contains organisms lacking a closed circulatory system?\nA: Polychaetes\nB: Cephalopoda\nC: Insecta\nD: Reptilia\nE: Eutheria\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following classes contains organisms lacking a closed circulatory system?\n\nA: Polychaetes\nB: Cephalopoda\nC: Insecta\nD: Reptilia\nE: Eutheria\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1281",
        "problem": "Portions of roots were immersed in culture solutions containing radioactive sulfur as ${ }^{35} \\mathrm{SO}_{4}{ }^{2-}$. One set was aerated and a second set had nitrogen bubbled through it. The amount of ${ }^{35} \\mathrm{SO}_{4}{ }^{2-}$ in the roots was measured at intervals and the results are shown in the graph below.\n\n[figure1]\n\nFrom these results the best conclusion is that\nA: the nitrogen interferes with ATP production\nB: passive diffusion of ${ }^{35} \\mathrm{SO}_{4}{ }^{2-}$ occurs between the solution and the roots\nC: the uptake of ${ }^{35} \\mathrm{SO}_{4}{ }^{2-}$ is an active process\nD: the uptake of ${ }^{35} \\mathrm{SO}_{4}{ }^{2-}$ involves both active and passive processes\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPortions of roots were immersed in culture solutions containing radioactive sulfur as ${ }^{35} \\mathrm{SO}_{4}{ }^{2-}$. One set was aerated and a second set had nitrogen bubbled through it. The amount of ${ }^{35} \\mathrm{SO}_{4}{ }^{2-}$ in the roots was measured at intervals and the results are shown in the graph below.\n\n[figure1]\n\nFrom these results the best conclusion is that\n\nA: the nitrogen interferes with ATP production\nB: passive diffusion of ${ }^{35} \\mathrm{SO}_{4}{ }^{2-}$ occurs between the solution and the roots\nC: the uptake of ${ }^{35} \\mathrm{SO}_{4}{ }^{2-}$ is an active process\nD: the uptake of ${ }^{35} \\mathrm{SO}_{4}{ }^{2-}$ involves both active and passive processes\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-08.jpg?height=472&width=770&top_left_y=607&top_left_x=431"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_546",
        "problem": "若某哺乳动物毛色由 3 对位于常染色体上的、独立分配的等位基因决定，其中， $\\mathrm{A}$ 基因编码的酶可使黄色素转化为褐色素: B 基因编码的酶可使该褐色素转化为黑色素; $\\mathrm{D}$ 基因的表达产物能完全抑制 $\\mathrm{A}$ 基因的表达: 相应的隐性等位基因 $a 、 b 、 d$ 的表达产物没有上述功能。若用两个纯合黄色品种的动物作为亲本进行杂交, $F_{1}$ 均为黄色, $F_{2}$ 中毛色表现型出现了黄: 褐: 黑=52: 3: 9 的数量比, 则杂交亲本的组合是 ( )\nA: AABBDD $\\times$ aaBBdd 或 $A A b b D D \\times$ aabbdd\nB: aaBBDD $\\times$ aabbdd 或 $A A b b D D \\times a a B B D D$\nC: aabbDD $\\times$ aabbdd 或 AAbbDD $\\times$ aabbdd\nD: AAbbDD $\\times$ aaBBdd 或 $A A B B D D \\times a a b b d d$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n若某哺乳动物毛色由 3 对位于常染色体上的、独立分配的等位基因决定，其中， $\\mathrm{A}$ 基因编码的酶可使黄色素转化为褐色素: B 基因编码的酶可使该褐色素转化为黑色素; $\\mathrm{D}$ 基因的表达产物能完全抑制 $\\mathrm{A}$ 基因的表达: 相应的隐性等位基因 $a 、 b 、 d$ 的表达产物没有上述功能。若用两个纯合黄色品种的动物作为亲本进行杂交, $F_{1}$ 均为黄色, $F_{2}$ 中毛色表现型出现了黄: 褐: 黑=52: 3: 9 的数量比, 则杂交亲本的组合是 ( )\n\nA: AABBDD $\\times$ aaBBdd 或 $A A b b D D \\times$ aabbdd\nB: aaBBDD $\\times$ aabbdd 或 $A A b b D D \\times a a B B D D$\nC: aabbDD $\\times$ aabbdd 或 AAbbDD $\\times$ aabbdd\nD: AAbbDD $\\times$ aaBBdd 或 $A A B B D D \\times a a b b d d$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1437",
        "problem": "[figure1]\n\nCell 1 is best described as what type of cell?\nA: red blood cell\nB: prokaryotic cell\nC: plant cell\nD: eukaryotic cell\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nCell 1 is best described as what type of cell?\n\nA: red blood cell\nB: prokaryotic cell\nC: plant cell\nD: eukaryotic cell\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-03.jpg?height=614&width=920&top_left_y=304&top_left_x=548"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_412",
        "problem": "烟草和许多果树如苹果、梨等存在不亲和基因的作用机制。其中烟草相应作用机制为: 基因型 $\\mathrm{S}_{1} \\mathrm{~S}_{2}$ 植株的花粉受到基因型 $\\mathrm{S}_{1} \\mathrm{~S}_{2}$ 植株的花柱抑阻, 不能参加受精,基因型 $\\mathrm{S}_{1} \\mathrm{~S}_{3}$的花粉落在 $S_{1} S_{2}$ 的柱头上时, $S_{1}$ 的花粉受到抑阻, 而 $S_{3}$ 的花粉不被抑阻, 可以参加受精, 生成 $\\mathrm{S}_{1} \\mathrm{~S}_{3}$ 和 $\\mathrm{S}_{2} \\mathrm{~S}_{3}$ 的合子，相关叙述错误的是（）\n\n[图1]\nA: 复等位基因 $\\mathrm{S}_{1} 、 \\mathrm{~S}_{2} 、 \\mathrm{~S}_{3}$ 的遗传遵循基因的分离定律\nB: 父本 $\\mathrm{S}_{1} \\mathrm{~S}_{2}$ 与母本 $\\mathrm{S}_{2} \\mathrm{~S}_{3}$ 杂交后代的基因型为 $\\mathrm{S}_{1} \\mathrm{~S}_{2} 、 \\mathrm{~S}_{1} \\mathrm{~S}_{3}$\nC: 这种机制不利于植物生存，终将在进化中被淘汰\nD: 在苹果园中添种不同基因型的授粉苹果树, 可提高结实率\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n烟草和许多果树如苹果、梨等存在不亲和基因的作用机制。其中烟草相应作用机制为: 基因型 $\\mathrm{S}_{1} \\mathrm{~S}_{2}$ 植株的花粉受到基因型 $\\mathrm{S}_{1} \\mathrm{~S}_{2}$ 植株的花柱抑阻, 不能参加受精,基因型 $\\mathrm{S}_{1} \\mathrm{~S}_{3}$的花粉落在 $S_{1} S_{2}$ 的柱头上时, $S_{1}$ 的花粉受到抑阻, 而 $S_{3}$ 的花粉不被抑阻, 可以参加受精, 生成 $\\mathrm{S}_{1} \\mathrm{~S}_{3}$ 和 $\\mathrm{S}_{2} \\mathrm{~S}_{3}$ 的合子，相关叙述错误的是（）\n\n[图1]\n\nA: 复等位基因 $\\mathrm{S}_{1} 、 \\mathrm{~S}_{2} 、 \\mathrm{~S}_{3}$ 的遗传遵循基因的分离定律\nB: 父本 $\\mathrm{S}_{1} \\mathrm{~S}_{2}$ 与母本 $\\mathrm{S}_{2} \\mathrm{~S}_{3}$ 杂交后代的基因型为 $\\mathrm{S}_{1} \\mathrm{~S}_{2} 、 \\mathrm{~S}_{1} \\mathrm{~S}_{3}$\nC: 这种机制不利于植物生存，终将在进化中被淘汰\nD: 在苹果园中添种不同基因型的授粉苹果树, 可提高结实率\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-18.jpg?height=611&width=579&top_left_y=146&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_460",
        "problem": "某植物控制叶形 (A/a) 和叶色（B/b）的基因分别位于两对同源染色体上。宽叶紫色植株和窄叶绿色植株杂交获得 $F_{1}, F_{1}$ 中的宽叶紫色植株自交获得 $F_{2}, F_{2}$ 中宽叶紫色: 宽叶绿色: 窄叶紫色: 窄叶绿色=10:2: 5: 1. 已知出现上述比例的原因是某性状中显性基因纯合植株致死，同时另一性状中某种基因型的花粉部分致死。下列叙述正确的是 ( )\nA: $\\mathrm{F}_{1}$ 产生的四种花粉比例相同\nB: 叶色性状中存在显性基因纯合致死现象\nC: 若 $F_{2}$ 中宽叶植株自交, 后代中窄叶植株占 $1 / 2$\nD: 若 $F_{2}$ 中紫色植株自由交配, 后代中绿色植株占 $9 / 170$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某植物控制叶形 (A/a) 和叶色（B/b）的基因分别位于两对同源染色体上。宽叶紫色植株和窄叶绿色植株杂交获得 $F_{1}, F_{1}$ 中的宽叶紫色植株自交获得 $F_{2}, F_{2}$ 中宽叶紫色: 宽叶绿色: 窄叶紫色: 窄叶绿色=10:2: 5: 1. 已知出现上述比例的原因是某性状中显性基因纯合植株致死，同时另一性状中某种基因型的花粉部分致死。下列叙述正确的是 ( )\n\nA: $\\mathrm{F}_{1}$ 产生的四种花粉比例相同\nB: 叶色性状中存在显性基因纯合致死现象\nC: 若 $F_{2}$ 中宽叶植株自交, 后代中窄叶植株占 $1 / 2$\nD: 若 $F_{2}$ 中紫色植株自由交配, 后代中绿色植株占 $9 / 170$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1060",
        "problem": "Consider a patchy environment where the habitat patches differ in ecological parameters. Two such habitats were found which differed in the size of seeds available to consumers. Habitat 1 had trees that bore small sized seeds. Thus, smallbilled birds would have the highest food intake and the habitat was thus populated by small-billed bird species. Habitat 2 had trees that bore large-sized seeds and was populated by large-billed birds of the same species.\n\nWe assume that birds with medium-bill size perform worse over both habitats than either small- or large-billed birds. Also there is no hindrance for birds to fly from one habitat to another.\n\nHabitat 1\n\n[figure1]\n\nHabitat 2\n\n[figure2]\n\nWhich of the following is correct?\nA: Stabilizing selection is operating within each habitat.\nB: Disruptive selection is operating for the entire bird population.\nC: Disruptive selection cannot occur as there is possible gene flow between the populations of birds occupying both habitats.\nD: After several generations, speciation is most likely to result because selection will constantly remove the intermediate phenotypes (medium-bill size).\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nConsider a patchy environment where the habitat patches differ in ecological parameters. Two such habitats were found which differed in the size of seeds available to consumers. Habitat 1 had trees that bore small sized seeds. Thus, smallbilled birds would have the highest food intake and the habitat was thus populated by small-billed bird species. Habitat 2 had trees that bore large-sized seeds and was populated by large-billed birds of the same species.\n\nWe assume that birds with medium-bill size perform worse over both habitats than either small- or large-billed birds. Also there is no hindrance for birds to fly from one habitat to another.\n\nHabitat 1\n\n[figure1]\n\nHabitat 2\n\n[figure2]\n\nWhich of the following is correct?\n\nA: Stabilizing selection is operating within each habitat.\nB: Disruptive selection is operating for the entire bird population.\nC: Disruptive selection cannot occur as there is possible gene flow between the populations of birds occupying both habitats.\nD: After several generations, speciation is most likely to result because selection will constantly remove the intermediate phenotypes (medium-bill size).\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-26.jpg?height=642&width=715&top_left_y=942&top_left_x=301",
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-26.jpg?height=609&width=451&top_left_y=972&top_left_x=1379"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_72",
        "problem": "Two groups of students separately performed an experiment on kidney function. Thirty minutes before the experiment, each student in one group was instructed to drink $500 \\mathrm{~mL}$ of water, while each student in the other group was instructed to drink $100 \\mathrm{~mL}$ of water. At $\\mathrm{t}=0$ minute, each student in both groups was instructed to drink $750 \\mathrm{~mL}$ of water. Each student was then asked to urinate as he or she would normally do without attempting to manipulate the speed or flow in any way at the different time points shown in Figure Q.26. An electronic uroflowmeter was used to measure the urine flow rate. The $\\mathrm{Cl}^{-}$concentration in each urine sample was measured. Figure Q. 26 shows the data of the experiment.\n[figure1]\nA: The reabsorption of water by nephrons of the students in Group II at $t=60 \\mathrm{~min}$ was lower than that of the students in Group I.\nB: The plasma aldosterone concentration of the students in Group I was the highest at $\\mathrm{t}=60 \\mathrm{~min}$.\nC: The students in Group I drank $500 \\mathrm{~mL}$ of water 30 minutes before the experiment.\nD: The students in Group II produced about $140 \\mathrm{~mL}$ of urine during the period between $\\mathrm{t}=40 \\mathrm{~min}$ and $\\mathrm{t}=60 \\mathrm{~min}$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nTwo groups of students separately performed an experiment on kidney function. Thirty minutes before the experiment, each student in one group was instructed to drink $500 \\mathrm{~mL}$ of water, while each student in the other group was instructed to drink $100 \\mathrm{~mL}$ of water. At $\\mathrm{t}=0$ minute, each student in both groups was instructed to drink $750 \\mathrm{~mL}$ of water. Each student was then asked to urinate as he or she would normally do without attempting to manipulate the speed or flow in any way at the different time points shown in Figure Q.26. An electronic uroflowmeter was used to measure the urine flow rate. The $\\mathrm{Cl}^{-}$concentration in each urine sample was measured. Figure Q. 26 shows the data of the experiment.\n[figure1]\n\nA: The reabsorption of water by nephrons of the students in Group II at $t=60 \\mathrm{~min}$ was lower than that of the students in Group I.\nB: The plasma aldosterone concentration of the students in Group I was the highest at $\\mathrm{t}=60 \\mathrm{~min}$.\nC: The students in Group I drank $500 \\mathrm{~mL}$ of water 30 minutes before the experiment.\nD: The students in Group II produced about $140 \\mathrm{~mL}$ of urine during the period between $\\mathrm{t}=40 \\mathrm{~min}$ and $\\mathrm{t}=60 \\mathrm{~min}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-059.jpg?height=526&width=1660&top_left_y=1179&top_left_x=197"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_591",
        "problem": "用基因型为 $\\mathrm{Aa}$ 的小麦分别进行连续自交、随机交配、连续自交并逐代淘汰隐性个体、随机交配并逐代淘汰隐性个体，根据各代 $\\mathrm{Aa}$ 基因型频率绘制曲线如图所示。下列分析正确的是 ( )\n\n[图1]\nA: 曲线 I 和 III 的各子代间 $\\mathrm{A}$ 和 $\\mathrm{a}$ 的基因频率始终相等\nB: 曲线 II 的 $F_{3}$ 代中纯合体的比例是 $3 / 5$\nC: 曲线III的 $\\mathrm{F}_{3}$ 代中纯合体的比例是 $7 / 10$\nD: 曲线 IV 的 $F_{n}$ 中纯合体的比例比上一代增加 $(1 / 2)^{n+1}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n用基因型为 $\\mathrm{Aa}$ 的小麦分别进行连续自交、随机交配、连续自交并逐代淘汰隐性个体、随机交配并逐代淘汰隐性个体，根据各代 $\\mathrm{Aa}$ 基因型频率绘制曲线如图所示。下列分析正确的是 ( )\n\n[图1]\n\nA: 曲线 I 和 III 的各子代间 $\\mathrm{A}$ 和 $\\mathrm{a}$ 的基因频率始终相等\nB: 曲线 II 的 $F_{3}$ 代中纯合体的比例是 $3 / 5$\nC: 曲线III的 $\\mathrm{F}_{3}$ 代中纯合体的比例是 $7 / 10$\nD: 曲线 IV 的 $F_{n}$ 中纯合体的比例比上一代增加 $(1 / 2)^{n+1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-015.jpg?height=494&width=711&top_left_y=1175&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_180",
        "problem": "Polycystic ovarian syndrome (PCOS) is a common disorder of women characterized by increased levels of testosterone and by chronic failure in ovulation. The ovary can be stimulated to produce more testosterone when insulin levels in the blood are high.\nA: PCOS patients are more likely to have acne than healthy people.\nB: PCOS patients have progesterone level higher than healthy people.\nC: Obese women have a higher risk of PCOS than normal-weight women.\nD: Follicle-stimulating hormone (FSH) and Luteinizing hormone (LH) can be used to increase the probability of PCOS women to conceive.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPolycystic ovarian syndrome (PCOS) is a common disorder of women characterized by increased levels of testosterone and by chronic failure in ovulation. The ovary can be stimulated to produce more testosterone when insulin levels in the blood are high.\n\nA: PCOS patients are more likely to have acne than healthy people.\nB: PCOS patients have progesterone level higher than healthy people.\nC: Obese women have a higher risk of PCOS than normal-weight women.\nD: Follicle-stimulating hormone (FSH) and Luteinizing hormone (LH) can be used to increase the probability of PCOS women to conceive.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_743",
        "problem": "DNA 分子经过诱变, 某位点上一个正常碱基（设为 $\\mathrm{P}$ ）变成了尿嘧啶。该 DNA 连续复制两次, 得到 4 个子代 DNA 分子相应位点上的碱基对分别为 U-A、A-T、G-C、C-G，推测“P”可能是（ ）\nA: 胸腺嘧啶\nB: 腺嘌呤\nC: 胸腺嘧啶或腺嘌呤\nD: 胞嘧啶\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nDNA 分子经过诱变, 某位点上一个正常碱基（设为 $\\mathrm{P}$ ）变成了尿嘧啶。该 DNA 连续复制两次, 得到 4 个子代 DNA 分子相应位点上的碱基对分别为 U-A、A-T、G-C、C-G，推测“P”可能是（ ）\n\nA: 胸腺嘧啶\nB: 腺嘌呤\nC: 胸腺嘧啶或腺嘌呤\nD: 胞嘧啶\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_401",
        "problem": "蜜蜂的工蜂和蜂王都是二倍体, 而雄蜂为单倍体。有一种黄蜂的群体中存在这样一种现象: 一些雄蜂体内含有一种特殊的染色体 psr, 在受精时它能破坏来自同一染色体组的其他染色体，导致后代表型变化，其作用机制如图所示。下列叙述正确的是（）\n\n[图1]\nA: psr 含有雄蜂生长发育所需的全部基因\nB: 卵细胞经过受精形成的都是雌性\nC: psr 能从父方传给雄性和䧳性子代\nD: psr 会在后代雄蜂中代代相传\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n蜜蜂的工蜂和蜂王都是二倍体, 而雄蜂为单倍体。有一种黄蜂的群体中存在这样一种现象: 一些雄蜂体内含有一种特殊的染色体 psr, 在受精时它能破坏来自同一染色体组的其他染色体，导致后代表型变化，其作用机制如图所示。下列叙述正确的是（）\n\n[图1]\n\nA: psr 含有雄蜂生长发育所需的全部基因\nB: 卵细胞经过受精形成的都是雌性\nC: psr 能从父方传给雄性和䧳性子代\nD: psr 会在后代雄蜂中代代相传\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-79.jpg?height=682&width=1042&top_left_y=573&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_427",
        "problem": "鸡的性别决定方式属于 ZW 型, 现有一只纯种雌性芦花鸡与一只纯种雄性非芦花鸡交配多次, $F_{1}$ 中雄的均为芦花形, 雌的均为非芦花形。据此推测错误的是\nA: 控制芦花和非芦花性状的基因在 Z 染色体上，属于性染色体遗传\nB: 雄鸡中芦花鸡的比例比雌鸡中的相应比例大\nC: 让 $F_{1}$ 中的雌雄鸡自由交配, 产生的 $F_{2}$ 雄鸡表现型有两种, 雌鸡也有两种\nD: 将 $F_{2}$ 中的芦花鸡雌雄交配, 产生的 $F_{3}$ 中非芦花鸡占 $3 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n鸡的性别决定方式属于 ZW 型, 现有一只纯种雌性芦花鸡与一只纯种雄性非芦花鸡交配多次, $F_{1}$ 中雄的均为芦花形, 雌的均为非芦花形。据此推测错误的是\n\nA: 控制芦花和非芦花性状的基因在 Z 染色体上，属于性染色体遗传\nB: 雄鸡中芦花鸡的比例比雌鸡中的相应比例大\nC: 让 $F_{1}$ 中的雌雄鸡自由交配, 产生的 $F_{2}$ 雄鸡表现型有两种, 雌鸡也有两种\nD: 将 $F_{2}$ 中的芦花鸡雌雄交配, 产生的 $F_{3}$ 中非芦花鸡占 $3 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_138",
        "problem": "Taxonomy has traditionally been based on morphology. DNA Barcoding is a new approach that aims to allow accurate and relatively simple species identification based on the nucleotide sequence of a 650 -bp fragment of the mitochondrial COI gene. The Kimura-2 parameter (K2P) distance is a widely accepted index that reflects the divergence between two DNA sequences.\n\nIn a study on fish inhabiting the Persian Gulf, over 150 individuals were studied that, based on morphology, represented 83 species. The average of intraspecific K2P distances based on COI sequence was calculated to be $1.15 \\%$. The average distance reported in various reputable fish studies in the world has been $0.25 \\%-0.45 \\%$.\n\nIn a study on fish inhabiting the Gulf of Mexico, again over 150 individuals were studied. Based on morphology, these fish represented 76 species, 56 genera, 32 families, 11 orders, and 2 classes. The table below presents K2P distances between specimens at different taxonomic levels.\n\n|  | No. of <br> comparisons | Min. <br> distance (\\%) | Mean <br> distance (\\%) | Max <br> distance (\\%) | Standard error <br> distance (\\%) |\n| :--- | :---: | :---: | :---: | :---: | :---: |\n| Within species | 185 | 0 | 0.18 | 1.66 | 0.02 |\n| Within genus | 76 | 6.19 | 12 | 20.23 | 0.42 |\n| Within family | 888 | 10.88 | 17.43 | 24.56 | 0.08 |\n| Within order | 9274 | 14.57 | 21.51 | 28.9 | 0.02 |\n| Within class | 3439 | 16.2 | 22.77 | 34.41 | 0.04 |\nA: The evolutionary rate of sequence change in a DNA fragment chosen for DNA barcoding should certainly be more rapid than the evolutionary rate of change\nB: The high average COI intraspecific K2P distance of the Persian gulf study compared to other fish studies could be explained by existence of relatively\nC: The data in table support the proposal that species identification of fish can be made on the basis of COI fragment sequences.\nD: Table 1 suggests that COI based barcoding is not appropriate for genus identification.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nTaxonomy has traditionally been based on morphology. DNA Barcoding is a new approach that aims to allow accurate and relatively simple species identification based on the nucleotide sequence of a 650 -bp fragment of the mitochondrial COI gene. The Kimura-2 parameter (K2P) distance is a widely accepted index that reflects the divergence between two DNA sequences.\n\nIn a study on fish inhabiting the Persian Gulf, over 150 individuals were studied that, based on morphology, represented 83 species. The average of intraspecific K2P distances based on COI sequence was calculated to be $1.15 \\%$. The average distance reported in various reputable fish studies in the world has been $0.25 \\%-0.45 \\%$.\n\nIn a study on fish inhabiting the Gulf of Mexico, again over 150 individuals were studied. Based on morphology, these fish represented 76 species, 56 genera, 32 families, 11 orders, and 2 classes. The table below presents K2P distances between specimens at different taxonomic levels.\n\n|  | No. of <br> comparisons | Min. <br> distance (\\%) | Mean <br> distance (\\%) | Max <br> distance (\\%) | Standard error <br> distance (\\%) |\n| :--- | :---: | :---: | :---: | :---: | :---: |\n| Within species | 185 | 0 | 0.18 | 1.66 | 0.02 |\n| Within genus | 76 | 6.19 | 12 | 20.23 | 0.42 |\n| Within family | 888 | 10.88 | 17.43 | 24.56 | 0.08 |\n| Within order | 9274 | 14.57 | 21.51 | 28.9 | 0.02 |\n| Within class | 3439 | 16.2 | 22.77 | 34.41 | 0.04 |\n\nA: The evolutionary rate of sequence change in a DNA fragment chosen for DNA barcoding should certainly be more rapid than the evolutionary rate of change\nB: The high average COI intraspecific K2P distance of the Persian gulf study compared to other fish studies could be explained by existence of relatively\nC: The data in table support the proposal that species identification of fish can be made on the basis of COI fragment sequences.\nD: Table 1 suggests that COI based barcoding is not appropriate for genus identification.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_571",
        "problem": "果蝇的长翅与残翅、圆眼与棒眼是两对相对性状，分别由等位基因 $\\mathrm{B} / \\mathrm{b} 、 \\mathrm{D} / \\mathrm{d}$ 控制。\n\n研究人员利用纯合残翅圆哏雄果蝇甲与两只雌果蝇（残翅棒眼乙、长翅圆眼丙）分别交配，结果如下表所示。下列叙述错误的是（）\n\n| 组 | 亲本 |\n| :--- | :--- |\n| 别 | 组合 |\n\n\n| 1 | 乙 | 残翅圆眼 ( |\n| :---: | :---: | :---: |\n| 2 | 甲× <br> 丙 | [图1] |\nA: 控制果蝇翅形的基因位于常染色体上\nB: 控制果蝇眼形的基因位于 $\\mathrm{X}$ 染色体上\nC: 乙果蝇的基因型是 $b b X^{d} X^{d}$\nD: 组别 2 实验 $\\mathrm{F}_{1}$ 中的长翅圆眼雌雄果蝇杂交, 子代的长翅圆眼雌果蝇中, 杂合子所占的比例为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的长翅与残翅、圆眼与棒眼是两对相对性状，分别由等位基因 $\\mathrm{B} / \\mathrm{b} 、 \\mathrm{D} / \\mathrm{d}$ 控制。\n\n研究人员利用纯合残翅圆哏雄果蝇甲与两只雌果蝇（残翅棒眼乙、长翅圆眼丙）分别交配，结果如下表所示。下列叙述错误的是（）\n\n| 组 | 亲本 |\n| :--- | :--- |\n| 别 | 组合 |\n\n\n| 1 | 乙 | 残翅圆眼 ( |\n| :---: | :---: | :---: |\n| 2 | 甲× <br> 丙 | [图1] |\n\nA: 控制果蝇翅形的基因位于常染色体上\nB: 控制果蝇眼形的基因位于 $\\mathrm{X}$ 染色体上\nC: 乙果蝇的基因型是 $b b X^{d} X^{d}$\nD: 组别 2 实验 $\\mathrm{F}_{1}$ 中的长翅圆眼雌雄果蝇杂交, 子代的长翅圆眼雌果蝇中, 杂合子所占的比例为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-84.jpg?height=133&width=1189&top_left_y=373&top_left_x=526"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_961",
        "problem": "By convincing soldiers that they are part of a \"brotherhood,\" and thus increasing the likelihood that they will protect, and even die for each other, the military is-consciously or unconsciously-tapping into genes that were adaptive in our ancestors due to\nA: Individual selection\nB: Kin selection\nC: Neoteny\nD: Inclusive selection\nE: Heterochrony\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBy convincing soldiers that they are part of a \"brotherhood,\" and thus increasing the likelihood that they will protect, and even die for each other, the military is-consciously or unconsciously-tapping into genes that were adaptive in our ancestors due to\n\nA: Individual selection\nB: Kin selection\nC: Neoteny\nD: Inclusive selection\nE: Heterochrony\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_210",
        "problem": "Figure Q. 80 demonstrates the relationship between oxygen concentration and oxygen partial pressure $\\left(\\mathrm{PO}_{2}\\right)$ in blood of two species of vertebrates (species $a$ and $b$ ). Each sample was subjected to two levels of carbon dioxide pressure $\\left(\\mathrm{PcO}_{2}\\right)$ : curve I represents the values measured at normal $\\mathrm{PcO}_{2}$ and curve II represents the values measured at elevated $\\mathrm{PCO}_{2}$. The blood having passed through the lungs of the two species normally has a $\\mathrm{Po}_{2}$ of $100 \\mathrm{~mm} \\mathrm{Hg}$ and the deoxygenated blood leaving the tissues has a $\\mathrm{Po}_{2}$ of 40 $\\mathrm{mm} \\mathrm{Hg}$.\n[figure1]\n\nFigure Q.80.\nA: While comparing curve I of species $a$ with curve I of species $b$, you can predict that the $\\mathrm{O}_{2}$ concentration of the blood in the lungs of species $a$ will be higher than that of species $b$.\nB: If you expose deoxygenated blood of the two species at the same level of $\\mathrm{PCO}_{2}$ to increasing $\\mathrm{Po}_{2}$, the first blood to become saturated with $\\mathrm{O}_{2}$ would be that of species a.\nC: In species $b$, if curves $\\mathrm{I}$ and II represent the $\\mathrm{PcO}_{2}$ of oxygenated and deoxygenated blood, respectively, there will be less than $160 \\mathrm{~mL}$ of $\\mathrm{O}_{2}$ released from a litre of blood as it passes through the tissues.\nD: In species $a$, an increase in $\\mathrm{PcO}_{2}$ in the blood reduces the affinity of hemoglobin for oxygen but has no effect on the maximum oxygen-carrying capacity in the blood.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFigure Q. 80 demonstrates the relationship between oxygen concentration and oxygen partial pressure $\\left(\\mathrm{PO}_{2}\\right)$ in blood of two species of vertebrates (species $a$ and $b$ ). Each sample was subjected to two levels of carbon dioxide pressure $\\left(\\mathrm{PcO}_{2}\\right)$ : curve I represents the values measured at normal $\\mathrm{PcO}_{2}$ and curve II represents the values measured at elevated $\\mathrm{PCO}_{2}$. The blood having passed through the lungs of the two species normally has a $\\mathrm{Po}_{2}$ of $100 \\mathrm{~mm} \\mathrm{Hg}$ and the deoxygenated blood leaving the tissues has a $\\mathrm{Po}_{2}$ of 40 $\\mathrm{mm} \\mathrm{Hg}$.\n[figure1]\n\nFigure Q.80.\n\nA: While comparing curve I of species $a$ with curve I of species $b$, you can predict that the $\\mathrm{O}_{2}$ concentration of the blood in the lungs of species $a$ will be higher than that of species $b$.\nB: If you expose deoxygenated blood of the two species at the same level of $\\mathrm{PCO}_{2}$ to increasing $\\mathrm{Po}_{2}$, the first blood to become saturated with $\\mathrm{O}_{2}$ would be that of species a.\nC: In species $b$, if curves $\\mathrm{I}$ and II represent the $\\mathrm{PcO}_{2}$ of oxygenated and deoxygenated blood, respectively, there will be less than $160 \\mathrm{~mL}$ of $\\mathrm{O}_{2}$ released from a litre of blood as it passes through the tissues.\nD: In species $a$, an increase in $\\mathrm{PcO}_{2}$ in the blood reduces the affinity of hemoglobin for oxygen but has no effect on the maximum oxygen-carrying capacity in the blood.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-070.jpg?height=768&width=1628&top_left_y=960&top_left_x=178"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_954",
        "problem": "Replacement of a lysine with a glycine in a protein could result in all of the following EXCEPT a:\nA: Change in the quaternary structure of the protein\nB: Change in the secondary structure of the protein\nC: Loss of catalytic activity of the protein\nD: Loss of a negatively charged side chain\nE: Loss of the protein's ability to interact with other proteins\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nReplacement of a lysine with a glycine in a protein could result in all of the following EXCEPT a:\n\nA: Change in the quaternary structure of the protein\nB: Change in the secondary structure of the protein\nC: Loss of catalytic activity of the protein\nD: Loss of a negatively charged side chain\nE: Loss of the protein's ability to interact with other proteins\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1431",
        "problem": "[figure1]\n\nWhat does the organelle $\\underline{K}$ produce when functioning normally in a cell?\nA: protein\nB: mRNA\nC: ATP\nD: endoplasmic reticulum\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nWhat does the organelle $\\underline{K}$ produce when functioning normally in a cell?\n\nA: protein\nB: mRNA\nC: ATP\nD: endoplasmic reticulum\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-03.jpg?height=614&width=920&top_left_y=304&top_left_x=548"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1476",
        "problem": "When someone loses weight by \"burning fat\", where does most of the mass of the fat go?\nA: It is excreted in the faeces\nB: It is converted into energy according to $\\mathrm{E}=\\mathrm{mc}^{2}$, where $\\mathrm{E}$ is energy, $\\mathrm{m}$ is mass and $c$ is the speed of light\nC: It is exhaled as carbon dioxide\nD: It is lost in sweat and urine\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen someone loses weight by \"burning fat\", where does most of the mass of the fat go?\n\nA: It is excreted in the faeces\nB: It is converted into energy according to $\\mathrm{E}=\\mathrm{mc}^{2}$, where $\\mathrm{E}$ is energy, $\\mathrm{m}$ is mass and $c$ is the speed of light\nC: It is exhaled as carbon dioxide\nD: It is lost in sweat and urine\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_511",
        "problem": "DNA 甲基化是常见的基因组印记方式。MAGEL2 是一种印记基因, 其致病性变异 （致病基因）会导致新生儿患某综合征。母源 MAGEL2 致病基因传递给子代时其印记区域由于被甲基化而沉默，而由父源 MAGEL2 致病基因传递给子代时可表达并致病。下图为该基因引起的遗传病的系谱图, $\\mathrm{II}_{4}$ 不携带致病基因。下列有关叙述正确的是 ( )\n\n[图1]\nA: 该病的遗传方式是常染色体隐性遗传\nB: $\\mathrm{II}_{3}$ 携带有 MAGEL2 致病基因, $\\mathrm{II}_{1}$ 和 $\\mathrm{II}_{2}$ 则不一定携带有 MAGEL2 致病基因\nC: MAGEL2 基因从 $\\mathrm{I}_{2} \\rightarrow \\mathrm{II}_{3} \\rightarrow \\mathrm{III}_{1}$ 的传递过程中碱基序列会发生改变\nD: 已被甲基化的基因在亲代传递给子代的过程中一般不会发生去甲基化\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nDNA 甲基化是常见的基因组印记方式。MAGEL2 是一种印记基因, 其致病性变异 （致病基因）会导致新生儿患某综合征。母源 MAGEL2 致病基因传递给子代时其印记区域由于被甲基化而沉默，而由父源 MAGEL2 致病基因传递给子代时可表达并致病。下图为该基因引起的遗传病的系谱图, $\\mathrm{II}_{4}$ 不携带致病基因。下列有关叙述正确的是 ( )\n\n[图1]\n\nA: 该病的遗传方式是常染色体隐性遗传\nB: $\\mathrm{II}_{3}$ 携带有 MAGEL2 致病基因, $\\mathrm{II}_{1}$ 和 $\\mathrm{II}_{2}$ 则不一定携带有 MAGEL2 致病基因\nC: MAGEL2 基因从 $\\mathrm{I}_{2} \\rightarrow \\mathrm{II}_{3} \\rightarrow \\mathrm{III}_{1}$ 的传递过程中碱基序列会发生改变\nD: 已被甲基化的基因在亲代传递给子代的过程中一般不会发生去甲基化\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-77.jpg?height=269&width=688&top_left_y=965&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_182",
        "problem": "Centromere position can be mapped using linear tetrads in some fungi. If there is no cross-over between gene $\\mathrm{E}$ and the centromere, four spores are arranged in sequence of eeEE or EEee (Fig.Q.35A). If there is a cross-over between gene E and the centromere, four spores are arranged in sequence of eEeE or EeEe (Fig.Q.35B).\n[figure1]\n\nB\n[figure2]\n\nFigure Q. 35\n\nStrain $a b$ was crossed to strain $a^{+} b^{+}$and 100 linear tetrads were isolated. Six classes in the following proportions were identified:\n\n| Class 1 | Class 2 | Class 3 | Class 4 | Class 5 | Class 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\mathrm{ab}$ | $\\mathrm{ab}$ | $\\mathrm{ab}$ | $\\mathrm{ab}$ | $\\mathrm{ab}^{+}$ | $\\mathrm{ab}$ |\n| $\\mathrm{a}^{+} \\mathrm{b}$ | $\\mathrm{ab}^{+}$ | $\\mathrm{ab}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ |\n| $\\mathrm{ab}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{ab}$ | $\\mathrm{ab}^{+}$ | $\\mathrm{ab}^{+}$ |\n| $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}$ | $\\mathrm{a}^{+} \\mathrm{b}$ |\n| 15 | 29 | 47 | 2 | 2 | 5 |\nA: Locus $\\mathrm{a}$ and $\\mathrm{b}$ are located on the same arm of chromosome.\nB: Class 4 involves a 4 -strand double crossover.\nC: Class 6 involves a 3 -strand double crossover, with one between gene a and the centromere and between the centromere and gene $b$.\nD: Class 5 involves a 2 -strand double crossover.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCentromere position can be mapped using linear tetrads in some fungi. If there is no cross-over between gene $\\mathrm{E}$ and the centromere, four spores are arranged in sequence of eeEE or EEee (Fig.Q.35A). If there is a cross-over between gene E and the centromere, four spores are arranged in sequence of eEeE or EeEe (Fig.Q.35B).\n[figure1]\n\nB\n[figure2]\n\nFigure Q. 35\n\nStrain $a b$ was crossed to strain $a^{+} b^{+}$and 100 linear tetrads were isolated. Six classes in the following proportions were identified:\n\n| Class 1 | Class 2 | Class 3 | Class 4 | Class 5 | Class 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\mathrm{ab}$ | $\\mathrm{ab}$ | $\\mathrm{ab}$ | $\\mathrm{ab}$ | $\\mathrm{ab}^{+}$ | $\\mathrm{ab}$ |\n| $\\mathrm{a}^{+} \\mathrm{b}$ | $\\mathrm{ab}^{+}$ | $\\mathrm{ab}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ |\n| $\\mathrm{ab}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{ab}$ | $\\mathrm{ab}^{+}$ | $\\mathrm{ab}^{+}$ |\n| $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}^{+}$ | $\\mathrm{a}^{+} \\mathrm{b}$ | $\\mathrm{a}^{+} \\mathrm{b}$ |\n| 15 | 29 | 47 | 2 | 2 | 5 |\n\nA: Locus $\\mathrm{a}$ and $\\mathrm{b}$ are located on the same arm of chromosome.\nB: Class 4 involves a 4 -strand double crossover.\nC: Class 6 involves a 3 -strand double crossover, with one between gene a and the centromere and between the centromere and gene $b$.\nD: Class 5 involves a 2 -strand double crossover.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-080.jpg?height=386&width=1202&top_left_y=582&top_left_x=434",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-080.jpg?height=336&width=1144&top_left_y=1010&top_left_x=502",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-081.jpg?height=245&width=780&top_left_y=1171&top_left_x=798",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-081.jpg?height=526&width=780&top_left_y=1596&top_left_x=798",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-081.jpg?height=246&width=780&top_left_y=2290&top_left_x=798"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_57",
        "problem": "The diagram below illustrates the $\\mathrm{ABCE}$ model of flower development, in which the identity structure of floral organs of eudicots is regulated by floral homeotic genes divided into ABCE-classes, based on function: A- and E-class genes specify sepal identity; A-, B-, and E-class genes petal identity; B-, C-, and E-class genes stamen identity; and $\\mathrm{C}$ - and E-class genes carpel identity.\n\nABCE-class MADS domain transcription factors (MTFs) are key regulators of floral organ development in eudicots. Aberrant expression of these genes can result in abnormal floral traits, such as phyllody. Certain plant pathogenic bacteria could trigger phyllody in particular species, such as Petunia, which is shown in the figures below.\n[figure1]\n\nA-E are parts of a control flower and $\\mathrm{F}-\\mathrm{I}$ are parts of a bacteria-treated flower. The same parts of the flowers are shown in $B$ and $G, C$ and $H, D$ and $I$, and $E$ and $J$.\nA: The bacteria caused perianth fusion of Petunia.\nB: The bacteria caused the expression of the protein coded by Gene E.\nC: The bacteria decreased the frequency of moths landing on the flower.\nD: The bacteria treatment proved that the flower reproductive parts are leaf modifications.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe diagram below illustrates the $\\mathrm{ABCE}$ model of flower development, in which the identity structure of floral organs of eudicots is regulated by floral homeotic genes divided into ABCE-classes, based on function: A- and E-class genes specify sepal identity; A-, B-, and E-class genes petal identity; B-, C-, and E-class genes stamen identity; and $\\mathrm{C}$ - and E-class genes carpel identity.\n\nABCE-class MADS domain transcription factors (MTFs) are key regulators of floral organ development in eudicots. Aberrant expression of these genes can result in abnormal floral traits, such as phyllody. Certain plant pathogenic bacteria could trigger phyllody in particular species, such as Petunia, which is shown in the figures below.\n[figure1]\n\nA-E are parts of a control flower and $\\mathrm{F}-\\mathrm{I}$ are parts of a bacteria-treated flower. The same parts of the flowers are shown in $B$ and $G, C$ and $H, D$ and $I$, and $E$ and $J$.\n\nA: The bacteria caused perianth fusion of Petunia.\nB: The bacteria caused the expression of the protein coded by Gene E.\nC: The bacteria decreased the frequency of moths landing on the flower.\nD: The bacteria treatment proved that the flower reproductive parts are leaf modifications.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-47.jpg?height=1674&width=1338&top_left_y=778&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1423",
        "problem": "The control of thyroid hormone secretion follows the pathway depicted in the figure. Thyroid releasing hormone (TRH) is secreted by the hypothalamus, which stimulates the production of thyroid stimulating hormone (TSH) from the anterior pituitary. TSH then stimulates the production of the hormones thyroxine (T4) and triiodothyronine (T3) from the thyroid gland.\n\nThe thyroid gland requires dietary iodine to form T4 and T3. T4 contains four iodine atoms while T3 contains three.\n\n[figure1]\n\nGraves' disease is an autoimmune disease where the body produces thyroid-stimulating immunoglobulins that activate TSH receptors on the thyroid gland. What is the likely consequence of Graves' disease?\nA: Decreased TRH, increased TSH, increased T4 and increased T3 levels\nB: Decreased TRH, decreased TSH, increased T4 and increased T3 levels\nC: Increased TRH, decreased TSH, decreased T4 and decreased T3 levels\nD: Increased TRH, increased TSH, increased T4 and increased T3 levels\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe control of thyroid hormone secretion follows the pathway depicted in the figure. Thyroid releasing hormone (TRH) is secreted by the hypothalamus, which stimulates the production of thyroid stimulating hormone (TSH) from the anterior pituitary. TSH then stimulates the production of the hormones thyroxine (T4) and triiodothyronine (T3) from the thyroid gland.\n\nThe thyroid gland requires dietary iodine to form T4 and T3. T4 contains four iodine atoms while T3 contains three.\n\n[figure1]\n\nGraves' disease is an autoimmune disease where the body produces thyroid-stimulating immunoglobulins that activate TSH receptors on the thyroid gland. What is the likely consequence of Graves' disease?\n\nA: Decreased TRH, increased TSH, increased T4 and increased T3 levels\nB: Decreased TRH, decreased TSH, increased T4 and increased T3 levels\nC: Increased TRH, decreased TSH, decreased T4 and decreased T3 levels\nD: Increased TRH, increased TSH, increased T4 and increased T3 levels\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-27.jpg?height=1714&width=1017&top_left_y=725&top_left_x=631"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_790",
        "problem": "果蝇的长翅和残翅受一对等位基因（Aa）控制, 灰体和黄体受另外一对等位基因\n\n$(\\mathrm{B} / \\mathrm{b})$ 控制。其中只有 1 对等位基因位于 $\\mathrm{X}$ 染色体上，且存在某一种配子致死的现象 (不考虑突变和交叉互换)。表型为长翅灰体的两雌雄果蝇杂交, 得到的雌果蝇中长翅灰体：残翅灰体 $=3: 1$, 无黄体性状出现, 雄果蝇中灰体：黄体： $=1: 1$, 无残翅出现。下列说法错误的是（ ）\nA: 上述两对等位基因中，位于 $\\mathrm{X}$ 染色体上的基因是 $\\mathrm{B} / \\mathrm{b}$\nB: 亲本产生的致死配子的基因型是 AY\nC: $F_{1}$ 中雄性长翅灰体个体的基因型都与父本相同\nD: $F_{1}$ 中长翅灰体雄果蝇与残翅灰体雌果蝇杂交, 子代中纯合残翅灰体雌果蝇占 $3 / 32$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n果蝇的长翅和残翅受一对等位基因（Aa）控制, 灰体和黄体受另外一对等位基因\n\n$(\\mathrm{B} / \\mathrm{b})$ 控制。其中只有 1 对等位基因位于 $\\mathrm{X}$ 染色体上，且存在某一种配子致死的现象 (不考虑突变和交叉互换)。表型为长翅灰体的两雌雄果蝇杂交, 得到的雌果蝇中长翅灰体：残翅灰体 $=3: 1$, 无黄体性状出现, 雄果蝇中灰体：黄体： $=1: 1$, 无残翅出现。下列说法错误的是（ ）\n\nA: 上述两对等位基因中，位于 $\\mathrm{X}$ 染色体上的基因是 $\\mathrm{B} / \\mathrm{b}$\nB: 亲本产生的致死配子的基因型是 AY\nC: $F_{1}$ 中雄性长翅灰体个体的基因型都与父本相同\nD: $F_{1}$ 中长翅灰体雄果蝇与残翅灰体雌果蝇杂交, 子代中纯合残翅灰体雌果蝇占 $3 / 32$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_196",
        "problem": "The diagram below represents the development of a human zygote from fertilization to the late blastocyst stage.\n\n[figure1]\n\nChoose a correct statement from the following choices.\nA: If two sperm penetrate the oocyte membrane at the time of fertilization, conjoined twins with shared body parts will be born.\nB: During the process of in vitro fertilization with embryo transfer (IVF-ET), the embryo is transferred at the 2-cell stage to the mother's uterus.\nC: The most appropriate stage for the collection of 'Embryonic stem cells' for regenerative-therapeutic purposes is the 8-cell stage.\nD: The outer cells (structure $a$ ) of the early-blastocyst embryo will eventually form the fetus.\nE: During the late blastocyst stage, the embryo is implanted in the uterine endometrium.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram below represents the development of a human zygote from fertilization to the late blastocyst stage.\n\n[figure1]\n\nChoose a correct statement from the following choices.\n\nA: If two sperm penetrate the oocyte membrane at the time of fertilization, conjoined twins with shared body parts will be born.\nB: During the process of in vitro fertilization with embryo transfer (IVF-ET), the embryo is transferred at the 2-cell stage to the mother's uterus.\nC: The most appropriate stage for the collection of 'Embryonic stem cells' for regenerative-therapeutic purposes is the 8-cell stage.\nD: The outer cells (structure $a$ ) of the early-blastocyst embryo will eventually form the fetus.\nE: During the late blastocyst stage, the embryo is implanted in the uterine endometrium.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-33.jpg?height=825&width=1360&top_left_y=593&top_left_x=448"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1152",
        "problem": "THE RENA OIL SPILL - BIOLOGY WHEN DISASTER STRIKES\n\n[figure1]\n\n\nThe grounding of the CV Rena on October 5, 2011 was New Zealand's worst maritime environmental disaster when the container ship spilled approximately 350 tonnes of heavy fuel oil into the ocean. The container ship ran aground on Otaiti (Astrolabe Reef) just off Mōtitīi Island in the Bay of Plenty, releasing oil into a relatively pristine marine environment. Te Mauri Moana, a group of scientists led by the University of Waikato as part of the Government's \\$ 4.2 million Rena Long-term Environmental Recovery Plan, undertook one of the most comprehensive, multi-disciplinary studies ever done following a marine pollution event. Their findings are summarised in the report; \"Rena Environmental Recovery Monitoring Programme 2011-2013.\"\n\nBy international standards the oil spill was relatively minor but it occurred in an otherwise uncontaminated coastline in an area renowned for its beauty and highly valued for tourism, recreation and fisheries. Iwi, government, commercial stakeholders and the public were rightly concerned there would be long-lasting negative impacts on beaches, reefs and fisheries. Te Mauri Moana reports on fifteen monitoring and research programmes focused on the range of ecosystems and habitats within the Bay of Plenty.\n\nChemical contamination of the sediments and fauna of Otaiti Reef and Môtitī Island and the surrounding soft sediments was examined to determine if the discharge of fuel oil (and later release of container debris) from the Rena resulted in significant chemical contamination of the marine environment and marine organisms. Amongst other contaminants the presence of polycyclic aromatic hydrocarbons (PAHs) from fuel oil was examined in sediments and from selected organisms.\n\nThe maps below shows the PAH levels in sediments collected from A. on Otaiti Reef and B. off-reef.\n\n\nA.\n\n[figure2]\n\nB.\n\n[figure3]\n\n[figure4]\n\nThe graph at left shows the PAH levels in sea urchins collected from Otaiti Reef and nearby islands. The numbers in brackets indicate the number of samples collected at each site. All sea urchins on Otaiti reef showed elevated PAH levels in the gonad with an average level of $0.057 \\mathrm{mg} \\mathrm{kg}^{-1}$. Gut levels averaged $1.58 \\mathrm{mg} \\mathrm{kg}^{-1}$ but at Astro 2 and 3 were orders of magnitude higher than at other sites on Otaiti Reef. PAH levels in the gonad and gut from urchins from Mōtîti Island averaged $0.026 \\mathrm{mg} \\mathrm{kg}^{-1}$ and $0.03 \\mathrm{mg} \\mathrm{kg}^{-1}$ respectively and were similar to levels found elsewhere, including East Cape.Off-reef sediment levels of PAH are greatest?\nA: North of the Rena.\nB: East of the Rena.\nC: South of the Rena.\nD: West of the Rena.\nE: East and West of the Rena.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nTHE RENA OIL SPILL - BIOLOGY WHEN DISASTER STRIKES\n\n[figure1]\n\n\nThe grounding of the CV Rena on October 5, 2011 was New Zealand's worst maritime environmental disaster when the container ship spilled approximately 350 tonnes of heavy fuel oil into the ocean. The container ship ran aground on Otaiti (Astrolabe Reef) just off Mōtitīi Island in the Bay of Plenty, releasing oil into a relatively pristine marine environment. Te Mauri Moana, a group of scientists led by the University of Waikato as part of the Government's \\$ 4.2 million Rena Long-term Environmental Recovery Plan, undertook one of the most comprehensive, multi-disciplinary studies ever done following a marine pollution event. Their findings are summarised in the report; \"Rena Environmental Recovery Monitoring Programme 2011-2013.\"\n\nBy international standards the oil spill was relatively minor but it occurred in an otherwise uncontaminated coastline in an area renowned for its beauty and highly valued for tourism, recreation and fisheries. Iwi, government, commercial stakeholders and the public were rightly concerned there would be long-lasting negative impacts on beaches, reefs and fisheries. Te Mauri Moana reports on fifteen monitoring and research programmes focused on the range of ecosystems and habitats within the Bay of Plenty.\n\nChemical contamination of the sediments and fauna of Otaiti Reef and Môtitī Island and the surrounding soft sediments was examined to determine if the discharge of fuel oil (and later release of container debris) from the Rena resulted in significant chemical contamination of the marine environment and marine organisms. Amongst other contaminants the presence of polycyclic aromatic hydrocarbons (PAHs) from fuel oil was examined in sediments and from selected organisms.\n\nThe maps below shows the PAH levels in sediments collected from A. on Otaiti Reef and B. off-reef.\n\n\nA.\n\n[figure2]\n\nB.\n\n[figure3]\n\n[figure4]\n\nThe graph at left shows the PAH levels in sea urchins collected from Otaiti Reef and nearby islands. The numbers in brackets indicate the number of samples collected at each site. All sea urchins on Otaiti reef showed elevated PAH levels in the gonad with an average level of $0.057 \\mathrm{mg} \\mathrm{kg}^{-1}$. Gut levels averaged $1.58 \\mathrm{mg} \\mathrm{kg}^{-1}$ but at Astro 2 and 3 were orders of magnitude higher than at other sites on Otaiti Reef. PAH levels in the gonad and gut from urchins from Mōtîti Island averaged $0.026 \\mathrm{mg} \\mathrm{kg}^{-1}$ and $0.03 \\mathrm{mg} \\mathrm{kg}^{-1}$ respectively and were similar to levels found elsewhere, including East Cape.\n\nproblem:\nOff-reef sediment levels of PAH are greatest?\n\nA: North of the Rena.\nB: East of the Rena.\nC: South of the Rena.\nD: West of the Rena.\nE: East and West of the Rena.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=556&width=1834&top_left_y=276&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=657&width=894&top_left_y=1876&top_left_x=113",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=548&width=854&top_left_y=1942&top_left_x=1092",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-25.jpg?height=591&width=802&top_left_y=201&top_left_x=136"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_125",
        "problem": "Poly(3-hydroxybutyrate) (PHB) is a bacterial storage material which is accumulated by various bacteria, usually when grown under limitation of a nutrient such as oxygen, nitrogen, phosphate, sulphur, or magnesium and in the presence of excess carbon. Fig. Q7 shows the PHB synthesis pathway of Ralstonia eutropha from acetyl-CoA. In addition, acetyl-CoA can enter the citric acid cycle.\n\n[figure1]\n\nFig. Q7. PHB synthesis pathway\nA: Citrate synthase is an important regulation factor in the PHB synthesis process.\nB: When the intracellular concentration of HSCoA is high, the rate of PHB synthesis will increase.\nC: When the rate of PHB synthesis increases, the growth rate of Ralstonia eutropha cells will also increase.\nD: PHB synthesis is stimulated by low ratios of $\\mathrm{NADPH}+\\mathrm{H}^{+} / \\mathrm{NADP}$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPoly(3-hydroxybutyrate) (PHB) is a bacterial storage material which is accumulated by various bacteria, usually when grown under limitation of a nutrient such as oxygen, nitrogen, phosphate, sulphur, or magnesium and in the presence of excess carbon. Fig. Q7 shows the PHB synthesis pathway of Ralstonia eutropha from acetyl-CoA. In addition, acetyl-CoA can enter the citric acid cycle.\n\n[figure1]\n\nFig. Q7. PHB synthesis pathway\n\nA: Citrate synthase is an important regulation factor in the PHB synthesis process.\nB: When the intracellular concentration of HSCoA is high, the rate of PHB synthesis will increase.\nC: When the rate of PHB synthesis increases, the growth rate of Ralstonia eutropha cells will also increase.\nD: PHB synthesis is stimulated by low ratios of $\\mathrm{NADPH}+\\mathrm{H}^{+} / \\mathrm{NADP}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-018.jpg?height=1261&width=1175&top_left_y=679&top_left_x=465"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_340",
        "problem": "心肌细胞中存在许多非编码 RNA（如 miR-223、HRCR），基因 ARC 可在该细胞中特异性表达, 它们共同参与调控细胞调亡, 相关过程如图, (1)(2)(3)(4)表示生理过程,下列有关叙述错误的是（）\n\n[图1]\nA: 过程(2)(3)(4)遵循的碱基互补配对原则完全相同\nB: 过程(4)的发生, 不利于基因 ARC 的表达，促进心肌细胞的调亡\nC: 基因 ARC、miR-223、HRCR 所含的游离磷酸基团数量分别为 $2 、 1 、 0$\nD: 过程(1)中形成 DNA一蛋白质复合体，过程(2)中核糖体向右移动\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n心肌细胞中存在许多非编码 RNA（如 miR-223、HRCR），基因 ARC 可在该细胞中特异性表达, 它们共同参与调控细胞调亡, 相关过程如图, (1)(2)(3)(4)表示生理过程,下列有关叙述错误的是（）\n\n[图1]\n\nA: 过程(2)(3)(4)遵循的碱基互补配对原则完全相同\nB: 过程(4)的发生, 不利于基因 ARC 的表达，促进心肌细胞的调亡\nC: 基因 ARC、miR-223、HRCR 所含的游离磷酸基团数量分别为 $2 、 1 、 0$\nD: 过程(1)中形成 DNA一蛋白质复合体，过程(2)中核糖体向右移动\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-27.jpg?height=634&width=1282&top_left_y=206&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_656",
        "problem": "研究发现, AGPAT2 基因表达的下调会延缓脂肪生成。湖羊尾部蓄脂量小，而广灵大尾羊尾部蓄脂量大。研究人员以若干只两种羊的尾部脂肪组织为材料, 检测 AGPAT2 基因启动子区 7 个位点的甲基化程度及基因表达水平，结果如下图。下列叙述正确的有\n\n[图1]\n\n[图2]\nA: AGPAT2 基因的甲基化可遗传给后代, 并改变 DNA 分子中嘌呤与嘧啶的比值\nB: 甲基化程度的差异会导致两种羊脂肪组织中 AGPAT2 基因的碱基序列不同\nC: 第 33 和 63 位点上的甲基化影响 AGPAT2 基因的翻译从而导致基因表达异常\nD: 两种羊中 AGPAT2 基因的甲基化程度与其在脂肪组织中的表达量呈负相关\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究发现, AGPAT2 基因表达的下调会延缓脂肪生成。湖羊尾部蓄脂量小，而广灵大尾羊尾部蓄脂量大。研究人员以若干只两种羊的尾部脂肪组织为材料, 检测 AGPAT2 基因启动子区 7 个位点的甲基化程度及基因表达水平，结果如下图。下列叙述正确的有\n\n[图1]\n\n[图2]\n\nA: AGPAT2 基因的甲基化可遗传给后代, 并改变 DNA 分子中嘌呤与嘧啶的比值\nB: 甲基化程度的差异会导致两种羊脂肪组织中 AGPAT2 基因的碱基序列不同\nC: 第 33 和 63 位点上的甲基化影响 AGPAT2 基因的翻译从而导致基因表达异常\nD: 两种羊中 AGPAT2 基因的甲基化程度与其在脂肪组织中的表达量呈负相关\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-03.jpg?height=402&width=1436&top_left_y=2169&top_left_x=344",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-04.jpg?height=462&width=677&top_left_y=166&top_left_x=381"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_208",
        "problem": "A population of dragonfly larvae (Leucorrhinia intacta) is separated into two groups. In both groups, larvae populations are put inside a cage with no food limitation. The first group is exposed to a fish predator that can swim freely, but cannot enter the cage. The second group is a control with no fish. The proportion of larvae surviving and the proportion of live larvae failing to metamorphose in the two groups are shown below:\n[figure1]\n\nFigure Q. 43\nA: One of the causes of high failure rate of metamorphosis of the larvae upon exposure to a non-lethal predator is cannibalism.\nB: The high mortality of larvae in the first group is due to predator-induced stress.\nC: In the predator treatment, the percentage of individuals that survived the larval stage completed emergence to the adult stage is lower than the percentage of those in the fishless treatment.\nD: The survival of dragonfly before metamorphosis is dependent on the predator while those during metamorphosis is not.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA population of dragonfly larvae (Leucorrhinia intacta) is separated into two groups. In both groups, larvae populations are put inside a cage with no food limitation. The first group is exposed to a fish predator that can swim freely, but cannot enter the cage. The second group is a control with no fish. The proportion of larvae surviving and the proportion of live larvae failing to metamorphose in the two groups are shown below:\n[figure1]\n\nFigure Q. 43\n\nA: One of the causes of high failure rate of metamorphosis of the larvae upon exposure to a non-lethal predator is cannibalism.\nB: The high mortality of larvae in the first group is due to predator-induced stress.\nC: In the predator treatment, the percentage of individuals that survived the larval stage completed emergence to the adult stage is lower than the percentage of those in the fishless treatment.\nD: The survival of dragonfly before metamorphosis is dependent on the predator while those during metamorphosis is not.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-094.jpg?height=1004&width=760&top_left_y=706&top_left_x=702"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_46",
        "problem": "Peter Godfrey-Smith (2009) describes a parameter space in which a population can evolve using three parameters:\n\n$\\boldsymbol{H}$ : Fidelity of heredity\n\n$S$ : Dependence of reproductive differences on genetic differences\n\nC: Continuity (when $\\mathrm{C}$ is maximum, adding beneficial mutations to a genome results in the proportional betterment of the genotype. When $\\mathrm{C}$ is zero, the effect of each mutation is entirely dependent on all the other loci)\n\nWe can imagine this space as cube (described by Peter Godfrey-Smith as A Darwinian space), as seen in the figure.\n\n[figure1]\nA: Accumulation of excessive mutations can result in the extinction of a population. $(0,1,1)$ in the Darwinian space describes this situation.\nB: In the absence of selection, a population of organisms resides at $(1,0,0)$.\nC: If your attempts at optimizing a bacteria species to consume glucose almost always results in sub-optimized populations, you are exploring the space close\nD: A somatic cell in the human body can reside at $(0,1,1)$ (Here, heredity is defined for the human and not the cell.).\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPeter Godfrey-Smith (2009) describes a parameter space in which a population can evolve using three parameters:\n\n$\\boldsymbol{H}$ : Fidelity of heredity\n\n$S$ : Dependence of reproductive differences on genetic differences\n\nC: Continuity (when $\\mathrm{C}$ is maximum, adding beneficial mutations to a genome results in the proportional betterment of the genotype. When $\\mathrm{C}$ is zero, the effect of each mutation is entirely dependent on all the other loci)\n\nWe can imagine this space as cube (described by Peter Godfrey-Smith as A Darwinian space), as seen in the figure.\n\n[figure1]\n\nA: Accumulation of excessive mutations can result in the extinction of a population. $(0,1,1)$ in the Darwinian space describes this situation.\nB: In the absence of selection, a population of organisms resides at $(1,0,0)$.\nC: If your attempts at optimizing a bacteria species to consume glucose almost always results in sub-optimized populations, you are exploring the space close\nD: A somatic cell in the human body can reside at $(0,1,1)$ (Here, heredity is defined for the human and not the cell.).\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-53.jpg?height=534&width=686&top_left_y=887&top_left_x=685"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1039",
        "problem": "A patient presents the following symptoms: diarrhea, dermatitis, and dementia, as well as \"necklace\" lesions on the lower neck, hyperpigmentation, thickening of the skin, inflammation of the mouth and tongue, digestive disturbances, amnesia, and delirium. Which of the following is deficient in this person's system?\nA: Vitamin D3 (cholecalciferol)\nB: Vitamin B3 (niacin)\nC: Vitamin C (ascorbic acid)\nD: Vitamin B1 (thiamine)\nE: Vitamin E (tocopherol)\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA patient presents the following symptoms: diarrhea, dermatitis, and dementia, as well as \"necklace\" lesions on the lower neck, hyperpigmentation, thickening of the skin, inflammation of the mouth and tongue, digestive disturbances, amnesia, and delirium. Which of the following is deficient in this person's system?\n\nA: Vitamin D3 (cholecalciferol)\nB: Vitamin B3 (niacin)\nC: Vitamin C (ascorbic acid)\nD: Vitamin B1 (thiamine)\nE: Vitamin E (tocopherol)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1095",
        "problem": "The slow rate of oxygen diffusion in water limits the efficiency of oxygen distribution from gas exchange surface to sites of cellular respiration. This is true for:\nA: only aquatic animals.\nB: only air-breathing animals.\nC: only those animals in which no circulatory system has developed.\nD: all animals.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe slow rate of oxygen diffusion in water limits the efficiency of oxygen distribution from gas exchange surface to sites of cellular respiration. This is true for:\n\nA: only aquatic animals.\nB: only air-breathing animals.\nC: only those animals in which no circulatory system has developed.\nD: all animals.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1109",
        "problem": "Typically a skeletal muscle has origin in one bone and insertion in the other across a joint. Consider the insertion of $X$ and $Y$. Which one of the following statements is correct?\n\n[figure1]\nA: Muscle $Y$ is responsible for rotation across the joint.\nB: Muscle $X$ is responsible for extension and $Y$ for flexion.\nC: Muscle $X$ is for stronger flexion, while $Y$ brings about weaker flexion.\nD: Muscle $X$ is for quick flexion while $Y$ produces strong flexion.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTypically a skeletal muscle has origin in one bone and insertion in the other across a joint. Consider the insertion of $X$ and $Y$. Which one of the following statements is correct?\n\n[figure1]\n\nA: Muscle $Y$ is responsible for rotation across the joint.\nB: Muscle $X$ is responsible for extension and $Y$ for flexion.\nC: Muscle $X$ is for stronger flexion, while $Y$ brings about weaker flexion.\nD: Muscle $X$ is for quick flexion while $Y$ produces strong flexion.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-06.jpg?height=425&width=616&top_left_y=384&top_left_x=798"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_897",
        "problem": "某单基因遗传病女性患者的家系遗传结果如下所示, 以下说法正确的是（）\n\n[图1]\nA: 该遗传病的致病基因一定在常染色体上\nB: 该遗传病的致病基因有可能在 $\\mathrm{X}$ 和 Y 染色体的非同源区段\nC: 若父母再生一个孩子，为患病男孩的概率为 $1 / 4$\nD: 若父亲为纯合子，则外祖母为杂合子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某单基因遗传病女性患者的家系遗传结果如下所示, 以下说法正确的是（）\n\n[图1]\n\nA: 该遗传病的致病基因一定在常染色体上\nB: 该遗传病的致病基因有可能在 $\\mathrm{X}$ 和 Y 染色体的非同源区段\nC: 若父母再生一个孩子，为患病男孩的概率为 $1 / 4$\nD: 若父亲为纯合子，则外祖母为杂合子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-009.jpg?height=371&width=522&top_left_y=220&top_left_x=367"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_223",
        "problem": "An organism has four genes, A, B, C and D with two alleles each. An individual heterozygous for these genes was bred with one that is homozygous recessive. The cross produced 3288 offspring with phenotypes shown in the table below:\n\n[figure1]\nA: The four loci are genetically linked.\nB: The distance between gene $B$ and gene $D$ is $9 \\mathrm{cM}$.\nC: The distance between gene $D$ and gene $C$ is $10.5 \\mathrm{cM}$.\nD: Interference happened with a value less than 0.25\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn organism has four genes, A, B, C and D with two alleles each. An individual heterozygous for these genes was bred with one that is homozygous recessive. The cross produced 3288 offspring with phenotypes shown in the table below:\n\n[figure1]\n\nA: The four loci are genetically linked.\nB: The distance between gene $B$ and gene $D$ is $9 \\mathrm{cM}$.\nC: The distance between gene $D$ and gene $C$ is $10.5 \\mathrm{cM}$.\nD: Interference happened with a value less than 0.25\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://i.postimg.cc/gjrp6MrQ/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_590",
        "problem": "蚕豆根尖细胞在含 ${ }^{3} \\mathrm{H}$ 标记的胸腺嘧啶脱氧核苷培养基中完成一个细胞周期, 然后在不含放射性标记的培养基中继续分裂至中期，其染色体的放射性标记分布情况是（ ）\nA: 每条染色体中都只有一条单体被标记\nB: 每条染色体的两条单体都被标记\nC: 半数的染色体中只有一条单体被标记\nD: 每条染色体的两条单体都不被标记\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n蚕豆根尖细胞在含 ${ }^{3} \\mathrm{H}$ 标记的胸腺嘧啶脱氧核苷培养基中完成一个细胞周期, 然后在不含放射性标记的培养基中继续分裂至中期，其染色体的放射性标记分布情况是（ ）\n\nA: 每条染色体中都只有一条单体被标记\nB: 每条染色体的两条单体都被标记\nC: 半数的染色体中只有一条单体被标记\nD: 每条染色体的两条单体都不被标记\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1350",
        "problem": "Genetic adaptation is best described as\nA: a process by which an individual organism gradually changes itself to suit its environment\nB: a change in allele frequencies in a population as a result of changing environmental circumstances\nC: the emergence of new forms as a result of mutations in the genetic material\nD: a change in the genotype of an individual organism as a result of selection pressure\nE: the occurrence of advantageous mutations\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGenetic adaptation is best described as\n\nA: a process by which an individual organism gradually changes itself to suit its environment\nB: a change in allele frequencies in a population as a result of changing environmental circumstances\nC: the emergence of new forms as a result of mutations in the genetic material\nD: a change in the genotype of an individual organism as a result of selection pressure\nE: the occurrence of advantageous mutations\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_804",
        "problem": "某雄果蝇的一个精原细胞甲经过连续两次正常分裂依次得到细胞乙、细胞丙: 甲 $\\xrightarrow{1}$ 乙 $\\xrightarrow{2}$ 丙。下列叙述错误的是（）\nA: 1 过程中细胞核 DNA 数是甲细胞的 2 倍, 2 过程中无四分体\nB: 1 过程中一定有同源染色体, 1 和 2 过程中一定都出现纺锤体\nC: 若 1 和 2 过程中着丝粒只分裂 1 次，则 2 过程中细胞可含 2 个染色体组\nD: 若 1 过程中染色体组数与中心体数之比不变, 则丙细胞中不存在 Y 染色体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某雄果蝇的一个精原细胞甲经过连续两次正常分裂依次得到细胞乙、细胞丙: 甲 $\\xrightarrow{1}$ 乙 $\\xrightarrow{2}$ 丙。下列叙述错误的是（）\n\nA: 1 过程中细胞核 DNA 数是甲细胞的 2 倍, 2 过程中无四分体\nB: 1 过程中一定有同源染色体, 1 和 2 过程中一定都出现纺锤体\nC: 若 1 和 2 过程中着丝粒只分裂 1 次，则 2 过程中细胞可含 2 个染色体组\nD: 若 1 过程中染色体组数与中心体数之比不变, 则丙细胞中不存在 Y 染色体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_225",
        "problem": "Dendrogramma is a new animal species which has been collected at 400 and 1000 metres on the Australian continental slope off eastern Bass Strait and Tasmania during a cruise in 1986 and described as a new taxon in 2016. The new taxon is multicellular (metazoan), non-bilateralian, apparently diploblastic with a dense mesoglea between an outer epidermis and an inner gastrodermis.\n\nThis animal is composed of a body divided into a stalk with a mouth opening terminally, and a flattened disc. The mouth is set in a specialized, lobed epidermis field, leading into a gastrodermislined gastrovascular canal (pharynx) in the stalk which aborally branches dichotomously into numerous radiating canals in the disc. We can state with considerable certainty that the organisms lack cnidocytes, tentacles, marginal pore openings for the radiating canals, ring canal, sense organs in the form of statocysts or the rhopalia, or colloblasts, ctenes, or an apical organ.\n\nNo cilia have been located. They have simple mouth opening with specialized lobes secreting mucus.\n\n[figure1]\nA: According to the characteristics of Dendrogramma it could be placed in Cnidaria.\nB: It seems likely that Dendrogramma feed on micro-organisms.\nC: Lack of statocysts and rhopalia shows that this taxon does not have nervous system.\nD: Most likely Dendrogramma could have Dinoflagellate symbionts such as Symbiodinium.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDendrogramma is a new animal species which has been collected at 400 and 1000 metres on the Australian continental slope off eastern Bass Strait and Tasmania during a cruise in 1986 and described as a new taxon in 2016. The new taxon is multicellular (metazoan), non-bilateralian, apparently diploblastic with a dense mesoglea between an outer epidermis and an inner gastrodermis.\n\nThis animal is composed of a body divided into a stalk with a mouth opening terminally, and a flattened disc. The mouth is set in a specialized, lobed epidermis field, leading into a gastrodermislined gastrovascular canal (pharynx) in the stalk which aborally branches dichotomously into numerous radiating canals in the disc. We can state with considerable certainty that the organisms lack cnidocytes, tentacles, marginal pore openings for the radiating canals, ring canal, sense organs in the form of statocysts or the rhopalia, or colloblasts, ctenes, or an apical organ.\n\nNo cilia have been located. They have simple mouth opening with specialized lobes secreting mucus.\n\n[figure1]\n\nA: According to the characteristics of Dendrogramma it could be placed in Cnidaria.\nB: It seems likely that Dendrogramma feed on micro-organisms.\nC: Lack of statocysts and rhopalia shows that this taxon does not have nervous system.\nD: Most likely Dendrogramma could have Dinoflagellate symbionts such as Symbiodinium.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-34.jpg?height=1202&width=1145&top_left_y=924&top_left_x=455"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1002",
        "problem": "During starvation, steroid hormones trigger the transcription of genes for lipid metabolism in their target cells. This would be an example of control by\nA: Negative feedback\nB: Positive feedback\nC: Repressors\nD: Inducers\nE: Modulators\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDuring starvation, steroid hormones trigger the transcription of genes for lipid metabolism in their target cells. This would be an example of control by\n\nA: Negative feedback\nB: Positive feedback\nC: Repressors\nD: Inducers\nE: Modulators\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_727",
        "problem": "菠菜的有刺籽粒与无刺籽粒、抗病与不抗病分别由基因 A（a）、B（b）控制，两对等位基因独立遗传。为研究其遗传机制, 选取纯合有刺不抗病雌株与杂合无刺抗病雄株杂交，结果见下图。下列叙述错误的是\n\n## P q纯合有剌不抗病 $\\times$ 含杂合无剌抗病\n\n[图1]\n\n## $F_{1}$ 有剌抗病崔株 有剌不抗病焳株 有剌抗病崔株 有刺不抗病碓株 <br> 比例 $3: 1: 1: 3$\nA: 有刺籽粒对无刺籽粒显性, 抗病对不抗病显性\nB: 雄性亲本的基因型及比例为 $a X^{B} Y^{b}: a a X^{b} Y^{B}=3: 1$\nC: 让 $F_{1}$ 中雌雄植株随机杂交, 子代中无刺抗病雌株占 $5 / 64$\nD: 亲本产生的配子基因型及比例决定了 $\\mathrm{F}_{1}$ 的基因型及比例\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n菠菜的有刺籽粒与无刺籽粒、抗病与不抗病分别由基因 A（a）、B（b）控制，两对等位基因独立遗传。为研究其遗传机制, 选取纯合有刺不抗病雌株与杂合无刺抗病雄株杂交，结果见下图。下列叙述错误的是\n\n## P q纯合有剌不抗病 $\\times$ 含杂合无剌抗病\n\n[图1]\n\n## $F_{1}$ 有剌抗病崔株 有剌不抗病焳株 有剌抗病崔株 有刺不抗病碓株 <br> 比例 $3: 1: 1: 3$\n\nA: 有刺籽粒对无刺籽粒显性, 抗病对不抗病显性\nB: 雄性亲本的基因型及比例为 $a X^{B} Y^{b}: a a X^{b} Y^{B}=3: 1$\nC: 让 $F_{1}$ 中雌雄植株随机杂交, 子代中无刺抗病雌株占 $5 / 64$\nD: 亲本产生的配子基因型及比例决定了 $\\mathrm{F}_{1}$ 的基因型及比例\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-042.jpg?height=114&width=71&top_left_y=1896&top_left_x=684"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_755",
        "problem": "某种植物的 4 号染色体上的 A 基因可以指导植酸合成, 不能合成植酸的该种植物\n\n[图1]\n前者不影响植酸合成, 后者不具有合成植酸的功能。将一个 $\\mathrm{A}$ 基因导入基因型为 $\\mathrm{A}^{3-} \\mathrm{A}^{25}$ -的植株的 6 号染色体, 构成基因型为 $\\mathrm{A}^{3-} \\mathrm{A}^{25-} \\mathrm{A}$ 的植株。不考虑子代植株细胞中 DNA 的复制，该植株自交子代中含有 $\\mathrm{A}^{25-} \\mathrm{A}^{25-}$ 的个体所占比例是（ ）\nA: $1 / 3$\nB: $1 / 5$\nC: $2 / 5$\nD: $3 / 8$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种植物的 4 号染色体上的 A 基因可以指导植酸合成, 不能合成植酸的该种植物\n\n[图1]\n前者不影响植酸合成, 后者不具有合成植酸的功能。将一个 $\\mathrm{A}$ 基因导入基因型为 $\\mathrm{A}^{3-} \\mathrm{A}^{25}$ -的植株的 6 号染色体, 构成基因型为 $\\mathrm{A}^{3-} \\mathrm{A}^{25-} \\mathrm{A}$ 的植株。不考虑子代植株细胞中 DNA 的复制，该植株自交子代中含有 $\\mathrm{A}^{25-} \\mathrm{A}^{25-}$ 的个体所占比例是（ ）\n\nA: $1 / 3$\nB: $1 / 5$\nC: $2 / 5$\nD: $3 / 8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-47.jpg?height=51&width=1373&top_left_y=1305&top_left_x=346"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1205",
        "problem": "The following diagram represents the pathways conveying the sensory information in the human body to the sensory part of the brain.\n\n[figure1]\n\nIn brief, the pathway conveying touch and pressure (shown in grey) passes from the left side of the body to the spinal cord and ascends to the medulla oblongata, where it crosses over to the other side and reaches the sensory regions of the right brain. For the pain and temperature pathway (shown in black), the sensory information passes from the left side of the body to the right side of the spinal cord before ascending up into the medulla and finally onto the right part of the brain. The same process occurs for pain and temperature, and touch and pressure pathways from the right side of the body.From this information, what would happen if the right side of the spinal cord were cut (i.e. damaged) at the level indicated by the arrow?\nA: There would be a loss in touch and pressure on the right side of the body and a loss in pain and temperature sensation on the left side of the body below that level.\nB: There would be a loss in touch and pressure sensation on the left side of the body below that level only.\nC: There would be a loss in touch and pressure on the left side of the body and a loss in pain and temperature sensation on the right side of the body below that level.\nD: There would be a loss in touch and pressure on the left side of the body and a loss in pain and temperature sensation on the left side of the body.\nE: There would be a loss of all sensation below that level, both touch and pressure and pain and temperature.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nThe following diagram represents the pathways conveying the sensory information in the human body to the sensory part of the brain.\n\n[figure1]\n\nIn brief, the pathway conveying touch and pressure (shown in grey) passes from the left side of the body to the spinal cord and ascends to the medulla oblongata, where it crosses over to the other side and reaches the sensory regions of the right brain. For the pain and temperature pathway (shown in black), the sensory information passes from the left side of the body to the right side of the spinal cord before ascending up into the medulla and finally onto the right part of the brain. The same process occurs for pain and temperature, and touch and pressure pathways from the right side of the body.\n\nproblem:\nFrom this information, what would happen if the right side of the spinal cord were cut (i.e. damaged) at the level indicated by the arrow?\n\nA: There would be a loss in touch and pressure on the right side of the body and a loss in pain and temperature sensation on the left side of the body below that level.\nB: There would be a loss in touch and pressure sensation on the left side of the body below that level only.\nC: There would be a loss in touch and pressure on the left side of the body and a loss in pain and temperature sensation on the right side of the body below that level.\nD: There would be a loss in touch and pressure on the left side of the body and a loss in pain and temperature sensation on the left side of the body.\nE: There would be a loss of all sensation below that level, both touch and pressure and pain and temperature.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-14.jpg?height=845&width=1516&top_left_y=431&top_left_x=224"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_487",
        "problem": "脊髓延髓肌肉萎缩症（SBMA）是由雄激素受体基因（AR）突变导致的遗传病,\n\n图 1 为 SBMA 遗传病的家系图, 实验人员利用 PCR 技术扩增正常人和该家系成员的 AR 基因片段, 电泳结果如图 2 所示, 下列叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\nM:Marker标准条带\n\n图2\nA: 可推知 SBMA 病为伴 X 染色体隐性遗传病\nB: III-4, III-6, III-8 的致病基因来自I-2\nC: I-1 和I-2 生出一个患病孩子的概率是 $1 / 4$\nD: 图 2 中III-3 的电泳条带与III- 4 是一致的\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n脊髓延髓肌肉萎缩症（SBMA）是由雄激素受体基因（AR）突变导致的遗传病,\n\n图 1 为 SBMA 遗传病的家系图, 实验人员利用 PCR 技术扩增正常人和该家系成员的 AR 基因片段, 电泳结果如图 2 所示, 下列叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\nM:Marker标准条带\n\n图2\n\nA: 可推知 SBMA 病为伴 X 染色体隐性遗传病\nB: III-4, III-6, III-8 的致病基因来自I-2\nC: I-1 和I-2 生出一个患病孩子的概率是 $1 / 4$\nD: 图 2 中III-3 的电泳条带与III- 4 是一致的\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-62.jpg?height=457&width=1142&top_left_y=637&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-62.jpg?height=437&width=985&top_left_y=1135&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1191",
        "problem": "THE RENA OIL SPILL - BIOLOGY WHEN DISASTER STRIKES\n\n[figure1]\n\n\nThe grounding of the CV Rena on October 5, 2011 was New Zealand's worst maritime environmental disaster when the container ship spilled approximately 350 tonnes of heavy fuel oil into the ocean. The container ship ran aground on Otaiti (Astrolabe Reef) just off Mōtitīi Island in the Bay of Plenty, releasing oil into a relatively pristine marine environment. Te Mauri Moana, a group of scientists led by the University of Waikato as part of the Government's \\$ 4.2 million Rena Long-term Environmental Recovery Plan, undertook one of the most comprehensive, multi-disciplinary studies ever done following a marine pollution event. Their findings are summarised in the report; \"Rena Environmental Recovery Monitoring Programme 2011-2013.\"\n\nBy international standards the oil spill was relatively minor but it occurred in an otherwise uncontaminated coastline in an area renowned for its beauty and highly valued for tourism, recreation and fisheries. Iwi, government, commercial stakeholders and the public were rightly concerned there would be long-lasting negative impacts on beaches, reefs and fisheries. Te Mauri Moana reports on fifteen monitoring and research programmes focused on the range of ecosystems and habitats within the Bay of Plenty.\n\nChemical contamination of the sediments and fauna of Otaiti Reef and Môtitī Island and the surrounding soft sediments was examined to determine if the discharge of fuel oil (and later release of container debris) from the Rena resulted in significant chemical contamination of the marine environment and marine organisms. Amongst other contaminants the presence of polycyclic aromatic hydrocarbons (PAHs) from fuel oil was examined in sediments and from selected organisms.\n\nThe maps below shows the PAH levels in sediments collected from A. on Otaiti Reef and B. off-reef.\n\n\nA.\n\n[figure2]\n\nB.\n\n[figure3]\n\n[figure4]\n\nThe graph at left shows the PAH levels in sea urchins collected from Otaiti Reef and nearby islands. The numbers in brackets indicate the number of samples collected at each site. All sea urchins on Otaiti reef showed elevated PAH levels in the gonad with an average level of $0.057 \\mathrm{mg} \\mathrm{kg}^{-1}$. Gut levels averaged $1.58 \\mathrm{mg} \\mathrm{kg}^{-1}$ but at Astro 2 and 3 were orders of magnitude higher than at other sites on Otaiti Reef. PAH levels in the gonad and gut from urchins from Mōtîti Island averaged $0.026 \\mathrm{mg} \\mathrm{kg}^{-1}$ and $0.03 \\mathrm{mg} \\mathrm{kg}^{-1}$ respectively and were similar to levels found elsewhere, including East Cape.Considering these data what is the main, substantiated, conclusion that can be drawn about the effect of fuel oil contamination on the sediments and fauna of Otaiti Reef and Môtitī Island from the grounding of the Rena?\nA: On reef sediment PAH levels generally match PAH levels in sea urchins because sea urchins ingest sand as they graze.\nB: Significant contamination of sediments and accumulation within the food chain has occurred.\nC: Contamination of sediments and sea urchins on and around Otaiti, particularly in the close vicinity of the ship's hull, has occurred.\nD: Otaiti Reef and Mōtitī Island show contamination of sediments and sea urchins.\nE: Widespread contamination of sediments and some marine organisms has occurred.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nTHE RENA OIL SPILL - BIOLOGY WHEN DISASTER STRIKES\n\n[figure1]\n\n\nThe grounding of the CV Rena on October 5, 2011 was New Zealand's worst maritime environmental disaster when the container ship spilled approximately 350 tonnes of heavy fuel oil into the ocean. The container ship ran aground on Otaiti (Astrolabe Reef) just off Mōtitīi Island in the Bay of Plenty, releasing oil into a relatively pristine marine environment. Te Mauri Moana, a group of scientists led by the University of Waikato as part of the Government's \\$ 4.2 million Rena Long-term Environmental Recovery Plan, undertook one of the most comprehensive, multi-disciplinary studies ever done following a marine pollution event. Their findings are summarised in the report; \"Rena Environmental Recovery Monitoring Programme 2011-2013.\"\n\nBy international standards the oil spill was relatively minor but it occurred in an otherwise uncontaminated coastline in an area renowned for its beauty and highly valued for tourism, recreation and fisheries. Iwi, government, commercial stakeholders and the public were rightly concerned there would be long-lasting negative impacts on beaches, reefs and fisheries. Te Mauri Moana reports on fifteen monitoring and research programmes focused on the range of ecosystems and habitats within the Bay of Plenty.\n\nChemical contamination of the sediments and fauna of Otaiti Reef and Môtitī Island and the surrounding soft sediments was examined to determine if the discharge of fuel oil (and later release of container debris) from the Rena resulted in significant chemical contamination of the marine environment and marine organisms. Amongst other contaminants the presence of polycyclic aromatic hydrocarbons (PAHs) from fuel oil was examined in sediments and from selected organisms.\n\nThe maps below shows the PAH levels in sediments collected from A. on Otaiti Reef and B. off-reef.\n\n\nA.\n\n[figure2]\n\nB.\n\n[figure3]\n\n[figure4]\n\nThe graph at left shows the PAH levels in sea urchins collected from Otaiti Reef and nearby islands. The numbers in brackets indicate the number of samples collected at each site. All sea urchins on Otaiti reef showed elevated PAH levels in the gonad with an average level of $0.057 \\mathrm{mg} \\mathrm{kg}^{-1}$. Gut levels averaged $1.58 \\mathrm{mg} \\mathrm{kg}^{-1}$ but at Astro 2 and 3 were orders of magnitude higher than at other sites on Otaiti Reef. PAH levels in the gonad and gut from urchins from Mōtîti Island averaged $0.026 \\mathrm{mg} \\mathrm{kg}^{-1}$ and $0.03 \\mathrm{mg} \\mathrm{kg}^{-1}$ respectively and were similar to levels found elsewhere, including East Cape.\n\nproblem:\nConsidering these data what is the main, substantiated, conclusion that can be drawn about the effect of fuel oil contamination on the sediments and fauna of Otaiti Reef and Môtitī Island from the grounding of the Rena?\n\nA: On reef sediment PAH levels generally match PAH levels in sea urchins because sea urchins ingest sand as they graze.\nB: Significant contamination of sediments and accumulation within the food chain has occurred.\nC: Contamination of sediments and sea urchins on and around Otaiti, particularly in the close vicinity of the ship's hull, has occurred.\nD: Otaiti Reef and Mōtitī Island show contamination of sediments and sea urchins.\nE: Widespread contamination of sediments and some marine organisms has occurred.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=556&width=1834&top_left_y=276&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=657&width=894&top_left_y=1876&top_left_x=113",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=548&width=854&top_left_y=1942&top_left_x=1092",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-25.jpg?height=591&width=802&top_left_y=201&top_left_x=136"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_454",
        "problem": "某动物 $(2 \\mathrm{~N}=4)$ 的一个精原细胞所有核 DNA 分子的一条链均被 ${ }^{32} \\mathrm{P}$ 标记, 在含 ${ }^{31} \\mathrm{P}$的培养液中完成减数分裂, 得到 4 个精细胞。下列有关这 4 个精细胞放射性含量及其原因的分析，错误的是（ ）\nA: 若只有 1 个精细胞有放射性, 可能是 MI后期一对同源染色体移向同一极\nB: 若只有 2 个精细胞有放射性, 可能是 MII后期 2 个次级精母细胞含有 ${ }^{32} \\mathrm{P}$ 标记的 2 条染色体均移向同一极\nC: 若只有 3 个精细胞有放射性, 可能是 MII后期 1 个次级精母细胞含有 ${ }^{32} \\mathrm{P}$ 标记的 2 条染色体移向同一极\nD: 若 4 个精细胞均有放射性, 则每个精细胞的 1 个核 DNA 分子的一条链被 ${ }^{32} \\mathrm{P}$ 标记\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某动物 $(2 \\mathrm{~N}=4)$ 的一个精原细胞所有核 DNA 分子的一条链均被 ${ }^{32} \\mathrm{P}$ 标记, 在含 ${ }^{31} \\mathrm{P}$的培养液中完成减数分裂, 得到 4 个精细胞。下列有关这 4 个精细胞放射性含量及其原因的分析，错误的是（ ）\n\nA: 若只有 1 个精细胞有放射性, 可能是 MI后期一对同源染色体移向同一极\nB: 若只有 2 个精细胞有放射性, 可能是 MII后期 2 个次级精母细胞含有 ${ }^{32} \\mathrm{P}$ 标记的 2 条染色体均移向同一极\nC: 若只有 3 个精细胞有放射性, 可能是 MII后期 1 个次级精母细胞含有 ${ }^{32} \\mathrm{P}$ 标记的 2 条染色体移向同一极\nD: 若 4 个精细胞均有放射性, 则每个精细胞的 1 个核 DNA 分子的一条链被 ${ }^{32} \\mathrm{P}$ 标记\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1471",
        "problem": "With the rise in drug-resistant bacteria, it is crucial that antibacterial drugs are effective and prescribed appropriately to prevent selecting for resistance in the bacteria. We also need to understand their effect on other bacteria cohabiting the human body to prevent development of resistant strains of these other bacteria.\n\nThe Kirby-Bauer disk diffusion method is one technique used to assess the resistance of a bacterium to a drug. The bacteria are cultured on plates and small disks impregnated with specific antibiotics placed on the surface (coloured and labelled A-D below). The antibiotic from the disk diffuses into the growth medium as the bacteria on the surface of the plate replicates. Susceptible bacteria cannot grow in the presence of the antibiotic, resulting in a clear zone of inhibition around the disk. In contrast, resistant bacteria can replicate freely around the disk and there is no clear zone seen, just a 'lawn' of bacteria. The size of the inhibition zone is indicative of the effectiveness of the antibiotic impregnated in that disk. Zones of antibiotic diffusion should not overlap: if they do so a repeat of the experiment is required to determine their effectiveness.\n\n[figure1]\n\nWhich of the following statements is INCORRECT?\nA: Antibiotics $B$ and $C$ require a repeated experiment to determine their effectiveness\nB: Antibiotic $D$ is the best inhibitor of the growth of this bacterium\nC: The incubation process caused the differential growth of this bacterium species\nD: Antibiotic $A$ is effective against this bacterium, although antibiotics $B$ and $C$ are potentially more effective than $A$\nE: Antibiotic B has not diffused evenly from disk B\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWith the rise in drug-resistant bacteria, it is crucial that antibacterial drugs are effective and prescribed appropriately to prevent selecting for resistance in the bacteria. We also need to understand their effect on other bacteria cohabiting the human body to prevent development of resistant strains of these other bacteria.\n\nThe Kirby-Bauer disk diffusion method is one technique used to assess the resistance of a bacterium to a drug. The bacteria are cultured on plates and small disks impregnated with specific antibiotics placed on the surface (coloured and labelled A-D below). The antibiotic from the disk diffuses into the growth medium as the bacteria on the surface of the plate replicates. Susceptible bacteria cannot grow in the presence of the antibiotic, resulting in a clear zone of inhibition around the disk. In contrast, resistant bacteria can replicate freely around the disk and there is no clear zone seen, just a 'lawn' of bacteria. The size of the inhibition zone is indicative of the effectiveness of the antibiotic impregnated in that disk. Zones of antibiotic diffusion should not overlap: if they do so a repeat of the experiment is required to determine their effectiveness.\n\n[figure1]\n\nWhich of the following statements is INCORRECT?\n\nA: Antibiotics $B$ and $C$ require a repeated experiment to determine their effectiveness\nB: Antibiotic $D$ is the best inhibitor of the growth of this bacterium\nC: The incubation process caused the differential growth of this bacterium species\nD: Antibiotic $A$ is effective against this bacterium, although antibiotics $B$ and $C$ are potentially more effective than $A$\nE: Antibiotic B has not diffused evenly from disk B\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-28.jpg?height=738&width=1457&top_left_y=976&top_left_x=265"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_333",
        "problem": "在一个果蝇品系中出现了一只染色体变异的果蝇, 正常果蝇和变异果蝇的染色体及所含的基因如图所示。变异果蝇的三条染色体在减数分裂中会发生联会, 其中两条随机移向一极, 另一条移向另一极。在变异的染色体中, B 基因所在染色体高度螺旋化, 无法表达。将变异果蝇进行测交, 缺体 (2n-1) 果蝇可以存活。下列说法正确的是 ( )\n\n[图1]\n\n正常果蝇\n\n[图2]\n\n$\\mathrm{A}$ 和 $\\mathrm{a}$ 只在减数第二次分裂发生分离\nA: 不考虑染色体发生交换, A 和 a 只在减数第二次分裂发生分离\nB: 若染色体发生交换 (考虑所有可能), 变异果蝇产生的正常配子有 4 种基因型.\nC: 变异果蝇产生的配子中正常配子所占比例为 $1 / 3$\nD: 不考虑染色体发生交换, 测交子代多翅脉短刚毛个体所占比例为 $1 / 6$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在一个果蝇品系中出现了一只染色体变异的果蝇, 正常果蝇和变异果蝇的染色体及所含的基因如图所示。变异果蝇的三条染色体在减数分裂中会发生联会, 其中两条随机移向一极, 另一条移向另一极。在变异的染色体中, B 基因所在染色体高度螺旋化, 无法表达。将变异果蝇进行测交, 缺体 (2n-1) 果蝇可以存活。下列说法正确的是 ( )\n\n[图1]\n\n正常果蝇\n\n[图2]\n\n$\\mathrm{A}$ 和 $\\mathrm{a}$ 只在减数第二次分裂发生分离\n\nA: 不考虑染色体发生交换, A 和 a 只在减数第二次分裂发生分离\nB: 若染色体发生交换 (考虑所有可能), 变异果蝇产生的正常配子有 4 种基因型.\nC: 变异果蝇产生的配子中正常配子所占比例为 $1 / 3$\nD: 不考虑染色体发生交换, 测交子代多翅脉短刚毛个体所占比例为 $1 / 6$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-18.jpg?height=277&width=465&top_left_y=181&top_left_x=333",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-18.jpg?height=297&width=466&top_left_y=183&top_left_x=858"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_501",
        "problem": "图 $\\mathrm{I}$ 表示基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$ 的生物体某个细胞的分裂过程中某种物质或结构数量变化的相关模式图, 图 2 表示图 1 过程某些时期细胞内染色体、染色单体和核 DNA 的数量关系。下列相关叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 图 1 呈现的是细胞内染色体组数在细胞分裂不同时期的变化\nB: 图 1 的 $\\mathrm{CD}$ 段对应图 2 的I时期, 图 2 的 I 时期对应图 1 的 $\\mathrm{AB}$ 段\nC: 若该细胞核 DNA 双链均被 ${ }^{32} \\mathrm{P}$ 标记, 在普通培养基中完成减数分裂后, 至少 3 个子细胞带标记\nD: 该细胞经历图 1 过程最终形成一个 $\\mathrm{aX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$ 的子细胞, 则其他的子细胞基因型可能有 4 种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 $\\mathrm{I}$ 表示基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$ 的生物体某个细胞的分裂过程中某种物质或结构数量变化的相关模式图, 图 2 表示图 1 过程某些时期细胞内染色体、染色单体和核 DNA 的数量关系。下列相关叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 图 1 呈现的是细胞内染色体组数在细胞分裂不同时期的变化\nB: 图 1 的 $\\mathrm{CD}$ 段对应图 2 的I时期, 图 2 的 I 时期对应图 1 的 $\\mathrm{AB}$ 段\nC: 若该细胞核 DNA 双链均被 ${ }^{32} \\mathrm{P}$ 标记, 在普通培养基中完成减数分裂后, 至少 3 个子细胞带标记\nD: 该细胞经历图 1 过程最终形成一个 $\\mathrm{aX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$ 的子细胞, 则其他的子细胞基因型可能有 4 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-36.jpg?height=468&width=693&top_left_y=1054&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-36.jpg?height=457&width=460&top_left_y=1054&top_left_x=1049"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_998",
        "problem": "Progressive change toward a climax in a community is called:\nA: Evolutionary change\nB: Succession\nC: Energy flow\nD: Dynamic equilibrium\nE: None of the above\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nProgressive change toward a climax in a community is called:\n\nA: Evolutionary change\nB: Succession\nC: Energy flow\nD: Dynamic equilibrium\nE: None of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_861",
        "problem": "果蝇的II号染色体上的 $\\mathrm{E}$ 或 $\\mathrm{G}$ 基因纯合致死。利用图中的果蝇(PZ) 可以鉴定隐性突变体的突变基因是否位于II号染色体。现有一只正常翅脉正常翅型的常染色体隐性紫眼突变果蝇。将该紫眼果蝇与 $\\mathrm{PZ}$ 杂交, 挑选 $\\mathrm{F}_{1}$ 中的多翅脉果蝇自由交配, 统计 $\\mathrm{F}_{2}$ 的\n表型及比例(假设II号染色体上的基因不发生互换)。\n\n[图1]\n\n果蝇（PZ）的II号染色体\n\n下列叙述错误的是（ ）\nA: $F_{1}$ 中多翅脉正常翅型野生眼色和正常翅脉卷曲翅野生眼色比例为 1: 1\nB: 若紫眼基因位于II号染色体, 可观察到 $\\mathrm{F}_{2}$ 中多翅脉正常翅型野生眼色: 正常翅脉正常翅型紫眼 $=2: 1$\nC: 若紫眼基因不位于II号染色体, 可观察到 $\\mathrm{F}_{2}$ 中多翅脉正常翅型野生眼色: 多翅脉正常翅型紫眼：正常翅脉正常翅型野生眼色：正常翅脉正常翅型紫眼=9: 3: 3 : 1\nD: 若 PZ 果蝇自交, 子代 $\\mathrm{E} 、 \\mathrm{G}$ 的基因频率不变, II号染色体的组成也与亲本相同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的II号染色体上的 $\\mathrm{E}$ 或 $\\mathrm{G}$ 基因纯合致死。利用图中的果蝇(PZ) 可以鉴定隐性突变体的突变基因是否位于II号染色体。现有一只正常翅脉正常翅型的常染色体隐性紫眼突变果蝇。将该紫眼果蝇与 $\\mathrm{PZ}$ 杂交, 挑选 $\\mathrm{F}_{1}$ 中的多翅脉果蝇自由交配, 统计 $\\mathrm{F}_{2}$ 的\n表型及比例(假设II号染色体上的基因不发生互换)。\n\n[图1]\n\n果蝇（PZ）的II号染色体\n\n下列叙述错误的是（ ）\n\nA: $F_{1}$ 中多翅脉正常翅型野生眼色和正常翅脉卷曲翅野生眼色比例为 1: 1\nB: 若紫眼基因位于II号染色体, 可观察到 $\\mathrm{F}_{2}$ 中多翅脉正常翅型野生眼色: 正常翅脉正常翅型紫眼 $=2: 1$\nC: 若紫眼基因不位于II号染色体, 可观察到 $\\mathrm{F}_{2}$ 中多翅脉正常翅型野生眼色: 多翅脉正常翅型紫眼：正常翅脉正常翅型野生眼色：正常翅脉正常翅型紫眼=9: 3: 3 : 1\nD: 若 PZ 果蝇自交, 子代 $\\mathrm{E} 、 \\mathrm{G}$ 的基因频率不变, II号染色体的组成也与亲本相同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-43.jpg?height=354&width=511&top_left_y=231&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_748",
        "problem": "将全部 DNA 分子双链经 ${ }^{32} \\mathrm{P}$ 标记的雄性动物细胞（染色体数为 $2 \\mathrm{~N}$ ) 置于不含 ${ }^{32} \\mathrm{P}$的培养基中培养。经过连续 3 次细胞分裂后产生 8 个子代细胞, 检测子代细胞中的情况。下列推断正确的是（ ）\nA: 若某子代细胞中的染色体都含 ${ }^{32} \\mathrm{P}$, 则一定进行有丝分裂\nB: 若某子代细胞中的染色体都不含 ${ }^{32} \\mathrm{P}$, 则一定进行减数分裂\nC: 若只进行有丝分裂, 则含 ${ }^{32} \\mathrm{P}$ 染色体的子代细胞比例至少为 $1 / 2$\nD: 若进行一次有丝分裂再进行一次减数分裂, 则含 ${ }^{32} \\mathrm{P}$ 染色体的子代细胞比例至少占 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将全部 DNA 分子双链经 ${ }^{32} \\mathrm{P}$ 标记的雄性动物细胞（染色体数为 $2 \\mathrm{~N}$ ) 置于不含 ${ }^{32} \\mathrm{P}$的培养基中培养。经过连续 3 次细胞分裂后产生 8 个子代细胞, 检测子代细胞中的情况。下列推断正确的是（ ）\n\nA: 若某子代细胞中的染色体都含 ${ }^{32} \\mathrm{P}$, 则一定进行有丝分裂\nB: 若某子代细胞中的染色体都不含 ${ }^{32} \\mathrm{P}$, 则一定进行减数分裂\nC: 若只进行有丝分裂, 则含 ${ }^{32} \\mathrm{P}$ 染色体的子代细胞比例至少为 $1 / 2$\nD: 若进行一次有丝分裂再进行一次减数分裂, 则含 ${ }^{32} \\mathrm{P}$ 染色体的子代细胞比例至少占 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1447",
        "problem": "Capillary hydrostatic pressure is the pressure exerted by the blood on the walls of capillaries, forcing fluid out of the capillary. Colloid osmotic pressure is exerted by proteins within the blood and draws fluid back into the capillary via osmosis.\n\nAs blood travels from the arterial to the venous end of the capillary, capillary hydrostatic pressure decreases as fluid leaves, while colloid osmotic pressure increases as the concentration of plasma proteins increases due to the loss of fluid. Therefore, there is net filtration at the arterial end of the capillary but net absorption at the venous end. Overall, across the entire capillary, there is generally more filtration than absorption. The diagram below summarises this information.\n\n[figure1]\n\nPeripheral oedema or swelling occurs when excessive fluid exits the capillaries and accumulates in the interstitial spaces.\n\nWhich of the following options could cause peripheral oedema?\nA: Liver disease, which decreases production of the plasma protein albumin\nB: Heart failure, where the heart is unable to pump sufficient blood to the rest of the body, leading to increased venous pressure\nC: Deep vein thrombosis, where a clot forms in the veins of the legs and obstructs venous blood flow\nD: Lymphatic filariasis, caused by worms such as Wuchereria bancrofti which may obstruct lymphatic vessels\nE: All of the options above would cause peripheral oedema\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCapillary hydrostatic pressure is the pressure exerted by the blood on the walls of capillaries, forcing fluid out of the capillary. Colloid osmotic pressure is exerted by proteins within the blood and draws fluid back into the capillary via osmosis.\n\nAs blood travels from the arterial to the venous end of the capillary, capillary hydrostatic pressure decreases as fluid leaves, while colloid osmotic pressure increases as the concentration of plasma proteins increases due to the loss of fluid. Therefore, there is net filtration at the arterial end of the capillary but net absorption at the venous end. Overall, across the entire capillary, there is generally more filtration than absorption. The diagram below summarises this information.\n\n[figure1]\n\nPeripheral oedema or swelling occurs when excessive fluid exits the capillaries and accumulates in the interstitial spaces.\n\nWhich of the following options could cause peripheral oedema?\n\nA: Liver disease, which decreases production of the plasma protein albumin\nB: Heart failure, where the heart is unable to pump sufficient blood to the rest of the body, leading to increased venous pressure\nC: Deep vein thrombosis, where a clot forms in the veins of the legs and obstructs venous blood flow\nD: Lymphatic filariasis, caused by worms such as Wuchereria bancrofti which may obstruct lymphatic vessels\nE: All of the options above would cause peripheral oedema\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-14.jpg?height=614&width=1448&top_left_y=1920&top_left_x=250"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1411",
        "problem": "A group of scientists measured the distance covered over 24 hours by two samples of a common periwinkle (Littorina littorea)\n\n| Individual | Sample A (cm) | Sample B (cm) |\n| :---: | :---: | :---: |\n| 1 | 12 | 30 |\n| 2 | 13 | 9 |\n| 3 | 20 | 17 |\n| 4 | 3 | 11 |\n| 5 | 11 | 18 |\n| 6 | 21 | 5 |\n| 7 | 18 | 6 |\n| 8 | 22 | 13 |\n| 9 | 8 | 11 |\n| 10 | 0 | 19 |\n| Standard Deviation | $\\mathbf{7 . 5 8}$ | $\\mathbf{7 . 4 2}$ |\n\nThe mean of a sample is defined as\n\n$$\n\\bar{x}=\\frac{\\sum_{i=1}^{n} x_{i}}{n}\n$$\n\nThe mean of sample $A$ is $12.8 \\mathrm{~cm}$\n\nThe skew of a sample is defined as\n\n$$\n\\text { skew }=\\frac{n}{(n-1)(n-2)} \\sum_{i=1}^{n}\\left(\\frac{x_{i}-\\bar{x}}{s}\\right)^{3}\n$$\n\nwhere s represents standard deviation\n\nWhat is the skew of sample A?\nA: 0.000\nB: -0.4176\nC: 1.0353\nD: -0.0241\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA group of scientists measured the distance covered over 24 hours by two samples of a common periwinkle (Littorina littorea)\n\n| Individual | Sample A (cm) | Sample B (cm) |\n| :---: | :---: | :---: |\n| 1 | 12 | 30 |\n| 2 | 13 | 9 |\n| 3 | 20 | 17 |\n| 4 | 3 | 11 |\n| 5 | 11 | 18 |\n| 6 | 21 | 5 |\n| 7 | 18 | 6 |\n| 8 | 22 | 13 |\n| 9 | 8 | 11 |\n| 10 | 0 | 19 |\n| Standard Deviation | $\\mathbf{7 . 5 8}$ | $\\mathbf{7 . 4 2}$ |\n\nThe mean of a sample is defined as\n\n$$\n\\bar{x}=\\frac{\\sum_{i=1}^{n} x_{i}}{n}\n$$\n\nThe mean of sample $A$ is $12.8 \\mathrm{~cm}$\n\nThe skew of a sample is defined as\n\n$$\n\\text { skew }=\\frac{n}{(n-1)(n-2)} \\sum_{i=1}^{n}\\left(\\frac{x_{i}-\\bar{x}}{s}\\right)^{3}\n$$\n\nwhere s represents standard deviation\n\nWhat is the skew of sample A?\n\nA: 0.000\nB: -0.4176\nC: 1.0353\nD: -0.0241\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1436",
        "problem": "The following graph shows relative oxygen concentrations in an airtight container containing one spinach plant exposed to varying light intensities. The y axis indicates oxygen concentration and the $x$ axis indicates increasing light intensity from left to right. The units of both axes are arbitrary; they do not matter.\n\n[figure1]\n\nConsider point $X$ labelled on the graph. Select the CORRECT option:\nA: At point $X$, the concentration of carbon dioxide is increasing.\nB: At point $X$, the concentration of carbon dioxide is decreasing.\nC: At point $X$, the concentration of carbon dioxide is constant.\nD: Insufficient information to answer the question.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph shows relative oxygen concentrations in an airtight container containing one spinach plant exposed to varying light intensities. The y axis indicates oxygen concentration and the $x$ axis indicates increasing light intensity from left to right. The units of both axes are arbitrary; they do not matter.\n\n[figure1]\n\nConsider point $X$ labelled on the graph. Select the CORRECT option:\n\nA: At point $X$, the concentration of carbon dioxide is increasing.\nB: At point $X$, the concentration of carbon dioxide is decreasing.\nC: At point $X$, the concentration of carbon dioxide is constant.\nD: Insufficient information to answer the question.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-18.jpg?height=765&width=1354&top_left_y=563&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_959",
        "problem": "A new therapeutic drug can treat the symptoms of the demyelinating disease multiple sclerosis (MS) by blockage of voltage-gated potassium channels. Neuron action potential conduction improves because:\nA: More potassium is allowed to enter the axon which increases the amplitude of the action potential\nB: The high resting leak of potassium out of the axon is minimized\nC: Less potassium is allowed to leave the axon which increases the size and duration of the action potential\nD: Less potassium is allowed to leave the axon which prevents the inactivation of voltage-gated sodium channels increasing the amplitude of the action potential\nE: The resistance of the plasma membrane decreases which increases the amplitude of the action potential\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA new therapeutic drug can treat the symptoms of the demyelinating disease multiple sclerosis (MS) by blockage of voltage-gated potassium channels. Neuron action potential conduction improves because:\n\nA: More potassium is allowed to enter the axon which increases the amplitude of the action potential\nB: The high resting leak of potassium out of the axon is minimized\nC: Less potassium is allowed to leave the axon which increases the size and duration of the action potential\nD: Less potassium is allowed to leave the axon which prevents the inactivation of voltage-gated sodium channels increasing the amplitude of the action potential\nE: The resistance of the plasma membrane decreases which increases the amplitude of the action potential\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1378",
        "problem": "Every day, some 200 litres of blood are filtered by your kidneys. Most of that fluid is reabsorbed back into the body; the rest of it becomes urine. The basic unit of structure in the kidney is the nephron. Blood enters the nephron through the afferent arteriole which branches into a network of glomerular capillaries. A series of hydrostatic and osmotic pressures then draw fluid in or drive fluid out of these capillaries. The filtrate is collected in the capsular space which drains into the proximal convoluted tubule (PCT) of the nephron. This is the basis of renal filtration.\n\nGlomerular hydrostatic pressure (GHP) is the pressure exerted by blood on the walls of the glomerular capillaries. Capsular hydrostatic pressure (CHP) is the pressure exerted by the filtrate in the capsular space. Blood colloid osmotic pressure (BCOP) is the pressure exerted by the concentration of proteins in the blood. No proteins are expected to found in the capsular space as the glomerular capillary walls do not allow them to pass. The net filtration pressure (NFP) is the sum of all these pressures drawing fluids in or out of the glomerular capillaries.\n\nThe arrows in the diagram represent fluid moving in and out of the capillaries under the influence of these pressures.\n\n[figure1]\n\nIn which direction would blood colloid osmotic pressure (BCOP) drive fluid?\nA: From the glomerular capillaries into the capsular space.\nB: From the capsular space back into the glomerular capillaries.\nC: BCOP does not have an effect on the direction of fluid movement because no proteins are expected to be found in the capsular space.\nD: Insufficient information to tell.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEvery day, some 200 litres of blood are filtered by your kidneys. Most of that fluid is reabsorbed back into the body; the rest of it becomes urine. The basic unit of structure in the kidney is the nephron. Blood enters the nephron through the afferent arteriole which branches into a network of glomerular capillaries. A series of hydrostatic and osmotic pressures then draw fluid in or drive fluid out of these capillaries. The filtrate is collected in the capsular space which drains into the proximal convoluted tubule (PCT) of the nephron. This is the basis of renal filtration.\n\nGlomerular hydrostatic pressure (GHP) is the pressure exerted by blood on the walls of the glomerular capillaries. Capsular hydrostatic pressure (CHP) is the pressure exerted by the filtrate in the capsular space. Blood colloid osmotic pressure (BCOP) is the pressure exerted by the concentration of proteins in the blood. No proteins are expected to found in the capsular space as the glomerular capillary walls do not allow them to pass. The net filtration pressure (NFP) is the sum of all these pressures drawing fluids in or out of the glomerular capillaries.\n\nThe arrows in the diagram represent fluid moving in and out of the capillaries under the influence of these pressures.\n\n[figure1]\n\nIn which direction would blood colloid osmotic pressure (BCOP) drive fluid?\n\nA: From the glomerular capillaries into the capsular space.\nB: From the capsular space back into the glomerular capillaries.\nC: BCOP does not have an effect on the direction of fluid movement because no proteins are expected to be found in the capsular space.\nD: Insufficient information to tell.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-25.jpg?height=633&width=851&top_left_y=1314&top_left_x=497"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1049",
        "problem": "[figure1]\n\nFluorescence intensity\n\nPropidium iodide ( $\\mathrm{PI})$ is a dye that can intercalate and stain cellular genome upon cell fixation. The amount of DNA indicated by the fluorescence intensity of PI in a population of cells which are in different stages of the cell cycle is depicted in the following graph. P, Q and $\\mathrm{R}$ respectively indicate:\nA: (G2), (S \\& G1) and (M).\nB: (G1), (G2\\&M) and (S).\nC: (G1), (G2) and (S\\&M).\nD: (S \\& G2), (G1) and (M).\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nFluorescence intensity\n\nPropidium iodide ( $\\mathrm{PI})$ is a dye that can intercalate and stain cellular genome upon cell fixation. The amount of DNA indicated by the fluorescence intensity of PI in a population of cells which are in different stages of the cell cycle is depicted in the following graph. P, Q and $\\mathrm{R}$ respectively indicate:\n\nA: (G2), (S \\& G1) and (M).\nB: (G1), (G2\\&M) and (S).\nC: (G1), (G2) and (S\\&M).\nD: (S \\& G2), (G1) and (M).\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-04.jpg?height=642&width=658&top_left_y=606&top_left_x=1189"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_986",
        "problem": "[figure1]\n\nA valid taxonomic group for the cladogram shown above would include:\nA: Tamarin + Squirrel\nB: Tamarin + Squirrel + Howler\nC: Squirrel + Howler + Wooly\nD: Wooly + Spider + Howler\nE: All are valid groups\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nA valid taxonomic group for the cladogram shown above would include:\n\nA: Tamarin + Squirrel\nB: Tamarin + Squirrel + Howler\nC: Squirrel + Howler + Wooly\nD: Wooly + Spider + Howler\nE: All are valid groups\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_c11579df664240d6158ag-17.jpg?height=588&width=1157&top_left_y=297&top_left_x=495"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1316",
        "problem": "The graph shows the effect of wavelength of light on the rate of photosynthesis and on the amount of light absorbed by the pigments in a green seaweed.\n\n[figure1]\n\nThe difference between the to curves at $X$ is due to\nA: inefficient trapping of light energy by the chlorophyll\nB: no production of ATP at that wavelength\nC: photorespiration occurring at that wavelength\nD: carotenoids absorbing light that is not then used in photosynthesis\nE: fluorescence of carotenoids\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows the effect of wavelength of light on the rate of photosynthesis and on the amount of light absorbed by the pigments in a green seaweed.\n\n[figure1]\n\nThe difference between the to curves at $X$ is due to\n\nA: inefficient trapping of light energy by the chlorophyll\nB: no production of ATP at that wavelength\nC: photorespiration occurring at that wavelength\nD: carotenoids absorbing light that is not then used in photosynthesis\nE: fluorescence of carotenoids\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-10.jpg?height=450&width=871&top_left_y=1363&top_left_x=627"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_336",
        "problem": "图(1)-3)分别表示人体细胞中发生的 3 种生物大分子的合成过程下列选项错误的是 ( )\n[图1]\nA: 细胞中过程(2)与过程(1)相比，特有的碱基对为 A-U\nB: 过程(1)的发生的主要场所是细胞核, 过程(3)的发生场所是细胞质\nC: 在人体内成熟红细胞、神经细胞、造血干细胞中, 能发生过程(2)、(3)而不能发生过程(1)的细胞是神经细胞\nD: 已知过程(2)的 $\\alpha$ 链中腺嘌岭与尿嘧啶之和占碱基总数的 48\\%, 则与 $\\alpha$ 链对应的 DNA 区段中鸟嘌呤所占的碱基比例为 $52 \\%$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图(1)-3)分别表示人体细胞中发生的 3 种生物大分子的合成过程下列选项错误的是 ( )\n[图1]\n\nA: 细胞中过程(2)与过程(1)相比，特有的碱基对为 A-U\nB: 过程(1)的发生的主要场所是细胞核, 过程(3)的发生场所是细胞质\nC: 在人体内成熟红细胞、神经细胞、造血干细胞中, 能发生过程(2)、(3)而不能发生过程(1)的细胞是神经细胞\nD: 已知过程(2)的 $\\alpha$ 链中腺嘌岭与尿嘧啶之和占碱基总数的 48\\%, 则与 $\\alpha$ 链对应的 DNA 区段中鸟嘌呤所占的碱基比例为 $52 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-101.jpg?height=524&width=1444&top_left_y=1458&top_left_x=342"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1372",
        "problem": "When a gene is 'expressed' it is?\nA: Transported around the body to make proteins\nB: Replicated within the cell\nC: Used as a blueprint to assemble the protein it codes for\nD: Used as a blueprint to assemble the peptide it codes for\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen a gene is 'expressed' it is?\n\nA: Transported around the body to make proteins\nB: Replicated within the cell\nC: Used as a blueprint to assemble the protein it codes for\nD: Used as a blueprint to assemble the peptide it codes for\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_982",
        "problem": "A rat chewing the insulation from the wiring in your car is analogous to:\nA: The depolarization of the unmyelinated axons\nB: The nodes of Ranvier in the PNS\nC: Demyelination of the nervous system\nD: Schwann cells failing to myelinate axons in the CNS\nE: The deterioration of the brain-blood barrier\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA rat chewing the insulation from the wiring in your car is analogous to:\n\nA: The depolarization of the unmyelinated axons\nB: The nodes of Ranvier in the PNS\nC: Demyelination of the nervous system\nD: Schwann cells failing to myelinate axons in the CNS\nE: The deterioration of the brain-blood barrier\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1131",
        "problem": "The table shows body weight and food intake of a few selected animal species.\n\n| Animal species | Average body weight | Average food intake/day |\n| :--- | :--- | :--- |\n| Black Bear <br> Euarctus americanus | $135 \\mathrm{~kg}$ | $3.9 \\mathrm{~kg}$ |\n| Common Shrew <br> Sorese cinereus | $5 \\mathrm{gm}$ | $13 \\mathrm{gm}$ |\n| Pigeon <br> Columba liva | $300 \\mathrm{gm}$ | $100 \\mathrm{gm}$ |\n| Horse <br> Equus caballus | $500 \\mathrm{~kg}$ | $12 \\mathrm{~kg}$ |\n\nBased in the data provided, analyze the following statements.\n\n1) Smaller the animal, higher is the metabolic rate.\n2) Larger the animal, greater is the energy requirement per unit body weight leading to larger intake of food.\n3) Mammals consume larger proportion of their own weight as food daily compared to birds.\n4) Demand for food per unit of body mass increases as the metabolic rate increases.\n\nThe correct statements are:\nA: 1 and 2\nB: 1 and 4\nC: 2 and 3\nD: 3 and 4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe table shows body weight and food intake of a few selected animal species.\n\n| Animal species | Average body weight | Average food intake/day |\n| :--- | :--- | :--- |\n| Black Bear <br> Euarctus americanus | $135 \\mathrm{~kg}$ | $3.9 \\mathrm{~kg}$ |\n| Common Shrew <br> Sorese cinereus | $5 \\mathrm{gm}$ | $13 \\mathrm{gm}$ |\n| Pigeon <br> Columba liva | $300 \\mathrm{gm}$ | $100 \\mathrm{gm}$ |\n| Horse <br> Equus caballus | $500 \\mathrm{~kg}$ | $12 \\mathrm{~kg}$ |\n\nBased in the data provided, analyze the following statements.\n\n1) Smaller the animal, higher is the metabolic rate.\n2) Larger the animal, greater is the energy requirement per unit body weight leading to larger intake of food.\n3) Mammals consume larger proportion of their own weight as food daily compared to birds.\n4) Demand for food per unit of body mass increases as the metabolic rate increases.\n\nThe correct statements are:\n\nA: 1 and 2\nB: 1 and 4\nC: 2 and 3\nD: 3 and 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_866",
        "problem": "种子的种皮是由母体植株雌荵的珠被发育而来的, 种子的胚是由受精卵发育来的。已知豌豆种皮有黄、白两种颜色。为了解种皮颜色性状的遗传规律, 某研究小组第 1 年在甲地用纯种黄种皮豌豆 (母本) 和纯种白种皮豌豆（父本）杂交, 当年收获的种子 (记作 $F_{1}$ ) 全为黄种皮: 在乙地用纯种黄种皮 (父本) 和纯种白种皮 (母本) 杂交, $F_{1}$全为白种皮; 第 2 年他们又将两地的 $F_{1}$ 种子分别种下得到 $F_{1}$ 植株, $F_{1}$ 植株自交所结种子为 $F_{2}$, 统计发现两地 $F_{2}$ 种皮都为黄色; 第 3 年他们又将两地的 $F_{2}$ 种子分别种下得到\n\n$F_{2}$ 植株, $F_{2}$ 植株自交所结种子为 $F_{3}$, 统计发现两地 $F_{3}$ 种皮中黄色与白色的比例接近 3: 1. 下列说法正确的是 ( )\nA: 甲、乙两地第 1 年正反交实验结果不同, 即可判断踠豆种皮颜色的遗传是细胞质遗传\nB: $F_{2}$ 种皮均为黄色, 不符合孟德尔一对相对性状杂交实验的分离比, 所以踠豆种皮颜色的遗传不遵循分离定律\nC: 若将甲、乙两地杂交产生的 $F_{1}$ 种子种下, 长成 $F_{1}$ 植株, 连续自交 4 代, 所得的自交第 4 代种子中白种皮种子占的比例为 $7 / 16$\nD: 若让其连续自交，白种皮种子中的纯合子比例会随着自交代数增多而逐代增大\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n种子的种皮是由母体植株雌荵的珠被发育而来的, 种子的胚是由受精卵发育来的。已知豌豆种皮有黄、白两种颜色。为了解种皮颜色性状的遗传规律, 某研究小组第 1 年在甲地用纯种黄种皮豌豆 (母本) 和纯种白种皮豌豆（父本）杂交, 当年收获的种子 (记作 $F_{1}$ ) 全为黄种皮: 在乙地用纯种黄种皮 (父本) 和纯种白种皮 (母本) 杂交, $F_{1}$全为白种皮; 第 2 年他们又将两地的 $F_{1}$ 种子分别种下得到 $F_{1}$ 植株, $F_{1}$ 植株自交所结种子为 $F_{2}$, 统计发现两地 $F_{2}$ 种皮都为黄色; 第 3 年他们又将两地的 $F_{2}$ 种子分别种下得到\n\n$F_{2}$ 植株, $F_{2}$ 植株自交所结种子为 $F_{3}$, 统计发现两地 $F_{3}$ 种皮中黄色与白色的比例接近 3: 1. 下列说法正确的是 ( )\n\nA: 甲、乙两地第 1 年正反交实验结果不同, 即可判断踠豆种皮颜色的遗传是细胞质遗传\nB: $F_{2}$ 种皮均为黄色, 不符合孟德尔一对相对性状杂交实验的分离比, 所以踠豆种皮颜色的遗传不遵循分离定律\nC: 若将甲、乙两地杂交产生的 $F_{1}$ 种子种下, 长成 $F_{1}$ 植株, 连续自交 4 代, 所得的自交第 4 代种子中白种皮种子占的比例为 $7 / 16$\nD: 若让其连续自交，白种皮种子中的纯合子比例会随着自交代数增多而逐代增大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_56",
        "problem": "The behaviour of two similar-sized species of fiddler crabs (Uca latimanus and $U$. musica) that intermingle in the same habitat was studied. Males build hoods over their burrows for mate attraction. Mate searching is a dangerous activity for fiddler crab females, so these females may be forced to make suboptimal choices for their own safety, especially in areas where their conspecifics are in lower densities. The figures Fig.Q. 82 below show the approaches made by females of the two species to male crabs as well as to burrows (with and without hoods) of conspecific males.\n[figure1]\n\n(A) Mean $( \\pm \\mathrm{SE}$ ) proportion of resident Uca latimanus and U. musica males approached by wandering\n\n(a) U. latimanus females and (b) U. musica females.\n\n(B) Mean ( $\\pm$ SE) proportion of resident conspecific males with and without hoods approached by wandering (a) U. latimanus females and (b) $U$. musica females. ${ }^{*} p<0.05$.\nA: Females of both species approached a greater proportion of the conspecific males than the heterospecific males they encountered.\nB: Attraction of $U$. musica females to hoods is not as strong as that of $U$. latimanus females.\nC: A male fiddler crab's willingness to court all females, regardless of species, is made used of by females of both species for shelter-seeking and avoidance of predators.\nD: An overlap in habitat use between these two similar-sized fiddler crabs has no impact on both signalers and receivers.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe behaviour of two similar-sized species of fiddler crabs (Uca latimanus and $U$. musica) that intermingle in the same habitat was studied. Males build hoods over their burrows for mate attraction. Mate searching is a dangerous activity for fiddler crab females, so these females may be forced to make suboptimal choices for their own safety, especially in areas where their conspecifics are in lower densities. The figures Fig.Q. 82 below show the approaches made by females of the two species to male crabs as well as to burrows (with and without hoods) of conspecific males.\n[figure1]\n\n(A) Mean $( \\pm \\mathrm{SE}$ ) proportion of resident Uca latimanus and U. musica males approached by wandering\n\n(a) U. latimanus females and (b) U. musica females.\n\n(B) Mean ( $\\pm$ SE) proportion of resident conspecific males with and without hoods approached by wandering (a) U. latimanus females and (b) $U$. musica females. ${ }^{*} p<0.05$.\n\nA: Females of both species approached a greater proportion of the conspecific males than the heterospecific males they encountered.\nB: Attraction of $U$. musica females to hoods is not as strong as that of $U$. latimanus females.\nC: A male fiddler crab's willingness to court all females, regardless of species, is made used of by females of both species for shelter-seeking and avoidance of predators.\nD: An overlap in habitat use between these two similar-sized fiddler crabs has no impact on both signalers and receivers.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-074.jpg?height=1196&width=1670&top_left_y=876&top_left_x=204"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_59",
        "problem": "When isolated mitochondria are suspended in a buffer containing ADP, $\\mathrm{P}_{\\mathrm{i}}$, and an oxidizable substrate, three easily measured processes occur: the substrate is oxidized; $\\mathrm{O}_{2}$ is consumed; and ATP is synthesized. Cyanide $\\left(\\mathrm{CN}^{*}\\right)$ is an inhibitor of the passage of electrons to $\\mathrm{O}_{2}$. Oligomycin inhibits ATP synthase by interacting with subunit $\\mathrm{F}_{0}$. 2,4dinitrophenol (DNP) can diffuse readily across mitochondrial membranes and release a proton into the matrix, thus dissipating the proton gradient.\n\n[figure1]\n\nTime\n\n[figure2]\n\nTime\n\nFig.Q.4. Oxygen consumption and ATP synthesis in mitochondria.\n\nThe solid lines indicate the amount of oxygen consumed and the dash lines indicate the amount of ATP synthesized.\nA: $\\mathrm{x}$ is the oxidizable substrate.\nB: y may be oligomycin or $\\mathrm{CN}^{-}$.\nC: $z$ is DNP.\nD: If $z$ is a mixture of oligomycin and DNP, the trend of each line in the figure $B$ is not changed.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nWhen isolated mitochondria are suspended in a buffer containing ADP, $\\mathrm{P}_{\\mathrm{i}}$, and an oxidizable substrate, three easily measured processes occur: the substrate is oxidized; $\\mathrm{O}_{2}$ is consumed; and ATP is synthesized. Cyanide $\\left(\\mathrm{CN}^{*}\\right)$ is an inhibitor of the passage of electrons to $\\mathrm{O}_{2}$. Oligomycin inhibits ATP synthase by interacting with subunit $\\mathrm{F}_{0}$. 2,4dinitrophenol (DNP) can diffuse readily across mitochondrial membranes and release a proton into the matrix, thus dissipating the proton gradient.\n\n[figure1]\n\nTime\n\n[figure2]\n\nTime\n\nFig.Q.4. Oxygen consumption and ATP synthesis in mitochondria.\n\nThe solid lines indicate the amount of oxygen consumed and the dash lines indicate the amount of ATP synthesized.\n\nA: $\\mathrm{x}$ is the oxidizable substrate.\nB: y may be oligomycin or $\\mathrm{CN}^{-}$.\nC: $z$ is DNP.\nD: If $z$ is a mixture of oligomycin and DNP, the trend of each line in the figure $B$ is not changed.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-011.jpg?height=536&width=693&top_left_y=890&top_left_x=264",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-011.jpg?height=531&width=873&top_left_y=887&top_left_x=978"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_439",
        "problem": "某种生物细胞减数分裂过程中几个时期的显微照片如下。下列叙述正确的是\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\n\n[图4]\n\n丁\nA: 图甲中，细胞的同源染色体之间发生了基因重组\nB: 图乙中，移向细胞两极的染色体组成相同\nC: 图丙中, 染色体的复制正在进行，着丝点尚未分裂\nD: 图丁中, 细胞的同源染色体分离, 染色体数目减半\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种生物细胞减数分裂过程中几个时期的显微照片如下。下列叙述正确的是\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\n\n[图4]\n\n丁\n\nA: 图甲中，细胞的同源染色体之间发生了基因重组\nB: 图乙中，移向细胞两极的染色体组成相同\nC: 图丙中, 染色体的复制正在进行，着丝点尚未分裂\nD: 图丁中, 细胞的同源染色体分离, 染色体数目减半\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-12.jpg?height=203&width=209&top_left_y=664&top_left_x=181",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-12.jpg?height=200&width=208&top_left_y=668&top_left_x=410",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-12.jpg?height=182&width=169&top_left_y=680&top_left_x=681",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-12.jpg?height=188&width=206&top_left_y=677&top_left_x=888"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_205",
        "problem": "The enzymatic reaction below includes two steps.\n\n$$\n\\text { para-nitrophenyl-peptide } \\rightarrow \\text { para-nitrophenol }+ \\text { peptide }\n$$\n\nPara-nitrophenol has maximum absorbance at wavelength $405 \\mathrm{~nm}$. The progress curve of the reaction is shown below.\n\n[figure1]\nA: The release of para-nitrophenol occurs during the first step of enzymatic mechanism.\nB: The rate limiting step of this enzymatic mechanism is the second step.\nC: Enzyme activity can be determined from the slope of region \" $a$ \" of the curve.\nD: One of the reaction products activates the enzyme.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe enzymatic reaction below includes two steps.\n\n$$\n\\text { para-nitrophenyl-peptide } \\rightarrow \\text { para-nitrophenol }+ \\text { peptide }\n$$\n\nPara-nitrophenol has maximum absorbance at wavelength $405 \\mathrm{~nm}$. The progress curve of the reaction is shown below.\n\n[figure1]\n\nA: The release of para-nitrophenol occurs during the first step of enzymatic mechanism.\nB: The rate limiting step of this enzymatic mechanism is the second step.\nC: Enzyme activity can be determined from the slope of region \" $a$ \" of the curve.\nD: One of the reaction products activates the enzyme.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-17.jpg?height=728&width=1006&top_left_y=624&top_left_x=525"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_119",
        "problem": "Physiological studies have been accomplished on the adaptations of bromeliad plants to nutrient acquisition and plant utilization of potassium. The Figure shows biphasic kinetics of potassium uptake by foliar trichomes of a tank bromeliad plant in the presence of $0.01-90 \\mathrm{mM}$ substrate. Uptake rates were calculated during the first $1-2 \\mathrm{~h}$ of the experiment. The obtained results suggested that there were two transporter systems, and their Michaelis-Menten constants $\\left(\\mathrm{K}_{\\mathrm{m}}\\right)$ were calculated to be $41.3 \\pm 8.7 \\mu \\mathrm{M}$ and $56.5 \\pm 13.7 \\mathrm{mM}$.\n\n[figure1]\nA: The low affinity system accumulated $\\mathrm{K}^{+}$slower than the high affinity system\nB: $\\mathrm{Km}$ of $56.5 \\mathrm{mM}$ belongs to $\\mathrm{B}$ transporter system\nC: Applying ATPase inhibitor compounds inhibits B transporters\nD: B transporters are blocked more by potassium channel blockers, as compared to A transporters\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPhysiological studies have been accomplished on the adaptations of bromeliad plants to nutrient acquisition and plant utilization of potassium. The Figure shows biphasic kinetics of potassium uptake by foliar trichomes of a tank bromeliad plant in the presence of $0.01-90 \\mathrm{mM}$ substrate. Uptake rates were calculated during the first $1-2 \\mathrm{~h}$ of the experiment. The obtained results suggested that there were two transporter systems, and their Michaelis-Menten constants $\\left(\\mathrm{K}_{\\mathrm{m}}\\right)$ were calculated to be $41.3 \\pm 8.7 \\mu \\mathrm{M}$ and $56.5 \\pm 13.7 \\mathrm{mM}$.\n\n[figure1]\n\nA: The low affinity system accumulated $\\mathrm{K}^{+}$slower than the high affinity system\nB: $\\mathrm{Km}$ of $56.5 \\mathrm{mM}$ belongs to $\\mathrm{B}$ transporter system\nC: Applying ATPase inhibitor compounds inhibits B transporters\nD: B transporters are blocked more by potassium channel blockers, as compared to A transporters\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-41.jpg?height=757&width=1102&top_left_y=684&top_left_x=471"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_677",
        "problem": "某男子的一条 14 号和一条 21 号染色体相互连接形成一条异常染色体, 如图甲。减数分裂时异常染色体的联会如图乙, 配对的三条染色体中, 任意配对的两条染色体分离时，另一条染色体随机移向细胞任一极。下列叙述正确的是（）\n[图1]\nA: 该男子体细胞中染色体为 47 条, 患有三体综合征\nB: 图甲发生了染色体的结构变异和染色体的数目变异\nC: 如不考虑其他染色体，理论上该男子产生的精子类型有 5 种\nD: 该男子与正常女子婚配理论上生育染色体组成正常的后代的概率为 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某男子的一条 14 号和一条 21 号染色体相互连接形成一条异常染色体, 如图甲。减数分裂时异常染色体的联会如图乙, 配对的三条染色体中, 任意配对的两条染色体分离时，另一条染色体随机移向细胞任一极。下列叙述正确的是（）\n[图1]\n\nA: 该男子体细胞中染色体为 47 条, 患有三体综合征\nB: 图甲发生了染色体的结构变异和染色体的数目变异\nC: 如不考虑其他染色体，理论上该男子产生的精子类型有 5 种\nD: 该男子与正常女子婚配理论上生育染色体组成正常的后代的概率为 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-014.jpg?height=254&width=648&top_left_y=2246&top_left_x=330"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1317",
        "problem": "Find the four-digit number designated by asterisks, given the following:\n\n- All four digits of the unknown number are different.\n- None of the digits is zero.\n- Each \"0\" on the right of each four-digit number below indicates that the number has a matching digit in a nonmatching position with the unknown number.\n- Each \"+\" on the right of each four-digit number below indicates that the number has a matching digit in a matching position with the unknown number.\n\n$6152 +0$\n\n$4182 00$\n\n$5314 00$\n\n$5789 +$\n\n-------------\n****\nA: 6419\nB: 6741\nC: 5619\nD: 5641\nE: 5629\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFind the four-digit number designated by asterisks, given the following:\n\n- All four digits of the unknown number are different.\n- None of the digits is zero.\n- Each \"0\" on the right of each four-digit number below indicates that the number has a matching digit in a nonmatching position with the unknown number.\n- Each \"+\" on the right of each four-digit number below indicates that the number has a matching digit in a matching position with the unknown number.\n\n$6152 +0$\n\n$4182 00$\n\n$5314 00$\n\n$5789 +$\n\n-------------\n****\n\nA: 6419\nB: 6741\nC: 5619\nD: 5641\nE: 5629\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_687",
        "problem": "某自花传粉植物的花色由 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 两对等位基因控制, $\\mathrm{A}$ 基因控制花青素的合成,当花瓣中无花青素时为白色, $\\mathrm{B}$ 基因能抑制 $\\mathrm{A}$ 基因的表达, $\\mathrm{BB}$ 会让花瓣成为黄色, $\\mathrm{Bb}$会让花瓣成为浅紫色, 无 $\\mathrm{B}$ 时为蓝色。某浅紫色植株自交, $\\mathrm{F}_{1}$ 中出现蓝色: 浅紫色:黄色: 白色=1：4: 3: 4。下列相关判断正确的是（ ）\nA: $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 两对基因的遗传遵循自由组合定律，亲本基因型为 $\\mathrm{AABb}$\nB: $F_{1}$ 出现上述表现型及比例的原因是基因型为 $a B$ 的雄配子或雌配子致死\nC: 让 $\\mathrm{F}_{1}$ 中黄色植株随机受粉, 子代表现型及比例为黄色: 白色 $=3: 1$\nD: 让 $F_{1}$ 中的浅紫色植株与白色植株杂交，可判断浅紫色植株的基因型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某自花传粉植物的花色由 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 两对等位基因控制, $\\mathrm{A}$ 基因控制花青素的合成,当花瓣中无花青素时为白色, $\\mathrm{B}$ 基因能抑制 $\\mathrm{A}$ 基因的表达, $\\mathrm{BB}$ 会让花瓣成为黄色, $\\mathrm{Bb}$会让花瓣成为浅紫色, 无 $\\mathrm{B}$ 时为蓝色。某浅紫色植株自交, $\\mathrm{F}_{1}$ 中出现蓝色: 浅紫色:黄色: 白色=1：4: 3: 4。下列相关判断正确的是（ ）\n\nA: $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 两对基因的遗传遵循自由组合定律，亲本基因型为 $\\mathrm{AABb}$\nB: $F_{1}$ 出现上述表现型及比例的原因是基因型为 $a B$ 的雄配子或雌配子致死\nC: 让 $\\mathrm{F}_{1}$ 中黄色植株随机受粉, 子代表现型及比例为黄色: 白色 $=3: 1$\nD: 让 $F_{1}$ 中的浅紫色植株与白色植株杂交，可判断浅紫色植株的基因型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1143",
        "problem": "A male Siamese fighting fish will respond to a model fish by producing a threat display. With continued presentation of the model fish the response is reduced. This is an example of\nA: classical conditioning\nB: imprinting\nC: operant conditioning\nD: habituation\nE: a simple reflex\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA male Siamese fighting fish will respond to a model fish by producing a threat display. With continued presentation of the model fish the response is reduced. This is an example of\n\nA: classical conditioning\nB: imprinting\nC: operant conditioning\nD: habituation\nE: a simple reflex\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_911",
        "problem": "果蝇的卷翅 (A) 对正常翅（a）完全显性, 裂翅（B）对非裂翅（b）完全显性,两雌雄果蝇杂交过程如下图, 含有两条缺失染色体的个体不能存活, 不考虑染色体互换及基因突变等变异。下列说法正确的是（）\n\n[图1]\nA: 亲本雄果蝇发生了染色体的缺失\nB: $F_{1}$ 中卷翅裂翅雄果蝇约占 $1 / 4$\nC: $\\mathrm{F}_{1}$ 中雌性个体能产生 5 种配子\nD: $F_{2}$ 中卷翅裂翅个体约占 $3 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的卷翅 (A) 对正常翅（a）完全显性, 裂翅（B）对非裂翅（b）完全显性,两雌雄果蝇杂交过程如下图, 含有两条缺失染色体的个体不能存活, 不考虑染色体互换及基因突变等变异。下列说法正确的是（）\n\n[图1]\n\nA: 亲本雄果蝇发生了染色体的缺失\nB: $F_{1}$ 中卷翅裂翅雄果蝇约占 $1 / 4$\nC: $\\mathrm{F}_{1}$ 中雌性个体能产生 5 种配子\nD: $F_{2}$ 中卷翅裂翅个体约占 $3 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-59.jpg?height=666&width=677&top_left_y=495&top_left_x=404"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_523",
        "problem": "噬菌体侵染细菌时, 子代噬菌体可组入细菌 DNA 片段, 在进一步侵染其他细菌时将供体菌的遗传物质整合到受体菌的 DNA 中, 这种现象称为噬菌体转导。下列有关叙述正确的是 ( )\nA: 噬菌体可利用细菌的 DNA 为模板合成子代噬菌体的 DNA\nB: 通过噬菌体转导可实现任意细菌之间的基因重组\nC: 用 ${ }^{32} \\mathrm{P}$ 标记噬菌体的 DNA 可证明噬菌体转导的存在\nD: 噬菌体侵染导致供体菌裂解才能成功实现噬菌体转导\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n噬菌体侵染细菌时, 子代噬菌体可组入细菌 DNA 片段, 在进一步侵染其他细菌时将供体菌的遗传物质整合到受体菌的 DNA 中, 这种现象称为噬菌体转导。下列有关叙述正确的是 ( )\n\nA: 噬菌体可利用细菌的 DNA 为模板合成子代噬菌体的 DNA\nB: 通过噬菌体转导可实现任意细菌之间的基因重组\nC: 用 ${ }^{32} \\mathrm{P}$ 标记噬菌体的 DNA 可证明噬菌体转导的存在\nD: 噬菌体侵染导致供体菌裂解才能成功实现噬菌体转导\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1330",
        "problem": "The graph shows the results of an experiment in which adult beetles of a species of Tribolium were starved and desiccated and then offered a choice of moist and dry air in a choice chamber.\n\nThe BEST interpretation of these results is that\n\n[figure1]\nA: The treatment reduces the beetles' response to humidity differences.\nB: The treatment reverses the beetles' normal preference for dry air.\nC: The stimulus of hunger overrides the stimulus of humidity.\nD: The stimulus of humidity overrides the stimulus of hunger.\nE: Desiccation leads to ALL beetles choosing a moist environment.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows the results of an experiment in which adult beetles of a species of Tribolium were starved and desiccated and then offered a choice of moist and dry air in a choice chamber.\n\nThe BEST interpretation of these results is that\n\n[figure1]\n\nA: The treatment reduces the beetles' response to humidity differences.\nB: The treatment reverses the beetles' normal preference for dry air.\nC: The stimulus of hunger overrides the stimulus of humidity.\nD: The stimulus of humidity overrides the stimulus of hunger.\nE: Desiccation leads to ALL beetles choosing a moist environment.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-08.jpg?height=657&width=697&top_left_y=377&top_left_x=1088"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1270",
        "problem": "Understanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.The most important food type for orange-fronted parakeets on Maud Island was?\nA: Leaves.\nB: Flowers.\nC: Fruit.\nD: Invertebrates.\nE: None of the above.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nUnderstanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.\n\nproblem:\nThe most important food type for orange-fronted parakeets on Maud Island was?\n\nA: Leaves.\nB: Flowers.\nC: Fruit.\nD: Invertebrates.\nE: None of the above.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_474",
        "problem": "某雌雄同株植物的花色有紫色和蓝色两种。为了研究其遗传机制, 研究者利用纯系品种进行了杂交实验，结果见表，下列叙述错误的是\n\n| 杂交组合 | 父本植株数目 <br> （表现性） | 母本植株数目 <br> （表现性） | $\\mathrm{F}_{1}$ 植株数目 <br> (表现性) | $\\mathrm{F}_{2}$ 植株数目 <br> （表现性） |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| I | 10 (紫色) | 10 (紫色) | 81（紫色） | 260 (紫色) | 61 (蓝色) |\n| II | 10 (紫色) | 10（蓝色） | 79 (紫色) | 270 (紫色) | 89 (蓝色) |\nA: 通过 I 判断, 紫色和蓝色这对相对性状至少由两对等位基因控制\nB: 将 I、II 中的 $\\mathrm{F}_{2}$ 紫色植株相互杂交, 产生的后代中紫色和蓝色的比例为 36:5\nC: 取杂交 II 中的 $\\mathrm{F}_{2}$ 紫色植株随机交配, 产生的后代紫色和蓝色的比例为 8: 1\nD: 将两个杂交组合中的 $\\mathrm{F}_{1}$ 相互杂交, 产生的后代紫色和蓝色的比例为 3:1\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某雌雄同株植物的花色有紫色和蓝色两种。为了研究其遗传机制, 研究者利用纯系品种进行了杂交实验，结果见表，下列叙述错误的是\n\n| 杂交组合 | 父本植株数目 <br> （表现性） | 母本植株数目 <br> （表现性） | $\\mathrm{F}_{1}$ 植株数目 <br> (表现性) | $\\mathrm{F}_{2}$ 植株数目 <br> （表现性） |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| I | 10 (紫色) | 10 (紫色) | 81（紫色） | 260 (紫色) | 61 (蓝色) |\n| II | 10 (紫色) | 10（蓝色） | 79 (紫色) | 270 (紫色) | 89 (蓝色) |\n\nA: 通过 I 判断, 紫色和蓝色这对相对性状至少由两对等位基因控制\nB: 将 I、II 中的 $\\mathrm{F}_{2}$ 紫色植株相互杂交, 产生的后代中紫色和蓝色的比例为 36:5\nC: 取杂交 II 中的 $\\mathrm{F}_{2}$ 紫色植株随机交配, 产生的后代紫色和蓝色的比例为 8: 1\nD: 将两个杂交组合中的 $\\mathrm{F}_{1}$ 相互杂交, 产生的后代紫色和蓝色的比例为 3:1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1013",
        "problem": "Which of the following is NOT an adaptation to survival in a grassland biome?\nA: Grasses have basal meristems that allow them to grow from the base of the plant instead of the tip.\nB: Trees retain their moisture below the ground and have bark that lacks resin or cork.\nC: Grasses have deep roots to anchor them and provide starch reserves.\nD: Grasses have bulliform cells that can change in turgor pressure allowing the plant to furl its blade and shelter its stomata.\nE: Grazing animals have long distance vision with eyes placed well above their snouts.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is NOT an adaptation to survival in a grassland biome?\n\nA: Grasses have basal meristems that allow them to grow from the base of the plant instead of the tip.\nB: Trees retain their moisture below the ground and have bark that lacks resin or cork.\nC: Grasses have deep roots to anchor them and provide starch reserves.\nD: Grasses have bulliform cells that can change in turgor pressure allowing the plant to furl its blade and shelter its stomata.\nE: Grazing animals have long distance vision with eyes placed well above their snouts.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_442",
        "problem": "下列有关计算中, 不正确的是（ ）\nA: 用 ${ }^{32} \\mathrm{P}$ 标记的噬菌体在不含放射性的大肠杆菌内增殖 3 代, 具有放射性的噬菌体占总数为 $1 / 4$\nB: 某蛋白质分子中含有 120 个氨基酸，则控制合成该蛋白质的基因中至少有 720 个碱基\nC: 某DNA片段有 300 个碱基对, 其中一条链上 A+T 比例为 40\\%: 则复制三次该 DNA 片段时总共需要 720 个胞嘧啶脱氧核苷酸\nD: 用 ${ }^{15} \\mathrm{~N}$ 标记的精原细胞（含 8 条染色体）在含 ${ }^{14} \\mathrm{~N}$ 的培养基中培养, 减数第一次分裂后期和减数第二次分裂后期含 ${ }^{15} \\mathrm{~N}$ 的染色体数分别是 8 和 8\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列有关计算中, 不正确的是（ ）\n\nA: 用 ${ }^{32} \\mathrm{P}$ 标记的噬菌体在不含放射性的大肠杆菌内增殖 3 代, 具有放射性的噬菌体占总数为 $1 / 4$\nB: 某蛋白质分子中含有 120 个氨基酸，则控制合成该蛋白质的基因中至少有 720 个碱基\nC: 某DNA片段有 300 个碱基对, 其中一条链上 A+T 比例为 40\\%: 则复制三次该 DNA 片段时总共需要 720 个胞嘧啶脱氧核苷酸\nD: 用 ${ }^{15} \\mathrm{~N}$ 标记的精原细胞（含 8 条染色体）在含 ${ }^{14} \\mathrm{~N}$ 的培养基中培养, 减数第一次分裂后期和减数第二次分裂后期含 ${ }^{15} \\mathrm{~N}$ 的染色体数分别是 8 和 8\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_807",
        "problem": "下列有关大肠杆菌的叙述, 正确的是 ( )\nA. 大肠杆菌拟核的 DNA 中有控制性状的基因\nB. 大肠杆菌中 DNA 分子数目与基因数目相同\nC. 在普通光学显微镜下能观察到大肠杆菌的核糖体\nD. 大肠杆菌分泌的蛋白, 需要经过内质网加工\nA: 大肠杆菌拟核的 DNA 中有控制性状的基因\nB: 大肠杆菌中 DNA 分子数目与基因数目相同\nC: 在普通光学显微镜下能观察到大肠杆菌的核糖体\nD: 大肠杆菌分泌的蛋白, 需要经过内质网加工\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列有关大肠杆菌的叙述, 正确的是 ( )\nA. 大肠杆菌拟核的 DNA 中有控制性状的基因\nB. 大肠杆菌中 DNA 分子数目与基因数目相同\nC. 在普通光学显微镜下能观察到大肠杆菌的核糖体\nD. 大肠杆菌分泌的蛋白, 需要经过内质网加工\n\nA: 大肠杆菌拟核的 DNA 中有控制性状的基因\nB: 大肠杆菌中 DNA 分子数目与基因数目相同\nC: 在普通光学显微镜下能观察到大肠杆菌的核糖体\nD: 大肠杆菌分泌的蛋白, 需要经过内质网加工\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_176",
        "problem": "A protocol that can be used for identification of gene structure with respect to exons and introns involves isolation and denaturation of a dsDNA fragment that includes a gene of interest, isolation of mature mRNA pertaining to gene of interest, hybridization of complementary single stranded nucleic acid molecules, performance of three types of chemical or enzymatic reactions, and electrophoresis on non-denaturing gels. Relevant factors pertaining the chemical or enzymatic reactions are as follows:\n\n- S1: a nuclease that degrades single stranded (ss) nucleic acid molecules or regions that are single stranded. It does not degrade double stranded (ds) molecules or double stranded regions.\n- Exonuclease VII: an exonuclease that degrades ss nucleic acids or ss ends of hybridized nucleic acids in both $5^{\\prime}$ to $3^{\\prime}$ and $3^{\\prime}$ to $5^{\\prime}$ directions. It does not degrade ds molecules or ds portions of molecules.\n- Alkali conditions degrade RNA and do not degrade DNA.\n\nExperiments pertaining to a gene that does not experience alternative splicing were as follows; appropriate size markers were included in the electrophoresis steps:\n\nA. Hybridization of denatured DNA with excess mRNA, followed by S1 treatment, followed by alkali treatment, followed by electrophoresis\n\nB. Hybridization of denatured DNA with excess mRNA, followed by S1 treatment, followed by electrophoresis\n\nC. Hybridization of denatured DNA with excess mRNA, followed by Exonuclease VII treatment, followed by alkali treatment, followed by electrophoresis\nA: Observation of three bands in electrophoresis pattern of reaction A signifies that the gene of interest has three exons.\nB: The result of reaction A allows prediction of number and size of bands to be seen after electrophoresis in reaction $B$.\nC: The result of reaction B allows estimation of size of gene of interest.\nD: The combined results of reactions A and B allow prediction of the size of the introns of the gene.\nE: The combined results of the three reactions $\\mathrm{A}, \\mathrm{B}$, and $\\mathrm{C}$ will allow prediction of length of ORF (open reading frame) of gene of interest.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA protocol that can be used for identification of gene structure with respect to exons and introns involves isolation and denaturation of a dsDNA fragment that includes a gene of interest, isolation of mature mRNA pertaining to gene of interest, hybridization of complementary single stranded nucleic acid molecules, performance of three types of chemical or enzymatic reactions, and electrophoresis on non-denaturing gels. Relevant factors pertaining the chemical or enzymatic reactions are as follows:\n\n- S1: a nuclease that degrades single stranded (ss) nucleic acid molecules or regions that are single stranded. It does not degrade double stranded (ds) molecules or double stranded regions.\n- Exonuclease VII: an exonuclease that degrades ss nucleic acids or ss ends of hybridized nucleic acids in both $5^{\\prime}$ to $3^{\\prime}$ and $3^{\\prime}$ to $5^{\\prime}$ directions. It does not degrade ds molecules or ds portions of molecules.\n- Alkali conditions degrade RNA and do not degrade DNA.\n\nExperiments pertaining to a gene that does not experience alternative splicing were as follows; appropriate size markers were included in the electrophoresis steps:\n\nA. Hybridization of denatured DNA with excess mRNA, followed by S1 treatment, followed by alkali treatment, followed by electrophoresis\n\nB. Hybridization of denatured DNA with excess mRNA, followed by S1 treatment, followed by electrophoresis\n\nC. Hybridization of denatured DNA with excess mRNA, followed by Exonuclease VII treatment, followed by alkali treatment, followed by electrophoresis\n\nA: Observation of three bands in electrophoresis pattern of reaction A signifies that the gene of interest has three exons.\nB: The result of reaction A allows prediction of number and size of bands to be seen after electrophoresis in reaction $B$.\nC: The result of reaction B allows estimation of size of gene of interest.\nD: The combined results of reactions A and B allow prediction of the size of the introns of the gene.\nE: The combined results of the three reactions $\\mathrm{A}, \\mathrm{B}$, and $\\mathrm{C}$ will allow prediction of length of ORF (open reading frame) of gene of interest.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_950",
        "problem": "野生型拟南芥的叶片是光滑形边缘, 研究影响其叶片形状的基因时, 发现了 6 个不同的隐性突变, 每个隐性突变只涉及 1 个基因。这些突变都能使拟南芥的叶片表现为锯齿状边缘。利用上述突变培育成 6 个不同纯合突变体(1) (6), 每个突变体只有 1 种隐性\n突变。不考虑其他突变, 根据表中的杂交实验结果, 下列推断正确的是（）\n\n| 杂交组合 | 子代叶片边缘 |\n| :---: | :---: |\n| $1 \\times 2$ | 光滑形 |\n| $1 \\times \\times 3$ | 锯齿状 |\n| $1 \\times \\times 4$ | 锯齿状 |\n| $1 \\times \\times 5$ | 光滑形 |\n| (2) $\\times 6$ | 锯齿状 |\nA: 导致(1)锯齿的突变基因与导致(6)锯齿的突变基因为等位基因\nB: (3)和(4)的后代与(1)和(2)的后代杂交, 子代全为叶片边缘光滑形\nC: (1)和(2)的后代与(1)和(5)的后代杂交, 子代叶片边缘锯齿的概率为 $1 / 2$ 或 $1 / 4$ 或 $7 / 16$\nD: 若(2)和(5)杂交, 子代叶片边缘为光滑形, 说明突变体(5)的突变基因与其他突变体突变基因不在同源染色体上\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n野生型拟南芥的叶片是光滑形边缘, 研究影响其叶片形状的基因时, 发现了 6 个不同的隐性突变, 每个隐性突变只涉及 1 个基因。这些突变都能使拟南芥的叶片表现为锯齿状边缘。利用上述突变培育成 6 个不同纯合突变体(1) (6), 每个突变体只有 1 种隐性\n突变。不考虑其他突变, 根据表中的杂交实验结果, 下列推断正确的是（）\n\n| 杂交组合 | 子代叶片边缘 |\n| :---: | :---: |\n| $1 \\times 2$ | 光滑形 |\n| $1 \\times \\times 3$ | 锯齿状 |\n| $1 \\times \\times 4$ | 锯齿状 |\n| $1 \\times \\times 5$ | 光滑形 |\n| (2) $\\times 6$ | 锯齿状 |\n\nA: 导致(1)锯齿的突变基因与导致(6)锯齿的突变基因为等位基因\nB: (3)和(4)的后代与(1)和(2)的后代杂交, 子代全为叶片边缘光滑形\nC: (1)和(2)的后代与(1)和(5)的后代杂交, 子代叶片边缘锯齿的概率为 $1 / 2$ 或 $1 / 4$ 或 $7 / 16$\nD: 若(2)和(5)杂交, 子代叶片边缘为光滑形, 说明突变体(5)的突变基因与其他突变体突变基因不在同源染色体上\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1324",
        "problem": "Which diagram below does not fit the logical series?\n\n[figure1]\nA: A\nB: B\nC: C\nD: D\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich diagram below does not fit the logical series?\n\n[figure1]\n\nA: A\nB: B\nC: C\nD: D\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-17.jpg?height=291&width=1836&top_left_y=1351&top_left_x=116"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1405",
        "problem": "The genetic sequence of the virus that causes COVID-19 was compared with virus samples from different non-human hosts and from environmentally collected strains and all samples were plotted against the average similarity of the group. The COVID-19 virus found in the environment is very similar to that from humans, likely reflecting the current high levels of transmission within the human population and shedding into the environment.\n\n[figure1]\n\nFrom the list below, what is the best inference that can be made from these data? The COVID19 virus\nA: evolved spontaneously in the environment, with no clear host\nB: has its closest evolutionary link to a pangolin virus\nC: has its closest evolutionary link to a bat (Rhinolophine) virus\nD: has its closest evolutionary link to a Dengue virus\nE: evolved from the common cold virus (HCov - 229E)\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe genetic sequence of the virus that causes COVID-19 was compared with virus samples from different non-human hosts and from environmentally collected strains and all samples were plotted against the average similarity of the group. The COVID-19 virus found in the environment is very similar to that from humans, likely reflecting the current high levels of transmission within the human population and shedding into the environment.\n\n[figure1]\n\nFrom the list below, what is the best inference that can be made from these data? The COVID19 virus\n\nA: evolved spontaneously in the environment, with no clear host\nB: has its closest evolutionary link to a pangolin virus\nC: has its closest evolutionary link to a bat (Rhinolophine) virus\nD: has its closest evolutionary link to a Dengue virus\nE: evolved from the common cold virus (HCov - 229E)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-36.jpg?height=800&width=1399&top_left_y=465&top_left_x=317"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_170",
        "problem": "Glucose-6-phosphate dehydrogenase (G6PDH) is encoded by a single X-linked gene in humans. There are multiple functional alleles for this gene, such as $A_{1}, A_{2}$, etc. G6PDH dimers are made in cells and secreted into blood.\n\nIf a woman has both $\\mathrm{A}_{1}$ and $\\mathrm{A}_{2}$ allele for G6PDH, what type(s) of dimers is/are found in her blood?\nA: Only $\\mathrm{A}_{1} \\mathrm{~A}_{1}$\nB: Only $\\mathrm{A}_{2} \\mathrm{~A}_{2}$\nC: Only $\\mathrm{A}_{1} \\mathrm{~A}_{2}$\nD: Only $\\mathrm{A}_{1} \\mathrm{~A}_{1}$ and $\\mathrm{A}_{2} \\mathrm{~A}_{2}$\nE: $\\mathrm{A}_{1} \\mathrm{~A}_{1}, \\mathrm{~A}_{2} \\mathrm{~A}_{2}$ and $\\mathrm{A}_{1} \\mathrm{~A}_{2}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGlucose-6-phosphate dehydrogenase (G6PDH) is encoded by a single X-linked gene in humans. There are multiple functional alleles for this gene, such as $A_{1}, A_{2}$, etc. G6PDH dimers are made in cells and secreted into blood.\n\nIf a woman has both $\\mathrm{A}_{1}$ and $\\mathrm{A}_{2}$ allele for G6PDH, what type(s) of dimers is/are found in her blood?\n\nA: Only $\\mathrm{A}_{1} \\mathrm{~A}_{1}$\nB: Only $\\mathrm{A}_{2} \\mathrm{~A}_{2}$\nC: Only $\\mathrm{A}_{1} \\mathrm{~A}_{2}$\nD: Only $\\mathrm{A}_{1} \\mathrm{~A}_{1}$ and $\\mathrm{A}_{2} \\mathrm{~A}_{2}$\nE: $\\mathrm{A}_{1} \\mathrm{~A}_{1}, \\mathrm{~A}_{2} \\mathrm{~A}_{2}$ and $\\mathrm{A}_{1} \\mathrm{~A}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_432",
        "problem": "下图是甲、乙两种单基因遗传病系谱图, 4 号不携带甲病致病基因, 其双亲均携带一个单碱基替换导致的乙病基因, 且突变位点不同。对家庭部分成员一对同源染色体上\n控制乙病的基因进行测序, 非模板链测序结果见下表。不考虑 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段,以下分析不正确的是（ ）\n\n[图1]\n\n| 家庭成员 | 1 | 2 | 4 | 5 |\n| :---: | :---: | :---: | :---: | :---: |\n| 测序结果 | .....G.....A...... <br> .....A.....A...... <br> 第 412 位第 420 <br> 位 | .....G............ <br> 第 412 位 第 420 <br> 位 | ? | .....G.....A...... <br> .....G.....A...... <br> 第 412 位第 420 <br> 位 |\nA: 无法判断甲病的显隐性\nB: 乙病基因位于 X 染色体\nC: 4 号控制乙病的基因测序结果为 $\\cdots \\cdots \\cdots A \\cdots \\cdot A \\cdots \\cdot . \\cdots$ ## 第412位 第420位\nD: 1 号和 2 号生一个不患乙病孩子的概率是 $3 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图是甲、乙两种单基因遗传病系谱图, 4 号不携带甲病致病基因, 其双亲均携带一个单碱基替换导致的乙病基因, 且突变位点不同。对家庭部分成员一对同源染色体上\n控制乙病的基因进行测序, 非模板链测序结果见下表。不考虑 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段,以下分析不正确的是（ ）\n\n[图1]\n\n| 家庭成员 | 1 | 2 | 4 | 5 |\n| :---: | :---: | :---: | :---: | :---: |\n| 测序结果 | .....G.....A...... <br> .....A.....A...... <br> 第 412 位第 420 <br> 位 | .....G............ <br> 第 412 位 第 420 <br> 位 | ? | .....G.....A...... <br> .....G.....A...... <br> 第 412 位第 420 <br> 位 |\n\nA: 无法判断甲病的显隐性\nB: 乙病基因位于 X 染色体\nC: 4 号控制乙病的基因测序结果为 $\\cdots \\cdots \\cdots A \\cdots \\cdot A \\cdots \\cdot . \\cdots$ ## 第412位 第420位\nD: 1 号和 2 号生一个不患乙病孩子的概率是 $3 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-78.jpg?height=305&width=643&top_left_y=356&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-78.jpg?height=117&width=222&top_left_y=2462&top_left_x=1431"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_258",
        "problem": "Following is the phylogenetic tree based on the amino-acid sequences of all opsin genes in the human and zebrafish genomes.\n\n[figure1]\n\nFigure 1. A phylogenetic tree based on the amino-acid sequences of all opsin genes in the human and zebrafish genomes. LWS: Long Wavelength Sensitive opsin, MWS: Middle Wavelength Sensitive opsin, SWS: Short Wavelength Sensitive opsin, UVS: Ultra Violet Sensitive opsin, RH: Rhodopsin type opsin. Branch length is not proportional to sequence divergence nor time.\nA: In the phylogenetic tree, zebrafish SWS is most closely related to RH.\nB: The common ancestor of the human and zebrafish had four opsin genes in the genome.\nC: The opsin gene at node I was supposed to encode an LWS.\nD: The zebrafish has acquired five opsin genes after splitting from the human.\nE: The human has lost two opsin genes after splitting from the zebrafish.\nF: The common ancestor of the human and zebrafish did not have any SWS in the genome.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFollowing is the phylogenetic tree based on the amino-acid sequences of all opsin genes in the human and zebrafish genomes.\n\n[figure1]\n\nFigure 1. A phylogenetic tree based on the amino-acid sequences of all opsin genes in the human and zebrafish genomes. LWS: Long Wavelength Sensitive opsin, MWS: Middle Wavelength Sensitive opsin, SWS: Short Wavelength Sensitive opsin, UVS: Ultra Violet Sensitive opsin, RH: Rhodopsin type opsin. Branch length is not proportional to sequence divergence nor time.\n\nA: In the phylogenetic tree, zebrafish SWS is most closely related to RH.\nB: The common ancestor of the human and zebrafish had four opsin genes in the genome.\nC: The opsin gene at node I was supposed to encode an LWS.\nD: The zebrafish has acquired five opsin genes after splitting from the human.\nE: The human has lost two opsin genes after splitting from the zebrafish.\nF: The common ancestor of the human and zebrafish did not have any SWS in the genome.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E, F].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-51.jpg?height=948&width=1510&top_left_y=657&top_left_x=296"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_566",
        "problem": "二倍体酵母经过减数分裂形成的子囊孢子可直接发育为单倍体酵母。科学家在二倍体酵母的III 号染色体上发现了 $\\mathrm{A}$ 和 $\\mathrm{B}$ 两个基因, 它们会抑制没有这两个基因的子囊胞子的发育。科学家测定了一批双杂合酵母（基因型记为 $\\mathrm{A} 0 / \\mathrm{B} 0$ ，“ 0 ”表示该基因缺失）减数分裂形成的孢子的存活力, 结果如图所示。下列叙述正确的是（）\n\n[图1]\n\n孢子基因型\nA: A 和 B 两对基因的遗传均不符合基因的分离定律\nB: $\\mathrm{AB}$ 基因全部缺失的酵母不能产生正常存活力的子囊孢子\nC: $\\mathrm{AO} / \\mathrm{BB}$ 酵母的单倍体后代中两种基因型比例为 $1: 1$\nD: 以上机制有利于提高 $\\mathrm{A}$ 基因和 $\\mathrm{B}$ 基因其在后代中的频率\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n二倍体酵母经过减数分裂形成的子囊孢子可直接发育为单倍体酵母。科学家在二倍体酵母的III 号染色体上发现了 $\\mathrm{A}$ 和 $\\mathrm{B}$ 两个基因, 它们会抑制没有这两个基因的子囊胞子的发育。科学家测定了一批双杂合酵母（基因型记为 $\\mathrm{A} 0 / \\mathrm{B} 0$ ，“ 0 ”表示该基因缺失）减数分裂形成的孢子的存活力, 结果如图所示。下列叙述正确的是（）\n\n[图1]\n\n孢子基因型\n\nA: A 和 B 两对基因的遗传均不符合基因的分离定律\nB: $\\mathrm{AB}$ 基因全部缺失的酵母不能产生正常存活力的子囊孢子\nC: $\\mathrm{AO} / \\mathrm{BB}$ 酵母的单倍体后代中两种基因型比例为 $1: 1$\nD: 以上机制有利于提高 $\\mathrm{A}$ 基因和 $\\mathrm{B}$ 基因其在后代中的频率\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-018.jpg?height=334&width=920&top_left_y=1472&top_left_x=385"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_673",
        "problem": "下列关于 DNA 分子的结构和复制的叙述, 正确的是 ( )\nA: DNA 分子都含有两个游离的磷酸基团\nB: 碱基排列顺序的千变万化, 构成了 DNA 分子的特异性\nC: 若用同位素标记原料, DNA 分子复制 $\\mathrm{n}$ 次后，含标记的 DNA 分子占 $100 \\%$,含标记的单链占 $1-2 / 2^{n}$\nD: 某个 DNA 片段由 500 对碱基组成, A+T 占碱基总数的 34\\%, 若该 DNA 分子片段复制 3 次, 共需游离的胞嘧啶脱氧核苷酸个数为 2310 个\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列关于 DNA 分子的结构和复制的叙述, 正确的是 ( )\n\nA: DNA 分子都含有两个游离的磷酸基团\nB: 碱基排列顺序的千变万化, 构成了 DNA 分子的特异性\nC: 若用同位素标记原料, DNA 分子复制 $\\mathrm{n}$ 次后，含标记的 DNA 分子占 $100 \\%$,含标记的单链占 $1-2 / 2^{n}$\nD: 某个 DNA 片段由 500 对碱基组成, A+T 占碱基总数的 34\\%, 若该 DNA 分子片段复制 3 次, 共需游离的胞嘧啶脱氧核苷酸个数为 2310 个\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_453",
        "problem": "西瓜 (雌雄同株) 黄色果肉和白色果肉为一对相对性状, 为研究西瓜果肉颜色的遗传特点, 研究人员将黄色果肉西瓜 $\\left(\\mathrm{P}_{1}\\right)$ 与白色果肉西瓜 $\\left(\\mathrm{P}_{2}\\right)$ 纯合亲本进行杂交, $\\mathrm{F}_{1}$皆为黄色果肉, $F_{1}$ 自交得到的 $F_{2}$ 中, 黄色果肉 106 株, 白色果肉 31 株。染色体上有一些非编码重复序列 (SSR), 不同亲本来源的染色体上 SSR 通常不同。利用 PCR 扩增亲本 $\\mathrm{P}_{1} 、 \\mathrm{P}_{2}$ 与 $\\mathrm{F}_{2}$ 中的白肉个体的 $4 、 6$ 号染色体 DNA 中的一段 SSR, 电泳后结果分别如图甲、图乙。下列有关实验目的或结论的叙述合理的是（ ）\n\n[图1]\n\n图甲: 4 号染色体SSR电泳结果\n\n[图2]\n\n图乙: 6号染色体SSR电泳结果\nA: 还需进一步通过正反交实验以判断西瓜果肉颜色基因是否位于性染色体上\nB: 依据 $\\mathrm{F}_{1}$ 自交结果可以推断控制西瓜果肉颜色基因的遗传遵循自由组合定律\nC: 图中 8 号白色果肉西瓜的 4 号染色体都来自亲本 $\\mathrm{P}_{2}$\nD: 依据电泳结果可推断西瓜果肉颜色基因位于 6 号染色体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n西瓜 (雌雄同株) 黄色果肉和白色果肉为一对相对性状, 为研究西瓜果肉颜色的遗传特点, 研究人员将黄色果肉西瓜 $\\left(\\mathrm{P}_{1}\\right)$ 与白色果肉西瓜 $\\left(\\mathrm{P}_{2}\\right)$ 纯合亲本进行杂交, $\\mathrm{F}_{1}$皆为黄色果肉, $F_{1}$ 自交得到的 $F_{2}$ 中, 黄色果肉 106 株, 白色果肉 31 株。染色体上有一些非编码重复序列 (SSR), 不同亲本来源的染色体上 SSR 通常不同。利用 PCR 扩增亲本 $\\mathrm{P}_{1} 、 \\mathrm{P}_{2}$ 与 $\\mathrm{F}_{2}$ 中的白肉个体的 $4 、 6$ 号染色体 DNA 中的一段 SSR, 电泳后结果分别如图甲、图乙。下列有关实验目的或结论的叙述合理的是（ ）\n\n[图1]\n\n图甲: 4 号染色体SSR电泳结果\n\n[图2]\n\n图乙: 6号染色体SSR电泳结果\n\nA: 还需进一步通过正反交实验以判断西瓜果肉颜色基因是否位于性染色体上\nB: 依据 $\\mathrm{F}_{1}$ 自交结果可以推断控制西瓜果肉颜色基因的遗传遵循自由组合定律\nC: 图中 8 号白色果肉西瓜的 4 号染色体都来自亲本 $\\mathrm{P}_{2}$\nD: 依据电泳结果可推断西瓜果肉颜色基因位于 6 号染色体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-64.jpg?height=206&width=691&top_left_y=982&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-64.jpg?height=206&width=677&top_left_y=982&top_left_x=1038"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_71",
        "problem": "Using sequence differences to establish phylogenies has some advantages and possible dangers. An inappropriate choice of molecule could result in molecular trees that greatly distort true phylogenetic relationships. Hence, care must be taken in using this approach.\nA: Cytochrome c molecules are very useful for establishing evolutionary relationships between closely related species.\nB: Comparing subunit sequences from many different genes is better than comparing whole-gene sequences of the organisms under phylogenetic study.\nC: Rates of nucleotide substitution are faster in organisms with short generation times than in organisms with long generation times.\nD: The Neutral Theory requires that all polypeptide and DNA sequences evolve at the same rate.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nUsing sequence differences to establish phylogenies has some advantages and possible dangers. An inappropriate choice of molecule could result in molecular trees that greatly distort true phylogenetic relationships. Hence, care must be taken in using this approach.\n\nA: Cytochrome c molecules are very useful for establishing evolutionary relationships between closely related species.\nB: Comparing subunit sequences from many different genes is better than comparing whole-gene sequences of the organisms under phylogenetic study.\nC: Rates of nucleotide substitution are faster in organisms with short generation times than in organisms with long generation times.\nD: The Neutral Theory requires that all polypeptide and DNA sequences evolve at the same rate.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_655",
        "problem": "油菜是我国重要的油料作物, 油菜株高适当降低对抗倒伏及机械化收割均有重要意义。某研究小组利用纯种高秆油菜 $\\mathrm{Z}$, 通过诱变育种培育出一个纯种半矮秆突变体 $\\mathrm{S}$ 并进行了相关试验，如下图所示，下列说法正确的是（）\n[图1]\nA: 杂交组合(1)的 $F_{1}$ 自交时, 雌雄配子有 4 种结合方式, 且每种结合方式几率相等\nB: 杂交组合(2)的 $F_{2}$ 中所有高秆植株自交，后代不发生性状分离的个体占 $1 / 9$\nC: 杂交组合(3)的 $F_{2}$ 高秆：半矮秆=3：1, 表明控制该性状的基因不遵循自由组合定律\nD: 杂交组合(3)的 $\\mathrm{F}_{2}$ 中所有高秆分别与半矮秆植株杂交, 后代中半矮秆个体占 $5 / 12$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n油菜是我国重要的油料作物, 油菜株高适当降低对抗倒伏及机械化收割均有重要意义。某研究小组利用纯种高秆油菜 $\\mathrm{Z}$, 通过诱变育种培育出一个纯种半矮秆突变体 $\\mathrm{S}$ 并进行了相关试验，如下图所示，下列说法正确的是（）\n[图1]\n\nA: 杂交组合(1)的 $F_{1}$ 自交时, 雌雄配子有 4 种结合方式, 且每种结合方式几率相等\nB: 杂交组合(2)的 $F_{2}$ 中所有高秆植株自交，后代不发生性状分离的个体占 $1 / 9$\nC: 杂交组合(3)的 $F_{2}$ 高秆：半矮秆=3：1, 表明控制该性状的基因不遵循自由组合定律\nD: 杂交组合(3)的 $\\mathrm{F}_{2}$ 中所有高秆分别与半矮秆植株杂交, 后代中半矮秆个体占 $5 / 12$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-25.jpg?height=296&width=1224&top_left_y=936&top_left_x=380"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_792",
        "problem": "下图甲为某二倍体动物的细胞分裂图像, 图乙为图甲所示细胞分裂过程中细胞内相关物质或结构数量变化的部分曲线图。下列有关叙述正确的是（ ）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\nA: 图甲细胞发生的变异是基因突变, 细胞中含有两个四分体\nB: 图乙 I 和II时期中，细胞内都存在 A 基因\nC: 图乙中曲线变化可表示某时期染色单体数的变化\nD: 若图乙表示染色体组的数目变化, 则 $\\mathrm{a}$ 值为 2\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图甲为某二倍体动物的细胞分裂图像, 图乙为图甲所示细胞分裂过程中细胞内相关物质或结构数量变化的部分曲线图。下列有关叙述正确的是（ ）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\nA: 图甲细胞发生的变异是基因突变, 细胞中含有两个四分体\nB: 图乙 I 和II时期中，细胞内都存在 A 基因\nC: 图乙中曲线变化可表示某时期染色单体数的变化\nD: 若图乙表示染色体组的数目变化, 则 $\\mathrm{a}$ 值为 2\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-03.jpg?height=448&width=601&top_left_y=744&top_left_x=362",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-03.jpg?height=383&width=666&top_left_y=822&top_left_x=1112"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1204",
        "problem": "Measurements of the rate of diffusion through stomata ('stomatal conductance') were made under various $\\mathrm{CO}_{2}$ concentrations, and the following stereograph was drawn.\n\n[figure1]\n\nWhich one of the following is a valid deduction from the stereograph?\nA: Stomatal conductance is maximal when conditions for photosynthesis are optimal.\nB: Increasing light intensity increases stomatal conductance at all $\\mathrm{CO}_{2}$ levels shown.\nC: Stomatal conductance is unaffected by an increase in $\\mathrm{CO}_{2}$ concentration from 170 to 810 p.p.m.\nD: Changes in $\\mathrm{CO}_{2}$ concentration have a greater influence on stomatal conductance than changes in light intensity.\nE: Changes in light intensity have a greater influence on stomatal conductance than changes in $\\mathrm{CO}_{2}$ concentration.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMeasurements of the rate of diffusion through stomata ('stomatal conductance') were made under various $\\mathrm{CO}_{2}$ concentrations, and the following stereograph was drawn.\n\n[figure1]\n\nWhich one of the following is a valid deduction from the stereograph?\n\nA: Stomatal conductance is maximal when conditions for photosynthesis are optimal.\nB: Increasing light intensity increases stomatal conductance at all $\\mathrm{CO}_{2}$ levels shown.\nC: Stomatal conductance is unaffected by an increase in $\\mathrm{CO}_{2}$ concentration from 170 to 810 p.p.m.\nD: Changes in $\\mathrm{CO}_{2}$ concentration have a greater influence on stomatal conductance than changes in light intensity.\nE: Changes in light intensity have a greater influence on stomatal conductance than changes in $\\mathrm{CO}_{2}$ concentration.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-05.jpg?height=934&width=1131&top_left_y=424&top_left_x=474"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_437",
        "problem": "下列关于蛋白质工程的叙述, 正确的是（）\nA: 蛋白质工程是在分子水平上对蛋白质分子直接进行操作\nB: 蛋白质工程可以创造出自然界不存在的新类型蛋白质\nC: 蛋白质工程的操作, 摆脱了自然界关于遗传信息在生物大分子上传递的规律\nD: 蛋白质工程产生的蛋白质类药物比基因工程药物更安全, 可直接用于治疗患者\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列关于蛋白质工程的叙述, 正确的是（）\n\nA: 蛋白质工程是在分子水平上对蛋白质分子直接进行操作\nB: 蛋白质工程可以创造出自然界不存在的新类型蛋白质\nC: 蛋白质工程的操作, 摆脱了自然界关于遗传信息在生物大分子上传递的规律\nD: 蛋白质工程产生的蛋白质类药物比基因工程药物更安全, 可直接用于治疗患者\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_603",
        "problem": "5-溴尿嘧啶（5-BU）是一种碱基类似物, 其能与碱基 $\\mathrm{A}$ 和 $\\mathrm{G}$ 配对, 因此当 5-BU\n\n掺入 DNA 复制体系时就可能导致碱基转换。现有含 5-BU 在内的五种碱基的复制体系进行 DNA 复制，下列相关叙述错误的是（）\nA: 上述 DNA 复制过程中容易出现碱基对的替换\nB: DNA 复制一次所消耗 5-BU 的量与 $\\mathrm{T}$ 或 $\\mathrm{C}$ 可能不同\nC: 实现碱基 A-T 到 G-C 替换后不会改变所控制肽链长度\nD: 至少复制 3 轮才能实现某位点碱基对从 T-A 到 C-G 的替换\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n5-溴尿嘧啶（5-BU）是一种碱基类似物, 其能与碱基 $\\mathrm{A}$ 和 $\\mathrm{G}$ 配对, 因此当 5-BU\n\n掺入 DNA 复制体系时就可能导致碱基转换。现有含 5-BU 在内的五种碱基的复制体系进行 DNA 复制，下列相关叙述错误的是（）\n\nA: 上述 DNA 复制过程中容易出现碱基对的替换\nB: DNA 复制一次所消耗 5-BU 的量与 $\\mathrm{T}$ 或 $\\mathrm{C}$ 可能不同\nC: 实现碱基 A-T 到 G-C 替换后不会改变所控制肽链长度\nD: 至少复制 3 轮才能实现某位点碱基对从 T-A 到 C-G 的替换\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_830",
        "problem": "基因是有遗传效应的 DNA 片段，眼色基因（红眼基因 $\\mathrm{R}$ 、白眼基因 $\\mathrm{r}$ ）位于果蝇的 $\\mathrm{X}$ 染色体上，下列相关叙述不正确的是（ ）\nA: 雌雄果蝇细胞内的核基因随染色体的复制而复制的\nB: 正常情况下，同一 DNA 分子上不可能同时含有两个 R 基因\nC: 果蝇正常的精原细胞有丝分裂时白眼基因最多有 2 个\nD: 白眼基因表达时, 两条链都可作为模板转录出 mRNA\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n基因是有遗传效应的 DNA 片段，眼色基因（红眼基因 $\\mathrm{R}$ 、白眼基因 $\\mathrm{r}$ ）位于果蝇的 $\\mathrm{X}$ 染色体上，下列相关叙述不正确的是（ ）\n\nA: 雌雄果蝇细胞内的核基因随染色体的复制而复制的\nB: 正常情况下，同一 DNA 分子上不可能同时含有两个 R 基因\nC: 果蝇正常的精原细胞有丝分裂时白眼基因最多有 2 个\nD: 白眼基因表达时, 两条链都可作为模板转录出 mRNA\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_563",
        "problem": "某二倍体植物中, 抗病和感病这对相对性状由一对等位基因控制, 要确定这对性状的显隐性关系, 应该选用的杂交组合是( )\nA: 抗病株×感病株\nB: 抗病纯合体 $x$ 感病纯合体\nC: 抗病株×抗病株, 或感病株×感病株\nD: 抗病纯合体 $\\times$ 抗病纯合体，或感病纯合体 $x$ 感病纯合体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二倍体植物中, 抗病和感病这对相对性状由一对等位基因控制, 要确定这对性状的显隐性关系, 应该选用的杂交组合是( )\n\nA: 抗病株×感病株\nB: 抗病纯合体 $x$ 感病纯合体\nC: 抗病株×抗病株, 或感病株×感病株\nD: 抗病纯合体 $\\times$ 抗病纯合体，或感病纯合体 $x$ 感病纯合体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_76",
        "problem": "Scientists constructed models for four threatened tree species in sub-tropical forests in\n\nVietnam, and used these models to estimate tree ages (figure Q.94). Tree age is measured by ring count and trunk diameter at breast height (DBH). Rates of growth were categorised using changes in DBH from 10 to 1000 , with 10 at the finest-grain measure of change.\n\n[figure1]\n\nFigure Q. 94\n\nEstimated (lines) and observed (circles) ages for DBH categories of four tree species.\nA: Using the smallest category gives the most accurate information of tree age of $P$. kwangtungenesis.\nB: Age estimates increase particularly strongly from 100 to 10 -category model in $D$. elatum.\nC: Model with just $10 \\mathrm{DBH}$ categories underestimate the observed ages for three species.\nD: For D.elatum, measuring DBH using either 100 or 1000 will give an accurate estimate of tree age, whereas to estimate the age of C. macrolepis, only 100 gives a reliable estimate.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nScientists constructed models for four threatened tree species in sub-tropical forests in\n\nVietnam, and used these models to estimate tree ages (figure Q.94). Tree age is measured by ring count and trunk diameter at breast height (DBH). Rates of growth were categorised using changes in DBH from 10 to 1000 , with 10 at the finest-grain measure of change.\n\n[figure1]\n\nFigure Q. 94\n\nEstimated (lines) and observed (circles) ages for DBH categories of four tree species.\n\nA: Using the smallest category gives the most accurate information of tree age of $P$. kwangtungenesis.\nB: Age estimates increase particularly strongly from 100 to 10 -category model in $D$. elatum.\nC: Model with just $10 \\mathrm{DBH}$ categories underestimate the observed ages for three species.\nD: For D.elatum, measuring DBH using either 100 or 1000 will give an accurate estimate of tree age, whereas to estimate the age of C. macrolepis, only 100 gives a reliable estimate.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-097.jpg?height=1054&width=1442&top_left_y=803&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1282",
        "problem": "Many laboratory procedures involve the use of dilutions. If a solution has a 1/10 dilution the number represents 1 part of the sample added to 9 parts of diluent. The dilution factor equals the final volume divided by the sample volume.\n\nA serial dilution is any dilution in which the concentration decreases by the same quantity in each successive step. Serial dilutions are multiplicative. Multiple dilution series use different dilution factors at each step. An example of a serial dilution of wastewater with a dilution factor of 10 is given in the diagram below (DI, dilutant).\n\n[figure1]A multiple dilution series was performed on a sample from a dairy farm's effluent pond. The sample was diluted initially by placing $25 \\mathrm{~mL}$ of effluent into $75 \\mathrm{~mL}$ of water. This solution was serially diluted by; $1 / 2,1 / 5$, and $1 / 10$. The final sample had 80 faecal coliform cells per $\\mathrm{mL}$. How many faecal coliform cells were in the original sample from the effluent pond?\nA: 800,000 cells per $\\mathrm{mL}$\nB: 234,000 cells per $\\mathrm{mL}$\nC: 80,000 cells per $\\mathrm{mL}$\nD: 32,000 cells per $\\mathrm{mL}$\nE: 3,200 cells per $\\mathrm{mL}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nMany laboratory procedures involve the use of dilutions. If a solution has a 1/10 dilution the number represents 1 part of the sample added to 9 parts of diluent. The dilution factor equals the final volume divided by the sample volume.\n\nA serial dilution is any dilution in which the concentration decreases by the same quantity in each successive step. Serial dilutions are multiplicative. Multiple dilution series use different dilution factors at each step. An example of a serial dilution of wastewater with a dilution factor of 10 is given in the diagram below (DI, dilutant).\n\n[figure1]\n\nproblem:\nA multiple dilution series was performed on a sample from a dairy farm's effluent pond. The sample was diluted initially by placing $25 \\mathrm{~mL}$ of effluent into $75 \\mathrm{~mL}$ of water. This solution was serially diluted by; $1 / 2,1 / 5$, and $1 / 10$. The final sample had 80 faecal coliform cells per $\\mathrm{mL}$. How many faecal coliform cells were in the original sample from the effluent pond?\n\nA: 800,000 cells per $\\mathrm{mL}$\nB: 234,000 cells per $\\mathrm{mL}$\nC: 80,000 cells per $\\mathrm{mL}$\nD: 32,000 cells per $\\mathrm{mL}$\nE: 3,200 cells per $\\mathrm{mL}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-10.jpg?height=994&width=1830&top_left_y=545&top_left_x=113"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_649",
        "problem": "果蝇 $(2 \\mathrm{~N}=8)$ 的精巢中, $\\mathrm{a} 、 \\mathrm{~b} 、 \\mathrm{c}$ 三个细胞中的染色体组数、四分体个数和染色单体数如表所示 (不考虑基因突变和染色体变异)。下列相关叙述错误的是（）\n\n| 细胞 | 染色体组数 | 四分体个数 | 染色单体数 |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{a}$ | 2 | 0 | 16 |\n| $\\mathrm{b}$ | 2 | 4 | 16 |\n| $\\mathrm{c}$ | 2 | 0 | 0 |\nA: $\\mathrm{a}$ 细胞中各对同源染色体将要排列在赤道板两侧\nB: b 细胞可能发生了非姐妹染色单体间片段的互换\nC: $\\mathrm{b}$ 细胞中染色体数目与 $\\mathrm{c}$ 细胞中染色体数目相同\nD: 若 $\\mathrm{c}$ 细胞的细胞膜内陷, 则细胞内可能有 2 条 Y 染色体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇 $(2 \\mathrm{~N}=8)$ 的精巢中, $\\mathrm{a} 、 \\mathrm{~b} 、 \\mathrm{c}$ 三个细胞中的染色体组数、四分体个数和染色单体数如表所示 (不考虑基因突变和染色体变异)。下列相关叙述错误的是（）\n\n| 细胞 | 染色体组数 | 四分体个数 | 染色单体数 |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{a}$ | 2 | 0 | 16 |\n| $\\mathrm{b}$ | 2 | 4 | 16 |\n| $\\mathrm{c}$ | 2 | 0 | 0 |\n\nA: $\\mathrm{a}$ 细胞中各对同源染色体将要排列在赤道板两侧\nB: b 细胞可能发生了非姐妹染色单体间片段的互换\nC: $\\mathrm{b}$ 细胞中染色体数目与 $\\mathrm{c}$ 细胞中染色体数目相同\nD: 若 $\\mathrm{c}$ 细胞的细胞膜内陷, 则细胞内可能有 2 条 Y 染色体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_368",
        "problem": "研究发现, 水稻蜡质基因（Wx）编码直链淀粉合成酶。若 $\\mathrm{Wx}$ 基因中第 226 位碱基是正常的 $\\mathrm{G}$, 该位点所在的内含子能被正常剪接, 胚乳中直链淀粉含量最高, 基因型记做 GG; 若该位点突变成 $T$, 则不能被正常剪接, 胚乳中直链淀粉的合成水平会降低,基因型记做 $\\mathrm{TT}$ 。为检测 $\\mathrm{Wx}$ 基因该位点碱基是 $\\mathrm{G}$ 或 $\\mathrm{T}$, 研究人员以待测水稻叶片总 DNA 为材料, 以 $\\mathrm{Wx}$ 基因片段设计引物如图所示, 对 PCR 扩增产物经 AccI酶切后电泳。已知引物 1 为 $5^{\\prime}$-GCTTCACTTCTCTGCTTGTG-3', 两引物之间无另外的 AccI 酶的识别位点。下列叙述正确的是 ( )\n\n[图1]\n\n高直链淀粉含量(GG)和中等直链淀粉含量(TT)水稻品种中蜡质基因部分DNA序列\nA: 组成上述两种淀粉合成酶的氨基酸的种类不同, 数目相同\nB: GT 杂合型水稻品种酶切电泳结果为 2 条条带\nC: 若酶切产物只能观察到大约 460bp 的 DNA 条带, 则表明该水稻品种为 TT 型\nD: 该实验中需设计的引物 2 为 $5^{\\prime}$-TTCAACTCTCGTTAAATCAT-3'\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究发现, 水稻蜡质基因（Wx）编码直链淀粉合成酶。若 $\\mathrm{Wx}$ 基因中第 226 位碱基是正常的 $\\mathrm{G}$, 该位点所在的内含子能被正常剪接, 胚乳中直链淀粉含量最高, 基因型记做 GG; 若该位点突变成 $T$, 则不能被正常剪接, 胚乳中直链淀粉的合成水平会降低,基因型记做 $\\mathrm{TT}$ 。为检测 $\\mathrm{Wx}$ 基因该位点碱基是 $\\mathrm{G}$ 或 $\\mathrm{T}$, 研究人员以待测水稻叶片总 DNA 为材料, 以 $\\mathrm{Wx}$ 基因片段设计引物如图所示, 对 PCR 扩增产物经 AccI酶切后电泳。已知引物 1 为 $5^{\\prime}$-GCTTCACTTCTCTGCTTGTG-3', 两引物之间无另外的 AccI 酶的识别位点。下列叙述正确的是 ( )\n\n[图1]\n\n高直链淀粉含量(GG)和中等直链淀粉含量(TT)水稻品种中蜡质基因部分DNA序列\n\nA: 组成上述两种淀粉合成酶的氨基酸的种类不同, 数目相同\nB: GT 杂合型水稻品种酶切电泳结果为 2 条条带\nC: 若酶切产物只能观察到大约 460bp 的 DNA 条带, 则表明该水稻品种为 TT 型\nD: 该实验中需设计的引物 2 为 $5^{\\prime}$-TTCAACTCTCGTTAAATCAT-3'\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-17.jpg?height=286&width=991&top_left_y=1162&top_left_x=378"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1491",
        "problem": "Quillwort is an amphibious plant that can live in both aerial and submerged conditions. When submerged, quillwort uses CAM metabolism; $\\mathrm{CO}_{2}$ is fixed into malate at night and released again in the day to be used in photosynthesis. CAM metabolism is not seen unless the plant is submerged. There is strong competition from other photosynthetic organisms during the day.\n\n[figure1]\n\nWhich is true?\nA: Malate concentration in the leaves is highest just before sunset.\nB: CAM metabolism is used by quillwort because it reduces water loss.\nC: $\\mathrm{CO}_{2}$ levels are much higher in air than water during the day.\nD: $\\quad \\mathrm{CO}_{2}$ levels are much higher in air than water during the night.\nE: Quillwort produces oxygen at night when submerged.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nQuillwort is an amphibious plant that can live in both aerial and submerged conditions. When submerged, quillwort uses CAM metabolism; $\\mathrm{CO}_{2}$ is fixed into malate at night and released again in the day to be used in photosynthesis. CAM metabolism is not seen unless the plant is submerged. There is strong competition from other photosynthetic organisms during the day.\n\n[figure1]\n\nWhich is true?\n\nA: Malate concentration in the leaves is highest just before sunset.\nB: CAM metabolism is used by quillwort because it reduces water loss.\nC: $\\mathrm{CO}_{2}$ levels are much higher in air than water during the day.\nD: $\\quad \\mathrm{CO}_{2}$ levels are much higher in air than water during the night.\nE: Quillwort produces oxygen at night when submerged.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-32.jpg?height=765&width=1011&top_left_y=554&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_756",
        "problem": "脱氧核苷三磷酸 (dNTP) 和双脱氧核苷三磷酸 (ddNTP, ddN $-P_{\\alpha} \\sim P_{\\beta} \\sim P_{\\gamma}$ ) 的结构均与核苷三磷酸（NTP）类似，其中 dNTP 核糖的第 2 位碳原子上的羟基（- $\\mathrm{OH}$ ）被氢原子取代而 ddNTP 核糖第 2 位和第 3 位碳原子上的羟基均被氢原子取代。DNA 复制时, ddNTP 可以与 dNTP 竞争核苷酸链延长位点, 并终止 DNA 片段的延伸。现有一些序列为 $5^{\\prime}$ - GCCTAAGATCGTA-3'的 DNA 分子单链片段, 拟通过 PCR 获得以 ${ }^{32} \\mathrm{P}$ 标记的碱基“ $\\mathrm{A}$ ”为末端 (3'为碱基 $\\mathrm{A}$ )、不同长度的子链 DNA. 在反应管中加入单链模板、引物、底物、 TaqDNA 聚合酶、 $\\mathrm{Mg}^{2+}$ 及缓冲溶液。下列叙述正确的是（）\nA: 实验结束后最多可得到 4 种被 ${ }^{32} \\mathrm{P}$ 标记的子链 DNA\nB: 反应底物是 dCTP、dGTP、dTTP 和 $\\gamma$ 位 ${ }^{32} \\mathrm{P}$ 标记的 ddATP\nC: ddNTP 与 dNTP 竞争的延长位点是脱氧核苷酸链的 5 '末端\nD: 获得以碱基“A”为末端子链 DNA 最多需脱去 9 分子水\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n脱氧核苷三磷酸 (dNTP) 和双脱氧核苷三磷酸 (ddNTP, ddN $-P_{\\alpha} \\sim P_{\\beta} \\sim P_{\\gamma}$ ) 的结构均与核苷三磷酸（NTP）类似，其中 dNTP 核糖的第 2 位碳原子上的羟基（- $\\mathrm{OH}$ ）被氢原子取代而 ddNTP 核糖第 2 位和第 3 位碳原子上的羟基均被氢原子取代。DNA 复制时, ddNTP 可以与 dNTP 竞争核苷酸链延长位点, 并终止 DNA 片段的延伸。现有一些序列为 $5^{\\prime}$ - GCCTAAGATCGTA-3'的 DNA 分子单链片段, 拟通过 PCR 获得以 ${ }^{32} \\mathrm{P}$ 标记的碱基“ $\\mathrm{A}$ ”为末端 (3'为碱基 $\\mathrm{A}$ )、不同长度的子链 DNA. 在反应管中加入单链模板、引物、底物、 TaqDNA 聚合酶、 $\\mathrm{Mg}^{2+}$ 及缓冲溶液。下列叙述正确的是（）\n\nA: 实验结束后最多可得到 4 种被 ${ }^{32} \\mathrm{P}$ 标记的子链 DNA\nB: 反应底物是 dCTP、dGTP、dTTP 和 $\\gamma$ 位 ${ }^{32} \\mathrm{P}$ 标记的 ddATP\nC: ddNTP 与 dNTP 竞争的延长位点是脱氧核苷酸链的 5 '末端\nD: 获得以碱基“A”为末端子链 DNA 最多需脱去 9 分子水\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_301",
        "problem": "组蛋白乙酰化可破坏染色质中组蛋白和 DNA 之间的紧密结合。下列相关叙述正确的是 ( )\nA: 组蛋白乙酰化程度与基因转录活性负相关\nB: 组蛋白去乙酰化酶抑制剂可用于抑制肿瘤生长\nC: 组蛋白去乙酰化酶在启动子上的富集通常与转录激活有关\nD: 组蛋白乙酰化是原核生物中一种重要的蛋白质翻译后修饰方式\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n组蛋白乙酰化可破坏染色质中组蛋白和 DNA 之间的紧密结合。下列相关叙述正确的是 ( )\n\nA: 组蛋白乙酰化程度与基因转录活性负相关\nB: 组蛋白去乙酰化酶抑制剂可用于抑制肿瘤生长\nC: 组蛋白去乙酰化酶在启动子上的富集通常与转录激活有关\nD: 组蛋白乙酰化是原核生物中一种重要的蛋白质翻译后修饰方式\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1234",
        "problem": "The graph below shows how the oxygen uptake of an estuarine crustacean changes with the salinity of the water.\n\n[figure1]\n\nFrom this information the most acceptable conclusion is that:\nA: Oxygen diffuses faster in more concentrated salt solutions.\nB: Oxygen diffuses faster in more dilute salt solutions.\nC: Respiration rate is greater in more concentrated salt solutions.\nD: Removal of excess water from the body is an energy-requiring process.\nE: Uptake of water by the body is an energy-requiring process.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph below shows how the oxygen uptake of an estuarine crustacean changes with the salinity of the water.\n\n[figure1]\n\nFrom this information the most acceptable conclusion is that:\n\nA: Oxygen diffuses faster in more concentrated salt solutions.\nB: Oxygen diffuses faster in more dilute salt solutions.\nC: Respiration rate is greater in more concentrated salt solutions.\nD: Removal of excess water from the body is an energy-requiring process.\nE: Uptake of water by the body is an energy-requiring process.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-13.jpg?height=580&width=882&top_left_y=1395&top_left_x=587"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_723",
        "problem": "某科研小组将抗虫相关基因 $A 、 B$ 导入水稻后发现，当这两个基因同时存在时，植株才能表现出抗虫性状。现将导入抗虫基因的水稻 $\\mathrm{P}$ 与普通水稻杂交，子代中抗虫: 不抗虫 $=1: 3$. 不考虑基因突变和同源染色体中非姐妹染色单体的互换, 推测 $\\mathrm{A}$ 和 $\\mathrm{B}$ 在抗虫水稻 $\\mathrm{P}$ 染色体上的位置关系是（）\nA: 在同一条染色体上 [图1]\nB: 在一对同源染色体上 [图2]\nC: 在两对同源染色体上 [图3]\nD: 在一对同源染色体上 [图4]\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某科研小组将抗虫相关基因 $A 、 B$ 导入水稻后发现，当这两个基因同时存在时，植株才能表现出抗虫性状。现将导入抗虫基因的水稻 $\\mathrm{P}$ 与普通水稻杂交，子代中抗虫: 不抗虫 $=1: 3$. 不考虑基因突变和同源染色体中非姐妹染色单体的互换, 推测 $\\mathrm{A}$ 和 $\\mathrm{B}$ 在抗虫水稻 $\\mathrm{P}$ 染色体上的位置关系是（）\n\nA: 在同一条染色体上 [图1]\nB: 在一对同源染色体上 [图2]\nC: 在两对同源染色体上 [图3]\nD: 在一对同源染色体上 [图4]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-014.jpg?height=243&width=259&top_left_y=170&top_left_x=750",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-014.jpg?height=246&width=260&top_left_y=168&top_left_x=1458",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-014.jpg?height=237&width=266&top_left_y=527&top_left_x=792",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-014.jpg?height=245&width=263&top_left_y=523&top_left_x=1453"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1522",
        "problem": "Dynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\n\nIt is more common for a pyrimidine nucleotide ( $\\mathrm{C}$ or $\\mathrm{T}$ ) to mutate into another pyrimidine, than into a purine ( $\\mathrm{G}$ or A). The sequence studied above also turned out to be in a protein coding sequence.\n\nProtein sequences can also be compared. What are the differences between protein and DNA alignments?\nA: DNA sequence changes at a faster rate overtime\nB: DNA is easier to sequence\nC: DNA is better for comparing more distantly related species\nD: DNA has a more complicated scoring system\nE: A and B only\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\n\nIt is more common for a pyrimidine nucleotide ( $\\mathrm{C}$ or $\\mathrm{T}$ ) to mutate into another pyrimidine, than into a purine ( $\\mathrm{G}$ or A). The sequence studied above also turned out to be in a protein coding sequence.\n\nProtein sequences can also be compared. What are the differences between protein and DNA alignments?\n\nA: DNA sequence changes at a faster rate overtime\nB: DNA is easier to sequence\nC: DNA is better for comparing more distantly related species\nD: DNA has a more complicated scoring system\nE: A and B only\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=925&width=928&top_left_y=354&top_left_x=238",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=922&width=928&top_left_y=1458&top_left_x=238"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_885",
        "problem": "下图为某生物一个细胞的分裂图像, 着丝点均在染色体端部, 图中 1、2、3、4 各表示一条染色体。下列表述正确的是( )\n\n[图1]\nA: 图中细胞处于减数第二次分裂前期\nB: 图中细胞的染色体数是体细胞的 2 倍\nC: 染色体 1 与 2 在后续的分裂过程中会相互分离\nD: 染色体 1 与 3 必定会出现在同一子细胞中\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为某生物一个细胞的分裂图像, 着丝点均在染色体端部, 图中 1、2、3、4 各表示一条染色体。下列表述正确的是( )\n\n[图1]\n\nA: 图中细胞处于减数第二次分裂前期\nB: 图中细胞的染色体数是体细胞的 2 倍\nC: 染色体 1 与 2 在后续的分裂过程中会相互分离\nD: 染色体 1 与 3 必定会出现在同一子细胞中\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-23.jpg?height=536&width=397&top_left_y=1254&top_left_x=178"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1001",
        "problem": "A child has overdosed on acetaminophen, causing acute liver failure. He displays symptoms of severe jaundice and swelling of the abdomen. Which of the following processes is not directly impacted?\nA: Bile synthesis\nB: Bilirubin glucuronidation\nC: Production of serum albumin\nD: Erythropoietin synthesis\nE: Glycogenesis\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA child has overdosed on acetaminophen, causing acute liver failure. He displays symptoms of severe jaundice and swelling of the abdomen. Which of the following processes is not directly impacted?\n\nA: Bile synthesis\nB: Bilirubin glucuronidation\nC: Production of serum albumin\nD: Erythropoietin synthesis\nE: Glycogenesis\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_476",
        "problem": "下图为四个细胞分裂过程中所有染色体的行为变化示意图。下列相关分析错误的是\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\n\n[图4]\n\n丁\nA: 甲细胞具有 4 对同源染色体\nB: 以上四个细胞可能存在于同一个动物的睪丸中\nC: 若乙细胞发生减数分裂, 染色体(1)(2)和(1)(3)出现在某一子细胞中的概率相同\nD: 丁细胞是一个没有同源染色体的次级精母细胞, 其可以是丙的一个子细胞\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为四个细胞分裂过程中所有染色体的行为变化示意图。下列相关分析错误的是\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\n\n[图4]\n\n丁\n\nA: 甲细胞具有 4 对同源染色体\nB: 以上四个细胞可能存在于同一个动物的睪丸中\nC: 若乙细胞发生减数分裂, 染色体(1)(2)和(1)(3)出现在某一子细胞中的概率相同\nD: 丁细胞是一个没有同源染色体的次级精母细胞, 其可以是丙的一个子细胞\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-085.jpg?height=302&width=308&top_left_y=192&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-085.jpg?height=274&width=286&top_left_y=203&top_left_x=631",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-085.jpg?height=308&width=283&top_left_y=183&top_left_x=909",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-085.jpg?height=291&width=234&top_left_y=186&top_left_x=1205"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_823",
        "problem": "某种植物的羽裂叶和全缘叶是一对相对性状。某同学用全缘叶植株 (植株甲) 进行了下列四个实验。(1)植株甲进行自花传粉, 子代出现性状分离\n\n(2)用植株甲给另一全缘叶植株授粉, 子代均为全缘叶\n\n(3)用植株甲给羽裂叶植株授粉, 子代中全缘叶与羽裂叶的比例为 $1: 1$\n\n(4)用植株甲给另一全缘叶植株授粉, 子代中全缘叶与羽裂叶的比例为 $3: 1$\n\n其中能够判定植株甲为杂合子的实验是( )\nA: (1)或(2)\nB: (1)或(4)\nC: (2)或(3)\nD: (3)或(4)\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种植物的羽裂叶和全缘叶是一对相对性状。某同学用全缘叶植株 (植株甲) 进行了下列四个实验。(1)植株甲进行自花传粉, 子代出现性状分离\n\n(2)用植株甲给另一全缘叶植株授粉, 子代均为全缘叶\n\n(3)用植株甲给羽裂叶植株授粉, 子代中全缘叶与羽裂叶的比例为 $1: 1$\n\n(4)用植株甲给另一全缘叶植株授粉, 子代中全缘叶与羽裂叶的比例为 $3: 1$\n\n其中能够判定植株甲为杂合子的实验是( )\n\nA: (1)或(2)\nB: (1)或(4)\nC: (2)或(3)\nD: (3)或(4)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_253",
        "problem": "The principle of magnetic resonance imaging (MRI) is based on the fact that the nuclei of certain elements align with the magnetic force when placed in a strong magnetic field. At the field strengths currently used in medical imaging, hydrogen nuclei (protons) in water molecules and lipids are responsible for producing anatomical images.\n\nIf a radiofrequency pulse at the resonant frequency of hydrogen is applied, a proportion of the protons change alignment, flipping through a present angle, and rotate in phase with one another. Following this radiofrequency pulse, the protons realign (relax), they induce a signal which, although very weak, can be detected and localized by copper coils placed around the patient. An image representing the distribution of the hydrogen protons can be built up.\n\nThe strength of the signal depends not only on proton density but also on two relaxation times, T1 and $\\mathrm{T} 2$; $\\mathrm{T} 1$ depends on the time the protons take to return to the axis of magnetic field, and $\\mathrm{T} 2$ depends on the time the protons take to dephase. A T1-weighted image is one in which the contrast between tissues is due mainly to their T1 relaxation properties, while in a T2-weighted image the contrast is due to the $\\mathrm{T} 2$ relaxation properties.\n\nT1-weighted image: Fat signal intensity $>$ Water signal intensity\n\nT2-weighted image: Fat signal intensity $<$ water signal intensity\n[figure1]\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe principle of magnetic resonance imaging (MRI) is based on the fact that the nuclei of certain elements align with the magnetic force when placed in a strong magnetic field. At the field strengths currently used in medical imaging, hydrogen nuclei (protons) in water molecules and lipids are responsible for producing anatomical images.\n\nIf a radiofrequency pulse at the resonant frequency of hydrogen is applied, a proportion of the protons change alignment, flipping through a present angle, and rotate in phase with one another. Following this radiofrequency pulse, the protons realign (relax), they induce a signal which, although very weak, can be detected and localized by copper coils placed around the patient. An image representing the distribution of the hydrogen protons can be built up.\n\nThe strength of the signal depends not only on proton density but also on two relaxation times, T1 and $\\mathrm{T} 2$; $\\mathrm{T} 1$ depends on the time the protons take to return to the axis of magnetic field, and $\\mathrm{T} 2$ depends on the time the protons take to dephase. A T1-weighted image is one in which the contrast between tissues is due mainly to their T1 relaxation properties, while in a T2-weighted image the contrast is due to the $\\mathrm{T} 2$ relaxation properties.\n\nT1-weighted image: Fat signal intensity $>$ Water signal intensity\n\nT2-weighted image: Fat signal intensity $<$ water signal intensity\n[figure1]\n\nA: A\nB: B\nC: C\nD: D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://i.postimg.cc/769XpgQD/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_232",
        "problem": "Lactic fermented vegetables are traditional food in many Asian cuisines. Microorganisms commonly found in the fermentation broth are lactic acid bacteria, yeast and filamentous fungi.\n\nFig.Q. 56 below shows the flowchart of viable cell counts (log CFU/mL) of three different microbial groups and the $\\mathrm{pH}$ value during the lactic fermentation course of cabbage. Oxygen dissolved in fermentation broth decreased with time and was completely consumed after the $22^{\\text {nd }}$ day.\n\n[figure1]\n\nFig.Q.56. Changes in microflora during lactic acid fermentation of cabbage.\nA: The drop in $\\mathrm{pH}$ value from day 1 to day 3 was caused by only organic acids produced by lactic acid bacteria.\nB: Lactic acid produced by lactic acid bacteria favours the growth of yeast cells from day 10 till day 26 .\nC: Yeast cells shifted from fermentation to respiration after day 22.\nD: Some filamentous fungi showed high tolerance at low $\\mathrm{pH}$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nLactic fermented vegetables are traditional food in many Asian cuisines. Microorganisms commonly found in the fermentation broth are lactic acid bacteria, yeast and filamentous fungi.\n\nFig.Q. 56 below shows the flowchart of viable cell counts (log CFU/mL) of three different microbial groups and the $\\mathrm{pH}$ value during the lactic fermentation course of cabbage. Oxygen dissolved in fermentation broth decreased with time and was completely consumed after the $22^{\\text {nd }}$ day.\n\n[figure1]\n\nFig.Q.56. Changes in microflora during lactic acid fermentation of cabbage.\n\nA: The drop in $\\mathrm{pH}$ value from day 1 to day 3 was caused by only organic acids produced by lactic acid bacteria.\nB: Lactic acid produced by lactic acid bacteria favours the growth of yeast cells from day 10 till day 26 .\nC: Yeast cells shifted from fermentation to respiration after day 22.\nD: Some filamentous fungi showed high tolerance at low $\\mathrm{pH}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-015.jpg?height=914&width=1171&top_left_y=819&top_left_x=474"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_101",
        "problem": "The following graph shows the relationship between the frequency or strength of disturbance and species diversity.\n\n[figure1]\n\nWhich of the following statements is not correct?\nA: The low species diversity of community (a) is due to the presence of dominant species in the community.\nB: In community (c), species diversity is low because there is not enough time between disturbances for a wide variety of species to colonize.\nC: The degree of competitive exclusion among species is highest in community $(b)$.\nD: In community (c), late successional species will be replaced rapidly by early successional species.\nE: Community (c) consists of environmental stress-tolerant species.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph shows the relationship between the frequency or strength of disturbance and species diversity.\n\n[figure1]\n\nWhich of the following statements is not correct?\n\nA: The low species diversity of community (a) is due to the presence of dominant species in the community.\nB: In community (c), species diversity is low because there is not enough time between disturbances for a wide variety of species to colonize.\nC: The degree of competitive exclusion among species is highest in community $(b)$.\nD: In community (c), late successional species will be replaced rapidly by early successional species.\nE: Community (c) consists of environmental stress-tolerant species.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-55.jpg?height=546&width=760&top_left_y=475&top_left_x=768"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1336",
        "problem": "The Census of Marine Life also examined the global threats to marine biodiversity and these are summarized in the table below.\n\n|  | Overfishing | Habitat loss | Pollution | Alien species | Temperature | Hypoxia | Acidification | Total | Median |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Mediterranean | 5 | 5 | 4 | 5 | 5 | 2 | 1 | 27 | 5.0 |\n| Gulf of Mexico | 5 | 5 | 5 | 4 | 2 | 3 | 1 | 25 | 4.0 |\n| China | 5 | 5 | 5 | 2 | 2 | 3 | 1 | 23 | 3.0 |\n| Baltic | 4 | 3 | 4 | 3 | 4 | 3 | 1 | 22 | 3.0 |\n| Caribbean | 4 | 4 | 4 | 4 | 2 | 2 | 2 | 22 | 4.0 |\n| USA Southeast | 4 | 4 | 3 | 3 | 3 | 2 | 3 | 22 | 3.0 |\n| Brazil and Tropical West Atlantic | 4 | 4 | 3 | 3 | 3 | 3 | 2 | 22 | 3.0 |\n| Humboldt Current and Patagonian Shelf | 4 | 3 | 3 | 3 | 2 | 4 | 2 | 21 | 3.0 |\n| North Indian Ocean | 3 | 4 | 4 | 3 | 3 | 2 | 2 | 21 | 3.0 |\n| Tropical East Pacific | 3 | 3 | 3 | 3 | 3 | 3 | 2 | 20 | 3.0 |\n| South Africa | 3 | 2 | 4 | 4 | 2 | 4 | 1 | 20 | 3.0 |\n| New Zealand | 4 | 3 | 2 | 4 | 2 | 1 | 3 | 19 | 3.0 |\n| Atlantic Europe | 4 | 2 | 4 | 2 | 4 | 1 | 2 | 19 | 2.0 |\n| USA Northeast | 4 | 3 | 3 | 2 | 3 | 2 | 1 | 18 | 3.0 |\n| Japan | 3 | 3 | 3 | 2 | 3 | 1 | 2 | 17 | 3.0 |\n| Canada (all) | 2 | 4 | 2 | 2 | 5 | 0 | 1 | 16 | 2.0 |\n| Australia | 3 | 3 | 2 | 3 | 2 | 0 | 1 | 14 | 2.0 |\n| Antarctica | 2 | 2 | 2 | 0 | 1 | 0 | 2 | 9 | 2.0 |\n| Total | 66 | 62 | 60 | 52 | 51 | 36 | 30 |  |  |\n| Median | 4.0 | 3.0 | 3.0 | 3.0 | 3.0 | 2.0 | 2.0 |  |  |\n\nEach threat was scored from 1 to 5 (minimum to maximum) across a comparative scale among different regions. Some regions (e.g., Australia) reported only known threats rather than predicted threats. Table is sorted by reported greatest threats and areas with greatest impacts. Median values of each threat and for each region are also reported. oi:10.1371/journal.pone.0012110.t005Worldwide, the greatest threat to marine biodiversity is?\nA: Overfishing\nB: Habitat loss\nC: Pollution\nD: Alien species\nE: Temperature\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nThe Census of Marine Life also examined the global threats to marine biodiversity and these are summarized in the table below.\n\n|  | Overfishing | Habitat loss | Pollution | Alien species | Temperature | Hypoxia | Acidification | Total | Median |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Mediterranean | 5 | 5 | 4 | 5 | 5 | 2 | 1 | 27 | 5.0 |\n| Gulf of Mexico | 5 | 5 | 5 | 4 | 2 | 3 | 1 | 25 | 4.0 |\n| China | 5 | 5 | 5 | 2 | 2 | 3 | 1 | 23 | 3.0 |\n| Baltic | 4 | 3 | 4 | 3 | 4 | 3 | 1 | 22 | 3.0 |\n| Caribbean | 4 | 4 | 4 | 4 | 2 | 2 | 2 | 22 | 4.0 |\n| USA Southeast | 4 | 4 | 3 | 3 | 3 | 2 | 3 | 22 | 3.0 |\n| Brazil and Tropical West Atlantic | 4 | 4 | 3 | 3 | 3 | 3 | 2 | 22 | 3.0 |\n| Humboldt Current and Patagonian Shelf | 4 | 3 | 3 | 3 | 2 | 4 | 2 | 21 | 3.0 |\n| North Indian Ocean | 3 | 4 | 4 | 3 | 3 | 2 | 2 | 21 | 3.0 |\n| Tropical East Pacific | 3 | 3 | 3 | 3 | 3 | 3 | 2 | 20 | 3.0 |\n| South Africa | 3 | 2 | 4 | 4 | 2 | 4 | 1 | 20 | 3.0 |\n| New Zealand | 4 | 3 | 2 | 4 | 2 | 1 | 3 | 19 | 3.0 |\n| Atlantic Europe | 4 | 2 | 4 | 2 | 4 | 1 | 2 | 19 | 2.0 |\n| USA Northeast | 4 | 3 | 3 | 2 | 3 | 2 | 1 | 18 | 3.0 |\n| Japan | 3 | 3 | 3 | 2 | 3 | 1 | 2 | 17 | 3.0 |\n| Canada (all) | 2 | 4 | 2 | 2 | 5 | 0 | 1 | 16 | 2.0 |\n| Australia | 3 | 3 | 2 | 3 | 2 | 0 | 1 | 14 | 2.0 |\n| Antarctica | 2 | 2 | 2 | 0 | 1 | 0 | 2 | 9 | 2.0 |\n| Total | 66 | 62 | 60 | 52 | 51 | 36 | 30 |  |  |\n| Median | 4.0 | 3.0 | 3.0 | 3.0 | 3.0 | 2.0 | 2.0 |  |  |\n\nEach threat was scored from 1 to 5 (minimum to maximum) across a comparative scale among different regions. Some regions (e.g., Australia) reported only known threats rather than predicted threats. Table is sorted by reported greatest threats and areas with greatest impacts. Median values of each threat and for each region are also reported. oi:10.1371/journal.pone.0012110.t005\n\nproblem:\nWorldwide, the greatest threat to marine biodiversity is?\n\nA: Overfishing\nB: Habitat loss\nC: Pollution\nD: Alien species\nE: Temperature\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1140",
        "problem": "Many laboratory procedures involve the use of dilutions. If a solution has a 1/10 dilution the number represents 1 part of the sample added to 9 parts of diluent. The dilution factor equals the final volume divided by the sample volume.\n\nA serial dilution is any dilution in which the concentration decreases by the same quantity in each successive step. Serial dilutions are multiplicative. Multiple dilution series use different dilution factors at each step. An example of a serial dilution of wastewater with a dilution factor of 10 is given in the diagram below (DI, dilutant).\n\n[figure1]In a haematology laboratory, a blood glucose of $800 \\mathrm{mg} / \\mathrm{dL}$ was obtained. (Note: $\\mathrm{dL}$ is a SI unit of volume, the deciliter. $1 \\mathrm{dL}=100 \\mathrm{~mL}$ ). According to the manufacturer the highest glucose result that can be measured accurately on this particular instrument is $500 \\mathrm{mg} / \\mathrm{dL}$. The sample must therefore be diluted. The serum was diluted by taking $5 \\mathrm{~mL}$ of serum and adding to $95 \\mathrm{~mL}$ of blood diluting fluid and retested. The result obtained was $35 \\mathrm{mg} / \\mathrm{dL}$. What was the blood glucose level of this patient?\nA: $800 \\mathrm{mg} / \\mathrm{dL}$\nB: $700 \\mathrm{mg} / \\mathrm{dL}$\nC: $665 \\mathrm{mg} / \\mathrm{dL}$\nD: $400 \\mathrm{mg} / \\mathrm{dL}$\nE: $350 \\mathrm{mg} / \\mathrm{dL}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nMany laboratory procedures involve the use of dilutions. If a solution has a 1/10 dilution the number represents 1 part of the sample added to 9 parts of diluent. The dilution factor equals the final volume divided by the sample volume.\n\nA serial dilution is any dilution in which the concentration decreases by the same quantity in each successive step. Serial dilutions are multiplicative. Multiple dilution series use different dilution factors at each step. An example of a serial dilution of wastewater with a dilution factor of 10 is given in the diagram below (DI, dilutant).\n\n[figure1]\n\nproblem:\nIn a haematology laboratory, a blood glucose of $800 \\mathrm{mg} / \\mathrm{dL}$ was obtained. (Note: $\\mathrm{dL}$ is a SI unit of volume, the deciliter. $1 \\mathrm{dL}=100 \\mathrm{~mL}$ ). According to the manufacturer the highest glucose result that can be measured accurately on this particular instrument is $500 \\mathrm{mg} / \\mathrm{dL}$. The sample must therefore be diluted. The serum was diluted by taking $5 \\mathrm{~mL}$ of serum and adding to $95 \\mathrm{~mL}$ of blood diluting fluid and retested. The result obtained was $35 \\mathrm{mg} / \\mathrm{dL}$. What was the blood glucose level of this patient?\n\nA: $800 \\mathrm{mg} / \\mathrm{dL}$\nB: $700 \\mathrm{mg} / \\mathrm{dL}$\nC: $665 \\mathrm{mg} / \\mathrm{dL}$\nD: $400 \\mathrm{mg} / \\mathrm{dL}$\nE: $350 \\mathrm{mg} / \\mathrm{dL}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-10.jpg?height=994&width=1830&top_left_y=545&top_left_x=113"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_303",
        "problem": "图 1 为某动物体内某一细胞分裂图像中部分染色体情况, 图 2 为同一动物体内细胞分裂过程中染色体组数目变化。下列叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 图 1 代表的细胞为极体, 出现在图 $2 \\mathrm{EF}$ 段\nB: 图 1 细胞所在的器官中只能观察到染色体组数为 $1 、 2$ 的不同细胞分裂图像\nC: 若该动物基因型为 Ddee, 图 1 中出现 $\\mathrm{D} 、 \\mathrm{~d}$ 的原因是减数分裂I过程中发生基因突变或基因重组\nD: 该动物体内细胞 $\\mathrm{D}$ 和 $\\mathrm{d}$ 基因的分离只发生在图 2 中 $\\mathrm{AB}$ 段\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 1 为某动物体内某一细胞分裂图像中部分染色体情况, 图 2 为同一动物体内细胞分裂过程中染色体组数目变化。下列叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 图 1 代表的细胞为极体, 出现在图 $2 \\mathrm{EF}$ 段\nB: 图 1 细胞所在的器官中只能观察到染色体组数为 $1 、 2$ 的不同细胞分裂图像\nC: 若该动物基因型为 Ddee, 图 1 中出现 $\\mathrm{D} 、 \\mathrm{~d}$ 的原因是减数分裂I过程中发生基因突变或基因重组\nD: 该动物体内细胞 $\\mathrm{D}$ 和 $\\mathrm{d}$ 基因的分离只发生在图 2 中 $\\mathrm{AB}$ 段\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-09.jpg?height=479&width=323&top_left_y=480&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-09.jpg?height=343&width=714&top_left_y=611&top_left_x=682"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_965",
        "problem": "Amino acids and nucleotides form complex polymers. Select ALL the statements that are TRUE\nA: Both contain peptide bonds\nB: Both contain nitrogen\nC: Both may form helical structures\nD: Nucleotides and amino acids form branched polymers\nE: Both contain phosphate groups\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAmino acids and nucleotides form complex polymers. Select ALL the statements that are TRUE\n\nA: Both contain peptide bonds\nB: Both contain nitrogen\nC: Both may form helical structures\nD: Nucleotides and amino acids form branched polymers\nE: Both contain phosphate groups\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1261",
        "problem": "The research team sequenced the genome of 11 species of New Zealand snails and compared this molecular information with detailed mathematical description of the shape of the shell and outline of the shell opening. The relationships between these species are shown in the cladograms below. A cladogram uses branching lines that end at groups of organisms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMolecular phylogeny\n[figure1]The results indicate significant congruence (agreement) between molecular and structural inferences. Which species is incongruent when these two phylogenies are compared?\nA: Alcithoe benthicola\nB: Alcithoe wilsoni\nC: Alcithoe arabica\nD: Amoria fusus\nE: Amoria hunteri\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nThe research team sequenced the genome of 11 species of New Zealand snails and compared this molecular information with detailed mathematical description of the shape of the shell and outline of the shell opening. The relationships between these species are shown in the cladograms below. A cladogram uses branching lines that end at groups of organisms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMolecular phylogeny\n[figure1]\n\nproblem:\nThe results indicate significant congruence (agreement) between molecular and structural inferences. Which species is incongruent when these two phylogenies are compared?\n\nA: Alcithoe benthicola\nB: Alcithoe wilsoni\nC: Alcithoe arabica\nD: Amoria fusus\nE: Amoria hunteri\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-22.jpg?height=971&width=1128&top_left_y=1428&top_left_x=772"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_219",
        "problem": "Sam is an imaginary single-celled organism. His body is covered with cilia which enable him to move. He has two nuclei: housekeeping genes are expressed in the macronucleus, while gene-expression tasks relevant to reproduction are undertaken by the micronucleus. Sam divides via a process called \"schizogony\", whereby nuclei undergo multiple fission and then the cell itself divides. Sam can eat food particles and even bacteria through his mouth, a characteristic that helps him to live freely as a saprophyte. Sam has a monomorphic shape throughout its life cycle and experience enlargement and division, consecutively. Sam has decided to abandon his free-living lifestyle and transform into a merry parasite (with limited harm to the host).\n\nJack is also a parasite and is more aggressive (damages its host) than Sam and his single-celled body is covered by an armour made out of protein. This protective layer is jagged and gives Jack a rugged look. Jack also has two nuclei. He has two flagella that allows him to move, attack, and penetrate. He is able to secrete enzymes that can degrade many animal tissues. immature individuals who belong to Jacks species lack traits mentioned above; they are small and proliferate rapidly. In addition, Immature individuals of Jack needs to find host to grow and cannot live independently.\n\nIndicate for each statement which strategy will be more suitable for Sam compared to Jack:\nA: Utilizing antigen shuffling, even though it is a costly trait.\nB: Having a life cycle consisting of two hosts.\nC: Choosing an r-selected host.\nD: Secreting Samilosporin (an antibiotic which destroys many of bacterial species that are virulent for the host).\nE: Transmitting to the next generation by infecting the gametes.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nSam is an imaginary single-celled organism. His body is covered with cilia which enable him to move. He has two nuclei: housekeeping genes are expressed in the macronucleus, while gene-expression tasks relevant to reproduction are undertaken by the micronucleus. Sam divides via a process called \"schizogony\", whereby nuclei undergo multiple fission and then the cell itself divides. Sam can eat food particles and even bacteria through his mouth, a characteristic that helps him to live freely as a saprophyte. Sam has a monomorphic shape throughout its life cycle and experience enlargement and division, consecutively. Sam has decided to abandon his free-living lifestyle and transform into a merry parasite (with limited harm to the host).\n\nJack is also a parasite and is more aggressive (damages its host) than Sam and his single-celled body is covered by an armour made out of protein. This protective layer is jagged and gives Jack a rugged look. Jack also has two nuclei. He has two flagella that allows him to move, attack, and penetrate. He is able to secrete enzymes that can degrade many animal tissues. immature individuals who belong to Jacks species lack traits mentioned above; they are small and proliferate rapidly. In addition, Immature individuals of Jack needs to find host to grow and cannot live independently.\n\nIndicate for each statement which strategy will be more suitable for Sam compared to Jack:\n\nA: Utilizing antigen shuffling, even though it is a costly trait.\nB: Having a life cycle consisting of two hosts.\nC: Choosing an r-selected host.\nD: Secreting Samilosporin (an antibiotic which destroys many of bacterial species that are virulent for the host).\nE: Transmitting to the next generation by infecting the gametes.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_941",
        "problem": "干扰素是动物细胞受到病毒侵染后产生的一种糖蛋白, 可用于对抗病毒的感染和癌症，但体外保存相当困难。下图是利用蛋白质工程设计生产干扰素的流程图，错误的是\n\n[图1]\nA: 图中构建新的干扰素模型的主要依据是蛋白质的预期功能\nB: 图中新的干扰素基因必须插入质粒上的起始密码子和终止密码子之间才能表达\nC: 图中改造干扰素结构的实质是改造干扰素基因的结构\nD: 图中各项技术环节中，有些需要通过基因工程实现\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n干扰素是动物细胞受到病毒侵染后产生的一种糖蛋白, 可用于对抗病毒的感染和癌症，但体外保存相当困难。下图是利用蛋白质工程设计生产干扰素的流程图，错误的是\n\n[图1]\n\nA: 图中构建新的干扰素模型的主要依据是蛋白质的预期功能\nB: 图中新的干扰素基因必须插入质粒上的起始密码子和终止密码子之间才能表达\nC: 图中改造干扰素结构的实质是改造干扰素基因的结构\nD: 图中各项技术环节中，有些需要通过基因工程实现\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-06.jpg?height=334&width=1416&top_left_y=1501&top_left_x=363"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_104",
        "problem": "A rocky shore contains many shallow rock pools dominated by macroalgae and grazing gastropods, comprising primarily Patella ulyssiponensis (P), Littorina littorea (L) and Gibbula umbilicalis $(\\mathrm{G})$. The experiment is designed to test the interaction between grazer species and the additive interaction with nutrient enrichment. Pools contained either none, one, two or all three of grazer species at realistic densities (Patella, Littorina and Gibbula). Another complete set of all the manipulation grazer treatments was also established concurrently where nutrient concentrations were enhanced to compare the simultaneous effects of grazer treatments at ambient and enriched nutrient conditions. Gross ecosystem productivity (GEP), number of algal taxa, and the biomass (dry weight) of all algal species were measured.\n[figure1]\n\nFigure Q. 97\nA: Gross ecosystem productivity is enhanced by nutrient enrichment and is greater in pools where Littorina is present.\nB: The effects of grazer species loss on accumulated algal biomass are regulated by nutrient conditions, grazer identity and grazer diversity.\nC: The effects of loss of grazer species on ecosystem functioning depend upon both the diversity and identity of the species present.\nD: The presence of all grazers results in lower algal diversity and biomass in both nutrient conditions.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA rocky shore contains many shallow rock pools dominated by macroalgae and grazing gastropods, comprising primarily Patella ulyssiponensis (P), Littorina littorea (L) and Gibbula umbilicalis $(\\mathrm{G})$. The experiment is designed to test the interaction between grazer species and the additive interaction with nutrient enrichment. Pools contained either none, one, two or all three of grazer species at realistic densities (Patella, Littorina and Gibbula). Another complete set of all the manipulation grazer treatments was also established concurrently where nutrient concentrations were enhanced to compare the simultaneous effects of grazer treatments at ambient and enriched nutrient conditions. Gross ecosystem productivity (GEP), number of algal taxa, and the biomass (dry weight) of all algal species were measured.\n[figure1]\n\nFigure Q. 97\n\nA: Gross ecosystem productivity is enhanced by nutrient enrichment and is greater in pools where Littorina is present.\nB: The effects of grazer species loss on accumulated algal biomass are regulated by nutrient conditions, grazer identity and grazer diversity.\nC: The effects of loss of grazer species on ecosystem functioning depend upon both the diversity and identity of the species present.\nD: The presence of all grazers results in lower algal diversity and biomass in both nutrient conditions.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-103.jpg?height=1560&width=834&top_left_y=1042&top_left_x=674"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1153",
        "problem": "Chickens of the variety 'Wyandotte' are white because they possess the dominant allele $(I)$ of a colour inhibitor gene in addition to the dominant allele $(F)$ of a colour factor gene. Hens of the variety 'leghorn' are white because they are homozygous recessive for the colour factor gene.\n\nThe ratio of white to coloured offspring in the second generation of a cross between white Wyandotte (IIFF) and white leghorn (iiff), if the genes assort independently, would be\nA: 15 white : 1 coloured\nB: 13 white : 3 coloured\nC: 12 white : 4 coloured\nD: 10 white : 6 coloured\nE: 9 white : 7 coloured\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nChickens of the variety 'Wyandotte' are white because they possess the dominant allele $(I)$ of a colour inhibitor gene in addition to the dominant allele $(F)$ of a colour factor gene. Hens of the variety 'leghorn' are white because they are homozygous recessive for the colour factor gene.\n\nThe ratio of white to coloured offspring in the second generation of a cross between white Wyandotte (IIFF) and white leghorn (iiff), if the genes assort independently, would be\n\nA: 15 white : 1 coloured\nB: 13 white : 3 coloured\nC: 12 white : 4 coloured\nD: 10 white : 6 coloured\nE: 9 white : 7 coloured\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1443",
        "problem": "[figure1]\n\nWhat does organelle A produce when functional in a cell?\nA: Carbon dioxide\nB: Starch\nC: Oxygen\nD: ATP\nE: Glucose\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nWhat does organelle A produce when functional in a cell?\n\nA: Carbon dioxide\nB: Starch\nC: Oxygen\nD: ATP\nE: Glucose\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-03.jpg?height=434&width=485&top_left_y=314&top_left_x=777"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_564",
        "problem": "鸡的卷羽 (F) 对片羽 (f) 为不完全显性, 基因位于 1 常染色体上, $\\mathrm{Ff}$ 表现为半卷羽; 体型正常（D）对矮小（d）为完全显性, 基因位于 Z 染色体上。卷羽鸡适应高温环境, 矮小鸡饲料利用率高。为培育耐热节粮型种鸡以实现规模化生产, 研究人员拟通过杂交将 $\\mathrm{d}$ 基因引入某特色肉鸡体内，育种过程如图所示。已知亲本均为纯合子，下列分析错误的是（ ）\n\n[图1]\nA: 符合要求的新品种个体的基因型为 $F F Z^{d} W 、 F F Z^{d} Z^{d}$\nB: 群体I和群体II个体的表型及基因型与亲本的都不同\nC: 为缩短育种时间, 应从群体 I、II中分别选择母本和父本进行杂交\nD: 若群体I、II的雌雄个体随机交配, 子代中半卷羽矮小鸡占 $3 / 8$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n鸡的卷羽 (F) 对片羽 (f) 为不完全显性, 基因位于 1 常染色体上, $\\mathrm{Ff}$ 表现为半卷羽; 体型正常（D）对矮小（d）为完全显性, 基因位于 Z 染色体上。卷羽鸡适应高温环境, 矮小鸡饲料利用率高。为培育耐热节粮型种鸡以实现规模化生产, 研究人员拟通过杂交将 $\\mathrm{d}$ 基因引入某特色肉鸡体内，育种过程如图所示。已知亲本均为纯合子，下列分析错误的是（ ）\n\n[图1]\n\nA: 符合要求的新品种个体的基因型为 $F F Z^{d} W 、 F F Z^{d} Z^{d}$\nB: 群体I和群体II个体的表型及基因型与亲本的都不同\nC: 为缩短育种时间, 应从群体 I、II中分别选择母本和父本进行杂交\nD: 若群体I、II的雌雄个体随机交配, 子代中半卷羽矮小鸡占 $3 / 8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-30.jpg?height=677&width=1268&top_left_y=173&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-30.jpg?height=63&width=1379&top_left_y=2167&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_914",
        "problem": "一对表现型正常的夫妇生了一个患半乳糖血症的女儿和一个正常的儿子。若这个儿子与一个半乳糖血症携带者的女性结婚, 他们所生子女中, 理论上患半乳糖血症女儿的可能性是( )\nA: $1 / 12$\nB: $1 / 8$\nC: $1 / 6$\nD: $1 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一对表现型正常的夫妇生了一个患半乳糖血症的女儿和一个正常的儿子。若这个儿子与一个半乳糖血症携带者的女性结婚, 他们所生子女中, 理论上患半乳糖血症女儿的可能性是( )\n\nA: $1 / 12$\nB: $1 / 8$\nC: $1 / 6$\nD: $1 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1138",
        "problem": "## AVOIDING FREEZING IN ANTARCTICA\n\nThe research group of Associate Professor Clive Evans at the University of Auckland has an active Antarctic Programme investigating how fish avoid freezing in the frigid waters of the Southern Ocean surrounding Antarctica.\n\nSeawater temperature hovers close to its freezing point of around $-1.93^{\\circ} \\mathrm{C}$ throughout the year but most marine fish freeze at about $-0.7^{\\circ} \\mathrm{C}$ so they cannot survive in the icy Antarctic waters. Notothenioid fishes (icefish) thrive in this freezing environment because they are able to produce antifreeze glycoproteins (AFGPs). AFGPs bind to and inhibit the growth of minute ice crystals that occasionally enter the fish, thus preventing their body fluids from freezing. This key evolutionary innovation allowed the icefish to colonize the frigid waters of the Southern Ocean some 5-15 million years ago.\n\n[figure1]\n\nIcefish risk freezing of the intestinal tract by swallowing ice in ingested seawater or food. This suggests that AFGPs should be present in the stomach and intestinal fluids to decrease the risk of freezing initiated by ingested ice. In the diagram below, the activity of antifreeze in these fluids is examined by measuring the difference between melting and freezing points in degrees Celsius, a measure of thermal hysteresis (TH). TH is represented by the gray box.\n\n[figure2]\n\nCheng C C et al. PNAS 2006;103:10491-10496Examining this data, in which fluid is the $\\mathrm{TH}$ particularly large\nA: Intestinal fluid only\nB: Stomach fluid only\nC: Pancreatic fluid only\nD: Both intestinal and stomach fluid\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## AVOIDING FREEZING IN ANTARCTICA\n\nThe research group of Associate Professor Clive Evans at the University of Auckland has an active Antarctic Programme investigating how fish avoid freezing in the frigid waters of the Southern Ocean surrounding Antarctica.\n\nSeawater temperature hovers close to its freezing point of around $-1.93^{\\circ} \\mathrm{C}$ throughout the year but most marine fish freeze at about $-0.7^{\\circ} \\mathrm{C}$ so they cannot survive in the icy Antarctic waters. Notothenioid fishes (icefish) thrive in this freezing environment because they are able to produce antifreeze glycoproteins (AFGPs). AFGPs bind to and inhibit the growth of minute ice crystals that occasionally enter the fish, thus preventing their body fluids from freezing. This key evolutionary innovation allowed the icefish to colonize the frigid waters of the Southern Ocean some 5-15 million years ago.\n\n[figure1]\n\nIcefish risk freezing of the intestinal tract by swallowing ice in ingested seawater or food. This suggests that AFGPs should be present in the stomach and intestinal fluids to decrease the risk of freezing initiated by ingested ice. In the diagram below, the activity of antifreeze in these fluids is examined by measuring the difference between melting and freezing points in degrees Celsius, a measure of thermal hysteresis (TH). TH is represented by the gray box.\n\n[figure2]\n\nCheng C C et al. PNAS 2006;103:10491-10496\n\nproblem:\nExamining this data, in which fluid is the $\\mathrm{TH}$ particularly large\n\nA: Intestinal fluid only\nB: Stomach fluid only\nC: Pancreatic fluid only\nD: Both intestinal and stomach fluid\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-30.jpg?height=554&width=813&top_left_y=440&top_left_x=1084",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-30.jpg?height=1011&width=868&top_left_y=1413&top_left_x=114"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_814",
        "problem": "亨廷顿症(HD)是由于异常 Htt 蛋白堆积, 从而损伤神经系统功能的一种单基因遗传病,相关基因不位于 $\\mathrm{Y}$ 染色体上，平均发病年龄为 45 岁左右。\n某家庭 HD 的系谱图及对该家庭中部分个体相关基因分析的电泳图如下所示。下列叙述正确的是（ ）\n[图1]\nA: $\\mathrm{II}_{3}$ 的致病基因一定来自 $\\mathrm{I}_{2}$ ，并遗传给 $\\mathrm{III}_{2}$ 和 $\\mathrm{III}_{3}$\nB: $\\mathrm{III}_{3}$ 与正常女性婚配, 所生孩子不会患 HD\nC: 由 $\\mathrm{III}_{2}$ 与 $\\mathrm{III}_{3}$ 的表型差异推测 $\\mathrm{HD}$ 的发病与环境相关\nD: 可向体细胞内随机导入正常基因用于治疗 HD\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n亨廷顿症(HD)是由于异常 Htt 蛋白堆积, 从而损伤神经系统功能的一种单基因遗传病,相关基因不位于 $\\mathrm{Y}$ 染色体上，平均发病年龄为 45 岁左右。\n某家庭 HD 的系谱图及对该家庭中部分个体相关基因分析的电泳图如下所示。下列叙述正确的是（ ）\n[图1]\n\nA: $\\mathrm{II}_{3}$ 的致病基因一定来自 $\\mathrm{I}_{2}$ ，并遗传给 $\\mathrm{III}_{2}$ 和 $\\mathrm{III}_{3}$\nB: $\\mathrm{III}_{3}$ 与正常女性婚配, 所生孩子不会患 HD\nC: 由 $\\mathrm{III}_{2}$ 与 $\\mathrm{III}_{3}$ 的表型差异推测 $\\mathrm{HD}$ 的发病与环境相关\nD: 可向体细胞内随机导入正常基因用于治疗 HD\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-70.jpg?height=732&width=1192&top_left_y=152&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_858",
        "problem": "若利用根瘤农杆菌转基因技术将抗虫基因和抗除草剂基因转入大豆, 获得若干转基因植物 ( $T_{0}$ 代), 从中选择抗虫抗除草剂的单株 $S_{1} 、 S_{2}$ 和 $S_{3}$ 分别进行自交获得 $T_{1}$ 代, $T_{1}$代性状表现如图所示。已知目的基因能 1 次或多次插入并整合到受体细胞染色体上。下列叙述正确的是 ( )\n\n[图1]\nA: 抗虫对不抗虫表现为完全显性, 抗除草剂对不抗除草剂表现为不完全显性\nB: 若给 $S_{1}$ 后代 $T_{1}$ 植株喷施适量的除草剂, 让存活植株自交, 得到的自交一代群体中不抗虫抗除草剂的基因型频率为 $1 / 2$\nC: 根瘤衣杆菌 $\\mathrm{Ti}$ 质粒携带的抗虫和抗除草剂基因分别插入到了 $\\mathrm{S}_{2}$ 的 2 条非同源染色体上, 并正常表达\nD: 若取 $S_{3}$ 后代 $T_{1}$ 纯合抗虫不抗除草剂与纯合不抗虫抗除草剂单株杂交, 得到的子二代中抗虫抗除草剂的纯合子占 $1 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n若利用根瘤农杆菌转基因技术将抗虫基因和抗除草剂基因转入大豆, 获得若干转基因植物 ( $T_{0}$ 代), 从中选择抗虫抗除草剂的单株 $S_{1} 、 S_{2}$ 和 $S_{3}$ 分别进行自交获得 $T_{1}$ 代, $T_{1}$代性状表现如图所示。已知目的基因能 1 次或多次插入并整合到受体细胞染色体上。下列叙述正确的是 ( )\n\n[图1]\n\nA: 抗虫对不抗虫表现为完全显性, 抗除草剂对不抗除草剂表现为不完全显性\nB: 若给 $S_{1}$ 后代 $T_{1}$ 植株喷施适量的除草剂, 让存活植株自交, 得到的自交一代群体中不抗虫抗除草剂的基因型频率为 $1 / 2$\nC: 根瘤衣杆菌 $\\mathrm{Ti}$ 质粒携带的抗虫和抗除草剂基因分别插入到了 $\\mathrm{S}_{2}$ 的 2 条非同源染色体上, 并正常表达\nD: 若取 $S_{3}$ 后代 $T_{1}$ 纯合抗虫不抗除草剂与纯合不抗虫抗除草剂单株杂交, 得到的子二代中抗虫抗除草剂的纯合子占 $1 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-080.jpg?height=483&width=1008&top_left_y=158&top_left_x=321"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_82",
        "problem": "Bacteria regulate gene expression through transcription factors that sense environmental changes in order to adapt to the ever-changing environment. One transcription factor often controls multiple genes. Since the expression of a gene consumes energy, the selection of the gene group to be expressed is important for the survival strategy of the bacterium. It is often observed that bacteria move vigorously in search of nutrients in the aquatic environment, while bacteria in biofilms rarely move.\nA: Generally, transcription factors, which induce the expression of glucose utilizing genes, suppress the expression of lactose-metabolizing genes.\nB: Transcription factors activated by phosphate depletion activate the expression of glycogen-utilizing genes.\nC: Transcription factors that activate the expression of fatty acids-metabolizing genes are generally activated under oxygen depleted conditions.\nD: Transcription factors that activate the expression of biofilm-forming genes usually suppress the expression of the genes of flagella formation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBacteria regulate gene expression through transcription factors that sense environmental changes in order to adapt to the ever-changing environment. One transcription factor often controls multiple genes. Since the expression of a gene consumes energy, the selection of the gene group to be expressed is important for the survival strategy of the bacterium. It is often observed that bacteria move vigorously in search of nutrients in the aquatic environment, while bacteria in biofilms rarely move.\n\nA: Generally, transcription factors, which induce the expression of glucose utilizing genes, suppress the expression of lactose-metabolizing genes.\nB: Transcription factors activated by phosphate depletion activate the expression of glycogen-utilizing genes.\nC: Transcription factors that activate the expression of fatty acids-metabolizing genes are generally activated under oxygen depleted conditions.\nD: Transcription factors that activate the expression of biofilm-forming genes usually suppress the expression of the genes of flagella formation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1477",
        "problem": "moebas are found in many different environments. An amoeba found living in a lake formed by a melting glacier was placed in a bucket of water taken from a river estuary. An amoeba taken from the ocean was also placed in the bucket.\n\n[figure1]\n\n\nWhat happened to the amoeba taken from the ocean?\nA: It expanded.\nB: It shrank.\nC: It stayed the same.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nmoebas are found in many different environments. An amoeba found living in a lake formed by a melting glacier was placed in a bucket of water taken from a river estuary. An amoeba taken from the ocean was also placed in the bucket.\n\n[figure1]\n\n\nWhat happened to the amoeba taken from the ocean?\n\nA: It expanded.\nB: It shrank.\nC: It stayed the same.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-03.jpg?height=751&width=1099&top_left_y=752&top_left_x=244"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1164",
        "problem": "New Zealand is the terrestrial part of a submerged continent called Zealandia, formed over a period of 500 million years. It was once part of the Gondwana supercontinent, and split off 80-60 million years ago when the Tasman Sea formed. New Zealand's complex geological history is mirrored in the complex patterns of evolution seen in New Zealand's flora and fauna.$$\n\\begin{aligned}\n& \\text { Time - million } \\\\\n& \\text { years (log scale) }\n\\end{aligned}\n$$\n\n[figure1]\n\nUsing the information from the geological timeline at left, place the paleo-geographical maps (A-E) of New Zealand below into the correct order.\n[figure2]\nA: $B C A D E$\nB: $B A C D E$\nC: $C B D A E$\nD: $C A B D E$\nE: $C D B A E$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nNew Zealand is the terrestrial part of a submerged continent called Zealandia, formed over a period of 500 million years. It was once part of the Gondwana supercontinent, and split off 80-60 million years ago when the Tasman Sea formed. New Zealand's complex geological history is mirrored in the complex patterns of evolution seen in New Zealand's flora and fauna.\n\nproblem:\n$$\n\\begin{aligned}\n& \\text { Time - million } \\\\\n& \\text { years (log scale) }\n\\end{aligned}\n$$\n\n[figure1]\n\nUsing the information from the geological timeline at left, place the paleo-geographical maps (A-E) of New Zealand below into the correct order.\n[figure2]\n\nA: $B C A D E$\nB: $B A C D E$\nC: $C B D A E$\nD: $C A B D E$\nE: $C D B A E$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-21.jpg?height=1856&width=743&top_left_y=814&top_left_x=128",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-21.jpg?height=1276&width=986&top_left_y=701&top_left_x=980"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_644",
        "problem": "2022 年 10 月, 我国科学家在小鼠的单倍体胚胎干细胞上首次实现 4 号和 5 号染色体的断裂和重新连接, 培育得到 $\\operatorname{ch} 14+5$ 单倍体小鼠。用野生型小鼠与 $\\operatorname{ch} 14+5$ 单倍体小鼠为材料可得到 ch14+5 二倍体纯合子小鼠, 但是成功的机率相对于自然状态下产生的试卷第 18 页，共 88 页\n后代低很多。下列叙述错误的是\nA: chl4+5 单倍体小鼠可能会表现出一些未知隐性基因所控制的表型\nB: 培育 ch14+5 单倍体小鼠过程涉及染色体结构变异和染色体数目变异\nC: 融合染色体无法与正常染色体联会降低 ch14+5 二倍体纯合子小鼠出现机率\nD: ch14+5 单倍体小鼠可用于研究染色体融合导致的不孕不育等疾病\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2022 年 10 月, 我国科学家在小鼠的单倍体胚胎干细胞上首次实现 4 号和 5 号染色体的断裂和重新连接, 培育得到 $\\operatorname{ch} 14+5$ 单倍体小鼠。用野生型小鼠与 $\\operatorname{ch} 14+5$ 单倍体小鼠为材料可得到 ch14+5 二倍体纯合子小鼠, 但是成功的机率相对于自然状态下产生的试卷第 18 页，共 88 页\n后代低很多。下列叙述错误的是\n\nA: chl4+5 单倍体小鼠可能会表现出一些未知隐性基因所控制的表型\nB: 培育 ch14+5 单倍体小鼠过程涉及染色体结构变异和染色体数目变异\nC: 融合染色体无法与正常染色体联会降低 ch14+5 二倍体纯合子小鼠出现机率\nD: ch14+5 单倍体小鼠可用于研究染色体融合导致的不孕不育等疾病\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_54",
        "problem": "Anthocyanins are vacuolar pigments that consist of two phenyl rings and one heterocyclic ring. They are the source of red to blue colours of flowers and fruits. Changes of vacuolar and soil $\\mathrm{pH}$, formation of complexes with metal ions, and introduction of $-\\mathrm{OH}$ groups particularly at $3^{\\prime}$ and $5^{\\prime}$ positions and subsequent methylations at these positions all result in production of variations in the colours of the pigments.\n\nFor example, increase in the number of hydroxyl groups or complex formation with metal ions shifts absorption of the pigments to longer wavelengths, whereas methoxyl groups $\\left(\\mathrm{OCH}_{3}\\right)$ shift absorption to a slightly shorter wavelengths. Given that plargonidin (1) is the most abundant pigment in the variety A of Tropical Water Lily flowers (Nymphaea spp.) indicate if each of the following statement is true or false.\n[figure1]\n\n\n[figure2]\nA: Pigment 3 is the most abundant pigment in flower E.\nB: In Nymphaea variety C, pigment 4 is more abundant than 5 .\nC: Compared to the other flowers, pigment 6 is mostly abundant in flower D.\nD: The colour of flower $\\mathrm{B}$ is explained by presence of higher levels of pigment 2 as compared to pigment 1.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAnthocyanins are vacuolar pigments that consist of two phenyl rings and one heterocyclic ring. They are the source of red to blue colours of flowers and fruits. Changes of vacuolar and soil $\\mathrm{pH}$, formation of complexes with metal ions, and introduction of $-\\mathrm{OH}$ groups particularly at $3^{\\prime}$ and $5^{\\prime}$ positions and subsequent methylations at these positions all result in production of variations in the colours of the pigments.\n\nFor example, increase in the number of hydroxyl groups or complex formation with metal ions shifts absorption of the pigments to longer wavelengths, whereas methoxyl groups $\\left(\\mathrm{OCH}_{3}\\right)$ shift absorption to a slightly shorter wavelengths. Given that plargonidin (1) is the most abundant pigment in the variety A of Tropical Water Lily flowers (Nymphaea spp.) indicate if each of the following statement is true or false.\n[figure1]\n\n\n[figure2]\n\nA: Pigment 3 is the most abundant pigment in flower E.\nB: In Nymphaea variety C, pigment 4 is more abundant than 5 .\nC: Compared to the other flowers, pigment 6 is mostly abundant in flower D.\nD: The colour of flower $\\mathrm{B}$ is explained by presence of higher levels of pigment 2 as compared to pigment 1.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-31.jpg?height=570&width=924&top_left_y=865&top_left_x=564",
            "https://i.postimg.cc/pXjY8CG1/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1442",
        "problem": "Julie noticed an introduced species of weed disrupted the growth of native plant seedlings in the local park. She also recalled that symbiotic bacteria in the soil play a role in native plant seedling growth. Julie took two soil samples from an area where the weed species were growing ('Affected' soil samples) and sterilised ONE sample using ultraviolet light. She then obtained another two soil samples from an area without any evidence of the weed species ('Unaffected' soil samples) and sterilised one sample as well. She then planted some native seedlings in each soil sample and let them grow for 3 months. At the end of 3 months, she removed all the roots, stems, and leaves of the seedlings and weighed the biomass of the collected plant material.\n\nShe repeated this experiment 5 times with five different sets of soils and documented the results in a table:\n\n| Soil category | Weighed seedling biomass (g) |  |  |  |  |\n| :--- | :---: | :---: | :---: | :---: | :---: |\n|  | First set | Second set | Third set | Fourth set | Fifth set |\n| Affected | 6 | 11 | 1 | 2 | 4 |\n| Unaffected | 141 | 149 | 139 | 128 | 140 |\n| Sterilised affected | 3 | 7 | 5 | 2 | 5 |\n| Sterilised unaffected | 10 | 21 | 9 | 7 | 14 |\n\nJulie also examined the soil samples with and without the weed species under a light microscope to examine the symbiotic bacteria present in the soil.\n\n| Soil category | Number of symbiotic bacteria viewed per $\\mathbf{m m}^{2}$ under the microscope |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | First set | Second set | Third set | Fourth set | Fifth set |\n| Affected | 6 | 7 | 11 | 6 | 8 |\n| Unaffected | 515 | 534 | 501 | 456 | 466 |\n\nHow could Julie best increase the validity of her experiment?\nA: Increase the number of times she repeated the experiment and collect more numerical data, and then average the results\nB: Introduce a friend to repeat her experiment and check Julie's data collection\nC: Dry out the seedling biomass samples in a dehydrator before recording the weight\nD: Count the number of seedlings instead of weighing them\nE: Recording biomass weights to two decimal points, rather than rounding them to the nearest gram\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nJulie noticed an introduced species of weed disrupted the growth of native plant seedlings in the local park. She also recalled that symbiotic bacteria in the soil play a role in native plant seedling growth. Julie took two soil samples from an area where the weed species were growing ('Affected' soil samples) and sterilised ONE sample using ultraviolet light. She then obtained another two soil samples from an area without any evidence of the weed species ('Unaffected' soil samples) and sterilised one sample as well. She then planted some native seedlings in each soil sample and let them grow for 3 months. At the end of 3 months, she removed all the roots, stems, and leaves of the seedlings and weighed the biomass of the collected plant material.\n\nShe repeated this experiment 5 times with five different sets of soils and documented the results in a table:\n\n| Soil category | Weighed seedling biomass (g) |  |  |  |  |\n| :--- | :---: | :---: | :---: | :---: | :---: |\n|  | First set | Second set | Third set | Fourth set | Fifth set |\n| Affected | 6 | 11 | 1 | 2 | 4 |\n| Unaffected | 141 | 149 | 139 | 128 | 140 |\n| Sterilised affected | 3 | 7 | 5 | 2 | 5 |\n| Sterilised unaffected | 10 | 21 | 9 | 7 | 14 |\n\nJulie also examined the soil samples with and without the weed species under a light microscope to examine the symbiotic bacteria present in the soil.\n\n| Soil category | Number of symbiotic bacteria viewed per $\\mathbf{m m}^{2}$ under the microscope |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | First set | Second set | Third set | Fourth set | Fifth set |\n| Affected | 6 | 7 | 11 | 6 | 8 |\n| Unaffected | 515 | 534 | 501 | 456 | 466 |\n\nHow could Julie best increase the validity of her experiment?\n\nA: Increase the number of times she repeated the experiment and collect more numerical data, and then average the results\nB: Introduce a friend to repeat her experiment and check Julie's data collection\nC: Dry out the seedling biomass samples in a dehydrator before recording the weight\nD: Count the number of seedlings instead of weighing them\nE: Recording biomass weights to two decimal points, rather than rounding them to the nearest gram\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_863",
        "problem": "$\\mathrm{T}_{2}$ 噬菌体由 $\\mathrm{DNA}$ 和蛋白质组成。 $\\mathrm{T}_{2}$ 噬菌体在侵染大肠杆菌时, 注入大肠杆菌体内的是什么物质, 某研究者的假设有三种: 注入 DNA、注入蛋白质、注入整个病毒。该研究者对实验结果进行预测：(1)放射性物质主要分布在上清液中、(2)放射性物质主要分布在沉淀物中、(3)放射性物质在上清液和沉淀物中均有分布。用 ${ }^{35} \\mathrm{~S}$ 和 ${ }^{32} \\mathrm{P}$ 分别标记 $\\mathrm{T}_{2}$噬菌体的蛋白质和 DNA, 让其分别去侵染无标记的大肠杆菌, 则上述假设对应的结果正确的是（ ）\nA: (2)(1)(3), (1)(2)(3)\nB: (1)(2)(2), (2)(1)(2)\nC: (3)(2)(1), (3)(2)(2)\nD: (2)(1)(2), (2)(1)\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{T}_{2}$ 噬菌体由 $\\mathrm{DNA}$ 和蛋白质组成。 $\\mathrm{T}_{2}$ 噬菌体在侵染大肠杆菌时, 注入大肠杆菌体内的是什么物质, 某研究者的假设有三种: 注入 DNA、注入蛋白质、注入整个病毒。该研究者对实验结果进行预测：(1)放射性物质主要分布在上清液中、(2)放射性物质主要分布在沉淀物中、(3)放射性物质在上清液和沉淀物中均有分布。用 ${ }^{35} \\mathrm{~S}$ 和 ${ }^{32} \\mathrm{P}$ 分别标记 $\\mathrm{T}_{2}$噬菌体的蛋白质和 DNA, 让其分别去侵染无标记的大肠杆菌, 则上述假设对应的结果正确的是（ ）\n\nA: (2)(1)(3), (1)(2)(3)\nB: (1)(2)(2), (2)(1)(2)\nC: (3)(2)(1), (3)(2)(2)\nD: (2)(1)(2), (2)(1)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_951",
        "problem": "一个基因型为 $\\mathrm{DdTt}$ 的精原细胞产生了四个精细胞, 其基因与染色体的位置关系见下图。导致该结果最可能的原因是[图1][图2][图3]\nA: 基因突变\nB: 染色体变异\nC: 同源染色体非姐妹染色单体交叉互换\nD: 非同源染色体自由组合\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一个基因型为 $\\mathrm{DdTt}$ 的精原细胞产生了四个精细胞, 其基因与染色体的位置关系见下图。导致该结果最可能的原因是[图1][图2][图3]\n\nA: 基因突变\nB: 染色体变异\nC: 同源染色体非姐妹染色单体交叉互换\nD: 非同源染色体自由组合\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-13.jpg?height=197&width=188&top_left_y=341&top_left_x=180",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-13.jpg?height=194&width=188&top_left_y=343&top_left_x=403",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-13.jpg?height=200&width=188&top_left_y=340&top_left_x=614"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_625",
        "problem": "果蝇中性染色体异常个体 XXX、YO、YY、XYY 致死, XXY 为雌果蝇、 $X O$ 为雄果蝇, 且都能正常存活。XXY 产生的配子随机含有一条或两条性染色体, XO 能够正常产生配子, 所有配子育性均正常。一只染色体组成为 XXY 的果蝇与一只染色体组成为 $\\mathrm{XO}$ 的果蝇杂交产生 $\\mathrm{F}_{1}, \\mathrm{~F}_{1}$ 随机交配产生 $\\mathrm{F}_{2}$ (假设后代足够多)。不考虑其他突变, 下列相关说法错误的是 ( )\nA: $F_{1}$ 中雌果蝇可能携带 $Y$ 染色体\nB: $F_{1}$ 中雌果蝇与雄果蝇的比例为 $1: 1$\nC: $\\mathrm{F}_{1}$ 存活个体中性染色体组成类型共有 4 种\nD: $F_{1} 、 F_{2}$ 中的致死个体的比例分别为 $1 / 6$ 和 $1 / 15$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇中性染色体异常个体 XXX、YO、YY、XYY 致死, XXY 为雌果蝇、 $X O$ 为雄果蝇, 且都能正常存活。XXY 产生的配子随机含有一条或两条性染色体, XO 能够正常产生配子, 所有配子育性均正常。一只染色体组成为 XXY 的果蝇与一只染色体组成为 $\\mathrm{XO}$ 的果蝇杂交产生 $\\mathrm{F}_{1}, \\mathrm{~F}_{1}$ 随机交配产生 $\\mathrm{F}_{2}$ (假设后代足够多)。不考虑其他突变, 下列相关说法错误的是 ( )\n\nA: $F_{1}$ 中雌果蝇可能携带 $Y$ 染色体\nB: $F_{1}$ 中雌果蝇与雄果蝇的比例为 $1: 1$\nC: $\\mathrm{F}_{1}$ 存活个体中性染色体组成类型共有 4 种\nD: $F_{1} 、 F_{2}$ 中的致死个体的比例分别为 $1 / 6$ 和 $1 / 15$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_962",
        "problem": "Among plants, long-term survival depends mostly on competition for:\nA: Water\nB: Air\nC: Minerals\nD: Light\nE: All or none of the above\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAmong plants, long-term survival depends mostly on competition for:\n\nA: Water\nB: Air\nC: Minerals\nD: Light\nE: All or none of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_616",
        "problem": "下图 1 为孟买型血型家系图, 该血型的形成过程如下图 2 所示。已知 $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$ 分别是 $\\mathrm{O}$ 型血和 $\\mathrm{B}$ 型血, $\\mathrm{III}_{1}$ 和 $\\mathrm{III}_{2}$ 分别为 $\\mathrm{AB}$ 型和 $\\mathrm{O}$ 型血, 下列叙述正确的是 ( )\n\n[图1]\nA: $\\mathrm{II}_{5}$ 的基因型为 $\\mathrm{HHI}^{\\mathrm{A}} \\mathrm{i}$ 或 $\\mathrm{HhI}^{\\mathrm{A}} \\mathrm{i}$\nB: 亲本再生一个 $\\mathrm{O}$ 型血的概率为 $3 / 4$\nC: $\\mathrm{II}_{3}$ 产生含有 i 基因的配子概率为 $3 / 16$\nD: $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 的表型受到环境的影响且可遗传\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图 1 为孟买型血型家系图, 该血型的形成过程如下图 2 所示。已知 $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$ 分别是 $\\mathrm{O}$ 型血和 $\\mathrm{B}$ 型血, $\\mathrm{III}_{1}$ 和 $\\mathrm{III}_{2}$ 分别为 $\\mathrm{AB}$ 型和 $\\mathrm{O}$ 型血, 下列叙述正确的是 ( )\n\n[图1]\n\nA: $\\mathrm{II}_{5}$ 的基因型为 $\\mathrm{HHI}^{\\mathrm{A}} \\mathrm{i}$ 或 $\\mathrm{HhI}^{\\mathrm{A}} \\mathrm{i}$\nB: 亲本再生一个 $\\mathrm{O}$ 型血的概率为 $3 / 4$\nC: $\\mathrm{II}_{3}$ 产生含有 i 基因的配子概率为 $3 / 16$\nD: $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 的表型受到环境的影响且可遗传\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-13.jpg?height=306&width=1393&top_left_y=1709&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_522",
        "problem": "假如下图是某生物体 $(2 n=4)$ 正常的细胞分裂示意图, 下列有关叙述错误的是 $(\\quad)$\n\n[图1]\nA: 该细胞处于减数第二次分裂后期\nB: 若染色体(1)有基因 A，则(4)有基因 A 或 $\\mathrm{a}$\nC: 若(2)表示 X 染色体，则(3)表示 Y 染色体\nD: 该细胞的子细胞有 2 对同源染色体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n假如下图是某生物体 $(2 n=4)$ 正常的细胞分裂示意图, 下列有关叙述错误的是 $(\\quad)$\n\n[图1]\n\nA: 该细胞处于减数第二次分裂后期\nB: 若染色体(1)有基因 A，则(4)有基因 A 或 $\\mathrm{a}$\nC: 若(2)表示 X 染色体，则(3)表示 Y 染色体\nD: 该细胞的子细胞有 2 对同源染色体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-09.jpg?height=328&width=259&top_left_y=2343&top_left_x=196"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_524",
        "problem": "为研究鸡 ( $Z W$ 型性别决定) 的性反转是否与 $\\mathrm{D}$ 基因相关, 研究人员分别检测了公鸡胚、母鸡胚和实验组（添加雄激素合成酶抑制剂的母鸡胚）睬化过程中 D 基因的相对表达量, 此过程中实验组最终发育成公鸡, 部分研究数据如下图所示。下列叙述错误的是 ( )\n\n[图1]\nA: D 基因在公鸡胚中的高表达与雄性性别决定相关\nB: 实验组最终发育成的公鸡的性染色体组成是 ZW\nC: 实验结果可以说明 D 基因的遗传遵循伴性遗传规律\nD: 实验表明鸡的性别是基因和环境共同作用的结果\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n为研究鸡 ( $Z W$ 型性别决定) 的性反转是否与 $\\mathrm{D}$ 基因相关, 研究人员分别检测了公鸡胚、母鸡胚和实验组（添加雄激素合成酶抑制剂的母鸡胚）睬化过程中 D 基因的相对表达量, 此过程中实验组最终发育成公鸡, 部分研究数据如下图所示。下列叙述错误的是 ( )\n\n[图1]\n\nA: D 基因在公鸡胚中的高表达与雄性性别决定相关\nB: 实验组最终发育成的公鸡的性染色体组成是 ZW\nC: 实验结果可以说明 D 基因的遗传遵循伴性遗传规律\nD: 实验表明鸡的性别是基因和环境共同作用的结果\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-59.jpg?height=474&width=768&top_left_y=1945&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_163",
        "problem": "Arabidopsis thaliana was used in a study on photosynthesis. A. thaliana plants of same age, with leaf blades of equal shape and size, were placed under various light sources that emitted different colours. Three plants were exposed to each light source. The groupings of were as follows:\n\n$$\n\\begin{array}{ll}\n\\text { Group a: blue light } & \\text { Group d: blue and red } \\\\\n\\text { Group b: green light } & \\text { Group e: yellow and green } \\\\\n\\text { Group c: red light } &\n\\end{array}\n$$\n\nAfter 5 days of illumination at equal intensity comparable to normal day light, in terms of total amount of photons and duration, the plant's physiological parameters were compared. The compensation point was measured using sunlight.\nA: The plants of group \"a\" showed the highest photosynthesis rate.\nB: Among the various groups, the light compensation point for $\\mathrm{CO}_{2}$ fixation was the lowest for group \"b\".\nC: The biomass of group \"d\" plants was higher than group \"c\" plants.\nD: At light intensities just above the compensation point, the rate of photosynthesis of group \"e\" is expected to increase linearly with light.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nArabidopsis thaliana was used in a study on photosynthesis. A. thaliana plants of same age, with leaf blades of equal shape and size, were placed under various light sources that emitted different colours. Three plants were exposed to each light source. The groupings of were as follows:\n\n$$\n\\begin{array}{ll}\n\\text { Group a: blue light } & \\text { Group d: blue and red } \\\\\n\\text { Group b: green light } & \\text { Group e: yellow and green } \\\\\n\\text { Group c: red light } &\n\\end{array}\n$$\n\nAfter 5 days of illumination at equal intensity comparable to normal day light, in terms of total amount of photons and duration, the plant's physiological parameters were compared. The compensation point was measured using sunlight.\n\nA: The plants of group \"a\" showed the highest photosynthesis rate.\nB: Among the various groups, the light compensation point for $\\mathrm{CO}_{2}$ fixation was the lowest for group \"b\".\nC: The biomass of group \"d\" plants was higher than group \"c\" plants.\nD: At light intensities just above the compensation point, the rate of photosynthesis of group \"e\" is expected to increase linearly with light.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_939",
        "problem": "玉米为雌雄同株异花的二倍体植株，利用转基因技术将抗虫基因和抗除草剂基因导入玉米植株, 选育出抗虫抗除草剂玉米植株。取该植株进行自交, $\\mathrm{F}_{1}$ 中抗虫抗除草剂:抗虫不抗除草剂: 不抗虫抗除草剂=14: 1：1. 已知外源基因能 1 次或多次插入并整合到受体细胞染色体上。下列叙述错误的是（）\nA: 玉米植株进行自交时, 不需要进行人工去雄, 但需要对雌荵进行套袋操作\nB: 该玉米植株细胞中，可能抗虫基因和抗除草剂基因同时位于一条染色体上\nC: 该玉米植株细胞中，可能抗虫基因和抗除草剂基因分别位于同源染色体的不同染色体上\nD: 取 $\\mathrm{F}_{1}$ 中抗虫不抗除草剂植株和不抗虫抗除草剂植株随机传粉, $\\mathrm{F}_{2}$ 中抗虫抗除草剂植株所占比例为 $8 / 9$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n玉米为雌雄同株异花的二倍体植株，利用转基因技术将抗虫基因和抗除草剂基因导入玉米植株, 选育出抗虫抗除草剂玉米植株。取该植株进行自交, $\\mathrm{F}_{1}$ 中抗虫抗除草剂:抗虫不抗除草剂: 不抗虫抗除草剂=14: 1：1. 已知外源基因能 1 次或多次插入并整合到受体细胞染色体上。下列叙述错误的是（）\n\nA: 玉米植株进行自交时, 不需要进行人工去雄, 但需要对雌荵进行套袋操作\nB: 该玉米植株细胞中，可能抗虫基因和抗除草剂基因同时位于一条染色体上\nC: 该玉米植株细胞中，可能抗虫基因和抗除草剂基因分别位于同源染色体的不同染色体上\nD: 取 $\\mathrm{F}_{1}$ 中抗虫不抗除草剂植株和不抗虫抗除草剂植株随机传粉, $\\mathrm{F}_{2}$ 中抗虫抗除草剂植株所占比例为 $8 / 9$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-85.jpg?height=306&width=1380&top_left_y=1383&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_976",
        "problem": "Pollen grains were treated with colchicine in culture. The resulting plants were\nA: No more than large calluses\nB: Dihaploid and fertile\nC: Dihaploid and sterile\nD: Haploid and sterile\nE: Haploid and fertile\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPollen grains were treated with colchicine in culture. The resulting plants were\n\nA: No more than large calluses\nB: Dihaploid and fertile\nC: Dihaploid and sterile\nD: Haploid and sterile\nE: Haploid and fertile\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1178",
        "problem": "Which one of the following steps represents the essential link between the light-dependent and light-independent stages of photosynthesis?\nA: photolysis of water and photophosphorylation\nB: carbon dioxide fixation\nC: activation of chlorophyll molecules\nD: production of reduced electron carriers and photophosphorylation\nE: reduction of phosphoglycerate\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following steps represents the essential link between the light-dependent and light-independent stages of photosynthesis?\n\nA: photolysis of water and photophosphorylation\nB: carbon dioxide fixation\nC: activation of chlorophyll molecules\nD: production of reduced electron carriers and photophosphorylation\nE: reduction of phosphoglycerate\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_712",
        "problem": "研究发现, 细胞中染色体的正确排列、分离与染色单体之间的粘连蛋白有关。动物\n细胞内存在一种 SGO 蛋白, 其主要集中在染色体的着丝粒位置, 保护粘连蛋白不被水解酶破坏，粘连蛋白如下图所示。相关叙述错误的是（ ）\n\n[图1]\nA: 粘连蛋白和 SGO 蛋白结构不同根本原因是控制它们合成的基因不同\nB: 该水解酶能水解粘连蛋白但对染色体蛋白无影响, 这体现酶专一性\nC: 如果阻断正在分裂的动物体细胞内 SGO 蛋白的合成, 则图示过程会滞后进行\nD: 图中染色体行为变化可发生在有丝分裂后期或减数第二次分裂后期\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究发现, 细胞中染色体的正确排列、分离与染色单体之间的粘连蛋白有关。动物\n细胞内存在一种 SGO 蛋白, 其主要集中在染色体的着丝粒位置, 保护粘连蛋白不被水解酶破坏，粘连蛋白如下图所示。相关叙述错误的是（ ）\n\n[图1]\n\nA: 粘连蛋白和 SGO 蛋白结构不同根本原因是控制它们合成的基因不同\nB: 该水解酶能水解粘连蛋白但对染色体蛋白无影响, 这体现酶专一性\nC: 如果阻断正在分裂的动物体细胞内 SGO 蛋白的合成, 则图示过程会滞后进行\nD: 图中染色体行为变化可发生在有丝分裂后期或减数第二次分裂后期\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-44.jpg?height=345&width=459&top_left_y=330&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_718",
        "problem": "豌豆的红花对白花是显性, 长花粉对圆花粉是显性。现有红花长花粉与白花圆花粉植株杂交, $F_{1}$ 都是红花长花粉植株。若 $F_{1}$ 自交获得 $F_{2}$ 共 200 株植株, 其中白花圆花粉个体为 32 株, 下列说法不正确的是（）\nA: 控制豌豆花色和花粉形状的两对等位基因不遵循自由组合定律\nB: 踠豆杂交实验时需要进行去雄 $\\rightarrow$ 套袋 $\\rightarrow$ 授粉 $\\rightarrow$ 套袋的过程\nC: 若 $F_{1}$ 与白花圆花粉植株杂交, 则子代表型比例为 $4: 1: 1: 4$\nD: $F_{2}$ 豌豆植株中杂合的红花圆花粉植株所占比例为 $4 \\%$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n豌豆的红花对白花是显性, 长花粉对圆花粉是显性。现有红花长花粉与白花圆花粉植株杂交, $F_{1}$ 都是红花长花粉植株。若 $F_{1}$ 自交获得 $F_{2}$ 共 200 株植株, 其中白花圆花粉个体为 32 株, 下列说法不正确的是（）\n\nA: 控制豌豆花色和花粉形状的两对等位基因不遵循自由组合定律\nB: 踠豆杂交实验时需要进行去雄 $\\rightarrow$ 套袋 $\\rightarrow$ 授粉 $\\rightarrow$ 套袋的过程\nC: 若 $F_{1}$ 与白花圆花粉植株杂交, 则子代表型比例为 $4: 1: 1: 4$\nD: $F_{2}$ 豌豆植株中杂合的红花圆花粉植株所占比例为 $4 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_969",
        "problem": "Polarity in the developing Drosophila embryo is determined by\nA: A protein gradient of the segmentation protein engrailed\nB: A protein gradient of the bicoid protein expressed from maternal mRNA\nC: A protein gradient of the gap protein bunchback\nD: Expression of the segmentation protein engrailed throughout the embryo\nE: Expression of the gap protein hunchback throughout the embryo\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPolarity in the developing Drosophila embryo is determined by\n\nA: A protein gradient of the segmentation protein engrailed\nB: A protein gradient of the bicoid protein expressed from maternal mRNA\nC: A protein gradient of the gap protein bunchback\nD: Expression of the segmentation protein engrailed throughout the embryo\nE: Expression of the gap protein hunchback throughout the embryo\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1504",
        "problem": "Dynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\n\nIt is more common for a pyrimidine nucleotide ( $\\mathrm{C}$ or $\\mathrm{T}$ ) to mutate into another pyrimidine, than into a purine ( $\\mathrm{G}$ or A). The sequence studied above also turned out to be in a protein coding sequence.\n\nHow should the scoring system be adapted to account for these findings?\nA: Penalty for $\\mathrm{C}$ to $\\mathrm{T}$ mutations increased\nB: Penalty for gaps in alignment increased\nC: Penalty for $\\mathrm{C}$ to $\\mathrm{A}$ mutations increased\nD: Penalty for $\\mathrm{C}$ to $\\mathrm{A}$ mutations reduced\nE: B and C only\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\n\nIt is more common for a pyrimidine nucleotide ( $\\mathrm{C}$ or $\\mathrm{T}$ ) to mutate into another pyrimidine, than into a purine ( $\\mathrm{G}$ or A). The sequence studied above also turned out to be in a protein coding sequence.\n\nHow should the scoring system be adapted to account for these findings?\n\nA: Penalty for $\\mathrm{C}$ to $\\mathrm{T}$ mutations increased\nB: Penalty for gaps in alignment increased\nC: Penalty for $\\mathrm{C}$ to $\\mathrm{A}$ mutations increased\nD: Penalty for $\\mathrm{C}$ to $\\mathrm{A}$ mutations reduced\nE: B and C only\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=925&width=928&top_left_y=354&top_left_x=238",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=922&width=928&top_left_y=1458&top_left_x=238"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_541",
        "problem": "水稻 $(2 \\mathrm{~N}=24)$ 雌雄同花, 自然状态下自花传粉。水稻的无香味 $(\\mathrm{Y})$ 对有香味\n\n（y）为显性。无香味人工杂交水稻（如图 1) 经过处理得到三种类型的水稻（如图 2)。若类型二的植株 $\\mathrm{M}$ 可正常产生配子，各种配子活力相同且可育，但含 $\\mathrm{Y}$ 或 $\\mathrm{y}$ 基因的个体才能存活。类型三的植株 $\\mathrm{N}$ 自交, 子代无香味: 有香味 $=2$ : 1 。下列分析错误的是\n\n[图1]\n\n图 1\n\n[图2]\n\n类型一\n\n[图3]\n\n类型二\n\n[图4]\n\n类型三\nA: M 自交产生的子代有 8 种基因型\nB: M 自交后代中无香味：有香味 $=4$ : 1\nC: $\\mathrm{N}$ 自交子代出现 2: 1 的原因可能是缺失染色体影响了花粉的受精能力\nD: 由图 1 水稻获得图 2 三种类型水稻是用理化学方法诱发基因突变\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n水稻 $(2 \\mathrm{~N}=24)$ 雌雄同花, 自然状态下自花传粉。水稻的无香味 $(\\mathrm{Y})$ 对有香味\n\n（y）为显性。无香味人工杂交水稻（如图 1) 经过处理得到三种类型的水稻（如图 2)。若类型二的植株 $\\mathrm{M}$ 可正常产生配子，各种配子活力相同且可育，但含 $\\mathrm{Y}$ 或 $\\mathrm{y}$ 基因的个体才能存活。类型三的植株 $\\mathrm{N}$ 自交, 子代无香味: 有香味 $=2$ : 1 。下列分析错误的是\n\n[图1]\n\n图 1\n\n[图2]\n\n类型一\n\n[图3]\n\n类型二\n\n[图4]\n\n类型三\n\nA: M 自交产生的子代有 8 种基因型\nB: M 自交后代中无香味：有香味 $=4$ : 1\nC: $\\mathrm{N}$ 自交子代出现 2: 1 的原因可能是缺失染色体影响了花粉的受精能力\nD: 由图 1 水稻获得图 2 三种类型水稻是用理化学方法诱发基因突变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-15.jpg?height=296&width=300&top_left_y=2282&top_left_x=364",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-15.jpg?height=329&width=296&top_left_y=2283&top_left_x=837",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-15.jpg?height=242&width=303&top_left_y=2289&top_left_x=1162",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-15.jpg?height=246&width=303&top_left_y=2284&top_left_x=1482"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_570",
        "problem": "某单基因遗传病的系谱图如下，其中II-3 不携带该致病基因。不考虑基因突变和染色体变异。下列分析错误的是（）\n\n[图1]\nA: 若该致病基因位于常染色体，III-1 与正常女性婚配，子女患病概率相同\nB: 若该致病基因位于性染色体，III-1 患病的原因是性染色体间发生了交换\nC: 若该致病基因位于性染色体, III-1 与正常女性婚配, 女儿的患病概率高于儿子\nD: III-3 与正常男性婚配, 子代患病的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某单基因遗传病的系谱图如下，其中II-3 不携带该致病基因。不考虑基因突变和染色体变异。下列分析错误的是（）\n\n[图1]\n\nA: 若该致病基因位于常染色体，III-1 与正常女性婚配，子女患病概率相同\nB: 若该致病基因位于性染色体，III-1 患病的原因是性染色体间发生了交换\nC: 若该致病基因位于性染色体, III-1 与正常女性婚配, 女儿的患病概率高于儿子\nD: III-3 与正常男性婚配, 子代患病的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-18.jpg?height=456&width=734&top_left_y=1551&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1097",
        "problem": "Photosynthesis in plants occurs due to the presence of chlorophyll and other accessory pigments present in leaves. When absorbance of chlorophyll pigments was\nstudied in different layers of tissues in a dorsiventral leaf, following graph was obtained. The lines indicate data obtained when the experiments was carried out in duplicate.\n\n[figure1]\n\nWhich of the following is correct?\nA: The graph indicates that palisade cells are richer in chlorophyll content as compared to spongy mesophyll cells.\nB: Spongy mesophyll cells are more adapted to gather available light which is then passed on to the reaction centre.\nC: If instead of chlorophyll estimation, invitro reduction of redox indicator is studied, it will show the same trend along the two surfaces.\nD: (Shade-sun) plant pair will show the same trend that is observed for the graph of (abaxial adaxial) surface.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPhotosynthesis in plants occurs due to the presence of chlorophyll and other accessory pigments present in leaves. When absorbance of chlorophyll pigments was\nstudied in different layers of tissues in a dorsiventral leaf, following graph was obtained. The lines indicate data obtained when the experiments was carried out in duplicate.\n\n[figure1]\n\nWhich of the following is correct?\n\nA: The graph indicates that palisade cells are richer in chlorophyll content as compared to spongy mesophyll cells.\nB: Spongy mesophyll cells are more adapted to gather available light which is then passed on to the reaction centre.\nC: If instead of chlorophyll estimation, invitro reduction of redox indicator is studied, it will show the same trend along the two surfaces.\nD: (Shade-sun) plant pair will show the same trend that is observed for the graph of (abaxial adaxial) surface.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-21.jpg?height=588&width=710&top_left_y=365&top_left_x=748"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_429",
        "problem": "加拿大底鳟是一种小型鱼类，其体内与细胞呼吸有关的基因 $L D H-B$ 有多种等位基\n\n因, 其中等位基因 $L D H-B^{b}$ 出现的比例表现出随温度和纬度变化的规律，如图所示。下列叙述正确的是（ ）\n\n[图1]\nA: $L D H-B$ 的多种等位基因的基因频率之和大于 1\nB: 冷水环境中突变产生 $L D H-B^{b}$ 的频率比温水环境中高\nC: $L D H-B^{b}$ 基因控制的性状与低温下快速供能有关\nD: 底鳟个体从温水环境游至冷水环境中会发生进化\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n加拿大底鳟是一种小型鱼类，其体内与细胞呼吸有关的基因 $L D H-B$ 有多种等位基\n\n因, 其中等位基因 $L D H-B^{b}$ 出现的比例表现出随温度和纬度变化的规律，如图所示。下列叙述正确的是（ ）\n\n[图1]\n\nA: $L D H-B$ 的多种等位基因的基因频率之和大于 1\nB: 冷水环境中突变产生 $L D H-B^{b}$ 的频率比温水环境中高\nC: $L D H-B^{b}$ 基因控制的性状与低温下快速供能有关\nD: 底鳟个体从温水环境游至冷水环境中会发生进化\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-66.jpg?height=572&width=820&top_left_y=1630&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1207",
        "problem": "[figure1]\n\nWhat is the best conclusion(s) that can be made from the diagram at left of measles cases and immunisation rates in Wales?\nA: A vaccination programme does not always stop outbreaks of disease.\nB: Higher infection rates are linked with lower vaccination rates.\nC: Outbreaks can occur in geographical clusters.\nD: Only $B$ and $C$ are valid conclusions.\nE: A, B and C are valid conclusions.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nWhat is the best conclusion(s) that can be made from the diagram at left of measles cases and immunisation rates in Wales?\n\nA: A vaccination programme does not always stop outbreaks of disease.\nB: Higher infection rates are linked with lower vaccination rates.\nC: Outbreaks can occur in geographical clusters.\nD: Only $B$ and $C$ are valid conclusions.\nE: A, B and C are valid conclusions.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-18.jpg?height=837&width=914&top_left_y=267&top_left_x=114"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1179",
        "problem": "THE RENA OIL SPILL - BIOLOGY WHEN DISASTER STRIKES\n\n[figure1]\n\n\nThe grounding of the CV Rena on October 5, 2011 was New Zealand's worst maritime environmental disaster when the container ship spilled approximately 350 tonnes of heavy fuel oil into the ocean. The container ship ran aground on Otaiti (Astrolabe Reef) just off Mōtitīi Island in the Bay of Plenty, releasing oil into a relatively pristine marine environment. Te Mauri Moana, a group of scientists led by the University of Waikato as part of the Government's \\$ 4.2 million Rena Long-term Environmental Recovery Plan, undertook one of the most comprehensive, multi-disciplinary studies ever done following a marine pollution event. Their findings are summarised in the report; \"Rena Environmental Recovery Monitoring Programme 2011-2013.\"\n\nBy international standards the oil spill was relatively minor but it occurred in an otherwise uncontaminated coastline in an area renowned for its beauty and highly valued for tourism, recreation and fisheries. Iwi, government, commercial stakeholders and the public were rightly concerned there would be long-lasting negative impacts on beaches, reefs and fisheries. Te Mauri Moana reports on fifteen monitoring and research programmes focused on the range of ecosystems and habitats within the Bay of Plenty.\n\nChemical contamination of the sediments and fauna of Otaiti Reef and Môtitī Island and the surrounding soft sediments was examined to determine if the discharge of fuel oil (and later release of container debris) from the Rena resulted in significant chemical contamination of the marine environment and marine organisms. Amongst other contaminants the presence of polycyclic aromatic hydrocarbons (PAHs) from fuel oil was examined in sediments and from selected organisms.\n\nThe maps below shows the PAH levels in sediments collected from A. on Otaiti Reef and B. off-reef.\n\n\nA.\n\n[figure2]\n\nB.\n\n[figure3]\n\n[figure4]\n\nThe graph at left shows the PAH levels in sea urchins collected from Otaiti Reef and nearby islands. The numbers in brackets indicate the number of samples collected at each site. All sea urchins on Otaiti reef showed elevated PAH levels in the gonad with an average level of $0.057 \\mathrm{mg} \\mathrm{kg}^{-1}$. Gut levels averaged $1.58 \\mathrm{mg} \\mathrm{kg}^{-1}$ but at Astro 2 and 3 were orders of magnitude higher than at other sites on Otaiti Reef. PAH levels in the gonad and gut from urchins from Mōtîti Island averaged $0.026 \\mathrm{mg} \\mathrm{kg}^{-1}$ and $0.03 \\mathrm{mg} \\mathrm{kg}^{-1}$ respectively and were similar to levels found elsewhere, including East Cape.The levels of PAH in sea urchin guts from the most contaminated Otaiti reef site are how much higher than those from Mōtitī Island?\nA: 61 times higher.\nB: 53 times higher.\nC: 2 times higher.\nD: 3 times higher.\nE: 4 times higher.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nTHE RENA OIL SPILL - BIOLOGY WHEN DISASTER STRIKES\n\n[figure1]\n\n\nThe grounding of the CV Rena on October 5, 2011 was New Zealand's worst maritime environmental disaster when the container ship spilled approximately 350 tonnes of heavy fuel oil into the ocean. The container ship ran aground on Otaiti (Astrolabe Reef) just off Mōtitīi Island in the Bay of Plenty, releasing oil into a relatively pristine marine environment. Te Mauri Moana, a group of scientists led by the University of Waikato as part of the Government's \\$ 4.2 million Rena Long-term Environmental Recovery Plan, undertook one of the most comprehensive, multi-disciplinary studies ever done following a marine pollution event. Their findings are summarised in the report; \"Rena Environmental Recovery Monitoring Programme 2011-2013.\"\n\nBy international standards the oil spill was relatively minor but it occurred in an otherwise uncontaminated coastline in an area renowned for its beauty and highly valued for tourism, recreation and fisheries. Iwi, government, commercial stakeholders and the public were rightly concerned there would be long-lasting negative impacts on beaches, reefs and fisheries. Te Mauri Moana reports on fifteen monitoring and research programmes focused on the range of ecosystems and habitats within the Bay of Plenty.\n\nChemical contamination of the sediments and fauna of Otaiti Reef and Môtitī Island and the surrounding soft sediments was examined to determine if the discharge of fuel oil (and later release of container debris) from the Rena resulted in significant chemical contamination of the marine environment and marine organisms. Amongst other contaminants the presence of polycyclic aromatic hydrocarbons (PAHs) from fuel oil was examined in sediments and from selected organisms.\n\nThe maps below shows the PAH levels in sediments collected from A. on Otaiti Reef and B. off-reef.\n\n\nA.\n\n[figure2]\n\nB.\n\n[figure3]\n\n[figure4]\n\nThe graph at left shows the PAH levels in sea urchins collected from Otaiti Reef and nearby islands. The numbers in brackets indicate the number of samples collected at each site. All sea urchins on Otaiti reef showed elevated PAH levels in the gonad with an average level of $0.057 \\mathrm{mg} \\mathrm{kg}^{-1}$. Gut levels averaged $1.58 \\mathrm{mg} \\mathrm{kg}^{-1}$ but at Astro 2 and 3 were orders of magnitude higher than at other sites on Otaiti Reef. PAH levels in the gonad and gut from urchins from Mōtîti Island averaged $0.026 \\mathrm{mg} \\mathrm{kg}^{-1}$ and $0.03 \\mathrm{mg} \\mathrm{kg}^{-1}$ respectively and were similar to levels found elsewhere, including East Cape.\n\nproblem:\nThe levels of PAH in sea urchin guts from the most contaminated Otaiti reef site are how much higher than those from Mōtitī Island?\n\nA: 61 times higher.\nB: 53 times higher.\nC: 2 times higher.\nD: 3 times higher.\nE: 4 times higher.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=556&width=1834&top_left_y=276&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=657&width=894&top_left_y=1876&top_left_x=113",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=548&width=854&top_left_y=1942&top_left_x=1092",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-25.jpg?height=591&width=802&top_left_y=201&top_left_x=136"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_89",
        "problem": "As depicted in the following figure, an oat seedling was germinated in the dark. A blue light was given unilaterallyto the right side of the coleoptile, and an agar block containing $\\mathrm{Ca}^{2+}$ was attached to the right side of root tip below the elongation zone.\n\n[figure1]\n\nWhat do you expect the bending responses of the oat seedling will be in a few days?\n|  |  |  |\n| :---: | :--- | :--- |\n| A | Bending towards the light. | Boot |\n| B | Growing upright. | Bending towards the $\\mathrm{Ca}^{2+}$ block. |\n| C | Bending away from the light. | Bending towards the $\\mathrm{Ca}^{2+}$ block. |\n| D | Bending towards the light. | Growing downwards. |\n| E | Growing upright. | Bending away from the $\\mathrm{Ca}^{2+}$ block. |\nA: A\nB: B\nC: C\nD: D\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAs depicted in the following figure, an oat seedling was germinated in the dark. A blue light was given unilaterallyto the right side of the coleoptile, and an agar block containing $\\mathrm{Ca}^{2+}$ was attached to the right side of root tip below the elongation zone.\n\n[figure1]\n\nWhat do you expect the bending responses of the oat seedling will be in a few days?\n|  |  |  |\n| :---: | :--- | :--- |\n| A | Bending towards the light. | Boot |\n| B | Growing upright. | Bending towards the $\\mathrm{Ca}^{2+}$ block. |\n| C | Bending away from the light. | Bending towards the $\\mathrm{Ca}^{2+}$ block. |\n| D | Bending towards the light. | Growing downwards. |\n| E | Growing upright. | Bending away from the $\\mathrm{Ca}^{2+}$ block. |\n\nA: A\nB: B\nC: C\nD: D\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-19.jpg?height=697&width=623&top_left_y=591&top_left_x=571"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1385",
        "problem": "The left ventricle of the heart pumps blood around the body. The amount of blood pumped out of the heart during one contraction is called the cardiac output. The cardiac output varies according to end diastolic volume (see diagram below). The end diastolic volume, also known as the preload, is the volume of blood in the left ventricle before contraction. The cardiac output is also influenced by contractility, which is the ability of the heart muscle to contract.\n\n[figure1]\n\nThis graph depicts the Frank-Starling law. The Frank-Starling Law is the description of cardiac function as it relates to myocyte stretch and contractility.\n\nWhich of the following best describes the Frank-Starling law of the heart?\nA: Increase in end-diastolic volume results in an increased cardiac output.\nB: Increase in end-diastolic volume results in an increased heart rate.\nC: Increase in end-systolic volume indicates an increased cardiac output.\nD: End-systolic volume equals end-diastolic volume.\nE: Increase in cardiac output results in an increase in end-diastolic volume.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe left ventricle of the heart pumps blood around the body. The amount of blood pumped out of the heart during one contraction is called the cardiac output. The cardiac output varies according to end diastolic volume (see diagram below). The end diastolic volume, also known as the preload, is the volume of blood in the left ventricle before contraction. The cardiac output is also influenced by contractility, which is the ability of the heart muscle to contract.\n\n[figure1]\n\nThis graph depicts the Frank-Starling law. The Frank-Starling Law is the description of cardiac function as it relates to myocyte stretch and contractility.\n\nWhich of the following best describes the Frank-Starling law of the heart?\n\nA: Increase in end-diastolic volume results in an increased cardiac output.\nB: Increase in end-diastolic volume results in an increased heart rate.\nC: Increase in end-systolic volume indicates an increased cardiac output.\nD: End-systolic volume equals end-diastolic volume.\nE: Increase in cardiac output results in an increase in end-diastolic volume.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-19.jpg?height=939&width=1245&top_left_y=661&top_left_x=405"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1286",
        "problem": "## AVOIDING FREEZING IN ANTARCTICA\n\nThe research group of Associate Professor Clive Evans at the University of Auckland has an active Antarctic Programme investigating how fish avoid freezing in the frigid waters of the Southern Ocean surrounding Antarctica.\n\nSeawater temperature hovers close to its freezing point of around $-1.93^{\\circ} \\mathrm{C}$ throughout the year but most marine fish freeze at about $-0.7^{\\circ} \\mathrm{C}$ so they cannot survive in the icy Antarctic waters. Notothenioid fishes (icefish) thrive in this freezing environment because they are able to produce antifreeze glycoproteins (AFGPs). AFGPs bind to and inhibit the growth of minute ice crystals that occasionally enter the fish, thus preventing their body fluids from freezing. This key evolutionary innovation allowed the icefish to colonize the frigid waters of the Southern Ocean some 5-15 million years ago.\n\n[figure1]\n\nIcefish risk freezing of the intestinal tract by swallowing ice in ingested seawater or food. This suggests that AFGPs should be present in the stomach and intestinal fluids to decrease the risk of freezing initiated by ingested ice. In the diagram below, the activity of antifreeze in these fluids is examined by measuring the difference between melting and freezing points in degrees Celsius, a measure of thermal hysteresis (TH). TH is represented by the gray box.\n\n[figure2]\n\nCheng C C et al. PNAS 2006;103:10491-10496Which species has the lowest intestinal fluid freezing point (FP)?\nA: D. mawsoni\nB: P. borchgrevinki\nC: T. bernacchii\nD: G. acuticeps\nE: L. dearborni\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## AVOIDING FREEZING IN ANTARCTICA\n\nThe research group of Associate Professor Clive Evans at the University of Auckland has an active Antarctic Programme investigating how fish avoid freezing in the frigid waters of the Southern Ocean surrounding Antarctica.\n\nSeawater temperature hovers close to its freezing point of around $-1.93^{\\circ} \\mathrm{C}$ throughout the year but most marine fish freeze at about $-0.7^{\\circ} \\mathrm{C}$ so they cannot survive in the icy Antarctic waters. Notothenioid fishes (icefish) thrive in this freezing environment because they are able to produce antifreeze glycoproteins (AFGPs). AFGPs bind to and inhibit the growth of minute ice crystals that occasionally enter the fish, thus preventing their body fluids from freezing. This key evolutionary innovation allowed the icefish to colonize the frigid waters of the Southern Ocean some 5-15 million years ago.\n\n[figure1]\n\nIcefish risk freezing of the intestinal tract by swallowing ice in ingested seawater or food. This suggests that AFGPs should be present in the stomach and intestinal fluids to decrease the risk of freezing initiated by ingested ice. In the diagram below, the activity of antifreeze in these fluids is examined by measuring the difference between melting and freezing points in degrees Celsius, a measure of thermal hysteresis (TH). TH is represented by the gray box.\n\n[figure2]\n\nCheng C C et al. PNAS 2006;103:10491-10496\n\nproblem:\nWhich species has the lowest intestinal fluid freezing point (FP)?\n\nA: D. mawsoni\nB: P. borchgrevinki\nC: T. bernacchii\nD: G. acuticeps\nE: L. dearborni\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-30.jpg?height=554&width=813&top_left_y=440&top_left_x=1084",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-30.jpg?height=1011&width=868&top_left_y=1413&top_left_x=114"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1139",
        "problem": "Conservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.Trawl fisheries contributes 1 death per year. This exceeds the estimate PBR by:\nA: Between $10-22.72$ times.\nB: 10 times.\nC: 22.72 times.\nD: 0.1 times.\nE: 0.04 times.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nConservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.\n\nproblem:\nTrawl fisheries contributes 1 death per year. This exceeds the estimate PBR by:\n\nA: Between $10-22.72$ times.\nB: 10 times.\nC: 22.72 times.\nD: 0.1 times.\nE: 0.04 times.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1454",
        "problem": "The three recognised colours of Labradors are black, chocolate and golden. These coat colours result from differences in two genetic loci which have alleles B/b and E/e respectively.\n\nSupplied is a simplified flowchart representing pigment deposition in labradors. $\\mathrm{E}$ is the allele for functional MRC1 protein, while e is a mutated allele resulting in no MRC1 protein. B is the allele for functional TYRP1 enzyme, while $b$ is a mutated allele resulting in no TYRP1 enzyme activity. If there is no pigment deposition, the labrador is considered golden.\n\n[figure1]\n\nWhich one of the following could be the genotype of a black female Labrador?\nA: bbEE\nB: BBee\nC: BbEe\nD: Bbee\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe three recognised colours of Labradors are black, chocolate and golden. These coat colours result from differences in two genetic loci which have alleles B/b and E/e respectively.\n\nSupplied is a simplified flowchart representing pigment deposition in labradors. $\\mathrm{E}$ is the allele for functional MRC1 protein, while e is a mutated allele resulting in no MRC1 protein. B is the allele for functional TYRP1 enzyme, while $b$ is a mutated allele resulting in no TYRP1 enzyme activity. If there is no pigment deposition, the labrador is considered golden.\n\n[figure1]\n\nWhich one of the following could be the genotype of a black female Labrador?\n\nA: bbEE\nB: BBee\nC: BbEe\nD: Bbee\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-19.jpg?height=780&width=926&top_left_y=1312&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_660",
        "problem": "如图显示某种鸟类羽毛的毛色\n（B、b）遗传图解，下列相关表述错误的是（）\n\n[图1]\nA: 该种鸟类的毛色遗传属于性染色体连锁遗传\nB: 芦花形状为显性性状，基因 B 对 $\\mathrm{b}$ 完全显现\nC: 非芦花雄鸟和芦花雌鸟的子代雌鸟均为非芦花\nD: 芦花雄鸟和非芦花雌鸟的子代雌鸟均为非芦花\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图显示某种鸟类羽毛的毛色\n（B、b）遗传图解，下列相关表述错误的是（）\n\n[图1]\n\nA: 该种鸟类的毛色遗传属于性染色体连锁遗传\nB: 芦花形状为显性性状，基因 B 对 $\\mathrm{b}$ 完全显现\nC: 非芦花雄鸟和芦花雌鸟的子代雌鸟均为非芦花\nD: 芦花雄鸟和非芦花雌鸟的子代雌鸟均为非芦花\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-12.jpg?height=277&width=531&top_left_y=1746&top_left_x=177"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_499",
        "problem": "苯丙氨酸是人体必需的氨基酸, 被摄入后其代谢路径如图 a. 苯丙氨酸的代谢异常可能使人患苯丙酮酸症 (苯丙酮酸积累过多阻碍脑的发育, 造成智力低下) 、白化病（黑色素合成受阻, 肤色异常）等。图 $\\mathrm{b}$ 为某个患有甲（由 $\\mathrm{A} / \\mathrm{a}$ 基因控制）、乙（由 $\\mathrm{B} / \\mathrm{b}$ 基因控制）两种遗传病的家族遗传系谱图: 已知其中一种为白化病且假设人群中白化病的发病率为 1/6400, II-4 无乙病的致病基因。下列有关说法错误的是（ ）\n\n[图1]\n\n(a)\n\n(b)\nA: I-2 个体细胞中正常最多有 4 个与甲、乙相关的致病基因\nB: II-5 与人群中某正常男子结婚, 所生孩子表型正常的概率为 $121 / 162$\nC: 据图推测, 缺乏正常酶(1)基因的个体表型为只患白化病且智力正常\nD: 与白化病相比，原发性高血压等多基因遗传病在群体中的发病率通常较高\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n苯丙氨酸是人体必需的氨基酸, 被摄入后其代谢路径如图 a. 苯丙氨酸的代谢异常可能使人患苯丙酮酸症 (苯丙酮酸积累过多阻碍脑的发育, 造成智力低下) 、白化病（黑色素合成受阻, 肤色异常）等。图 $\\mathrm{b}$ 为某个患有甲（由 $\\mathrm{A} / \\mathrm{a}$ 基因控制）、乙（由 $\\mathrm{B} / \\mathrm{b}$ 基因控制）两种遗传病的家族遗传系谱图: 已知其中一种为白化病且假设人群中白化病的发病率为 1/6400, II-4 无乙病的致病基因。下列有关说法错误的是（ ）\n\n[图1]\n\n(a)\n\n(b)\n\nA: I-2 个体细胞中正常最多有 4 个与甲、乙相关的致病基因\nB: II-5 与人群中某正常男子结婚, 所生孩子表型正常的概率为 $121 / 162$\nC: 据图推测, 缺乏正常酶(1)基因的个体表型为只患白化病且智力正常\nD: 与白化病相比，原发性高血压等多基因遗传病在群体中的发病率通常较高\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-30.jpg?height=283&width=1468&top_left_y=1800&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_831",
        "problem": "下图家族中的甲、乙两种病分别由基因 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 控制, 已知 $\\mathrm{II}_{4}$ 不携带乙病的致病基因。下列相关分析正确的（）\n[图1]\nA: $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$ 再生一个孩子, 从优生优育的角度建议生男孩\nB: $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$ 生一个女孩, 两病都不患的概率是 $3 / 8$\nC: $\\mathrm{II}_{3}$ 的基因型及概率为: $1 / 2 \\mathrm{aaX} X^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$ 或 $1 / 2 \\mathrm{aa} X^{\\mathrm{B}} X^{\\mathrm{B}}$\nD: $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生一个只患一种病的男孩的概率是 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图家族中的甲、乙两种病分别由基因 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 控制, 已知 $\\mathrm{II}_{4}$ 不携带乙病的致病基因。下列相关分析正确的（）\n[图1]\n\nA: $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$ 再生一个孩子, 从优生优育的角度建议生男孩\nB: $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$ 生一个女孩, 两病都不患的概率是 $3 / 8$\nC: $\\mathrm{II}_{3}$ 的基因型及概率为: $1 / 2 \\mathrm{aaX} X^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$ 或 $1 / 2 \\mathrm{aa} X^{\\mathrm{B}} X^{\\mathrm{B}}$\nD: $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生一个只患一种病的男孩的概率是 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-22.jpg?height=470&width=1148&top_left_y=2108&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_458",
        "problem": "如图是某家系甲、乙、丙三种单基因遗传病的系谱图, 其基因分别用 $A 、 a ， B 、 b$和 $\\mathrm{D} 、 \\mathrm{~d}$ 表示。甲病是伴性遗传病， $\\mathrm{II}_{7}$ 不携带乙病的致病基因。在不考虑家系内发生新的基因突变的情况下，下列叙述错误的是（ ）\n\nI\n\nII\n\nIII\n\n[图1]\n\n$\\square$ 正常男性正常女性患甲病男性患甲病女性患乙病男性患甲病、乙病男性患丙病女性\nA: 甲病和乙病的遗传方式分别是伴 X 染色体显性遗传和伴 X 染色体隐性遗传\nB: $\\mathrm{II}_{6}$ 基因型为 $\\mathrm{DDX}^{\\mathrm{AB}} \\mathrm{X}^{\\mathrm{ab}}$ 或 $\\mathrm{DdX} X^{\\mathrm{AB}} \\mathrm{X}^{\\mathrm{ab}}$\nC: $\\mathrm{III}_{13}$ 患两种遗传病的原因是 $\\mathrm{II}_{6}$ 在减数分裂第一次分裂前期, 两条 $\\mathrm{X}$ 染色体的非姐妹染色单体之间发生片段交换, 产生 $\\mathrm{X}^{\\mathrm{Ab}}$ 的配子\nD: 若 $\\mathrm{III}_{15}$ 为乙病致病基因的杂合子、为丙病致病基因携带者的概率是 $1 / 100, \\mathrm{III}_{15}$和 III $_{16}$ 结婚, 所生的子女只患一种病的概率是 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图是某家系甲、乙、丙三种单基因遗传病的系谱图, 其基因分别用 $A 、 a ， B 、 b$和 $\\mathrm{D} 、 \\mathrm{~d}$ 表示。甲病是伴性遗传病， $\\mathrm{II}_{7}$ 不携带乙病的致病基因。在不考虑家系内发生新的基因突变的情况下，下列叙述错误的是（ ）\n\nI\n\nII\n\nIII\n\n[图1]\n\n$\\square$ 正常男性正常女性患甲病男性患甲病女性患乙病男性患甲病、乙病男性患丙病女性\n\nA: 甲病和乙病的遗传方式分别是伴 X 染色体显性遗传和伴 X 染色体隐性遗传\nB: $\\mathrm{II}_{6}$ 基因型为 $\\mathrm{DDX}^{\\mathrm{AB}} \\mathrm{X}^{\\mathrm{ab}}$ 或 $\\mathrm{DdX} X^{\\mathrm{AB}} \\mathrm{X}^{\\mathrm{ab}}$\nC: $\\mathrm{III}_{13}$ 患两种遗传病的原因是 $\\mathrm{II}_{6}$ 在减数分裂第一次分裂前期, 两条 $\\mathrm{X}$ 染色体的非姐妹染色单体之间发生片段交换, 产生 $\\mathrm{X}^{\\mathrm{Ab}}$ 的配子\nD: 若 $\\mathrm{III}_{15}$ 为乙病致病基因的杂合子、为丙病致病基因携带者的概率是 $1 / 100, \\mathrm{III}_{15}$和 III $_{16}$ 结婚, 所生的子女只患一种病的概率是 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-007.jpg?height=520&width=771&top_left_y=2107&top_left_x=448"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1233",
        "problem": "A cladogram is a branching diagram in which groups of closely related species (i.e. those sharing many characteristics) are shown to have branched away from the common line of descent via the same ancestor. The table gives a number of characteristics which are found in a group of four species, W, X, Y and Z.\n\n| Characteristic | Species |  |  |  |\n| :--- | :--- | :--- | :--- | :--- |\n|  | W | X | Y | Z |\n| Fingers and toes |  |  |  |  |\n| Endothermy |  |  |  |  |\n| 3 ear ossicles |  |  |  |  |\n| Amnion |  |  |  |  |\n| Placenta |  |  |  |  |\n| Internal <br> fertilisation |  |  |  |  |\n| Mammary glands |  |  |  |  |\n| Oviparity (lays <br> eggs) |  |  |  |  |\n| Webbed feet |  |  |  |  |\n| Hair |  |  |  |  |\n| Feathers |  |  |  |  |\n\nWhich one of the cladograms below is the most likely hypothesis explaining the distribution of characters?\n[figure1]\nA: A\nB: B\nC: C\nD: D\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA cladogram is a branching diagram in which groups of closely related species (i.e. those sharing many characteristics) are shown to have branched away from the common line of descent via the same ancestor. The table gives a number of characteristics which are found in a group of four species, W, X, Y and Z.\n\n| Characteristic | Species |  |  |  |\n| :--- | :--- | :--- | :--- | :--- |\n|  | W | X | Y | Z |\n| Fingers and toes |  |  |  |  |\n| Endothermy |  |  |  |  |\n| 3 ear ossicles |  |  |  |  |\n| Amnion |  |  |  |  |\n| Placenta |  |  |  |  |\n| Internal <br> fertilisation |  |  |  |  |\n| Mammary glands |  |  |  |  |\n| Oviparity (lays <br> eggs) |  |  |  |  |\n| Webbed feet |  |  |  |  |\n| Hair |  |  |  |  |\n| Feathers |  |  |  |  |\n\nWhich one of the cladograms below is the most likely hypothesis explaining the distribution of characters?\n[figure1]\n\nA: A\nB: B\nC: C\nD: D\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-17.jpg?height=488&width=1040&top_left_y=2080&top_left_x=542"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1386",
        "problem": "A spectrophotometer is a machine that measures how much light is absorbed or transmitted when passed through a solution. The result depends on the colour and clarity of the solution as well as the wavelength (colour) of light passed through it. Zijlstra \\& Buursma (1997) examined the absorptivity of four derivatives of haemoglobin (the pigment that makes blood red) for humans and cows (bovines) (Figure 1).\n[figure1]\n\nFigure 1. Absorption spectra of the common derivatives of cow (-) and human (- - - -) haemoglobin in the visible range. (A) oxyhaemoglobin, $\\mathrm{O}_{2} \\mathrm{Hb}$ (1) and deoxyhaemoglobin, $\\mathrm{HHb}$ (2); (B) methaemoglobin, $\\mathrm{Hi}$ (1) and carboxyhaemoglobin, $\\mathrm{COHb}(2)$. The absorptivity is expressed in $\\mathrm{L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$.\n\nWhich of the following can be concluded from the data contained in the graphs?\nA: There is a statistically significant spectral difference between bovine and human $\\mathrm{Hi}$.\nB: For $\\mathrm{HHb}, \\mathrm{O}_{2} \\mathrm{Hb}$, and $\\mathrm{COHb}$ there are only minute spectral differences between bovine and human haemoglobin.\nC: In the $\\mathrm{O}_{2} \\mathrm{Hb}$ spectrum the minimum at $560 \\mathrm{~nm}$ is slightly higher for bovine than for human haemoglobin.\nD: At $508 \\mathrm{~nm}$ the absorptivity of bovine haemoglobin is slightly lower than that of human haemoglobin.\nE: In the $\\mathrm{COHb}$ spectrum the $\\alpha$-peak around $569 \\mathrm{~nm}$ is somewhat higher for bovine haemoglobin, and in the minimum near $496 \\mathrm{~nm}$ the spectrum of bovine haemoglobin lies a little above that of the human.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA spectrophotometer is a machine that measures how much light is absorbed or transmitted when passed through a solution. The result depends on the colour and clarity of the solution as well as the wavelength (colour) of light passed through it. Zijlstra \\& Buursma (1997) examined the absorptivity of four derivatives of haemoglobin (the pigment that makes blood red) for humans and cows (bovines) (Figure 1).\n[figure1]\n\nFigure 1. Absorption spectra of the common derivatives of cow (-) and human (- - - -) haemoglobin in the visible range. (A) oxyhaemoglobin, $\\mathrm{O}_{2} \\mathrm{Hb}$ (1) and deoxyhaemoglobin, $\\mathrm{HHb}$ (2); (B) methaemoglobin, $\\mathrm{Hi}$ (1) and carboxyhaemoglobin, $\\mathrm{COHb}(2)$. The absorptivity is expressed in $\\mathrm{L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$.\n\nWhich of the following can be concluded from the data contained in the graphs?\n\nA: There is a statistically significant spectral difference between bovine and human $\\mathrm{Hi}$.\nB: For $\\mathrm{HHb}, \\mathrm{O}_{2} \\mathrm{Hb}$, and $\\mathrm{COHb}$ there are only minute spectral differences between bovine and human haemoglobin.\nC: In the $\\mathrm{O}_{2} \\mathrm{Hb}$ spectrum the minimum at $560 \\mathrm{~nm}$ is slightly higher for bovine than for human haemoglobin.\nD: At $508 \\mathrm{~nm}$ the absorptivity of bovine haemoglobin is slightly lower than that of human haemoglobin.\nE: In the $\\mathrm{COHb}$ spectrum the $\\alpha$-peak around $569 \\mathrm{~nm}$ is somewhat higher for bovine haemoglobin, and in the minimum near $496 \\mathrm{~nm}$ the spectrum of bovine haemoglobin lies a little above that of the human.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-25.jpg?height=528&width=1558&top_left_y=513&top_left_x=244"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1127",
        "problem": "Two ecological pyramids are drawn below. If 'I' represents aquatic and 'II' represents terrestrial ecosystem, which of the following statements are true?\n\nI\n\n[figure1]\n\nII\n\n[figure2]\n\ni. Pyramid ' $\\mathrm{I}$ ' is based on biomass.\n\nii. Pyramid 'II' is based on numbers.\n\niii. The unshaded box in pyramid 'I' represents zooplankton and in pyramid 'II' it represents insects.\n\niv. Neither 'I' nor 'II' can be based on energy.\nA: ii and iii only\nB: i and iii only\nC: i, ii and iv only\nD: i, ii, iii and iv\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTwo ecological pyramids are drawn below. If 'I' represents aquatic and 'II' represents terrestrial ecosystem, which of the following statements are true?\n\nI\n\n[figure1]\n\nII\n\n[figure2]\n\ni. Pyramid ' $\\mathrm{I}$ ' is based on biomass.\n\nii. Pyramid 'II' is based on numbers.\n\niii. The unshaded box in pyramid 'I' represents zooplankton and in pyramid 'II' it represents insects.\n\niv. Neither 'I' nor 'II' can be based on energy.\n\nA: ii and iii only\nB: i and iii only\nC: i, ii and iv only\nD: i, ii, iii and iv\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-12.jpg?height=195&width=347&top_left_y=1171&top_left_x=607",
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-12.jpg?height=195&width=282&top_left_y=1171&top_left_x=1125"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1277",
        "problem": "Two species of plant, $X$ and $Y$, were grown in each of two regimes of day length, and their flowering response recorded.\n\nRegime 1: 8 hours daylight, 16 hours darkness;\n\nResult: $\\mathrm{X}$ flowered, $\\mathrm{Y}$ did not flower.\n\nRegime 2: 8 hours of daylight, 16 hours of darkness, interrupted after 8 hours by 10 minutes of bright light from an incandescent lamp;\n\nResult: $\\mathrm{X}$ and $\\mathrm{Y}$ both flowered.\n\nThe probable explanation of these results is that\nA: $\\mathrm{X}$ is a long-day plant, $\\mathrm{Y}$ is a short-day plant\nB: $X$ is a long-day plant, $Y$ is a day-neutral plant\nC: $X$ is a day-neutral plant, $Y$ is a long-day plant\nD: $X$ is a day-neutral plant, $Y$ is a short-day plant\nE: $\\mathrm{X}$ is a short-day plant, $\\mathrm{Y}$ is a long-day plant\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTwo species of plant, $X$ and $Y$, were grown in each of two regimes of day length, and their flowering response recorded.\n\nRegime 1: 8 hours daylight, 16 hours darkness;\n\nResult: $\\mathrm{X}$ flowered, $\\mathrm{Y}$ did not flower.\n\nRegime 2: 8 hours of daylight, 16 hours of darkness, interrupted after 8 hours by 10 minutes of bright light from an incandescent lamp;\n\nResult: $\\mathrm{X}$ and $\\mathrm{Y}$ both flowered.\n\nThe probable explanation of these results is that\n\nA: $\\mathrm{X}$ is a long-day plant, $\\mathrm{Y}$ is a short-day plant\nB: $X$ is a long-day plant, $Y$ is a day-neutral plant\nC: $X$ is a day-neutral plant, $Y$ is a long-day plant\nD: $X$ is a day-neutral plant, $Y$ is a short-day plant\nE: $\\mathrm{X}$ is a short-day plant, $\\mathrm{Y}$ is a long-day plant\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_829",
        "problem": "精原细胞中发生臂内倒位(如图)将导致染色体桥和断片的形成, 已知有缺失或重复的染色单体产生的配子不育。下列说法不正确的是（ ）\n[图1]\nA: 臂内倒位发生于减数第一次分裂前期的同源染色体非姐妹染色单体间\nB: 断片由于不含着丝点，分裂末期将游离于主核之外\nC: 染色体桥断裂后只能形成一条正常染色体\nD: 发生该现象的精原细胞产生的配子将有一半以上不可育\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n精原细胞中发生臂内倒位(如图)将导致染色体桥和断片的形成, 已知有缺失或重复的染色单体产生的配子不育。下列说法不正确的是（ ）\n[图1]\n\nA: 臂内倒位发生于减数第一次分裂前期的同源染色体非姐妹染色单体间\nB: 断片由于不含着丝点，分裂末期将游离于主核之外\nC: 染色体桥断裂后只能形成一条正常染色体\nD: 发生该现象的精原细胞产生的配子将有一半以上不可育\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-84.jpg?height=398&width=852&top_left_y=2256&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_581",
        "problem": "下图为某家族低磷酸酯酶症 (甲病)、先天性夜盲症（乙病）两种遗传病的系谱图。\n\n低磷酸酯酶症由等位基因 $\\mathrm{A} / \\mathrm{a}$ 控制, 先天性夜盲症由等位基因 $\\mathrm{B} / \\mathrm{b}$ 控制, 这两对等位基因独立遗传。已知 $\\mathrm{III}_{4}$ 携带低磷酸酯酶症的致病基因，但不携带先天性夜盲症的致病基因。下列叙述错误的是（ ）\n\n[图1]\nA: 先天性夜盲症在人群中男性的发病率高于女性\nB: $\\mathrm{III}_{3}$ 的基因型及概率为 $1 / 3 \\mathrm{AAX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}} 、 2 / 3 \\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$\nC: [图2]\nD: 若 $\\mathrm{IV}_{1}$ 与一个正常男性结婚, 则他们生一个患先天性夜盲症男孩的概率是 $1 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为某家族低磷酸酯酶症 (甲病)、先天性夜盲症（乙病）两种遗传病的系谱图。\n\n低磷酸酯酶症由等位基因 $\\mathrm{A} / \\mathrm{a}$ 控制, 先天性夜盲症由等位基因 $\\mathrm{B} / \\mathrm{b}$ 控制, 这两对等位基因独立遗传。已知 $\\mathrm{III}_{4}$ 携带低磷酸酯酶症的致病基因，但不携带先天性夜盲症的致病基因。下列叙述错误的是（ ）\n\n[图1]\n\nA: 先天性夜盲症在人群中男性的发病率高于女性\nB: $\\mathrm{III}_{3}$ 的基因型及概率为 $1 / 3 \\mathrm{AAX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}} 、 2 / 3 \\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$\nC: [图2]\nD: 若 $\\mathrm{IV}_{1}$ 与一个正常男性结婚, 则他们生一个患先天性夜盲症男孩的概率是 $1 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-65.jpg?height=394&width=782&top_left_y=908&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-65.jpg?height=57&width=1090&top_left_y=1491&top_left_x=403"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_956",
        "problem": "Which of the following features of flowers is NOT meant to facilitate its pollination by animals?\nA: The production of nectar at the base of the flower.\nB: The presence of leaf-like sepals.\nC: Brightly-colored and conspicuous petals.\nD: A strong odor of rotting meat.\nE: Patterns only visible under ultraviolet light.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following features of flowers is NOT meant to facilitate its pollination by animals?\n\nA: The production of nectar at the base of the flower.\nB: The presence of leaf-like sepals.\nC: Brightly-colored and conspicuous petals.\nD: A strong odor of rotting meat.\nE: Patterns only visible under ultraviolet light.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_798",
        "problem": "甲乙丙三种酶参与葡萄糖和糖原之间的转化, 过程如图 1 所示。任一酶的基因发生突变导致相应酶功能缺陷, 均会引发 GSD 病。图 2 为三种 GSD 亚型患者家系, 其中至少一种是伴性遗传。不考虑新的突变, 下列分析正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 若(1)同时患有红绿色盲，则其父母再生育健康孩子的概率是 $3 / 8$\nB: 若(2)长期表现为低血糖, 则一定不是乙酶功能缺陷所致\nC: 若丙酶缺陷 GSD 发病率是 $1 / 10000$, 则(3)患该病的概率为 $1 / 300$\nD: 三种 GSD 亚型患者体内的糖原含量都会异常升高\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n甲乙丙三种酶参与葡萄糖和糖原之间的转化, 过程如图 1 所示。任一酶的基因发生突变导致相应酶功能缺陷, 均会引发 GSD 病。图 2 为三种 GSD 亚型患者家系, 其中至少一种是伴性遗传。不考虑新的突变, 下列分析正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 若(1)同时患有红绿色盲，则其父母再生育健康孩子的概率是 $3 / 8$\nB: 若(2)长期表现为低血糖, 则一定不是乙酶功能缺陷所致\nC: 若丙酶缺陷 GSD 发病率是 $1 / 10000$, 则(3)患该病的概率为 $1 / 300$\nD: 三种 GSD 亚型患者体内的糖原含量都会异常升高\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-013.jpg?height=274&width=624&top_left_y=157&top_left_x=339",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-013.jpg?height=289&width=831&top_left_y=158&top_left_x=955"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_554",
        "problem": "油菜植株体内的中间代谢产物磷酸烯醇式丙酮酸 (PEP) 运向种子后有两条转变途\n\n径，如图甲所示，其中酶 $\\mathrm{a}$ 和酶 $\\mathrm{b}$ 分别由基因 $\\mathrm{A}$ 和基因 $\\mathrm{B}$ 控制合成。图乙表示基因 $\\mathrm{B}$, $a$ 链是转录链, 经诱导 $\\beta$ 链也能转录, 从而形成双链 mRNA。下列有关叙述错误的是\n\n[图1]\nA: 据图甲分析, 抑制酶 $\\mathrm{b}$ 合成, 促进酶 $\\mathrm{a}$ 合成可提高油菜产油量\nB: 转录出的双链 mRNA 与图乙基因在化学组成上的区别是 mRNA 中不含 $\\mathrm{T}$ 含 $\\mathrm{U}$,五碳糖为核糖\nC: 基因 B 经诱导后转录出双链 mRNA 不能提高产油量\nD: 该过程体现了基因通过控制酶的合成来控制代谢过程，进而控制生物体的性状。\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n油菜植株体内的中间代谢产物磷酸烯醇式丙酮酸 (PEP) 运向种子后有两条转变途\n\n径，如图甲所示，其中酶 $\\mathrm{a}$ 和酶 $\\mathrm{b}$ 分别由基因 $\\mathrm{A}$ 和基因 $\\mathrm{B}$ 控制合成。图乙表示基因 $\\mathrm{B}$, $a$ 链是转录链, 经诱导 $\\beta$ 链也能转录, 从而形成双链 mRNA。下列有关叙述错误的是\n\n[图1]\n\nA: 据图甲分析, 抑制酶 $\\mathrm{b}$ 合成, 促进酶 $\\mathrm{a}$ 合成可提高油菜产油量\nB: 转录出的双链 mRNA 与图乙基因在化学组成上的区别是 mRNA 中不含 $\\mathrm{T}$ 含 $\\mathrm{U}$,五碳糖为核糖\nC: 基因 B 经诱导后转录出双链 mRNA 不能提高产油量\nD: 该过程体现了基因通过控制酶的合成来控制代谢过程，进而控制生物体的性状。\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-09.jpg?height=301&width=1060&top_left_y=1777&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_297",
        "problem": "Figure I illustrates a skeletal muscle and its innervation. Figures II and III depict cross and longitudinal sections of the muscle, respectively. Figure IV shows an electron micrograph of neuromuscular junction.\n\n[figure1]\n\nWhich of the following statements gives the most accurate description of each structure?\nA: The number of muscle cells innervated by a single motor neuron is larger in a muscle that controls fine movement than in one that controls unskilled movement.\nB: During embryonic development, structure (a) is derived from a single cell.\nC: Within the same muscle, the population of small-diameter cells $(b)$ is increased after several weeks of intense exercise.\nD: Structure (c) is called a myofibril, which is the structural unit of the skeletal muscle.\nE: The main mechanism to terminate the action of secreted acetylcholine at the neuromuscular junction is neurotransmitter reuptake into the nerve terminal (d).\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFigure I illustrates a skeletal muscle and its innervation. Figures II and III depict cross and longitudinal sections of the muscle, respectively. Figure IV shows an electron micrograph of neuromuscular junction.\n\n[figure1]\n\nWhich of the following statements gives the most accurate description of each structure?\n\nA: The number of muscle cells innervated by a single motor neuron is larger in a muscle that controls fine movement than in one that controls unskilled movement.\nB: During embryonic development, structure (a) is derived from a single cell.\nC: Within the same muscle, the population of small-diameter cells $(b)$ is increased after several weeks of intense exercise.\nD: Structure (c) is called a myofibril, which is the structural unit of the skeletal muscle.\nE: The main mechanism to terminate the action of secreted acetylcholine at the neuromuscular junction is neurotransmitter reuptake into the nerve terminal (d).\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-34.jpg?height=885&width=1308&top_left_y=574&top_left_x=431"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1224",
        "problem": "This diagram shows the female hypothalamic-pituitarygonadal axis. Hormones are secreted from each of the organs. A \" $+\"$ \" indicates a stimulation of secretion from the organ and a \"-\" indicates a decrease in secretion.\n\nAccording to the diagram, a decrease in GnRH will cause:\n\n[figure1]\nA: A decrease in oestrogen secretion.\nB: An increase in inhibin secretion.\nC: A decrease in GnRH secretion.\nD: An increase in FSH secretion.\nE: An increase in LH secretion\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThis diagram shows the female hypothalamic-pituitarygonadal axis. Hormones are secreted from each of the organs. A \" $+\"$ \" indicates a stimulation of secretion from the organ and a \"-\" indicates a decrease in secretion.\n\nAccording to the diagram, a decrease in GnRH will cause:\n\n[figure1]\n\nA: A decrease in oestrogen secretion.\nB: An increase in inhibin secretion.\nC: A decrease in GnRH secretion.\nD: An increase in FSH secretion.\nE: An increase in LH secretion\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-05.jpg?height=636&width=720&top_left_y=1875&top_left_x=1116"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_745",
        "problem": "某种鸟类( $Z$ W 型)的翅膀羽毛红色(D)对灰色 (d)为显性, 羽毛有条纹(R)对无条纹( $r$ )为显性, 其中一对基因位于 $\\mathrm{Z}$ 染色体上。现有一只红色有条纹个体与一只灰色有条纹个体杂交, $F_{1}$ 雄鸟中约有 $1 / 8$ 为灰色无条纹。下列叙述错误的是（）\nA: 亲本产生的配子中含 $r Z^{d}$ 的配子均占 $1 / 4$\nB: $F_{1}$ 出现红色有条纹雌性个体的概率为 $3 / 16$\nC: $F_{1}$ 中无条纹个体随机交配，子代出现红色无条纹的概率为 $7 / 16$\nD: $F_{1}$ 中红色有条纹个体随机交配, 子代出现灰色有条纹的概率为 $1 / 9$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种鸟类( $Z$ W 型)的翅膀羽毛红色(D)对灰色 (d)为显性, 羽毛有条纹(R)对无条纹( $r$ )为显性, 其中一对基因位于 $\\mathrm{Z}$ 染色体上。现有一只红色有条纹个体与一只灰色有条纹个体杂交, $F_{1}$ 雄鸟中约有 $1 / 8$ 为灰色无条纹。下列叙述错误的是（）\n\nA: 亲本产生的配子中含 $r Z^{d}$ 的配子均占 $1 / 4$\nB: $F_{1}$ 出现红色有条纹雌性个体的概率为 $3 / 16$\nC: $F_{1}$ 中无条纹个体随机交配，子代出现红色无条纹的概率为 $7 / 16$\nD: $F_{1}$ 中红色有条纹个体随机交配, 子代出现灰色有条纹的概率为 $1 / 9$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_805",
        "problem": "在雌果蝇中, 胚胎发育所需要的部分养分、蛋白质和 mRNA 由卵母细胞旁边的营\n养细胞和滤泡细胞提供。有一个位于常染色体上的基因所产生的 mRNA 被运送到卵母细胞, 从而保证受精后形成的胚胎正常发育, 如果此基因发生突变将会导致胚胎畸形而且无法存活。以下叙述正确的是（）\nA: 如果此突变是显性的, 则突变杂合子雄果蝇和正常雌果蝇交配所生的雌性子代不可以存活\nB: 如果此突变是显性的, 则可观察到存活的突变纯合子个体\nC: 如果此突变是隐性的, 则对于突变杂合子母体所生的雌、雄性胚胎都有一半可正常发育\nD: 如果此突变是隐性的, 两个突变杂合子的个体杂交, 子二代中有 $1 / 6$ 是突变纯合子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在雌果蝇中, 胚胎发育所需要的部分养分、蛋白质和 mRNA 由卵母细胞旁边的营\n养细胞和滤泡细胞提供。有一个位于常染色体上的基因所产生的 mRNA 被运送到卵母细胞, 从而保证受精后形成的胚胎正常发育, 如果此基因发生突变将会导致胚胎畸形而且无法存活。以下叙述正确的是（）\n\nA: 如果此突变是显性的, 则突变杂合子雄果蝇和正常雌果蝇交配所生的雌性子代不可以存活\nB: 如果此突变是显性的, 则可观察到存活的突变纯合子个体\nC: 如果此突变是隐性的, 则对于突变杂合子母体所生的雌、雄性胚胎都有一半可正常发育\nD: 如果此突变是隐性的, 两个突变杂合子的个体杂交, 子二代中有 $1 / 6$ 是突变纯合子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_155",
        "problem": "HL-60 cells differentiate to neutrophils after eight days of incubation in the presence of All Trans retinoic acid (ATRA). siRNAs are synthetic molecules used to suppress expression of genes. In the experiment whose results are presented below, an siRNA against the PTEN gene was used. Annexin V is a marker used to show apoptosis, and CD11b is a neutrophil surface marker. Percentage of apoptotic and differentiated cells as well as combination of both are shown in panels A, B and C. In panels A and $B$, stars indicate statistical significant.\n\n(A)\n\n[figure1]\n\n(B)\n\n[figure2]\n\n(C)\n\n[figure3]\n\nDay 7\n\nDay 8\n\n[figure4]\n\nATRA-PTEN SIRNA\n[figure5]\n\nATRA\n\n(C) Horizontal line within each box reflects threshold value for apoptosis. Vertical line within each box reflects threshold value for neutrophil differentiation.\nA: The level of differentiation to neutrophils in the presence of PTEN siRNA is less than in the absence of PTEN siRNA at a statistically significant level.\nB: Panels a and b show that reduction of apoptosis causes an increase in the number of live neutrophil cells at all time points.\nC: Panel c shows an inverse correlation between apoptosis and the number of differentiated neutrophil cells at at least two time points.\nD: PTEN siRNA applications causes larger reduction of apoptosis in differentiated cells as compared to undifferentiated cells after 7 days of incubation with ATRA.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nHL-60 cells differentiate to neutrophils after eight days of incubation in the presence of All Trans retinoic acid (ATRA). siRNAs are synthetic molecules used to suppress expression of genes. In the experiment whose results are presented below, an siRNA against the PTEN gene was used. Annexin V is a marker used to show apoptosis, and CD11b is a neutrophil surface marker. Percentage of apoptotic and differentiated cells as well as combination of both are shown in panels A, B and C. In panels A and $B$, stars indicate statistical significant.\n\n(A)\n\n[figure1]\n\n(B)\n\n[figure2]\n\n(C)\n\n[figure3]\n\nDay 7\n\nDay 8\n\n[figure4]\n\nATRA-PTEN SIRNA\n[figure5]\n\nATRA\n\n(C) Horizontal line within each box reflects threshold value for apoptosis. Vertical line within each box reflects threshold value for neutrophil differentiation.\n\nA: The level of differentiation to neutrophils in the presence of PTEN siRNA is less than in the absence of PTEN siRNA at a statistically significant level.\nB: Panels a and b show that reduction of apoptosis causes an increase in the number of live neutrophil cells at all time points.\nC: Panel c shows an inverse correlation between apoptosis and the number of differentiated neutrophil cells at at least two time points.\nD: PTEN siRNA applications causes larger reduction of apoptosis in differentiated cells as compared to undifferentiated cells after 7 days of incubation with ATRA.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-10.jpg?height=614&width=665&top_left_y=795&top_left_x=296",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-10.jpg?height=557&width=688&top_left_y=838&top_left_x=1024",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-10.jpg?height=440&width=400&top_left_y=1453&top_left_x=377",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-10.jpg?height=346&width=616&top_left_y=1546&top_left_x=797",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-10.jpg?height=378&width=1034&top_left_y=1924&top_left_x=380"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_812",
        "problem": "某雌雄同株的二倍体植物中，控制抗病（A）与易感病（a）、高茎（B）与矮茎\n\n（b）的基因分别位于两对染色体上。让纯种抗病高茎植株与纯种易感病矮茎植株杂交, $F_{1}$ 全为抗病高茎植株, $F_{1}$ 自交获得的 $F_{2}$ 中, 抗病高茎: 抗病矮茎: 易感病高茎: 易感病矮茎 $=9: 3: 3: 1$. 下列有关叙述错误的是（）\nA: 等位基因 $A 、 a$ 与 $B 、 b$ 的遗传既遵循分离定律又遵循自由组合定律\nB: $F_{2}$ 中的抗病植株连续进行多代的自交和随机交配, 后代中抗病基因频率均不变\nC: $F_{2}$ 中的抗病高茎植株进行自交, 后代的性状比例为 25:5:5:1\nD: $F_{2}$ 中的抗病高茎植株随机交配, 后代的性状比例为 $64: 8: 8: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某雌雄同株的二倍体植物中，控制抗病（A）与易感病（a）、高茎（B）与矮茎\n\n（b）的基因分别位于两对染色体上。让纯种抗病高茎植株与纯种易感病矮茎植株杂交, $F_{1}$ 全为抗病高茎植株, $F_{1}$ 自交获得的 $F_{2}$ 中, 抗病高茎: 抗病矮茎: 易感病高茎: 易感病矮茎 $=9: 3: 3: 1$. 下列有关叙述错误的是（）\n\nA: 等位基因 $A 、 a$ 与 $B 、 b$ 的遗传既遵循分离定律又遵循自由组合定律\nB: $F_{2}$ 中的抗病植株连续进行多代的自交和随机交配, 后代中抗病基因频率均不变\nC: $F_{2}$ 中的抗病高茎植株进行自交, 后代的性状比例为 25:5:5:1\nD: $F_{2}$ 中的抗病高茎植株随机交配, 后代的性状比例为 $64: 8: 8: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1380",
        "problem": "A spectrophotometer is a machine that measures how much light is absorbed or transmitted when passed through a solution. The result depends on the colour and clarity of the solution as well as the wavelength (colour) of light passed through it. Zijlstra \\& Buursma (1997) examined the absorptivity of four derivatives of haemoglobin (the pigment that makes blood red) for humans and cows (bovines) (Figure 1).\n[figure1]\n\nFigure 1. Absorption spectra of the common derivatives of cow (-) and human (- - - -) haemoglobin in the visible range. (A) oxyhaemoglobin, $\\mathrm{O}_{2} \\mathrm{Hb}$ (1) and deoxyhaemoglobin, $\\mathrm{HHb}$ (2); (B) methaemoglobin, $\\mathrm{Hi}$ (1) and carboxyhaemoglobin, $\\mathrm{COHb}(2)$. The absorptivity is expressed in $\\mathrm{L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$.\n\nWhat is the approximate absorptivity of methaemoglobin for humans at $550 \\mathrm{~nm}$ ?\nA: $\\quad 4 \\mathrm{~L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$\nB: $\\quad 6 \\mathrm{~L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$\nC: $\\quad 8 \\mathrm{~L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$\nD: $\\quad 10 \\mathrm{~L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$\nE: $\\quad 12 \\mathrm{~L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA spectrophotometer is a machine that measures how much light is absorbed or transmitted when passed through a solution. The result depends on the colour and clarity of the solution as well as the wavelength (colour) of light passed through it. Zijlstra \\& Buursma (1997) examined the absorptivity of four derivatives of haemoglobin (the pigment that makes blood red) for humans and cows (bovines) (Figure 1).\n[figure1]\n\nFigure 1. Absorption spectra of the common derivatives of cow (-) and human (- - - -) haemoglobin in the visible range. (A) oxyhaemoglobin, $\\mathrm{O}_{2} \\mathrm{Hb}$ (1) and deoxyhaemoglobin, $\\mathrm{HHb}$ (2); (B) methaemoglobin, $\\mathrm{Hi}$ (1) and carboxyhaemoglobin, $\\mathrm{COHb}(2)$. The absorptivity is expressed in $\\mathrm{L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$.\n\nWhat is the approximate absorptivity of methaemoglobin for humans at $550 \\mathrm{~nm}$ ?\n\nA: $\\quad 4 \\mathrm{~L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$\nB: $\\quad 6 \\mathrm{~L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$\nC: $\\quad 8 \\mathrm{~L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$\nD: $\\quad 10 \\mathrm{~L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$\nE: $\\quad 12 \\mathrm{~L} \\mathrm{mmol}^{-1} \\mathrm{~cm}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-25.jpg?height=528&width=1558&top_left_y=513&top_left_x=244"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_672",
        "problem": "粗糙链孢霉是一种真核生物, 繁殖过程中通常由单倍体菌丝杂交形成二倍体合子。该霉菌的一个子囊母细胞(具有二倍体核, 基因型为 $\\mathrm{Aa}$ ), 通过细胞分裂形成 8 个孢子,它们在子囊中的排列顺序与减数分裂中期赤道板上染色单体的排列顺序完全一致, 如图所示。下列叙述错误的有（）\n\n[图1]\nA: 减数分裂过程中发生了染色体互换导致染色体变异\nB: 减数分裂I和有丝分裂过程中存在同源染色体\nC: 有丝分裂过程中有一个孢子的基因发生了隐性突变\nD: 子囊中孢子的染色体数目与合子的染色体数目一致\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n粗糙链孢霉是一种真核生物, 繁殖过程中通常由单倍体菌丝杂交形成二倍体合子。该霉菌的一个子囊母细胞(具有二倍体核, 基因型为 $\\mathrm{Aa}$ ), 通过细胞分裂形成 8 个孢子,它们在子囊中的排列顺序与减数分裂中期赤道板上染色单体的排列顺序完全一致, 如图所示。下列叙述错误的有（）\n\n[图1]\n\nA: 减数分裂过程中发生了染色体互换导致染色体变异\nB: 减数分裂I和有丝分裂过程中存在同源染色体\nC: 有丝分裂过程中有一个孢子的基因发生了隐性突变\nD: 子囊中孢子的染色体数目与合子的染色体数目一致\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-12.jpg?height=365&width=597&top_left_y=163&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_323",
        "problem": "科研人员用下图所示的果蝇作亲本进行了杂交实验。图示基因不在 $\\mathrm{Y}$ 染色体上,雌雄个体产生的配子均有活力, 下列叙述错误的是（）\n\n[图1]\nA: 亲本雌果蝇的两条 X 染色体均发生了结构变异\nB: 若亲本雌果蝇产生了四种基因型的卵细胞, 则亲本雌蝇发生了基因突变\nC: 若 $F_{1}$ 出现了 $X$ 染色体构型均异常的雌果蝇, 可能是亲本雄果蝇的生殖细胞发生了染色体结构变异\nD: 若 $F_{1}$ 出现了 $X^{\\mathrm{dB}} \\mathrm{X}^{\\mathrm{db}} \\mathrm{Y}$ 的个体, 可能是亲本雄性个体的初级精母细胞减数分裂 $\\mathrm{I}$时, 性染色体未分离\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n科研人员用下图所示的果蝇作亲本进行了杂交实验。图示基因不在 $\\mathrm{Y}$ 染色体上,雌雄个体产生的配子均有活力, 下列叙述错误的是（）\n\n[图1]\n\nA: 亲本雌果蝇的两条 X 染色体均发生了结构变异\nB: 若亲本雌果蝇产生了四种基因型的卵细胞, 则亲本雌蝇发生了基因突变\nC: 若 $F_{1}$ 出现了 $X$ 染色体构型均异常的雌果蝇, 可能是亲本雄果蝇的生殖细胞发生了染色体结构变异\nD: 若 $F_{1}$ 出现了 $X^{\\mathrm{dB}} \\mathrm{X}^{\\mathrm{db}} \\mathrm{Y}$ 的个体, 可能是亲本雄性个体的初级精母细胞减数分裂 $\\mathrm{I}$时, 性染色体未分离\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-042.jpg?height=363&width=1291&top_left_y=164&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_62",
        "problem": "For the three heritable features, Alfa, Baker, and Charlie, pedigree analysis was performed on pedigree A, pedigree B, and pedigree C, respectively, and the results in Figure 1 were obtained.\n[figure1]\n\nFigure 1\nA: An analysis of pedigree $\\mathbf{A}$ suggests that the inheritance pattern of characteristic Alfa could be due to a dominant allele.\nB: An analysis of pedigree $\\mathbf{C}$ suggests that the inheritance of the characteristic Charlie could be due to a dominant allele. \nC: B1 and B3 of family $\\mathbf{B}$ are definitely carriers.\nD: $\\mathbf{C} 1$ and $\\mathbf{C} 3$ of family $\\mathbf{C}$ are definitely carriers. \n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFor the three heritable features, Alfa, Baker, and Charlie, pedigree analysis was performed on pedigree A, pedigree B, and pedigree C, respectively, and the results in Figure 1 were obtained.\n[figure1]\n\nFigure 1\n\nA: An analysis of pedigree $\\mathbf{A}$ suggests that the inheritance pattern of characteristic Alfa could be due to a dominant allele.\nB: An analysis of pedigree $\\mathbf{C}$ suggests that the inheritance of the characteristic Charlie could be due to a dominant allele. \nC: B1 and B3 of family $\\mathbf{B}$ are definitely carriers.\nD: $\\mathbf{C} 1$ and $\\mathbf{C} 3$ of family $\\mathbf{C}$ are definitely carriers. \n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-15.jpg?height=958&width=1304&top_left_y=680&top_left_x=178"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1087",
        "problem": "The manner in which species are connected within a food web has implications for community structure and dynamics in addition to direct interactions between prey and predators. One of the ways in which the food webs of different communities differ is their degree of connectance, also known as 'connectance index'.\n\nConnectance is a way of describing how many possible links in a food web are present. They are simply the lines that link consumers and the consumed. A simple formula for calculating connectance index $(\\mathrm{C})$ is:\n\n$$\nC=L /\\left[\\frac{S(S-1)}{2}\\right]\n$$\n\nwhere $L$ is the actual number of links in the food chain and $S$ is the number of species in the food chain.\n\nWhat would be the number of possible links in a food web if the number of species was 5 ?",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe manner in which species are connected within a food web has implications for community structure and dynamics in addition to direct interactions between prey and predators. One of the ways in which the food webs of different communities differ is their degree of connectance, also known as 'connectance index'.\n\nConnectance is a way of describing how many possible links in a food web are present. They are simply the lines that link consumers and the consumed. A simple formula for calculating connectance index $(\\mathrm{C})$ is:\n\n$$\nC=L /\\left[\\frac{S(S-1)}{2}\\right]\n$$\n\nwhere $L$ is the actual number of links in the food chain and $S$ is the number of species in the food chain.\n\nWhat would be the number of possible links in a food web if the number of species was 5 ?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_932",
        "problem": "某果蝇种群眼色分为野生型和朱红色（由基因 A、a 控制）、野生型和棕色（由基因\n\n$\\mathrm{B} 、 \\mathrm{~b}$ 控制)，两对基因分别位于两对同源染色体上（不考虑 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段）。为研究其遗传机制, 进行了杂交实验, 结果如下表。下列说法正确的是 ( )\n\n| 杂交组合 | P |  | $\\mathrm{F}_{1}$ |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | o | $\\hat{0}$ | q | $\\hat{0}$ |\n\n\n| 甲 | 野生型 | 野生型 | 402 野生型 | 198 野生型、 201 朱红眼 |\n| :--- | :--- | :--- | :--- | :--- |\n| 乙 | 野生型 | 朱红眼 | 302 野生型、 99 <br> 棕眼 | 300 野生型、 101 棕眼 |\n| 丙 | 野生型 | 野生型 | 299 野生型、 51 <br> 棕眼 | 150 野生型、 149 朱红眼、 50 棕眼、 <br> 49 白眼 |\nA: 野生型果蝇的基因型有 6 种, 组合乙的亲本基因型分别是 $\\mathrm{AaX}^{\\mathrm{B}} X^{\\mathrm{B}} 、 \\mathrm{AaX}^{\\mathrm{b}} Y$\nB: 组合丙的 $\\mathrm{F}_{1}$ 野生型雌性与棕眼雌性比例原因可能是 $\\mathrm{b}$ 基因纯合致死\nC: 组合乙 $F_{1}$ 的棕眼果蝇自由交配, 后代中棕眼果蝇雌雄比为 $2: 1$\nD: 可通过与白眼雄果蝇杂交观察子代性状及比例判断某一野生型雌蝇的基因型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某果蝇种群眼色分为野生型和朱红色（由基因 A、a 控制）、野生型和棕色（由基因\n\n$\\mathrm{B} 、 \\mathrm{~b}$ 控制)，两对基因分别位于两对同源染色体上（不考虑 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段）。为研究其遗传机制, 进行了杂交实验, 结果如下表。下列说法正确的是 ( )\n\n| 杂交组合 | P |  | $\\mathrm{F}_{1}$ |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | o | $\\hat{0}$ | q | $\\hat{0}$ |\n\n\n| 甲 | 野生型 | 野生型 | 402 野生型 | 198 野生型、 201 朱红眼 |\n| :--- | :--- | :--- | :--- | :--- |\n| 乙 | 野生型 | 朱红眼 | 302 野生型、 99 <br> 棕眼 | 300 野生型、 101 棕眼 |\n| 丙 | 野生型 | 野生型 | 299 野生型、 51 <br> 棕眼 | 150 野生型、 149 朱红眼、 50 棕眼、 <br> 49 白眼 |\n\nA: 野生型果蝇的基因型有 6 种, 组合乙的亲本基因型分别是 $\\mathrm{AaX}^{\\mathrm{B}} X^{\\mathrm{B}} 、 \\mathrm{AaX}^{\\mathrm{b}} Y$\nB: 组合丙的 $\\mathrm{F}_{1}$ 野生型雌性与棕眼雌性比例原因可能是 $\\mathrm{b}$ 基因纯合致死\nC: 组合乙 $F_{1}$ 的棕眼果蝇自由交配, 后代中棕眼果蝇雌雄比为 $2: 1$\nD: 可通过与白眼雄果蝇杂交观察子代性状及比例判断某一野生型雌蝇的基因型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1506",
        "problem": "The 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nHere are some observations:\n\n1. The average speed with which allele frequency changes\n2. Some alleles takeover very large populations even when they have impacts too small to measure in laboratories\n3. The probability an allele will takeover a population\n4. The speed an allele's frequency changes in any one population at any one time\n5. The frequency at which an allele is initially introduced alters the probability it will takeover a population\n6. More fit alleles sometimes go extinct before less fit alleles\n\n\nWhich are caused by natural selection alone\nA: 1 and 2\nB: 3 and 4\nC: 5 and 6\nD: 1 only\nE: 2, 3 and 4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nHere are some observations:\n\n1. The average speed with which allele frequency changes\n2. Some alleles takeover very large populations even when they have impacts too small to measure in laboratories\n3. The probability an allele will takeover a population\n4. The speed an allele's frequency changes in any one population at any one time\n5. The frequency at which an allele is initially introduced alters the probability it will takeover a population\n6. More fit alleles sometimes go extinct before less fit alleles\n\n\nWhich are caused by natural selection alone\n\nA: 1 and 2\nB: 3 and 4\nC: 5 and 6\nD: 1 only\nE: 2, 3 and 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=970&width=704&top_left_y=778&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=966&width=706&top_left_y=778&top_left_x=1118"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1198",
        "problem": "Day-old chicks are difficult to sex but their down feathers will reveal the effects of a sex-linked dominant allele called 'barred'. In birds the female is the heterogametic sex. Which one of the following crosses allows correct sexing of the resultant chicks?\nA: Hen: barred, Cock: heterozygous barred\nB: Hen: unbarred, Cock: homozygous barred\nC: Hen: barred, Cock: unbarred\nD: Hen: unbarred, Cock: heterozygous barred\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDay-old chicks are difficult to sex but their down feathers will reveal the effects of a sex-linked dominant allele called 'barred'. In birds the female is the heterogametic sex. Which one of the following crosses allows correct sexing of the resultant chicks?\n\nA: Hen: barred, Cock: heterozygous barred\nB: Hen: unbarred, Cock: homozygous barred\nC: Hen: barred, Cock: unbarred\nD: Hen: unbarred, Cock: heterozygous barred\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1519",
        "problem": "Blue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nGive an estimate of the total whale population in 2010 .\nA: 200\nB: 2000\nC: 20000\nD: 200000\nE: 2000000\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBlue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nGive an estimate of the total whale population in 2010 .\n\nA: 200\nB: 2000\nC: 20000\nD: 200000\nE: 2000000\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-36.jpg?height=948&width=1673&top_left_y=657&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_451",
        "problem": "雄蜂由未受精的卵细胞发育而来, 雌蜂由受精卵发育而来, 受精卵发育至幼虫后,少数幼虫以蜂王浆为食而发育成蜂王, 而大多数幼虫以花粉和花蜜为食发育成工蜂 (生殖器官发育不全)。DNMT3 蛋白是 DNMT3 基因表达的一种 DNA 甲基化转移酶, 能使 DNA 特定结构添加甲基基团（如图所示）。敲除 DNMT3 基因后，取食花粉和花蜜的幼虫将发育成蜂王, 与取食蜂王浆有相同的效果。蜂群中蜜蜂个体之间有明确的分工, 又能通力合作，共同维持群体的生存和发展。下列有关叙述错误的是（ ）\n\n[图1]\n\n胞嘧啶\n5'甲基胞嘧啶\n\n[图2]\n\n部分被甲基化的DNA片段\nA: 雄蜂的体细胞中不含有同源染色体, 但它能产生正常的精子\nB: DNMT3 蛋白也能使腺嘌呤、胸腺嘧啶、鸟嘌呤添加甲基因而形成相应产物\nC: 基因敲除技术的应用有利于促进蜜蜂养殖业的发展，以提高养蜂人的收入\nD: 蜜蜂以跳舞的方式告知同伴蜜源的方向和位置, 传递的信息属于行为信息\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n雄蜂由未受精的卵细胞发育而来, 雌蜂由受精卵发育而来, 受精卵发育至幼虫后,少数幼虫以蜂王浆为食而发育成蜂王, 而大多数幼虫以花粉和花蜜为食发育成工蜂 (生殖器官发育不全)。DNMT3 蛋白是 DNMT3 基因表达的一种 DNA 甲基化转移酶, 能使 DNA 特定结构添加甲基基团（如图所示）。敲除 DNMT3 基因后，取食花粉和花蜜的幼虫将发育成蜂王, 与取食蜂王浆有相同的效果。蜂群中蜜蜂个体之间有明确的分工, 又能通力合作，共同维持群体的生存和发展。下列有关叙述错误的是（ ）\n\n[图1]\n\n胞嘧啶\n5'甲基胞嘧啶\n\n[图2]\n\n部分被甲基化的DNA片段\n\nA: 雄蜂的体细胞中不含有同源染色体, 但它能产生正常的精子\nB: DNMT3 蛋白也能使腺嘌呤、胸腺嘧啶、鸟嘌呤添加甲基因而形成相应产物\nC: 基因敲除技术的应用有利于促进蜜蜂养殖业的发展，以提高养蜂人的收入\nD: 蜜蜂以跳舞的方式告知同伴蜜源的方向和位置, 传递的信息属于行为信息\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-01.jpg?height=254&width=893&top_left_y=1004&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-01.jpg?height=189&width=509&top_left_y=1002&top_left_x=1276"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1313",
        "problem": "An experiment on gas exchange in a water plant, collected data about the rate of oxygen uptake and release. The plant was in darkness for 5 hours and then illuminated for 5 hours, while the temperature was constant. The results are shown in the graph.\n\nIf it is assumed that changes in light intensity have no effect on the rate of respiration, the best estimate of total oxygen produced by photosynthesis in the last two hours of the experiment is:\n\n[figure1]\nA: $600 \\mathrm{~cm}^{3}$\nB: $800 \\mathrm{~cm}^{3}$\nC: $1200 \\mathrm{~cm}^{3}$\nD: $1600 \\mathrm{~cm}^{3}$\nE: $1800 \\mathrm{~cm}^{3}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn experiment on gas exchange in a water plant, collected data about the rate of oxygen uptake and release. The plant was in darkness for 5 hours and then illuminated for 5 hours, while the temperature was constant. The results are shown in the graph.\n\nIf it is assumed that changes in light intensity have no effect on the rate of respiration, the best estimate of total oxygen produced by photosynthesis in the last two hours of the experiment is:\n\n[figure1]\n\nA: $600 \\mathrm{~cm}^{3}$\nB: $800 \\mathrm{~cm}^{3}$\nC: $1200 \\mathrm{~cm}^{3}$\nD: $1600 \\mathrm{~cm}^{3}$\nE: $1800 \\mathrm{~cm}^{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-03.jpg?height=648&width=991&top_left_y=441&top_left_x=1024"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_100",
        "problem": "The evolution of sexual reproduction has remained in certain aspects a puzzle which is yet to be fully pieced back together. Over the years many hypotheses have been proposed to explain this phenomenon.\nA: The effect of sexual reproduction on the association between alleles is expected to improve the fitness of co-adapted genes.\nB: Given that the majority of mutations are deleterious, sex is expected to counter the loss of mutation-free individuals in small populations.\nC: Avoiding inbreeding in sexually-reproducing organisms is expected to be advantageous when deleterious mutations are co-dominant.\nD: One would expect a greater proportion of sexually reproducing species in stable regions, as compared to unstable regions.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe evolution of sexual reproduction has remained in certain aspects a puzzle which is yet to be fully pieced back together. Over the years many hypotheses have been proposed to explain this phenomenon.\n\nA: The effect of sexual reproduction on the association between alleles is expected to improve the fitness of co-adapted genes.\nB: Given that the majority of mutations are deleterious, sex is expected to counter the loss of mutation-free individuals in small populations.\nC: Avoiding inbreeding in sexually-reproducing organisms is expected to be advantageous when deleterious mutations are co-dominant.\nD: One would expect a greater proportion of sexually reproducing species in stable regions, as compared to unstable regions.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_394",
        "problem": "黑腹果蝇 $\\mathrm{X}$ 染色体存在缺刻现象 (缺少某一片段)。缺刻红眼雌果蝇 $\\left(\\mathrm{X}^{\\mathrm{R}} \\mathrm{X}^{-}\\right.$- 与白眼雄果蝇 ( $\\left.X^{r} Y\\right)$ 杂交得 $F_{1}, F_{1}$ 雌雄个体杂交得 $F_{2}$ 。已知 $F_{1}$ 中雌雄个体数量比例为 2:1, 雄性全部为红眼, 雌性中既有红眼又有白眼。以下分析合理的是（ ）\nA: $F_{2}$ 中红眼个体的比例为 $1 / 2$\nB: $F_{1}$ 白眼的基因型为 $X^{r} X^{r}$\nC: $F_{2}$ 中雌雄个体数量比为 $3: 4$\nD: $\\mathrm{X}-$ 与 $\\mathrm{Y}$ 结合的子代会死亡\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n黑腹果蝇 $\\mathrm{X}$ 染色体存在缺刻现象 (缺少某一片段)。缺刻红眼雌果蝇 $\\left(\\mathrm{X}^{\\mathrm{R}} \\mathrm{X}^{-}\\right.$- 与白眼雄果蝇 ( $\\left.X^{r} Y\\right)$ 杂交得 $F_{1}, F_{1}$ 雌雄个体杂交得 $F_{2}$ 。已知 $F_{1}$ 中雌雄个体数量比例为 2:1, 雄性全部为红眼, 雌性中既有红眼又有白眼。以下分析合理的是（ ）\n\nA: $F_{2}$ 中红眼个体的比例为 $1 / 2$\nB: $F_{1}$ 白眼的基因型为 $X^{r} X^{r}$\nC: $F_{2}$ 中雌雄个体数量比为 $3: 4$\nD: $\\mathrm{X}-$ 与 $\\mathrm{Y}$ 结合的子代会死亡\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1211",
        "problem": "During the light-dependent reactions of photosynthesis, $\\mathrm{NADP}^{+}$is converted into NADPH. This requires the addition of\nA: one hydrogen atom\nB: one electron and $\\mathrm{H}^{+}$ion\nC: two electrons and a $\\mathrm{H}^{+}$ion\nD: two $\\mathrm{H}^{+}$ions and an electron\nE: two electrons and two $\\mathrm{H}^{+}$ions\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDuring the light-dependent reactions of photosynthesis, $\\mathrm{NADP}^{+}$is converted into NADPH. This requires the addition of\n\nA: one hydrogen atom\nB: one electron and $\\mathrm{H}^{+}$ion\nC: two electrons and a $\\mathrm{H}^{+}$ion\nD: two $\\mathrm{H}^{+}$ions and an electron\nE: two electrons and two $\\mathrm{H}^{+}$ions\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1348",
        "problem": "The research team sequenced the genome of 11 species of New Zealand snails and compared this molecular information with detailed mathematical description of the shape of the shell and outline of the shell opening. The relationships between these species are shown in the cladograms below. A cladogram uses branching lines that end at groups of organisms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMolecular phylogeny\n[figure1]Which species is inferred to be most closely related to Alcithoe benthicola?\nA: Alcithoe flemingi\nB: Alcithoe tigrina\nC: Alcithoe wilsoni\nD: Amoria hunteri\nE: Alcithoe fissurata\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nThe research team sequenced the genome of 11 species of New Zealand snails and compared this molecular information with detailed mathematical description of the shape of the shell and outline of the shell opening. The relationships between these species are shown in the cladograms below. A cladogram uses branching lines that end at groups of organisms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMolecular phylogeny\n[figure1]\n\nproblem:\nWhich species is inferred to be most closely related to Alcithoe benthicola?\n\nA: Alcithoe flemingi\nB: Alcithoe tigrina\nC: Alcithoe wilsoni\nD: Amoria hunteri\nE: Alcithoe fissurata\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-22.jpg?height=971&width=1128&top_left_y=1428&top_left_x=772"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_972",
        "problem": "The Costa Rican vampire bat is often not able to acquire blood from a mammal on a given night. Wilkinson (1984) trapped bats which were not allowed to feed for a night and found that they were given regurgitated blood by certain cave-mates. Based on this knowledge, which of the following observations are essential to confirm the occurrence of reciprocal altruism in this species?\n\nData showing that:\n\nI. Blood is exchanged only between kin\n\nII. Blood is exchanged between non-kin\n\nIII. Weak bats are frequently given blood even if they cannot give it to others\n\nIV. Bats who are given blood donate it to those who have given it to them previously\nA: I, II\nB: I, III\nC: II, III\nD: II, IV\nE: III, IV\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe Costa Rican vampire bat is often not able to acquire blood from a mammal on a given night. Wilkinson (1984) trapped bats which were not allowed to feed for a night and found that they were given regurgitated blood by certain cave-mates. Based on this knowledge, which of the following observations are essential to confirm the occurrence of reciprocal altruism in this species?\n\nData showing that:\n\nI. Blood is exchanged only between kin\n\nII. Blood is exchanged between non-kin\n\nIII. Weak bats are frequently given blood even if they cannot give it to others\n\nIV. Bats who are given blood donate it to those who have given it to them previously\n\nA: I, II\nB: I, III\nC: II, III\nD: II, IV\nE: III, IV\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_376",
        "problem": "如图是雄性哺乳动物体内处于分裂某时期的一个细胞的染色体示意图。相关叙述不正确的是\n\n[图1]\nA: 该个体的基因型为 $\\mathrm{AaBbDd}$\nB: 该细胞正在进行减数分裂\nC: 该细胞分裂完成后只产生 2 种基因型的精子\nD: A、 $\\mathrm{a}$ 和 D、 $\\mathrm{d}$ 基因的遗传遵循自由组合定律\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图是雄性哺乳动物体内处于分裂某时期的一个细胞的染色体示意图。相关叙述不正确的是\n\n[图1]\n\nA: 该个体的基因型为 $\\mathrm{AaBbDd}$\nB: 该细胞正在进行减数分裂\nC: 该细胞分裂完成后只产生 2 种基因型的精子\nD: A、 $\\mathrm{a}$ 和 D、 $\\mathrm{d}$ 基因的遗传遵循自由组合定律\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-11.jpg?height=448&width=460&top_left_y=1906&top_left_x=181"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_773",
        "problem": "在果蝇中，正常眼（A）对棘眼（a）是显性性状，长翅（B）对截翅（b）是显性性状, 体色褐色 (D) 对黄色 (d) 是显性性状。一只正常眼长翅褐色雌果蝇与欶眼截翅黄色雄果蝇进行杂交，子代个体数如下表所示。下列说法正确的是（ ）\n\n| 表 <br> 现 <br> 型 | 正常眼 <br> 翅褐 <br> 色 | 正常眼 <br> 长翅黄 <br> 色 | 棘眼截 <br> 翅褐色 | 欶眼截 <br> 翅黄色 | 正常眼 <br> 截翅褐 <br> 色 | 正常眼 <br> 截翅黄 <br> 色 | 欶眼长 <br> 翅褐色 | 棘眼长 <br> 翅黄色 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 数 <br> 量 | 97 | 98 | 97 | 98 | 3 | 3 | 2 | 2 |\nA: $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 基因位于同一对常染色体, $\\mathrm{D} / \\mathrm{d}$ 基因位于 $\\mathrm{X}$ 染色体\nB: 亲本雌果蝇发生交换的初级卵母细胞的比例约为 10\\%\nC: 上述杂交过程发生了基因突变、基因重组等可遗传变异\nD: 子代正常眼长翅褐色果蝇自由交配可获得三个性状都是隐性的个体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在果蝇中，正常眼（A）对棘眼（a）是显性性状，长翅（B）对截翅（b）是显性性状, 体色褐色 (D) 对黄色 (d) 是显性性状。一只正常眼长翅褐色雌果蝇与欶眼截翅黄色雄果蝇进行杂交，子代个体数如下表所示。下列说法正确的是（ ）\n\n| 表 <br> 现 <br> 型 | 正常眼 <br> 翅褐 <br> 色 | 正常眼 <br> 长翅黄 <br> 色 | 棘眼截 <br> 翅褐色 | 欶眼截 <br> 翅黄色 | 正常眼 <br> 截翅褐 <br> 色 | 正常眼 <br> 截翅黄 <br> 色 | 欶眼长 <br> 翅褐色 | 棘眼长 <br> 翅黄色 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 数 <br> 量 | 97 | 98 | 97 | 98 | 3 | 3 | 2 | 2 |\n\nA: $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 基因位于同一对常染色体, $\\mathrm{D} / \\mathrm{d}$ 基因位于 $\\mathrm{X}$ 染色体\nB: 亲本雌果蝇发生交换的初级卵母细胞的比例约为 10\\%\nC: 上述杂交过程发生了基因突变、基因重组等可遗传变异\nD: 子代正常眼长翅褐色果蝇自由交配可获得三个性状都是隐性的个体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_90",
        "problem": "Proteins encoded by Hox genes share a 60 -amino-acid DNA-binding motif, the homeodomain. In the fruit fly Drosophila melanogaster genome, eight Hox genes are assembled in one cluster on the same chromosome (Figure 1A). The segmental expression pattern of Hox genes along the anterior-posterior axis of the fruit fly embryo shows collinearity with the gene order on the chromosome (Figure 1B). Fruit flies normally possess a pair of wings that develop from the second thoracic segment (T2) of the embryo, and a pair of balance organs (halteres) that develop from the third thoracic segment (T3). When the gene expression of $U b x$ gene is lost by mutations, T3 transform into T2 and two pairs of wings are formed. Beetles and grasshoppers have two pairs of wings although the most anterior segment of the UBX protein expression of their embryos is found in T3, the same as that of the wild fruit fly (Figure 1C).\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\nC Fruit fly\n\n[figure3]\n\n[figure4]\n\n[figure5]\n\nFigure 1. (A) Eight Hox genes on the fruit fly genome: $l a b, p b, D f d, S c r, A n t p, U b x, a b d-A$, and $A b d-B$. (B) Segmental expression pattern of the Hox genes in the fruit fly embryo. Anterior is the left. The expression patterns for each gene are illustrated by labels that correspond to (A). Arced bars shown below indicate the range of the expression of each gene. (C) Schematic drawings for UBX protein expression in three species of embryos. Anterior is up. Red arrows indicate the boundary between T2 and T3. Area with gene expression is painted dark.\nA: Proteins encoded by Hox genes act as transcription factors that regulate gene expression.\nB: The segmental expression pattern of the Hox genes determines the identities of each segment in fruit fly embryos.\nC: In the $U b x$ gene mutants, the extension of the $a b d-A$ expression to the anterior region leads to the transformation of thoracic segment.\nD: Beetles and grasshoppers have two pairs of wings because their $U b x$ genes control a different set of genes in T3 from that of fruit flies.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nProteins encoded by Hox genes share a 60 -amino-acid DNA-binding motif, the homeodomain. In the fruit fly Drosophila melanogaster genome, eight Hox genes are assembled in one cluster on the same chromosome (Figure 1A). The segmental expression pattern of Hox genes along the anterior-posterior axis of the fruit fly embryo shows collinearity with the gene order on the chromosome (Figure 1B). Fruit flies normally possess a pair of wings that develop from the second thoracic segment (T2) of the embryo, and a pair of balance organs (halteres) that develop from the third thoracic segment (T3). When the gene expression of $U b x$ gene is lost by mutations, T3 transform into T2 and two pairs of wings are formed. Beetles and grasshoppers have two pairs of wings although the most anterior segment of the UBX protein expression of their embryos is found in T3, the same as that of the wild fruit fly (Figure 1C).\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\nC Fruit fly\n\n[figure3]\n\n[figure4]\n\n[figure5]\n\nFigure 1. (A) Eight Hox genes on the fruit fly genome: $l a b, p b, D f d, S c r, A n t p, U b x, a b d-A$, and $A b d-B$. (B) Segmental expression pattern of the Hox genes in the fruit fly embryo. Anterior is the left. The expression patterns for each gene are illustrated by labels that correspond to (A). Arced bars shown below indicate the range of the expression of each gene. (C) Schematic drawings for UBX protein expression in three species of embryos. Anterior is up. Red arrows indicate the boundary between T2 and T3. Area with gene expression is painted dark.\n\nA: Proteins encoded by Hox genes act as transcription factors that regulate gene expression.\nB: The segmental expression pattern of the Hox genes determines the identities of each segment in fruit fly embryos.\nC: In the $U b x$ gene mutants, the extension of the $a b d-A$ expression to the anterior region leads to the transformation of thoracic segment.\nD: Beetles and grasshoppers have two pairs of wings because their $U b x$ genes control a different set of genes in T3 from that of fruit flies.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-57.jpg?height=88&width=819&top_left_y=1184&top_left_x=276",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-57.jpg?height=411&width=731&top_left_y=1299&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-57.jpg?height=391&width=209&top_left_y=1249&top_left_x=1209",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-57.jpg?height=472&width=174&top_left_y=1189&top_left_x=1435",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-57.jpg?height=476&width=195&top_left_y=1187&top_left_x=1633"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1069",
        "problem": "The wild type (normal) fruit fly, Drosophila melanogaster has straight wings and long bristles. Mutant strains have been isolated that show either curled wings or short bristles. It is known that all genes for these phenotypes are located on separate chromosomes.\n\nWhen flies with straight wings and short bristles were self-crossed (Cross I), the following results were obtained:\n\n| Straight wings, long bristles | 30 |\n| :--- | :--- |\n| Straight wings, short bristles | 90 |\n| Curled wings, long bristles | 10 |\n| Curled wings, short bristles | 30 |\n| Total | 160 |\n\nNote: Assume $A$ and $a$ as alleles for wing structure and $B$ and $b$ as alleles for bristle length.\n\nWhen flies used for cross I were then used to cross with flies with curled wings short bristles, all flies in the progeny had short bristles. What is the genotype of the parental curled winged flies?",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThe wild type (normal) fruit fly, Drosophila melanogaster has straight wings and long bristles. Mutant strains have been isolated that show either curled wings or short bristles. It is known that all genes for these phenotypes are located on separate chromosomes.\n\nWhen flies with straight wings and short bristles were self-crossed (Cross I), the following results were obtained:\n\n| Straight wings, long bristles | 30 |\n| :--- | :--- |\n| Straight wings, short bristles | 90 |\n| Curled wings, long bristles | 10 |\n| Curled wings, short bristles | 30 |\n| Total | 160 |\n\nNote: Assume $A$ and $a$ as alleles for wing structure and $B$ and $b$ as alleles for bristle length.\n\nWhen flies used for cross I were then used to cross with flies with curled wings short bristles, all flies in the progeny had short bristles. What is the genotype of the parental curled winged flies?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_440",
        "problem": "科研人员用一种甜瓜 $(2 n)$ 的纯合亲本进行杂交得到 $F_{1}, F_{1}$ 经自交得到 $F_{2}$,结果如表所示。已知 $\\mathrm{A} 、 \\mathrm{E}$ 基因同在一条染色体上， $\\mathrm{a} 、 \\mathrm{e}$ 基因同在另一条染色体上，当 $\\mathrm{E}$ 和 $\\mathrm{F}$ 同时存在时果皮才表现出有覆纹性状。不考虑染色体互换、突变, 下列分析正确的是（）\n\n| 性状 | 控制基因及其所在染色体 | 母本 | 父本 | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 果皮底 <br> 色 | $\\mathrm{A} / \\mathrm{a}$ 位于 4 号染色体 | 黄绿 <br> 色 | 黄色 | 黄绿 <br> 色 | 黄绿色: 黄色 $\\approx 3: 1$ |\n| 果皮覆 <br> 纹 | $\\mathrm{E} / \\mathrm{e}$ 位于 4 号染色体, $\\mathrm{F} / \\mathrm{f}$ 位 <br> 于 2 号染色体 | 无覆 <br> 纹 | 无覆 <br> 纹 | 有覆 <br> 纹 | 有覆纹: 无覆纹 $\\approx 9:$ <br> 7 |\nA: 母本、父本的基因型分别是 aaeeFF、AAEEff\nB: $F_{1}$ 的基因型为 $\\mathrm{AaEeFf}$,产生的配子类型有 8 种\nC: $F_{2}$ 的表型有 4 种, 基因型有 9 种\nD: $F_{2}$ 中黄绿色无覆纹果皮植株所占的比例为 $3 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n科研人员用一种甜瓜 $(2 n)$ 的纯合亲本进行杂交得到 $F_{1}, F_{1}$ 经自交得到 $F_{2}$,结果如表所示。已知 $\\mathrm{A} 、 \\mathrm{E}$ 基因同在一条染色体上， $\\mathrm{a} 、 \\mathrm{e}$ 基因同在另一条染色体上，当 $\\mathrm{E}$ 和 $\\mathrm{F}$ 同时存在时果皮才表现出有覆纹性状。不考虑染色体互换、突变, 下列分析正确的是（）\n\n| 性状 | 控制基因及其所在染色体 | 母本 | 父本 | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 果皮底 <br> 色 | $\\mathrm{A} / \\mathrm{a}$ 位于 4 号染色体 | 黄绿 <br> 色 | 黄色 | 黄绿 <br> 色 | 黄绿色: 黄色 $\\approx 3: 1$ |\n| 果皮覆 <br> 纹 | $\\mathrm{E} / \\mathrm{e}$ 位于 4 号染色体, $\\mathrm{F} / \\mathrm{f}$ 位 <br> 于 2 号染色体 | 无覆 <br> 纹 | 无覆 <br> 纹 | 有覆 <br> 纹 | 有覆纹: 无覆纹 $\\approx 9:$ <br> 7 |\n\nA: 母本、父本的基因型分别是 aaeeFF、AAEEff\nB: $F_{1}$ 的基因型为 $\\mathrm{AaEeFf}$,产生的配子类型有 8 种\nC: $F_{2}$ 的表型有 4 种, 基因型有 9 种\nD: $F_{2}$ 中黄绿色无覆纹果皮植株所占的比例为 $3 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_920",
        "problem": "摩尔根等人做完突变体白眼雄蝇与红眼雌蝇杂交实验后, 还进行了这只白眼雄蝇与 $\\mathrm{F}_{1}$ 红眼雌蝇的回交实验, 圆满地说明了他的实验结论。为了进一步验证控制果蝇眼睛颜色的基因位于 $\\mathrm{X}$ 染色体上, 他们根据上述实验中的材料, 又设计了三个新的实验。这三个实验分别是：(1) $\\mathrm{F}_{2}$ 雌蝇 $\\times$ 白眼雄蝇; (2)白眼雌蝇 $\\times$ 红眼雄蝇; (3)白眼雌蝇 $\\times$ 白眼雄蝇。关于这三个实验，下列说法错误的是（ ）\nA: (1)实验所产的后代是 $1 / 4$ 红眼雌蝇、 $1 / 4$ 白眼雌蝇、 $1 / 4$ 红眼雄蝇、 $1 / 4$ 白眼雄蝇\nB: (2)实验子代中雌蝇都是红眼, 雄蝇都是白眼\nC: (3)实验子代能成为稳定的品系\nD: 三个实验中, (2)实验最为关键\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n摩尔根等人做完突变体白眼雄蝇与红眼雌蝇杂交实验后, 还进行了这只白眼雄蝇与 $\\mathrm{F}_{1}$ 红眼雌蝇的回交实验, 圆满地说明了他的实验结论。为了进一步验证控制果蝇眼睛颜色的基因位于 $\\mathrm{X}$ 染色体上, 他们根据上述实验中的材料, 又设计了三个新的实验。这三个实验分别是：(1) $\\mathrm{F}_{2}$ 雌蝇 $\\times$ 白眼雄蝇; (2)白眼雌蝇 $\\times$ 红眼雄蝇; (3)白眼雌蝇 $\\times$ 白眼雄蝇。关于这三个实验，下列说法错误的是（ ）\n\nA: (1)实验所产的后代是 $1 / 4$ 红眼雌蝇、 $1 / 4$ 白眼雌蝇、 $1 / 4$ 红眼雄蝇、 $1 / 4$ 白眼雄蝇\nB: (2)实验子代中雌蝇都是红眼, 雄蝇都是白眼\nC: (3)实验子代能成为稳定的品系\nD: 三个实验中, (2)实验最为关键\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1434",
        "problem": "The following table provides information about three different solutions, with respect to the plasma membrane.\n\n| Solution | Diffusible across plasma <br> membrane |\n| :--- | :---: |\n| 0.9\\% sodium chloride <br> solution | No |\n| $5.0 \\%$ dextrose (glucose) <br> solution | No |\n| $1.8 \\%$ urea solution | Yes |\n\nThese solutions can all be treated as iso-osmolar to human cells.\n\nWhich of the following is not a characteristic of the plasma membrane?\nA: It is selectively permeable.\nB: Cholesterol helps to maintain membrane fluidity at low temperatures.\nC: It is composed of a phospholipid bilayer.\nD: Peripheral proteins enable facilitated diffusion.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following table provides information about three different solutions, with respect to the plasma membrane.\n\n| Solution | Diffusible across plasma <br> membrane |\n| :--- | :---: |\n| 0.9\\% sodium chloride <br> solution | No |\n| $5.0 \\%$ dextrose (glucose) <br> solution | No |\n| $1.8 \\%$ urea solution | Yes |\n\nThese solutions can all be treated as iso-osmolar to human cells.\n\nWhich of the following is not a characteristic of the plasma membrane?\n\nA: It is selectively permeable.\nB: Cholesterol helps to maintain membrane fluidity at low temperatures.\nC: It is composed of a phospholipid bilayer.\nD: Peripheral proteins enable facilitated diffusion.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_648",
        "problem": "“母性效应”是指子代某一性状的表型由母体的核基因型决定, 而不受本身基因型的支配。椎实螺是一种雌雄同体的软体动物, 可异体受精, 也可自体受精, 其螺壳的旋转方向有左旋和右旋的区分, 旋转方向符合“母性效应”。母本右旋椎实螺 A (SS) 和父本左旋椎实螺 $B$ (ss) 杂交得到的 $F_{1}$ 企为右旋椎实螺, $F_{1}$ 随机交配获得 $F_{2}$ 。下列叙述错误的是 ( )\nA: 自然状态下 SS 个体的亲本可能都是左旋椎实螺\nB: $\\mathrm{F}_{2}$ 自交产生的 $\\mathrm{F}_{3}$ 全为右旋椎实螺\nC: 利用“母性效应”的个体可以验证基因分离定律\nD: 欲判断某左旋椎实螺的基因型, 可采用自交统计后代性状的方法确定\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n“母性效应”是指子代某一性状的表型由母体的核基因型决定, 而不受本身基因型的支配。椎实螺是一种雌雄同体的软体动物, 可异体受精, 也可自体受精, 其螺壳的旋转方向有左旋和右旋的区分, 旋转方向符合“母性效应”。母本右旋椎实螺 A (SS) 和父本左旋椎实螺 $B$ (ss) 杂交得到的 $F_{1}$ 企为右旋椎实螺, $F_{1}$ 随机交配获得 $F_{2}$ 。下列叙述错误的是 ( )\n\nA: 自然状态下 SS 个体的亲本可能都是左旋椎实螺\nB: $\\mathrm{F}_{2}$ 自交产生的 $\\mathrm{F}_{3}$ 全为右旋椎实螺\nC: 利用“母性效应”的个体可以验证基因分离定律\nD: 欲判断某左旋椎实螺的基因型, 可采用自交统计后代性状的方法确定\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_702",
        "problem": "如图表示细胞内遗传信息的传递过程, 下列有关叙述错误的是（ ）\n\n[图1]\nA: 过程(1)和(2)都需改变 DNA 的空间结构\nB: 相较于过程(2), 过程(3)特有的碱基配对方式为 U-A\nC: 过程(3)中核糖体在 mRNA 上的移动方向是 $\\mathrm{a}$ 到 $\\mathrm{b}$\nD: 图示 tRNA 搬运的氨基酸对应的密码子为 $\\mathrm{CCA}\\left(5^{\\prime} \\rightarrow 3^{\\prime}\\right)$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图表示细胞内遗传信息的传递过程, 下列有关叙述错误的是（ ）\n\n[图1]\n\nA: 过程(1)和(2)都需改变 DNA 的空间结构\nB: 相较于过程(2), 过程(3)特有的碱基配对方式为 U-A\nC: 过程(3)中核糖体在 mRNA 上的移动方向是 $\\mathrm{a}$ 到 $\\mathrm{b}$\nD: 图示 tRNA 搬运的氨基酸对应的密码子为 $\\mathrm{CCA}\\left(5^{\\prime} \\rightarrow 3^{\\prime}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-81.jpg?height=537&width=879&top_left_y=2050&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1149",
        "problem": "Which of the following is NOT a density-dependent factor that influences population density\nA: Competition\nB: Disease\nC: Drought\nD: Toxic waste\nE: Predation\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is NOT a density-dependent factor that influences population density\n\nA: Competition\nB: Disease\nC: Drought\nD: Toxic waste\nE: Predation\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1162",
        "problem": "Facial recognition has been show in the behaviour of many species. Cows (Bos taurus) of the breed Prim Holstein were studied to measure this trait. Cows were shown two pictures, one of a member of their own social group and one of an unfamiliar cow of the same breed. They were rewarded with food when they walked towards the unfamiliar cow's picture. The cows were first trained, using the same pictures of a familiar cow and an unfamiliar cow. Then, in 'generalisation' tests, experimental subjects were shown different angles of familiar cows' faces and unfamiliar cows' faces (examples are shown below). The number of sessions required for cows to reliably choose the correct picture was recorded. The * sign indicates a statistically significant difference.\n[figure1]\n\n◊ùPrim'Holstein Familiar Individual $\\square$ Prim'Holstein Unfamiliar Individual\n\n[figure2]\n\nhttp://www.plosone.org/article/info\\%3Adoi\\%2F10.1371\\%2Fjournal.pone.0004441\\#pone-0004441-t001\n\nWhat conclusion can be drawn from the data?\nA: Cows only have a short term memory.\nB: Cows show recognition of familiar individuals.\nC: Cows cannot recognise familiar individuals.\nD: Cows are colour-blind.\nE: Cows will only ever walk towards food.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFacial recognition has been show in the behaviour of many species. Cows (Bos taurus) of the breed Prim Holstein were studied to measure this trait. Cows were shown two pictures, one of a member of their own social group and one of an unfamiliar cow of the same breed. They were rewarded with food when they walked towards the unfamiliar cow's picture. The cows were first trained, using the same pictures of a familiar cow and an unfamiliar cow. Then, in 'generalisation' tests, experimental subjects were shown different angles of familiar cows' faces and unfamiliar cows' faces (examples are shown below). The number of sessions required for cows to reliably choose the correct picture was recorded. The * sign indicates a statistically significant difference.\n[figure1]\n\n◊ùPrim'Holstein Familiar Individual $\\square$ Prim'Holstein Unfamiliar Individual\n\n[figure2]\n\nhttp://www.plosone.org/article/info\\%3Adoi\\%2F10.1371\\%2Fjournal.pone.0004441\\#pone-0004441-t001\n\nWhat conclusion can be drawn from the data?\n\nA: Cows only have a short term memory.\nB: Cows show recognition of familiar individuals.\nC: Cows cannot recognise familiar individuals.\nD: Cows are colour-blind.\nE: Cows will only ever walk towards food.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-07.jpg?height=674&width=436&top_left_y=570&top_left_x=112",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-07.jpg?height=457&width=991&top_left_y=737&top_left_x=721"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_291",
        "problem": "The figure below depicts life-history strategies for three plant species $(a \\sim c)$ along 3 axes: strength of competition with other organisms, level of disturbance in the habitat, and level of environmental stress in the habitat. Species $a$ grows in habitats where competition among species is high but disturbance and stress are low. Species $b$ grows in habitats with high environmental stress but with low interspecies competition. Species $c$ grows in highly disturbed habitats with low environmental stress.\n\n[figure1]\n\nWhich of the statements below is/are correct?\n\nI Characteristics of \" $a$-type\" plants are slow growth rate and short-lived leaves.\n\nII Desert annual plants are \"b-type\" species. They have rapid growth and produce large amount of seeds in a short time after rains.\n\nIII Most plants belonging to \"c-type\" species would be herbaceous while \" $a$-type\" and \"b-types\" species are likely to be trees or shrubs.\nA: Only II\nB: Only I and II\nC: Only I and III\nD: Only II and III\nE: I, II, and III\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe figure below depicts life-history strategies for three plant species $(a \\sim c)$ along 3 axes: strength of competition with other organisms, level of disturbance in the habitat, and level of environmental stress in the habitat. Species $a$ grows in habitats where competition among species is high but disturbance and stress are low. Species $b$ grows in habitats with high environmental stress but with low interspecies competition. Species $c$ grows in highly disturbed habitats with low environmental stress.\n\n[figure1]\n\nWhich of the statements below is/are correct?\n\nI Characteristics of \" $a$-type\" plants are slow growth rate and short-lived leaves.\n\nII Desert annual plants are \"b-type\" species. They have rapid growth and produce large amount of seeds in a short time after rains.\n\nIII Most plants belonging to \"c-type\" species would be herbaceous while \" $a$-type\" and \"b-types\" species are likely to be trees or shrubs.\n\nA: Only II\nB: Only I and II\nC: Only I and III\nD: Only II and III\nE: I, II, and III\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-52.jpg?height=397&width=685&top_left_y=864&top_left_x=728"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1242",
        "problem": "Conservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.The greatest risk for the survival of Mauis dolphin remains?\nA: Trawl fishing\nB: Set netting\nC: Seismic surveying\nD: Mining and oil activities\nE: Noise leading to trauma\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nConservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.\n\nproblem:\nThe greatest risk for the survival of Mauis dolphin remains?\n\nA: Trawl fishing\nB: Set netting\nC: Seismic surveying\nD: Mining and oil activities\nE: Noise leading to trauma\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_899",
        "problem": "生物兴趣小组开展人类遗传病发病率和遗传方式的调查活动, 发现某家系同时有甲遗传病（基因为 $\\mathrm{A} 、 \\mathrm{a}$ ）和乙遗传病（基因为 $\\mathrm{B} 、 \\mathrm{~b}$ ）患者，系谱如图 1; 对部分家庭成员是否携带甲病基因进行核酸分子检测, 结果如图 2, 已知乙病为常染色体遗传病, 且在人群中的发病率为 $1 / 10000$ 。上述基因均不位于 $\\mathrm{Y}$ 染色体上,下列叙述错误的是 ( )\n\n[图1]\n\n四10.病男女目乙病男\n\n两病兼患男 $\\square 0$ 正常男女\n\n图1\n\n[图2]\n\n图2\nA: 甲病属于常染色体隐性遗传病\nB: III $_{9}$ 是患病男孩的概率是 $11 / 27$\nC: $\\mathrm{II}_{7}$ 同时携带两种致病基因的概率是 $4 / 9$\nD: $\\mathrm{II}_{7}$ 与另一健康男子再婚, 生一个患乙病孩子的概率约为 $1 / 303$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n生物兴趣小组开展人类遗传病发病率和遗传方式的调查活动, 发现某家系同时有甲遗传病（基因为 $\\mathrm{A} 、 \\mathrm{a}$ ）和乙遗传病（基因为 $\\mathrm{B} 、 \\mathrm{~b}$ ）患者，系谱如图 1; 对部分家庭成员是否携带甲病基因进行核酸分子检测, 结果如图 2, 已知乙病为常染色体遗传病, 且在人群中的发病率为 $1 / 10000$ 。上述基因均不位于 $\\mathrm{Y}$ 染色体上,下列叙述错误的是 ( )\n\n[图1]\n\n四10.病男女目乙病男\n\n两病兼患男 $\\square 0$ 正常男女\n\n图1\n\n[图2]\n\n图2\n\nA: 甲病属于常染色体隐性遗传病\nB: III $_{9}$ 是患病男孩的概率是 $11 / 27$\nC: $\\mathrm{II}_{7}$ 同时携带两种致病基因的概率是 $4 / 9$\nD: $\\mathrm{II}_{7}$ 与另一健康男子再婚, 生一个患乙病孩子的概率约为 $1 / 303$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-028.jpg?height=286&width=391&top_left_y=642&top_left_x=387",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-028.jpg?height=366&width=739&top_left_y=725&top_left_x=1047"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1383",
        "problem": "All vertebrates show some sort of bilateral symmetry. What advantage does this provide?\nA: Concentration of most sensory organs are in the head\nB: Walking on land\nC: Flying\nD: Development of bones\nE: Ability to give birth to live offspring\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAll vertebrates show some sort of bilateral symmetry. What advantage does this provide?\n\nA: Concentration of most sensory organs are in the head\nB: Walking on land\nC: Flying\nD: Development of bones\nE: Ability to give birth to live offspring\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_710",
        "problem": "为了使番茄成为乡村振兴的致富果, 科技工作者研究了番茄遗传方式。已知番茄果肉的颜色由基因 $\\mathrm{A} / \\mathrm{a}$ 与 $\\mathrm{B} / \\mathrm{b}$ 控制。甲、乙两种番茄杂交，结果如图所示; 用 $\\mathrm{A} 、 \\mathrm{a} 、 \\mathrm{~B} 、$ $\\mathrm{b}$ 四种基因的特异性引物对甲、乙番茄果肉细胞的 DNA 进行 PCR 扩增, 并用 A 基因特异性引物对红色番茄丙、用 B 基因特异性引物对红色番茄丁的 DNA 进 PCR 扩增作为标准参照, PCR 产物电泳结果如下图所示。在不考虑变异的情下, 下列叙述正确的是\n\n( )\n[图1]\nA: 甲番茄的基因型为 $a a B B$\nB: $F_{1}$ 番茄与乙番茄杂交, 子代可出现橙色番茄\nC: $F_{2}$ 中橙色番茄自交后代不会发生性状分离\nD: 理论上, $\\mathrm{F}_{2}$ 红色番茄中自交能产生橙色番茄的占 $2 / 9$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n为了使番茄成为乡村振兴的致富果, 科技工作者研究了番茄遗传方式。已知番茄果肉的颜色由基因 $\\mathrm{A} / \\mathrm{a}$ 与 $\\mathrm{B} / \\mathrm{b}$ 控制。甲、乙两种番茄杂交，结果如图所示; 用 $\\mathrm{A} 、 \\mathrm{a} 、 \\mathrm{~B} 、$ $\\mathrm{b}$ 四种基因的特异性引物对甲、乙番茄果肉细胞的 DNA 进行 PCR 扩增, 并用 A 基因特异性引物对红色番茄丙、用 B 基因特异性引物对红色番茄丁的 DNA 进 PCR 扩增作为标准参照, PCR 产物电泳结果如下图所示。在不考虑变异的情下, 下列叙述正确的是\n\n( )\n[图1]\n\nA: 甲番茄的基因型为 $a a B B$\nB: $F_{1}$ 番茄与乙番茄杂交, 子代可出现橙色番茄\nC: $F_{2}$ 中橙色番茄自交后代不会发生性状分离\nD: 理论上, $\\mathrm{F}_{2}$ 红色番茄中自交能产生橙色番茄的占 $2 / 9$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-63.jpg?height=472&width=1448&top_left_y=894&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1294",
        "problem": "yellow-crowned parakeet, orange-fronted parakeet, red-crowned parakeet, Forbes' parakeet and Antipodes Island parakeet. All are bright green in colour but with distinguishing coloured areas on the head. The yellow-crowned parakeet is rare but is found throughout forested areas of the North, South and Stewart Islands as well as the sub-Antarctic Auckland Islands. The orange-fronted parakeet is critically endangered with around 300 birds found in just three alpine beech forest valleys in Canterbury. The red-crowned parakeet was widespread throughout the mainland last century but today is very rare on the mainland and only common on islands free of mammalian predators. Forbes' parakeet is restricted to the Chatham Island and also critically endangered. The Antipodes Island parakeet is restricted to the Antipodes Islands.\n\nDeforestation, disease, introduced predators, and shooting by farmers have all contributed to the present restricted distribution of these species. An important conservation tool for the restoration of endangered birds is the translocation of species to habitats that have been restored and introduced predators eradicated.Luis Ortiz-Catedral at the Ecology and Conservation Lab, Institute of Natural \\& Mathematical Sciences, Massey University, has been studying kkriki on Raoul Island, a remote volcanic island approximately $995 \\mathrm{~km}$ north of New Zealand. There had not been a confirmed record of resident parakeets on Raoul Island since 1836 following the introduction of goats, cats and rats. Goats were removed from the island in 1986. Then, in the world's largest multispecies eradication project to date, the New Zealand Department of Conservation (DOC) successfully removed domestic cats, and Norway and Pacific rats (kiore) from Raoul Island using aerial drops of poisoned bait for rats between 2002 and 2004, and follow-up ground-based control for cats.\n\nPrior to the removal of these invasive species on Raoul, the last strongholds for Kermadec red-crowned parakeets were the Herald Islets (ca. 50 breeding pairs) and Macauley (ca. 10,000 breeding pairs) 2-4 km east and $108 \\mathrm{~km}$ south respectively off the coast of Raoul Island. Since 2000 (i.e. two years prior to initiation of the predator removal programme), staff from DOC have carried out bird surveys roughly once a year on Raoul to assess the effect of the removal of predators. No parakeets were detected prior to eradication of cats and rats. After the cat and rat eradication there were infrequent sightings of one to three parakeets. In 2008 during the parakeet survey, 100 parakeets were caught during a 13-day mist-netting period. Of these, 59 were female and 41 were male, of which 56 were adults and 44 sub-adults hatched in 2008 . One full pre-mating display followed by copulation was also observed and two nests were located in fallen logs of Kermadec pohutukawa.\n\nConsidering the data given above what conclusion can be drawn about the presence of red-crowned parakeets on Raoul Island?\nA: Red-crowned parakeets are now frequent visitors to Raoul Island from their breeding sites on the Herald Islets.\nB: Red-crowned parakeets have successfully recolonised Raoul Island after eradication of invasive predators.\nC: Red-crowned parakeets have been successfully translocated to Raoul Island.\nD: Eradication of invasive predators on Raoul Island has allowed remnant populations of red-crowned parakeets to increase.\nE: Red-crowned parakeets are now common on Raoul Island after eradication of invasive species.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nyellow-crowned parakeet, orange-fronted parakeet, red-crowned parakeet, Forbes' parakeet and Antipodes Island parakeet. All are bright green in colour but with distinguishing coloured areas on the head. The yellow-crowned parakeet is rare but is found throughout forested areas of the North, South and Stewart Islands as well as the sub-Antarctic Auckland Islands. The orange-fronted parakeet is critically endangered with around 300 birds found in just three alpine beech forest valleys in Canterbury. The red-crowned parakeet was widespread throughout the mainland last century but today is very rare on the mainland and only common on islands free of mammalian predators. Forbes' parakeet is restricted to the Chatham Island and also critically endangered. The Antipodes Island parakeet is restricted to the Antipodes Islands.\n\nDeforestation, disease, introduced predators, and shooting by farmers have all contributed to the present restricted distribution of these species. An important conservation tool for the restoration of endangered birds is the translocation of species to habitats that have been restored and introduced predators eradicated.\n\nproblem:\nLuis Ortiz-Catedral at the Ecology and Conservation Lab, Institute of Natural \\& Mathematical Sciences, Massey University, has been studying kkriki on Raoul Island, a remote volcanic island approximately $995 \\mathrm{~km}$ north of New Zealand. There had not been a confirmed record of resident parakeets on Raoul Island since 1836 following the introduction of goats, cats and rats. Goats were removed from the island in 1986. Then, in the world's largest multispecies eradication project to date, the New Zealand Department of Conservation (DOC) successfully removed domestic cats, and Norway and Pacific rats (kiore) from Raoul Island using aerial drops of poisoned bait for rats between 2002 and 2004, and follow-up ground-based control for cats.\n\nPrior to the removal of these invasive species on Raoul, the last strongholds for Kermadec red-crowned parakeets were the Herald Islets (ca. 50 breeding pairs) and Macauley (ca. 10,000 breeding pairs) 2-4 km east and $108 \\mathrm{~km}$ south respectively off the coast of Raoul Island. Since 2000 (i.e. two years prior to initiation of the predator removal programme), staff from DOC have carried out bird surveys roughly once a year on Raoul to assess the effect of the removal of predators. No parakeets were detected prior to eradication of cats and rats. After the cat and rat eradication there were infrequent sightings of one to three parakeets. In 2008 during the parakeet survey, 100 parakeets were caught during a 13-day mist-netting period. Of these, 59 were female and 41 were male, of which 56 were adults and 44 sub-adults hatched in 2008 . One full pre-mating display followed by copulation was also observed and two nests were located in fallen logs of Kermadec pohutukawa.\n\nConsidering the data given above what conclusion can be drawn about the presence of red-crowned parakeets on Raoul Island?\n\nA: Red-crowned parakeets are now frequent visitors to Raoul Island from their breeding sites on the Herald Islets.\nB: Red-crowned parakeets have successfully recolonised Raoul Island after eradication of invasive predators.\nC: Red-crowned parakeets have been successfully translocated to Raoul Island.\nD: Eradication of invasive predators on Raoul Island has allowed remnant populations of red-crowned parakeets to increase.\nE: Red-crowned parakeets are now common on Raoul Island after eradication of invasive species.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1073",
        "problem": "Batesian mimicry is a type of mimicry in which an edible animal is protected by its resemblance to one avoided by the predators. It typically includes three elements:\n\nI. Predator,\n\nII. Species which is dangerous or unpalatable to the predator- termed as \"model\" and\n\nIII. Palatable species termed as \"mimic\".\n\nA few statements about Batesian mimicry are given. Which of the following is correct?\nA: Batesian mimicry can be stably maintained only if the harm caused to the predator by eating a model overweighs the benefit of eating a mimic.\nB: Batesian mimicry can be stably maintained if the mimics become more abundant than the models.\nC: Young predator with first experience with mimic is likely to destabilize the evolution of Batesian mimicry in a population.\nD: In Batesian mimicry, there is always a directional selection that causes model species to evolve away from mimic.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBatesian mimicry is a type of mimicry in which an edible animal is protected by its resemblance to one avoided by the predators. It typically includes three elements:\n\nI. Predator,\n\nII. Species which is dangerous or unpalatable to the predator- termed as \"model\" and\n\nIII. Palatable species termed as \"mimic\".\n\nA few statements about Batesian mimicry are given. Which of the following is correct?\n\nA: Batesian mimicry can be stably maintained only if the harm caused to the predator by eating a model overweighs the benefit of eating a mimic.\nB: Batesian mimicry can be stably maintained if the mimics become more abundant than the models.\nC: Young predator with first experience with mimic is likely to destabilize the evolution of Batesian mimicry in a population.\nD: In Batesian mimicry, there is always a directional selection that causes model species to evolve away from mimic.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_402",
        "problem": "某男子患轻度的遗传性疾病, 其 1 条 14 号染色体和 1 条 21 号染色体连接成 1 条异\n\n常染色体，如图甲所示。减数分裂时, 异常染色体的联会如图乙所示, 配对的三条染色体中, 任意配对的两条染色体分离时, 另一条染色体随机移向细胞任意一极。下列叙述错误的是（ ）\n\n[图1]\nA: 该病的致病机理与原发性高血压的不同\nB: 染色体部分片段丢失会改变基因的数目\nC: 理论上，该男子能产生 6 种染色体组型的精子\nD: 该男子与正常女子婚配，子代染色体正常的概率为 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某男子患轻度的遗传性疾病, 其 1 条 14 号染色体和 1 条 21 号染色体连接成 1 条异\n\n常染色体，如图甲所示。减数分裂时, 异常染色体的联会如图乙所示, 配对的三条染色体中, 任意配对的两条染色体分离时, 另一条染色体随机移向细胞任意一极。下列叙述错误的是（ ）\n\n[图1]\n\nA: 该病的致病机理与原发性高血压的不同\nB: 染色体部分片段丢失会改变基因的数目\nC: 理论上，该男子能产生 6 种染色体组型的精子\nD: 该男子与正常女子婚配，子代染色体正常的概率为 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-45.jpg?height=329&width=674&top_left_y=1640&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_33",
        "problem": "The domain structure of protein Z, which is composed of 180 amino acids, is shown in the upper part of the figure below. Protein $\\mathrm{Z}$ is palmitoylated at a cysteine residue (the third amino acid) through the mechanism shown in the box.\n[figure1]\n\nWhich of the following diagrams shows the correct topology of protein $\\mathrm{Z}$ in the plasma membrane?\nA: [figure2]\nB: [figure3]\nC: [figure4]\nD: [figure5]\nE: [figure6]\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe domain structure of protein Z, which is composed of 180 amino acids, is shown in the upper part of the figure below. Protein $\\mathrm{Z}$ is palmitoylated at a cysteine residue (the third amino acid) through the mechanism shown in the box.\n[figure1]\n\nWhich of the following diagrams shows the correct topology of protein $\\mathrm{Z}$ in the plasma membrane?\n\nA: [figure2]\nB: [figure3]\nC: [figure4]\nD: [figure5]\nE: [figure6]\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-09.jpg?height=840&width=1104&top_left_y=593&top_left_x=538",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-09.jpg?height=212&width=454&top_left_y=1710&top_left_x=561",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-09.jpg?height=234&width=463&top_left_y=1993&top_left_x=548",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-09.jpg?height=234&width=494&top_left_y=2282&top_left_x=518",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-09.jpg?height=257&width=460&top_left_y=1688&top_left_x=1249",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-09.jpg?height=219&width=462&top_left_y=1975&top_left_x=1251"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_841",
        "problem": "普通小麦为异源六倍体 $(6 n=42$, 记为 $42 \\mathrm{~W})$, 染色体组成为 $\\mathrm{AABBDD}(\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{D}$ 分别代表不同物种的一个染色体组), 甲、乙、丙表示三个小麦的纯合“二体异附加系”(即普通小麦在有 $42 \\mathrm{~W}$ 的基础上, 还有来自偃麦草的染色体, 来自偃麦草的两条染色体称为二体)。如图所示, 减数分裂时, 染色体I和II不能联会, 但均可随机移向一极(各种配子的产生机会和可育性相等)。 $\\mathrm{L}$ 为抗叶锈病基因、 $\\mathrm{M}$ 为抗白粉病基因、 $\\mathrm{S}$ 为抗秆锈病基因, 均为显性。若甲和乙杂交的子代为丁, 甲和丙杂交的子代为戊, 乙和丙杂交的子代为己, 下列叙述错误的是（）\n\n[图1]\n\n甲 $(42 \\mathrm{~W}+$ I I $)$\n\n[图2]\n\n[图3]\nA: 甲、乙、丙品系的培育过程发生了染色体数目变异\nB: 甲和丙杂交所得 $F_{1}$ 在减数分裂时, 能形成 22 个四分体\nC: 丁自交，子代中只抗一种病性状的植株占 $3 / 8$\nD: 戊与已杂交, 子代同时有三种抗病性状的植株占 $3 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n普通小麦为异源六倍体 $(6 n=42$, 记为 $42 \\mathrm{~W})$, 染色体组成为 $\\mathrm{AABBDD}(\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{D}$ 分别代表不同物种的一个染色体组), 甲、乙、丙表示三个小麦的纯合“二体异附加系”(即普通小麦在有 $42 \\mathrm{~W}$ 的基础上, 还有来自偃麦草的染色体, 来自偃麦草的两条染色体称为二体)。如图所示, 减数分裂时, 染色体I和II不能联会, 但均可随机移向一极(各种配子的产生机会和可育性相等)。 $\\mathrm{L}$ 为抗叶锈病基因、 $\\mathrm{M}$ 为抗白粉病基因、 $\\mathrm{S}$ 为抗秆锈病基因, 均为显性。若甲和乙杂交的子代为丁, 甲和丙杂交的子代为戊, 乙和丙杂交的子代为己, 下列叙述错误的是（）\n\n[图1]\n\n甲 $(42 \\mathrm{~W}+$ I I $)$\n\n[图2]\n\n[图3]\n\nA: 甲、乙、丙品系的培育过程发生了染色体数目变异\nB: 甲和丙杂交所得 $F_{1}$ 在减数分裂时, 能形成 22 个四分体\nC: 丁自交，子代中只抗一种病性状的植株占 $3 / 8$\nD: 戊与已杂交, 子代同时有三种抗病性状的植株占 $3 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-066.jpg?height=237&width=240&top_left_y=755&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-066.jpg?height=268&width=231&top_left_y=754&top_left_x=770",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-066.jpg?height=258&width=242&top_left_y=756&top_left_x=1184"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_953",
        "problem": "An animal experiences an acid-base imbalance in the arterial blood that results in acidosis. To increase $\\mathrm{pH}$ toward normal, which direction would the ventilation rate be changed and what would be the corresponding change in arterial $\\mathbf{P}_{\\mathrm{CO}_{2}}$ ?\nA: Ventilation rate increases, arterial $\\mathrm{P}_{\\mathrm{CO} 2}$ increases\nB: Ventilation rate increases, arterial $\\mathrm{P}_{\\mathrm{CO} 2}$ decreases\nC: Ventilation rate decreases, arterial $\\mathrm{P}_{\\mathrm{CO} 2}$ increases\nD: Ventilation rate decreases, arterial $\\mathrm{P}_{\\mathrm{CO} 2}$ decreases\nE: None of the above\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn animal experiences an acid-base imbalance in the arterial blood that results in acidosis. To increase $\\mathrm{pH}$ toward normal, which direction would the ventilation rate be changed and what would be the corresponding change in arterial $\\mathbf{P}_{\\mathrm{CO}_{2}}$ ?\n\nA: Ventilation rate increases, arterial $\\mathrm{P}_{\\mathrm{CO} 2}$ increases\nB: Ventilation rate increases, arterial $\\mathrm{P}_{\\mathrm{CO} 2}$ decreases\nC: Ventilation rate decreases, arterial $\\mathrm{P}_{\\mathrm{CO} 2}$ increases\nD: Ventilation rate decreases, arterial $\\mathrm{P}_{\\mathrm{CO} 2}$ decreases\nE: None of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1285",
        "problem": "Dental arcades are a feature that is used to differentiate between Homo sapiens and their recent ancestors. Dental arcades of recently discovered Homo species have been compared with data on previously discovered species. The reconstructed upper arcade of KNM-ER 62000 (outlined in black; discovered 2012) is occluded with (a) the reconstructed lower arcades of KNM-ER 60000 (shown in grey; also discovered 2012) and (b) KNM-ER 1802 (shown in grey; discovered 1973). What is a logical conclusion that can be made from these data?\n[figure1]\n\nhttp://www.nature.com/nature/journal/v488/n7410/full/nature11322.html\nA: Early Homo species were scavengers.\nB: There was morphological variation in early Homo species.\nC: Modern Homo sapiens have taken large evolutionary leaps from early Homo species.\nD: Early Homo species had an 'overbite' in their jaws.\nE: Fossils that are discovered at the same time, are the same shape.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDental arcades are a feature that is used to differentiate between Homo sapiens and their recent ancestors. Dental arcades of recently discovered Homo species have been compared with data on previously discovered species. The reconstructed upper arcade of KNM-ER 62000 (outlined in black; discovered 2012) is occluded with (a) the reconstructed lower arcades of KNM-ER 60000 (shown in grey; also discovered 2012) and (b) KNM-ER 1802 (shown in grey; discovered 1973). What is a logical conclusion that can be made from these data?\n[figure1]\n\nhttp://www.nature.com/nature/journal/v488/n7410/full/nature11322.html\n\nA: Early Homo species were scavengers.\nB: There was morphological variation in early Homo species.\nC: Modern Homo sapiens have taken large evolutionary leaps from early Homo species.\nD: Early Homo species had an 'overbite' in their jaws.\nE: Fossils that are discovered at the same time, are the same shape.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-16.jpg?height=463&width=483&top_left_y=1479&top_left_x=975"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1553",
        "problem": "The NDMAR protein is a membrane channel found on neurones. It binds glutamate then opens to allow ions into the cell. Several drugs (NO, ethanol, ketamine) inhibit NDMAR and can cause powerful psychological dissociation. The effect of several inhibitors is shown.\n\n$$\n\\text { Ion entry }\n$$\n\n$$\n\\text { into neurone }\n$$\n[figure1]\n\nWhich inhibitor binds the same site as glutamate?\nA: A\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: All of them\nE: None of them\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe NDMAR protein is a membrane channel found on neurones. It binds glutamate then opens to allow ions into the cell. Several drugs (NO, ethanol, ketamine) inhibit NDMAR and can cause powerful psychological dissociation. The effect of several inhibitors is shown.\n\n$$\n\\text { Ion entry }\n$$\n\n$$\n\\text { into neurone }\n$$\n[figure1]\n\nWhich inhibitor binds the same site as glutamate?\n\nA: A\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: All of them\nE: None of them\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-10.jpg?height=960&width=1476&top_left_y=591&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1086",
        "problem": "The following pattern of venous blood drainage is typical of:\n\n[figure1]\nA: fish\nB: amphibian\nC: bird\nD: mammal\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following pattern of venous blood drainage is typical of:\n\n[figure1]\n\nA: fish\nB: amphibian\nC: bird\nD: mammal\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-06.jpg?height=994&width=1011&top_left_y=1346&top_left_x=600"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_991",
        "problem": "Some of the oxygen molecules given off in the process of photosynthesis are used by the plant:\nA: To form water by combining with hydrogen stored in the vacuoles\nB: In cellular respiration in the mitochondria\nC: To build enzymes in the Golgi complex\nD: In transcribing DNA in the lysosome\nE: In membrane-bound amyloplasts to aid in the storage of starch\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSome of the oxygen molecules given off in the process of photosynthesis are used by the plant:\n\nA: To form water by combining with hydrogen stored in the vacuoles\nB: In cellular respiration in the mitochondria\nC: To build enzymes in the Golgi complex\nD: In transcribing DNA in the lysosome\nE: In membrane-bound amyloplasts to aid in the storage of starch\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1507",
        "problem": "The main purpose of flower pigments is to create contrasting patterns which guide pollinators to pollen. However, scientists wanted to investigate whether pigments also protect pollen. Some flowers have anthers exposed in a dish of petals, whereas some flowers have anthers enveloped under petals.\n\n[figure1]\n\nThe change in pigmentation of these types of flowers was measured across the world over time. The change in temperature due to global warming, and the change in UV exposure due to depletion and recovery of the ozone layer varies between regions. The correlation between these variables is plotted below.\n\n[figure2]\n\nWhich of the following conclusions are NOT supported by these data?\nA: Flower pigments absorb UV light.\nB: Flower pigments absorb heat.\nC: Pollen in flowers with exposed anthers is at risk of overheating.\nD: Pollen in flowers with enveloped anthers is at risk of overheating.\nE: Pollen in flowers with exposed anthers is at risk of damage by UV radiation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe main purpose of flower pigments is to create contrasting patterns which guide pollinators to pollen. However, scientists wanted to investigate whether pigments also protect pollen. Some flowers have anthers exposed in a dish of petals, whereas some flowers have anthers enveloped under petals.\n\n[figure1]\n\nThe change in pigmentation of these types of flowers was measured across the world over time. The change in temperature due to global warming, and the change in UV exposure due to depletion and recovery of the ozone layer varies between regions. The correlation between these variables is plotted below.\n\n[figure2]\n\nWhich of the following conclusions are NOT supported by these data?\n\nA: Flower pigments absorb UV light.\nB: Flower pigments absorb heat.\nC: Pollen in flowers with exposed anthers is at risk of overheating.\nD: Pollen in flowers with enveloped anthers is at risk of overheating.\nE: Pollen in flowers with exposed anthers is at risk of damage by UV radiation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-34.jpg?height=616&width=1176&top_left_y=523&top_left_x=240",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-34.jpg?height=854&width=1625&top_left_y=1338&top_left_x=227"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1051",
        "problem": "A urine test was conducted for a person as a routine check-up programme. It showed the following results:\n\nI. Colour: light yellow\n\nII. pH: alkaline\n\nIII. Glucose: Nil\n\nIV. Amylase: 500 units\n\nWhich of the following is correct?\nA: The results do not indicate any abnormality and all four factors depend on diet consumed.\nB: The results are indicative of liver dysfunction as well as kidney homeostasis dysfunction.\nC: The results indicate that the person is on heavy carbohydrate diet.\nD: The presence of amylase is likely to indicate pancreatitis while alkaline $\\mathrm{pH}$ can correlate to diet consumed.\nE: The results are indicative that the person is dehydrated and starved which has led to defective glomerular filtration.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA urine test was conducted for a person as a routine check-up programme. It showed the following results:\n\nI. Colour: light yellow\n\nII. pH: alkaline\n\nIII. Glucose: Nil\n\nIV. Amylase: 500 units\n\nWhich of the following is correct?\n\nA: The results do not indicate any abnormality and all four factors depend on diet consumed.\nB: The results are indicative of liver dysfunction as well as kidney homeostasis dysfunction.\nC: The results indicate that the person is on heavy carbohydrate diet.\nD: The presence of amylase is likely to indicate pancreatitis while alkaline $\\mathrm{pH}$ can correlate to diet consumed.\nE: The results are indicative that the person is dehydrated and starved which has led to defective glomerular filtration.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_262",
        "problem": "Bats and birds are typical animal groups with flying form of locomotion. Compared with non-flying mammals of similar size and weight, birds and bats have a number of different physiological characteristics.\nA: Despite having higher daily energy needs, birds have statistically shorter intestines compared to those of the non-flying mammals.\nB: The ratio of lung weight per body weight of birds is greater than that of the nonflying mammals.\nC: The speed of water absorption and excretion via the digestive tracts of bats is faster than that of the non-flying mammals.\nD: Although bats and the non-flying mammals have similar blood volume and hemoglobin's affinity for oxygen, bats have higher hematocrit level.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBats and birds are typical animal groups with flying form of locomotion. Compared with non-flying mammals of similar size and weight, birds and bats have a number of different physiological characteristics.\n\nA: Despite having higher daily energy needs, birds have statistically shorter intestines compared to those of the non-flying mammals.\nB: The ratio of lung weight per body weight of birds is greater than that of the nonflying mammals.\nC: The speed of water absorption and excretion via the digestive tracts of bats is faster than that of the non-flying mammals.\nD: Although bats and the non-flying mammals have similar blood volume and hemoglobin's affinity for oxygen, bats have higher hematocrit level.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_750",
        "problem": "如图为一对近亲结婚的青年夫妇的遗传分析图, 其中白化病由基因 a 控制, 色盲由基因 b 控制（不考虑基因突变）。下列有关叙述错误的是（ ）\n\n[图1]\nA: 这对夫妇所生子女中, 患一种病的概率为 $7 / 16$\nB: D 细胞为次级精母细胞, 其中含有两个 $\\mathrm{B}$ 基因\nC: $\\mathrm{E}$ 细胞和 $\\mathrm{G}$ 细胞结合发育成的个体表现型为正常女性\nD: $\\mathrm{F}$ 细胞和 $\\mathrm{G}$ 细胞含有的相同致病基因可能来自于家族中的同一祖先\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图为一对近亲结婚的青年夫妇的遗传分析图, 其中白化病由基因 a 控制, 色盲由基因 b 控制（不考虑基因突变）。下列有关叙述错误的是（ ）\n\n[图1]\n\nA: 这对夫妇所生子女中, 患一种病的概率为 $7 / 16$\nB: D 细胞为次级精母细胞, 其中含有两个 $\\mathrm{B}$ 基因\nC: $\\mathrm{E}$ 细胞和 $\\mathrm{G}$ 细胞结合发育成的个体表现型为正常女性\nD: $\\mathrm{F}$ 细胞和 $\\mathrm{G}$ 细胞含有的相同致病基因可能来自于家族中的同一祖先\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-064.jpg?height=483&width=631&top_left_y=1049&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_883",
        "problem": "关于 DNA 分子的结构与复制的叙述中, 正确的是 ( )\nA: 1 个含有 $\\mathrm{m}$ 个腺嘌呤的 DNA 分子, 第 $\\mathrm{n}$ 次复制需要腺嘌呤脱氧核苷酸的数量为 $\\left(2^{\\mathrm{n}}-1\\right) \\times \\mathrm{m}$ 个\nB: DNA 双链被 ${ }^{32} \\mathrm{P}$ 标记后, 复制 $\\mathrm{n}$ 次, 子代 DNA 中含有 ${ }^{32} \\mathrm{P}$ 标记的 DNA 所占比例为 $(1 / 2)^{n}$\nC: 双链 DNA 分子中 $\\mathrm{G}+\\mathrm{C}$ 占碱基总数的 $\\mathrm{M} \\%$, 那么每条 $\\mathrm{DNA}$ 单链中 $\\mathrm{G}+\\mathrm{C}$ 都占该链碱基总数的 $\\mathrm{M} \\%$\nD: 细胞内 DNA 双链均被 ${ }^{32} \\mathrm{P}$ 标记后在不含 ${ }^{32} \\mathrm{P}$ 的环境中进行连续 2 次有丝分裂,每个子细胞的染色体均有一半有标记\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n关于 DNA 分子的结构与复制的叙述中, 正确的是 ( )\n\nA: 1 个含有 $\\mathrm{m}$ 个腺嘌呤的 DNA 分子, 第 $\\mathrm{n}$ 次复制需要腺嘌呤脱氧核苷酸的数量为 $\\left(2^{\\mathrm{n}}-1\\right) \\times \\mathrm{m}$ 个\nB: DNA 双链被 ${ }^{32} \\mathrm{P}$ 标记后, 复制 $\\mathrm{n}$ 次, 子代 DNA 中含有 ${ }^{32} \\mathrm{P}$ 标记的 DNA 所占比例为 $(1 / 2)^{n}$\nC: 双链 DNA 分子中 $\\mathrm{G}+\\mathrm{C}$ 占碱基总数的 $\\mathrm{M} \\%$, 那么每条 $\\mathrm{DNA}$ 单链中 $\\mathrm{G}+\\mathrm{C}$ 都占该链碱基总数的 $\\mathrm{M} \\%$\nD: 细胞内 DNA 双链均被 ${ }^{32} \\mathrm{P}$ 标记后在不含 ${ }^{32} \\mathrm{P}$ 的环境中进行连续 2 次有丝分裂,每个子细胞的染色体均有一半有标记\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_420",
        "problem": "水稻的育性由细胞核基因（Rf、rf）和细胞质基因（N、S）共同控制, 在细胞质基因为 $S$ 、细胞核基因为 $r f r f$ 时，水稻表现为雄性不育，基因型写作 $S$ （rfrf），其他情况下\n均表现为雄性可育。除雄性不育系 (简称不育系) 外还有雄性不育保持系 (简称保持系)和雄性不育恢复系 (简称恢复系)。保持系与不育系杂交后代为不育系; 恢复系与不育系杂交可使不育系的后代恢复育性。生产上常用不育系做母本和恢复系做父本杂交得到杂交种。下列叙述错误的是\nA: 因为杂交种的核基因型为杂合的 Rfrf,因而具有杂种优势\nB: 恢复系的基因型为 $\\mathrm{N}$ （RfRf）或 $S$ （RfRf）\nC: 杂交种自交，后代会出现性状分离，故需年年制种\nD: 以 $S$ (Rfrf) 为父本, $N$ (Rfrf) 为母本进行杂交, 后代均表现为雄性可育\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n水稻的育性由细胞核基因（Rf、rf）和细胞质基因（N、S）共同控制, 在细胞质基因为 $S$ 、细胞核基因为 $r f r f$ 时，水稻表现为雄性不育，基因型写作 $S$ （rfrf），其他情况下\n均表现为雄性可育。除雄性不育系 (简称不育系) 外还有雄性不育保持系 (简称保持系)和雄性不育恢复系 (简称恢复系)。保持系与不育系杂交后代为不育系; 恢复系与不育系杂交可使不育系的后代恢复育性。生产上常用不育系做母本和恢复系做父本杂交得到杂交种。下列叙述错误的是\n\nA: 因为杂交种的核基因型为杂合的 Rfrf,因而具有杂种优势\nB: 恢复系的基因型为 $\\mathrm{N}$ （RfRf）或 $S$ （RfRf）\nC: 杂交种自交，后代会出现性状分离，故需年年制种\nD: 以 $S$ (Rfrf) 为父本, $N$ (Rfrf) 为母本进行杂交, 后代均表现为雄性可育\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1212",
        "problem": "The SEM photomicrograph of a moth at right shows the compound eye in great detail. What is the best estimate of the maximum horizontal width of this eye, given that the scale bar represents $200 \\mu \\mathrm{m}$ ?\n[figure1]\nA: $550 \\mu \\mathrm{m}$\nB: $600 \\mu \\mathrm{m}$\nC: $650 \\mu \\mathrm{m}$\nD: $700 \\mu \\mathrm{m}$\nE: $750 \\mu \\mathrm{m}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe SEM photomicrograph of a moth at right shows the compound eye in great detail. What is the best estimate of the maximum horizontal width of this eye, given that the scale bar represents $200 \\mu \\mathrm{m}$ ?\n[figure1]\n\nA: $550 \\mu \\mathrm{m}$\nB: $600 \\mu \\mathrm{m}$\nC: $650 \\mu \\mathrm{m}$\nD: $700 \\mu \\mathrm{m}$\nE: $750 \\mu \\mathrm{m}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-15.jpg?height=796&width=1038&top_left_y=1126&top_left_x=932"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1019",
        "problem": "A plant cell with a solute potential of $-0.8 \\mathrm{MPa}$ maintains a constant volume when bathed in a solution that has a solute potential of $\\mathbf{- 0 . 2 5}$ $\\mathrm{MPa}$ and is in an open container. From this information, one knows that:\nA: The cell has a pressure potential of $+0.55 \\mathrm{MPa}$\nB: The cell has a pressure potential of $+0.25 \\mathrm{MPa}$\nC: The cell has a pressure potential of $+0.8 \\mathrm{MPa}$\nD: The cell has a water potential of $-0.8 \\mathrm{MPa}$\nE: None of the above answers is correct\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA plant cell with a solute potential of $-0.8 \\mathrm{MPa}$ maintains a constant volume when bathed in a solution that has a solute potential of $\\mathbf{- 0 . 2 5}$ $\\mathrm{MPa}$ and is in an open container. From this information, one knows that:\n\nA: The cell has a pressure potential of $+0.55 \\mathrm{MPa}$\nB: The cell has a pressure potential of $+0.25 \\mathrm{MPa}$\nC: The cell has a pressure potential of $+0.8 \\mathrm{MPa}$\nD: The cell has a water potential of $-0.8 \\mathrm{MPa}$\nE: None of the above answers is correct\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_733",
        "problem": "5-溴尿嘧啶 (BrdU) 与胸腺嘧啶脱氧核苷类似, 能够与腺嘌呤 (A) 配对, 掺入到新合成的 DNA 子链中。DNA 的一条链含有 BrdU 的染色单体为 $\\alpha$ 型, DNA 的两条链均含 $\\mathrm{BrdU}$ 的染色单体为 $\\beta$ 型。将洋葱根尖分生组织的某一细胞放在含 $\\mathrm{BrdU}$ 的培养液中进行培养时, 所有子细胞的分裂都同步进行。关于子细胞 $(2 \\mathrm{n}=16)$ 处于第二、三个细胞周期时的姐妹染色单体情况的叙述错误的是（）\nA: 第二次分裂中期每个细胞含 $\\alpha$ 型染色单体 16 条\nB: 第二次分裂中期所有细胞中的 $\\alpha$ 型染色单体共 32 条\nC: 第三次分裂中期每个细胞中的 $\\alpha$ 型染色单体为 $0 \\sim 16$ 条\nD: 第三次分裂中期所有细胞中的 $\\alpha$ 型染色单体共 48 条\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n5-溴尿嘧啶 (BrdU) 与胸腺嘧啶脱氧核苷类似, 能够与腺嘌呤 (A) 配对, 掺入到新合成的 DNA 子链中。DNA 的一条链含有 BrdU 的染色单体为 $\\alpha$ 型, DNA 的两条链均含 $\\mathrm{BrdU}$ 的染色单体为 $\\beta$ 型。将洋葱根尖分生组织的某一细胞放在含 $\\mathrm{BrdU}$ 的培养液中进行培养时, 所有子细胞的分裂都同步进行。关于子细胞 $(2 \\mathrm{n}=16)$ 处于第二、三个细胞周期时的姐妹染色单体情况的叙述错误的是（）\n\nA: 第二次分裂中期每个细胞含 $\\alpha$ 型染色单体 16 条\nB: 第二次分裂中期所有细胞中的 $\\alpha$ 型染色单体共 32 条\nC: 第三次分裂中期每个细胞中的 $\\alpha$ 型染色单体为 $0 \\sim 16$ 条\nD: 第三次分裂中期所有细胞中的 $\\alpha$ 型染色单体共 48 条\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1058",
        "problem": "Which one of the following characteristics found in birds contributes LEAST towards the adaptation for flight?\nA: Keratinized beak\nB: Hind limbs\nC: Endothermy\nD: High metabolic rate\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following characteristics found in birds contributes LEAST towards the adaptation for flight?\n\nA: Keratinized beak\nB: Hind limbs\nC: Endothermy\nD: High metabolic rate\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1410",
        "problem": "Review the following map and answer the 5 subsequent questions.\n\n[figure1]\n\nEcologists have been studying the small archipelago shown in the map above. They have been comparing the ecology of the mainland with those on the nearby islands and have come up with a few discoveries. The area has not been known to have been inhabited by humans, so it remains very natural. The mainland contains a low mountain range approximately $50 \\mathrm{~km}$ inland from the coast (highest peak approximately $400 \\mathrm{~m}$ above sea level) and the prevailing winds are westerlies.\n\nThe ecologists monitor the number of geckos on Island 3 and find that their birth rate equals 0.2 per annum. When the ecologist counted them this year there were 110 geckos on the island.\n\nHow many will there be at the same time next year?\nA: 55\nB: 132\nC: 22\nD: 165\nE: 172\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nReview the following map and answer the 5 subsequent questions.\n\n[figure1]\n\nEcologists have been studying the small archipelago shown in the map above. They have been comparing the ecology of the mainland with those on the nearby islands and have come up with a few discoveries. The area has not been known to have been inhabited by humans, so it remains very natural. The mainland contains a low mountain range approximately $50 \\mathrm{~km}$ inland from the coast (highest peak approximately $400 \\mathrm{~m}$ above sea level) and the prevailing winds are westerlies.\n\nThe ecologists monitor the number of geckos on Island 3 and find that their birth rate equals 0.2 per annum. When the ecologist counted them this year there were 110 geckos on the island.\n\nHow many will there be at the same time next year?\n\nA: 55\nB: 132\nC: 22\nD: 165\nE: 172\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-39.jpg?height=925&width=1479&top_left_y=451&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1446",
        "problem": "There are three types of muscle contraction, see diagram below. All these muscle contractions occur against a load. Concentric contraction occurs when a muscle shortens against a load. Eccentric contraction occurs when muscle lengthens against a load. Finally, isometric contraction occurs when muscle does not shorten or lengthen against a load.\n\n[figure1]\n\nCONCENTRIC (shortening)\n\n[figure2]\n\nECCENTRIC (lengthening)\n[figure3]\n\nISOMETRIC (no movement)\n\nWithin a muscle, there are two types of receptors: muscle spindles and golgi tendon organs, see diagram below. Muscle spindles are found within the belly of a muscle. These are activated when the muscle stretches. Golgi tendon organs are found in the tendon of a muscle. A tendon attaches muscle to bone. Golgi tendon organs are activated when the muscle contracts and shortens. This information is sent to the central nervous system via sensory neurons.\n\n[figure4]\n\nWhat happens to sensory neuron activity during concentric contraction?\nA: Muscle spindle nerve activity increase, Golgi tendon organ nerve activity decrease\nB: Muscle spindle nerve activity decrease, Golgi tendon organ nerve activity decrease\nC: Muscle spindle nerve activity increase, Golgi tendon organ nerve activity increase\nD: Muscle spindle nerve activity decrease, Golgi tendon organ nerve activity increase\nE: Muscle spindle nerve activity stays the same, Golgi tendon organ nerve activity stays the same\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThere are three types of muscle contraction, see diagram below. All these muscle contractions occur against a load. Concentric contraction occurs when a muscle shortens against a load. Eccentric contraction occurs when muscle lengthens against a load. Finally, isometric contraction occurs when muscle does not shorten or lengthen against a load.\n\n[figure1]\n\nCONCENTRIC (shortening)\n\n[figure2]\n\nECCENTRIC (lengthening)\n[figure3]\n\nISOMETRIC (no movement)\n\nWithin a muscle, there are two types of receptors: muscle spindles and golgi tendon organs, see diagram below. Muscle spindles are found within the belly of a muscle. These are activated when the muscle stretches. Golgi tendon organs are found in the tendon of a muscle. A tendon attaches muscle to bone. Golgi tendon organs are activated when the muscle contracts and shortens. This information is sent to the central nervous system via sensory neurons.\n\n[figure4]\n\nWhat happens to sensory neuron activity during concentric contraction?\n\nA: Muscle spindle nerve activity increase, Golgi tendon organ nerve activity decrease\nB: Muscle spindle nerve activity decrease, Golgi tendon organ nerve activity decrease\nC: Muscle spindle nerve activity increase, Golgi tendon organ nerve activity increase\nD: Muscle spindle nerve activity decrease, Golgi tendon organ nerve activity increase\nE: Muscle spindle nerve activity stays the same, Golgi tendon organ nerve activity stays the same\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-21.jpg?height=300&width=177&top_left_y=455&top_left_x=288",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-21.jpg?height=326&width=394&top_left_y=454&top_left_x=820",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-21.jpg?height=292&width=410&top_left_y=476&top_left_x=1369",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-21.jpg?height=985&width=1353&top_left_y=1501&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_482",
        "problem": "下图分别表示某动物 (2n) 精巢中正在分裂的甲细胞和乙细胞, 用红色荧光和绿色苂光分别标记其中两条染色体的着丝粒，在苂光显微镜下观察着丝粒附时间的变化，发现其依次出现在细胞(1)～(3)（或(1)～(4)）的不同位置处。下列叙述错误的是（）\n\n[图1]\n\n赤道板位置\n\n[图2]\n\n赤道板位置\nA: 甲细胞和乙细胞中均标记的是一对同源染色体的着丝粒\nB: 甲细胞的着丝粒到达(3)位置时, 细胞内的染色单体数为 0\nC: 乙细胞的着丝粒从(2)到(3)位置的过程中可能仍保留着四分体\nD: 乙细胞的着丝粒到达(4)位置时, 每条染色体上 DNA 含量为 2\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图分别表示某动物 (2n) 精巢中正在分裂的甲细胞和乙细胞, 用红色荧光和绿色苂光分别标记其中两条染色体的着丝粒，在苂光显微镜下观察着丝粒附时间的变化，发现其依次出现在细胞(1)～(3)（或(1)～(4)）的不同位置处。下列叙述错误的是（）\n\n[图1]\n\n赤道板位置\n\n[图2]\n\n赤道板位置\n\nA: 甲细胞和乙细胞中均标记的是一对同源染色体的着丝粒\nB: 甲细胞的着丝粒到达(3)位置时, 细胞内的染色单体数为 0\nC: 乙细胞的着丝粒从(2)到(3)位置的过程中可能仍保留着四分体\nD: 乙细胞的着丝粒到达(4)位置时, 每条染色体上 DNA 含量为 2\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-14.jpg?height=408&width=759&top_left_y=1121&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-14.jpg?height=419&width=631&top_left_y=1104&top_left_x=1084"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_636",
        "problem": "斑马鱼（2n=50）作为水生模式生物, 其幼鱼皮肤中的一群表面上皮细胞（SEC）会在生物体发育过程中出现一种非典型的细胞分裂机制--无合成分裂（不涉及基因组的复制), SEC 进行两轮分裂, 产生总体积与初级母细胞的体积一致的四个子细胞。生物\n体生长产生的组织张力通过积累可以促进无合成分裂, Piezo1 作为张力传感器, 是机械敏感性离子通道蛋白。下列说法正确的是（）\n\n[图1]\nA: SEC 进行无合成分裂过程中会出现姐妹染色单体\nB: 无合成分裂过程会出现同源染色体联会现象\nC: SEC 进行两轮分裂产生的 4 个子细胞染色体数均为 25 条\nD: 该分裂能迅速增加上皮表面积，从而保障对生长中幼鱼的体表覆盖\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n斑马鱼（2n=50）作为水生模式生物, 其幼鱼皮肤中的一群表面上皮细胞（SEC）会在生物体发育过程中出现一种非典型的细胞分裂机制--无合成分裂（不涉及基因组的复制), SEC 进行两轮分裂, 产生总体积与初级母细胞的体积一致的四个子细胞。生物\n体生长产生的组织张力通过积累可以促进无合成分裂, Piezo1 作为张力传感器, 是机械敏感性离子通道蛋白。下列说法正确的是（）\n\n[图1]\n\nA: SEC 进行无合成分裂过程中会出现姐妹染色单体\nB: 无合成分裂过程会出现同源染色体联会现象\nC: SEC 进行两轮分裂产生的 4 个子细胞染色体数均为 25 条\nD: 该分裂能迅速增加上皮表面积，从而保障对生长中幼鱼的体表覆盖\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-010.jpg?height=392&width=788&top_left_y=338&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1537",
        "problem": "The 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nFor which of these will genetic drift dominate over natural selection?\nA: Population split into isolated groups\nB: Large population\nC: High current allele frequency\nD: Large difference in relative fitness\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nFor which of these will genetic drift dominate over natural selection?\n\nA: Population split into isolated groups\nB: Large population\nC: High current allele frequency\nD: Large difference in relative fitness\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=970&width=704&top_left_y=778&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=966&width=706&top_left_y=778&top_left_x=1118"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_338",
        "problem": "苦瓜植株中含一对等位基因 $\\mathrm{D}$ 和 $\\mathrm{d}$ ，其中 $\\mathrm{D}$ 基因纯合的植株不能产生卵细胞,而 $\\mathrm{d}$ 基因纯合的植株花粉不能正常发育，杂合子植株完全正常。现有基因型为 Dd 的苦瓜植株若干做亲本，下列有关叙述错误的是（ ）\nA: 如果每代均自交直至 $F_{2}$, 则 $F_{2}$ 植株中基因型为 $\\mathrm{dd}$ 植株所占比例为 $1 / 4$\nB: 如果每代均自交直至 $F_{2}$, 则 $F_{2}$ 植株中正常植株所占比例为 $1 / 2$\nC: 如果每代均自由交配直至 $F_{2}$ ，则 $F_{2}$ 植株中 D 基因的频率为 $1 / 2$\nD: 如果每代均自由交配直至 $F_{2}$, 则 $F_{2}$ 植株中正常植株所占比例为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n苦瓜植株中含一对等位基因 $\\mathrm{D}$ 和 $\\mathrm{d}$ ，其中 $\\mathrm{D}$ 基因纯合的植株不能产生卵细胞,而 $\\mathrm{d}$ 基因纯合的植株花粉不能正常发育，杂合子植株完全正常。现有基因型为 Dd 的苦瓜植株若干做亲本，下列有关叙述错误的是（ ）\n\nA: 如果每代均自交直至 $F_{2}$, 则 $F_{2}$ 植株中基因型为 $\\mathrm{dd}$ 植株所占比例为 $1 / 4$\nB: 如果每代均自交直至 $F_{2}$, 则 $F_{2}$ 植株中正常植株所占比例为 $1 / 2$\nC: 如果每代均自由交配直至 $F_{2}$ ，则 $F_{2}$ 植株中 D 基因的频率为 $1 / 2$\nD: 如果每代均自由交配直至 $F_{2}$, 则 $F_{2}$ 植株中正常植株所占比例为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_385",
        "problem": "从雄果蝇 $(2 \\mathrm{~N}=8)$ 性腺获取甲、乙、丙、丁、戊五个细胞 (不考虑染色体畸变),记录细胞中核 DNA/染色体比值和染色体组数的关系如图所示, 下列说法错误的是 ( )\n\n[图1]\nA: 甲细胞和乙细胞中一定无同源染色体\nB: 丙细胞中一定有 4 个四分体\nC: 丁细胞可能为次级精母细胞\nD: 戊细胞一定处于有丝分裂过程中\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n从雄果蝇 $(2 \\mathrm{~N}=8)$ 性腺获取甲、乙、丙、丁、戊五个细胞 (不考虑染色体畸变),记录细胞中核 DNA/染色体比值和染色体组数的关系如图所示, 下列说法错误的是 ( )\n\n[图1]\n\nA: 甲细胞和乙细胞中一定无同源染色体\nB: 丙细胞中一定有 4 个四分体\nC: 丁细胞可能为次级精母细胞\nD: 戊细胞一定处于有丝分裂过程中\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-68.jpg?height=572&width=919&top_left_y=1630&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1250",
        "problem": "Conservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.Considering all the scientific evidence you have available about the Maui's dolphin, what would you recommend that the government does to protect this critically endangered dolphin?\nA: The extension to the set netting restricted areas together with the new codes of conduct for acoustic surveying and boat racing is sufficient.\nB: Ban set netting and mining and oil activities within the West Coast North Island Marine Mammal Sanctuary.\nC: Ban set netting throughout the range of Maui's dolphins including south to Whanganui, out to the $100 \\mathrm{~m}$ depth contour and within all harbours.\nD: Prevent all mining and oil activities within the West Coast North Island Marine Mammal Sanctuary.\nE: Ban trawl fishing, set netting, and mining and oil activities throughout the range of Maui's dolphins including south to Whanganui, out to the $100 \\mathrm{~m}$ depth contour and within all harbours.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nConservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.\n\nproblem:\nConsidering all the scientific evidence you have available about the Maui's dolphin, what would you recommend that the government does to protect this critically endangered dolphin?\n\nA: The extension to the set netting restricted areas together with the new codes of conduct for acoustic surveying and boat racing is sufficient.\nB: Ban set netting and mining and oil activities within the West Coast North Island Marine Mammal Sanctuary.\nC: Ban set netting throughout the range of Maui's dolphins including south to Whanganui, out to the $100 \\mathrm{~m}$ depth contour and within all harbours.\nD: Prevent all mining and oil activities within the West Coast North Island Marine Mammal Sanctuary.\nE: Ban trawl fishing, set netting, and mining and oil activities throughout the range of Maui's dolphins including south to Whanganui, out to the $100 \\mathrm{~m}$ depth contour and within all harbours.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1099",
        "problem": "' $A$ ' and ' $B$ ' are leaves belong to the same tree. The morphology of the two leaves vary in shape, size, thickness and in the number of lobes.\n[figure1]\n\nWhich of the following is correct?\nA: Leaf ' $A$ ' is from the top of canopy and leaf ' $B$ ' is from the bottom.\nB: Leaves in the upper tree canopy are thicker as they receive higher intensity of light.\nC: Leaf ' $B$ ' can lose heat by convection more easily than leaf ' $A$ ' as it is more lobed and small.\nD: Leaves lower in the canopy will have more chlorophyll a: b ratio as compared to leaves from the top of the tree.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\n' $A$ ' and ' $B$ ' are leaves belong to the same tree. The morphology of the two leaves vary in shape, size, thickness and in the number of lobes.\n[figure1]\n\nWhich of the following is correct?\n\nA: Leaf ' $A$ ' is from the top of canopy and leaf ' $B$ ' is from the bottom.\nB: Leaves in the upper tree canopy are thicker as they receive higher intensity of light.\nC: Leaf ' $B$ ' can lose heat by convection more easily than leaf ' $A$ ' as it is more lobed and small.\nD: Leaves lower in the canopy will have more chlorophyll a: b ratio as compared to leaves from the top of the tree.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-19.jpg?height=320&width=724&top_left_y=1694&top_left_x=778"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1290",
        "problem": "A new diamond graph method has been developed that can represent 3 variables equally on a 2-D graph. The diamond graph is essentially a 3-D bar graph viewed from above rather than from the side. Instead of using rising parallel bars, the diamond graph uses expanding polygons within a diamond-shaped grid to represent values. This diamond graph depicts the age-adjusted incidence rates of end-stage renal disease due to any cause per 100,000 person-years according to six categories of both systolic and diastolic blood pressure (Klag, Whelton, Randall, et al. 1996).\n\n\n[figure1]\n(Li, Buechner, Tarwater, and Muñoz, 2003)\n\nWhat is the incidence per 100,000 person-years of end-stage renal disease expected in individuals with a blood pressure of 170/100 $(\\mathrm{mm} \\mathrm{Hg})$\nA: 143.5\nB: 124.4\nC: 62.3\nD: 49.7\nE: None of the above.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA new diamond graph method has been developed that can represent 3 variables equally on a 2-D graph. The diamond graph is essentially a 3-D bar graph viewed from above rather than from the side. Instead of using rising parallel bars, the diamond graph uses expanding polygons within a diamond-shaped grid to represent values. This diamond graph depicts the age-adjusted incidence rates of end-stage renal disease due to any cause per 100,000 person-years according to six categories of both systolic and diastolic blood pressure (Klag, Whelton, Randall, et al. 1996).\n\n\n[figure1]\n(Li, Buechner, Tarwater, and Muñoz, 2003)\n\nWhat is the incidence per 100,000 person-years of end-stage renal disease expected in individuals with a blood pressure of 170/100 $(\\mathrm{mm} \\mathrm{Hg})$\n\nA: 143.5\nB: 124.4\nC: 62.3\nD: 49.7\nE: None of the above.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-15.jpg?height=877&width=1202&top_left_y=675&top_left_x=695"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1291",
        "problem": "In the fungus Sordaria fimicola, a diploid nucleus containing alleles for both black and for white spore colours can give rise to an ascus containing one of several spore patterns. The spores are arranged linearly in the ascus according to the order in which they are formed. Two possible asci, and their method of formation, are shown in the following diagram.\n\n[figure1]\n\nThe arrangement of spores in the two asci is explained by the alleles for spore colour undergoing\nA: crossing over\nB: independent assortment\nC: segregation during meiosis I\nD: segregation during meiosis II\nE: mutation\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the fungus Sordaria fimicola, a diploid nucleus containing alleles for both black and for white spore colours can give rise to an ascus containing one of several spore patterns. The spores are arranged linearly in the ascus according to the order in which they are formed. Two possible asci, and their method of formation, are shown in the following diagram.\n\n[figure1]\n\nThe arrangement of spores in the two asci is explained by the alleles for spore colour undergoing\n\nA: crossing over\nB: independent assortment\nC: segregation during meiosis I\nD: segregation during meiosis II\nE: mutation\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-15.jpg?height=399&width=623&top_left_y=386&top_left_x=748"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1387",
        "problem": "Rhizophores are leafless branches that arise from the stem and grow downward, producing roots at its tip when it reaches the soil. Rhizophores are commonly known as aerial roots. Scientists measured length and height of rhizophores of mangrove plant (Rhizophora mangle, Fig A). They also made cross sections of rhizophores and observed their anatomical characteristics. The rhizophore cross sections were classified in orders according to the number of arches away from the main stem. The results are shown in Fig B and Fig $\\mathrm{C}$.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\n[figure3]\n\nFig: Rhizophores of Rhizophora mangle plants.\n\nA: Rhizophore height and length measurement.\n\nB: Change in height (empty square) and in length/height proportion (full circle) in five sequential orders of rhizophores.\n\nC: Relative proportions of bark (including aerenchyma), xylem and pith along the length of individual first-order rhizophores (left), and at the base of rhizophores of sequential orders (right).\n\nHow does water support a mangrove plant?\nA: By transpiration, the process whereby the water is lost at the leaves, enabling constant cycling of water throughout the plant.\nB: By making the cells flaccid, enabling the mangrove plant to strengthen its structural integrity.\nC: By filling the stem with water, assisting the stem to resist gravitational as well as osmotic forces.\nD: By making the cells turgid, promoting rigidity of plant structures.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRhizophores are leafless branches that arise from the stem and grow downward, producing roots at its tip when it reaches the soil. Rhizophores are commonly known as aerial roots. Scientists measured length and height of rhizophores of mangrove plant (Rhizophora mangle, Fig A). They also made cross sections of rhizophores and observed their anatomical characteristics. The rhizophore cross sections were classified in orders according to the number of arches away from the main stem. The results are shown in Fig B and Fig $\\mathrm{C}$.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\n[figure3]\n\nFig: Rhizophores of Rhizophora mangle plants.\n\nA: Rhizophore height and length measurement.\n\nB: Change in height (empty square) and in length/height proportion (full circle) in five sequential orders of rhizophores.\n\nC: Relative proportions of bark (including aerenchyma), xylem and pith along the length of individual first-order rhizophores (left), and at the base of rhizophores of sequential orders (right).\n\nHow does water support a mangrove plant?\n\nA: By transpiration, the process whereby the water is lost at the leaves, enabling constant cycling of water throughout the plant.\nB: By making the cells flaccid, enabling the mangrove plant to strengthen its structural integrity.\nC: By filling the stem with water, assisting the stem to resist gravitational as well as osmotic forces.\nD: By making the cells turgid, promoting rigidity of plant structures.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-09.jpg?height=454&width=508&top_left_y=681&top_left_x=266",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-09.jpg?height=448&width=377&top_left_y=690&top_left_x=814",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-09.jpg?height=496&width=552&top_left_y=640&top_left_x=1226"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_719",
        "problem": "下列实验及结果中，能作为直接证据说明“核糖核酸是遗传物质”的是（）\nA: 红花植株与白花植株杂交, $F_{1}$ 为红花, $F_{2}$ 中红花：白花 $=3: 1$\nB: 病毒甲的 RNA 与病毒乙的蛋白质混合后感染烟草只能得到病毒甲\nC: 加热杀死的 $\\mathrm{S}$ 型肺炎双球菌与 $\\mathrm{R}$ 型活菌混合培养后可分离出 $\\mathrm{S}$ 型活菌\nD: 用放射性同位素标记 $\\mathrm{T}_{2}$ 噬菌体外壳蛋白，在子代噬菌体中检测不到放射性\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列实验及结果中，能作为直接证据说明“核糖核酸是遗传物质”的是（）\n\nA: 红花植株与白花植株杂交, $F_{1}$ 为红花, $F_{2}$ 中红花：白花 $=3: 1$\nB: 病毒甲的 RNA 与病毒乙的蛋白质混合后感染烟草只能得到病毒甲\nC: 加热杀死的 $\\mathrm{S}$ 型肺炎双球菌与 $\\mathrm{R}$ 型活菌混合培养后可分离出 $\\mathrm{S}$ 型活菌\nD: 用放射性同位素标记 $\\mathrm{T}_{2}$ 噬菌体外壳蛋白，在子代噬菌体中检测不到放射性\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1457",
        "problem": "Polypeptide chains are formed via condensation polymerisation, where multiple amino acids are joined via condensation reactions to form a polymer. The Figure below demonstrates how a peptide bond (shaded in red) is formed via a condensation reaction that removes one water molecule and joins two amino acids together.\n\n[figure1]\n\nGlycine is an amino acid with a molecular formula of $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{NO}_{2}$.\n\nWhat is the molecular formula of a polypeptide chain composed of one hundred (100) glycine molecules?\nA: $\\mathrm{C}_{200} \\mathrm{H}_{500} \\mathrm{~N}_{100} \\mathrm{O}_{200}$\nB: $\\mathrm{C}_{200} \\mathrm{H}_{300} \\mathrm{~N}_{100} \\mathrm{O}_{100}$\nC: $\\mathrm{C}_{200} \\mathrm{H}_{298} \\mathrm{~N}_{100} \\mathrm{O}_{99}$\nD: $\\mathrm{C}_{200} \\mathrm{H}_{302} \\mathrm{~N}_{100} \\mathrm{O}_{101}$\nE: None of the above\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPolypeptide chains are formed via condensation polymerisation, where multiple amino acids are joined via condensation reactions to form a polymer. The Figure below demonstrates how a peptide bond (shaded in red) is formed via a condensation reaction that removes one water molecule and joins two amino acids together.\n\n[figure1]\n\nGlycine is an amino acid with a molecular formula of $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{NO}_{2}$.\n\nWhat is the molecular formula of a polypeptide chain composed of one hundred (100) glycine molecules?\n\nA: $\\mathrm{C}_{200} \\mathrm{H}_{500} \\mathrm{~N}_{100} \\mathrm{O}_{200}$\nB: $\\mathrm{C}_{200} \\mathrm{H}_{300} \\mathrm{~N}_{100} \\mathrm{O}_{100}$\nC: $\\mathrm{C}_{200} \\mathrm{H}_{298} \\mathrm{~N}_{100} \\mathrm{O}_{99}$\nD: $\\mathrm{C}_{200} \\mathrm{H}_{302} \\mathrm{~N}_{100} \\mathrm{O}_{101}$\nE: None of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-16.jpg?height=469&width=625&top_left_y=448&top_left_x=704"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_274",
        "problem": "Three-spined stickleback Gasterosteus aculeatus are widely distributed in both marine and freshwater areas across the world. Adaptive radiation has led to morphological differences between marine and freshwater populations. Of such differences, all of the marine population have a pair of pelvic spines that evolved from the pelvic skeleton, while some freshwater populations of various localities have lost their spines (Figure 1). Genetic analyses revealed that the causative genomic region for this pelvic difference is located around the Pitxl gene. This Pitxl plays an important role in the development of the ventral spine, thymus, and neuromast. Although the amino acid sequences of Pitxl transcripts are identical in both populations, the expression patterns of Pitxl in the pelvic fin buds of embryos are different: Pitxl is expressed in the marine population (purple), while it is not in the freshwater population (Figure 1 insets).\n[figure1]\n\nFigure 1. Ventral view of marine (left) and freshwater (right) sticklebacks, showing the presence/absence of pelvic spines (shown by dark blue). Anterior is to up. (Boxes) Magnified ventral view of stickleback embryos showing Pitxl expressions in the pelvic fin buds.\nA: The freshwater population without pelvic spines independently have likely evolved from the marine population with pelvic spines.\nB: Pelvic spines can function to protect the marine population against predators\nC: The Pitxl-knockout individual of the marine population are likely to show similar phenotypes to those of the freshwater population.\nD: The presence/absence of Pitxl expression in the pelvic fin buds of embryos may result from the difference of enhancer sequences that control the gene expression.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThree-spined stickleback Gasterosteus aculeatus are widely distributed in both marine and freshwater areas across the world. Adaptive radiation has led to morphological differences between marine and freshwater populations. Of such differences, all of the marine population have a pair of pelvic spines that evolved from the pelvic skeleton, while some freshwater populations of various localities have lost their spines (Figure 1). Genetic analyses revealed that the causative genomic region for this pelvic difference is located around the Pitxl gene. This Pitxl plays an important role in the development of the ventral spine, thymus, and neuromast. Although the amino acid sequences of Pitxl transcripts are identical in both populations, the expression patterns of Pitxl in the pelvic fin buds of embryos are different: Pitxl is expressed in the marine population (purple), while it is not in the freshwater population (Figure 1 insets).\n[figure1]\n\nFigure 1. Ventral view of marine (left) and freshwater (right) sticklebacks, showing the presence/absence of pelvic spines (shown by dark blue). Anterior is to up. (Boxes) Magnified ventral view of stickleback embryos showing Pitxl expressions in the pelvic fin buds.\n\nA: The freshwater population without pelvic spines independently have likely evolved from the marine population with pelvic spines.\nB: Pelvic spines can function to protect the marine population against predators\nC: The Pitxl-knockout individual of the marine population are likely to show similar phenotypes to those of the freshwater population.\nD: The presence/absence of Pitxl expression in the pelvic fin buds of embryos may result from the difference of enhancer sequences that control the gene expression.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-59.jpg?height=590&width=1122&top_left_y=1170&top_left_x=450"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_925",
        "problem": "某哺乳动物雄性个体的基因型为 $\\mathrm{AaBb}$, 关于该哺乳动物减数分裂产生精细胞的叙述, 正确的是 ( )\nA: 若该哺乳动物的一个精原细胞经减数分裂产生了 $\\mathrm{AB}$ 和 $\\mathrm{ab}$ 两种精细胞, 则说明 $A 、 B$ 基因位于同一条染色体上， $a 、 b$ 基因位于同一条染色体上\nB: 若该哺乳动物的一个精原细胞在减数分裂前的间期一个 $\\mathrm{A}$ 基因突变成了 $\\mathrm{a}$ 基因,则该精原细胞将产生 4 种精细胞\nC: 若该哺乳动物的一个精原细胞经减数分裂产生的精细胞种类及比例为 $\\mathrm{AB}: \\mathrm{Ab}: \\mathrm{aB}$ : $a b=1: 1: 1: 1$, 则说明该两对基因位于一对同源染色体上\nD: 若该哺乳动物产生的精细胞种类及比例为 $A B: A b: a B: a b=4: 1: 1: 4$, 则说明 $A$ 与 $b$ 、 $\\mathrm{a}$ 与 $\\mathrm{B}$ 基因各位于一条染色体上，且部分初级精母细胞发生了姐妹染色单体的互换\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某哺乳动物雄性个体的基因型为 $\\mathrm{AaBb}$, 关于该哺乳动物减数分裂产生精细胞的叙述, 正确的是 ( )\n\nA: 若该哺乳动物的一个精原细胞经减数分裂产生了 $\\mathrm{AB}$ 和 $\\mathrm{ab}$ 两种精细胞, 则说明 $A 、 B$ 基因位于同一条染色体上， $a 、 b$ 基因位于同一条染色体上\nB: 若该哺乳动物的一个精原细胞在减数分裂前的间期一个 $\\mathrm{A}$ 基因突变成了 $\\mathrm{a}$ 基因,则该精原细胞将产生 4 种精细胞\nC: 若该哺乳动物的一个精原细胞经减数分裂产生的精细胞种类及比例为 $\\mathrm{AB}: \\mathrm{Ab}: \\mathrm{aB}$ : $a b=1: 1: 1: 1$, 则说明该两对基因位于一对同源染色体上\nD: 若该哺乳动物产生的精细胞种类及比例为 $A B: A b: a B: a b=4: 1: 1: 4$, 则说明 $A$ 与 $b$ 、 $\\mathrm{a}$ 与 $\\mathrm{B}$ 基因各位于一条染色体上，且部分初级精母细胞发生了姐妹染色单体的互换\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1279",
        "problem": "## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAAWe can investigate the composition of a bolus in detail and then compare how similar different bolus' are using a Percent Similarity Index (PSI). Total PSI values of 100 indicate $100 \\%$ overlap; values of $>80 \\%$ can be considered similar.\n\nBolus 3 and 4 have been analysed for you and the results recorded in the table. You are to calculate the PSI by choosing the smallest value of the percent numerical abundance (\\% NA) for each item and entering this value in the PSI column at right. The first PSI value for squid beaks has been done for you. The PSI values are then summed to obtain overall PSI (dark grey box at bottom right). Record your value for the overall PSI (rounded to the nearest whole number) on your answer sheet.\n\n|  | Bolc |  | Bol |  | PSI |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | Count | $\\% N A$ | Count | $\\% N A$ |  |\n| Squid beaks | 47 | 37.0 | 63 | 43.4 | 37.0 |\n| Fishing line | 4 | 3.1 | 3 | 2.1 |  |\n| Whole Plastic Items <br> eg bottle caps | 2 | 1.6 | 5 | 3.4 |  |\n| Extra large plastic fragments <br> $\\qquad(\\geq 50 \\mathrm{~mm})$ | 0 | 0 | 0 | 0 |  |\n| Large plastic fragments <br> $(\\geq .20 \\mathrm{~mm}$ and $<50 \\mathrm{~mm})$ | 4 | 3.1 | 6 | 4.1 |  |\n| Medium plastic fragments <br> $\\qquad(<20 \\mathrm{~mm}$ and $\\geq 10 \\mathrm{~mm})$ | 5 | 3.9 | 8 | 5.5 |  |\n| Small plastic fragments <br> $(<10 \\mathrm{~mm})$ | 62 | 48.8 | 55 | 37.9 |  |\n| Other | 4 | 3.1 | 4 | 2.8 |  |\n| TOTAL | 127 |  | 145 |  |  |\n\nNote: $\\%$ numerical abundance (NA) for each item in each bolus is calculated as follows: $\\%$ NA $=$ count/total $* 100$",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAA\n\nproblem:\nWe can investigate the composition of a bolus in detail and then compare how similar different bolus' are using a Percent Similarity Index (PSI). Total PSI values of 100 indicate $100 \\%$ overlap; values of $>80 \\%$ can be considered similar.\n\nBolus 3 and 4 have been analysed for you and the results recorded in the table. You are to calculate the PSI by choosing the smallest value of the percent numerical abundance (\\% NA) for each item and entering this value in the PSI column at right. The first PSI value for squid beaks has been done for you. The PSI values are then summed to obtain overall PSI (dark grey box at bottom right). Record your value for the overall PSI (rounded to the nearest whole number) on your answer sheet.\n\n|  | Bolc |  | Bol |  | PSI |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | Count | $\\% N A$ | Count | $\\% N A$ |  |\n| Squid beaks | 47 | 37.0 | 63 | 43.4 | 37.0 |\n| Fishing line | 4 | 3.1 | 3 | 2.1 |  |\n| Whole Plastic Items <br> eg bottle caps | 2 | 1.6 | 5 | 3.4 |  |\n| Extra large plastic fragments <br> $\\qquad(\\geq 50 \\mathrm{~mm})$ | 0 | 0 | 0 | 0 |  |\n| Large plastic fragments <br> $(\\geq .20 \\mathrm{~mm}$ and $<50 \\mathrm{~mm})$ | 4 | 3.1 | 6 | 4.1 |  |\n| Medium plastic fragments <br> $\\qquad(<20 \\mathrm{~mm}$ and $\\geq 10 \\mathrm{~mm})$ | 5 | 3.9 | 8 | 5.5 |  |\n| Small plastic fragments <br> $(<10 \\mathrm{~mm})$ | 62 | 48.8 | 55 | 37.9 |  |\n| Other | 4 | 3.1 | 4 | 2.8 |  |\n| TOTAL | 127 |  | 145 |  |  |\n\nNote: $\\%$ numerical abundance (NA) for each item in each bolus is calculated as follows: $\\%$ NA $=$ count/total $* 100$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=800&width=617&top_left_y=1299&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=817&width=1097&top_left_y=1299&top_left_x=845"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_825",
        "problem": "人类 B 珠蛋白基因位于常染色体上。其基因显性突变和隐性突变均引起地中海贫血。一对地中海贫血夫妻生了一个正常孩子。据此判断, 这对有下列哪些夫妻可能? ( )\nA: 都是杂合子，但基因型不同\nB: 各自携带不同的致病基因\nC: 各自含有不同的正常基因\nD: 基因型相同，且携带相同的致病基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n人类 B 珠蛋白基因位于常染色体上。其基因显性突变和隐性突变均引起地中海贫血。一对地中海贫血夫妻生了一个正常孩子。据此判断, 这对有下列哪些夫妻可能? ( )\n\nA: 都是杂合子，但基因型不同\nB: 各自携带不同的致病基因\nC: 各自含有不同的正常基因\nD: 基因型相同，且携带相同的致病基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_468",
        "problem": "双脱氧核苷三磷酸 (如图甲) 在人工合成 DNA 体系中, 可脱去两个磷酸基团形成焦磷酸和双脱氧核苷酸并释放能量, 双脱氧核苷酸可使 DNA 子链延伸终止。在人工合成 DNA 体系中, 有适量某单链模板、某一种双脱氧核件三磷酸 (ddNTP) 和四种正常脱氧核苷三磷酸 (dNTP), 反应终止时对合成的不同长度子链进行电泳 (结果如图乙)。下列说法正确的是（ ）\n\n[图1]\nA: ddNTP 和 dNTP 的区别是 ddNTP 的五碳糖上无氧原子\nB: 双脱氧核苷酸无法与模板链发生碱基互补配对, 导致子链延伸终止\nC: 人工合成 DNA 体系中必须要加入 ATP 以供能\nD: 据图乙推测, 模板链的碱基序列为 3'-ATGATGCGAT-5'\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n双脱氧核苷三磷酸 (如图甲) 在人工合成 DNA 体系中, 可脱去两个磷酸基团形成焦磷酸和双脱氧核苷酸并释放能量, 双脱氧核苷酸可使 DNA 子链延伸终止。在人工合成 DNA 体系中, 有适量某单链模板、某一种双脱氧核件三磷酸 (ddNTP) 和四种正常脱氧核苷三磷酸 (dNTP), 反应终止时对合成的不同长度子链进行电泳 (结果如图乙)。下列说法正确的是（ ）\n\n[图1]\n\nA: ddNTP 和 dNTP 的区别是 ddNTP 的五碳糖上无氧原子\nB: 双脱氧核苷酸无法与模板链发生碱基互补配对, 导致子链延伸终止\nC: 人工合成 DNA 体系中必须要加入 ATP 以供能\nD: 据图乙推测, 模板链的碱基序列为 3'-ATGATGCGAT-5'\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-44.jpg?height=505&width=1453&top_left_y=1392&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1268",
        "problem": "The graph refers to the surface layers of the Tasman Sea. It shows the seasonal changes in numbers of phytoplankton and zooplankton, and in inorganic nutrient concentration.\n\n[figure1]\n\nWhich one of the following sequences correctly represents the passage of nutrients through the ecosystem?\nA: $1 \\rightarrow 2 \\rightarrow 3$\nB: $1 \\rightarrow 3 \\rightarrow 2$\nC: $2 \\rightarrow 1 \\rightarrow 3$\nD: $2 \\rightarrow 3 \\rightarrow 1$\nE: $3 \\rightarrow 2 \\rightarrow 1$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph refers to the surface layers of the Tasman Sea. It shows the seasonal changes in numbers of phytoplankton and zooplankton, and in inorganic nutrient concentration.\n\n[figure1]\n\nWhich one of the following sequences correctly represents the passage of nutrients through the ecosystem?\n\nA: $1 \\rightarrow 2 \\rightarrow 3$\nB: $1 \\rightarrow 3 \\rightarrow 2$\nC: $2 \\rightarrow 1 \\rightarrow 3$\nD: $2 \\rightarrow 3 \\rightarrow 1$\nE: $3 \\rightarrow 2 \\rightarrow 1$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-05.jpg?height=510&width=1159&top_left_y=1363&top_left_x=475"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_165",
        "problem": "Figure Q.76A shows neuron $\\mathrm{X}$ receives signals directly from three separate nerve terminals $b, c$ and $d$. Neuron $Y$ receives signals from nerve terminal $a$.\n\nFigure Q.76B shows the various postsynaptic potentials recorded in neuron $\\mathrm{X}$ after receiving input signals directly from terminals $b, c$, and $d$.\n\nFigure Q.76C shows the action potential recorded in neuron $\\mathrm{Y}$ after receiving input signals from the presynaptic terminal $a$.\n\n[figure1]\n\nFigure Q.76A\n\n[figure2]\n\nFigure Q.76B.\n\n[figure3]\n\nFigure Q.76C.\nA: Action potentials could be generated in neuron $\\mathrm{X}$ if nerve terminal $c$ is stimulated rapidly.\nB: When three nerve terminals $a, c$ and $d$ are stimulated simultaneously, the postsynaptic potentials recorded in neuron $\\mathrm{X}$ are smaller than those when the nerve terminals $c$ and $d$ are stimulated simultaneously.\nC: Nerve terminal $a$ releases inhibitory neurotransmitter and nerve terminal $b$ releases excitatory neurotransmitter.\nD: In the mammalian body, there are many neurons $\\mathrm{X}, \\mathrm{Y}, \\mathrm{Z}$ and interneurons. Neurons Z, Y, X are sensory neurons, Renshaw cells (inhibitory neurons) and motor neurons, respectively. If a substance (e.g., Strychnine) injected in the body blocks glycine receptors, diaphragm contracts fully and remains contracted.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFigure Q.76A shows neuron $\\mathrm{X}$ receives signals directly from three separate nerve terminals $b, c$ and $d$. Neuron $Y$ receives signals from nerve terminal $a$.\n\nFigure Q.76B shows the various postsynaptic potentials recorded in neuron $\\mathrm{X}$ after receiving input signals directly from terminals $b, c$, and $d$.\n\nFigure Q.76C shows the action potential recorded in neuron $\\mathrm{Y}$ after receiving input signals from the presynaptic terminal $a$.\n\n[figure1]\n\nFigure Q.76A\n\n[figure2]\n\nFigure Q.76B.\n\n[figure3]\n\nFigure Q.76C.\n\nA: Action potentials could be generated in neuron $\\mathrm{X}$ if nerve terminal $c$ is stimulated rapidly.\nB: When three nerve terminals $a, c$ and $d$ are stimulated simultaneously, the postsynaptic potentials recorded in neuron $\\mathrm{X}$ are smaller than those when the nerve terminals $c$ and $d$ are stimulated simultaneously.\nC: Nerve terminal $a$ releases inhibitory neurotransmitter and nerve terminal $b$ releases excitatory neurotransmitter.\nD: In the mammalian body, there are many neurons $\\mathrm{X}, \\mathrm{Y}, \\mathrm{Z}$ and interneurons. Neurons Z, Y, X are sensory neurons, Renshaw cells (inhibitory neurons) and motor neurons, respectively. If a substance (e.g., Strychnine) injected in the body blocks glycine receptors, diaphragm contracts fully and remains contracted.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-061.jpg?height=845&width=1511&top_left_y=881&top_left_x=332",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-061.jpg?height=655&width=803&top_left_y=1915&top_left_x=293",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-061.jpg?height=564&width=714&top_left_y=1909&top_left_x=1180"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_218",
        "problem": "Drugs that render the inner mitochondrial membrane permeable to $\\mathrm{H}^{+}$are called \"uncouplers\nA: These drugs will increase oxygen consumption.\nB: These drugs will reduce body carbohydrate catabolism.\nC: These drugs will decrease body temperature.\nD: These drugs may cause death upon overdose because of severe weight loss.\nE: These drugs may cause death upon overdose because of severe ATP loss.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nDrugs that render the inner mitochondrial membrane permeable to $\\mathrm{H}^{+}$are called \"uncouplers\n\nA: These drugs will increase oxygen consumption.\nB: These drugs will reduce body carbohydrate catabolism.\nC: These drugs will decrease body temperature.\nD: These drugs may cause death upon overdose because of severe weight loss.\nE: These drugs may cause death upon overdose because of severe ATP loss.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_193",
        "problem": "The following figure shows a hypothetical evolutionary tree of species $a \\sim e$ along with the variability between pairs of these species.\n\n[figure1]\n\nChoose a statement that is correct.\nA: The speciation rate shows a linear relationship to evolutionary time.\nB: Species variation shows a linear relationship to evolutionary time.\nC: The species pair $a$ - $b$ and the pair $c$ - $d$ shows sister group relationship.\nD: The tree contains three monophyletic groups.\nE: Species $a$ can be used as an outgroup for the other four species.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following figure shows a hypothetical evolutionary tree of species $a \\sim e$ along with the variability between pairs of these species.\n\n[figure1]\n\nChoose a statement that is correct.\n\nA: The speciation rate shows a linear relationship to evolutionary time.\nB: Species variation shows a linear relationship to evolutionary time.\nC: The species pair $a$ - $b$ and the pair $c$ - $d$ shows sister group relationship.\nD: The tree contains three monophyletic groups.\nE: Species $a$ can be used as an outgroup for the other four species.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-60.jpg?height=502&width=528&top_left_y=480&top_left_x=821"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_754",
        "problem": "潜伏感染是指 HIV 病毒 DNA 整合到宿主细胞基因组中, 但不产生游离的病毒颗粒, 整合了 HIV 基因组的宿主细胞就成为了“病毒储存库”。HIV 病毒储存库通常存在于记忆 $\\mathrm{CD} 4+\\mathrm{T}$ 细胞中, 也可以存在于单核巨噬细胞和星形胶质细胞中。根除艾滋病毒感染的关键在于抑制或清除“病毒储存库”。下列关于艾滋病预防与治疗的叙述错误的是（）\nA: 抑制 HIV 病毒与细胞表面的 CD4 受体结合, 可有效防止 HIV 的感染\nB: 逆转录酶抑制剂和整合酶抑制剂能有效抑制“病毒储存库”的形成\nC: 调节特定基因表达的转录因子, 即可使 HIV 在储存细胞中保持沉默\nD: 通过推广行为干预, 避免高危行为, 最好的治疗永远是预防\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n潜伏感染是指 HIV 病毒 DNA 整合到宿主细胞基因组中, 但不产生游离的病毒颗粒, 整合了 HIV 基因组的宿主细胞就成为了“病毒储存库”。HIV 病毒储存库通常存在于记忆 $\\mathrm{CD} 4+\\mathrm{T}$ 细胞中, 也可以存在于单核巨噬细胞和星形胶质细胞中。根除艾滋病毒感染的关键在于抑制或清除“病毒储存库”。下列关于艾滋病预防与治疗的叙述错误的是（）\n\nA: 抑制 HIV 病毒与细胞表面的 CD4 受体结合, 可有效防止 HIV 的感染\nB: 逆转录酶抑制剂和整合酶抑制剂能有效抑制“病毒储存库”的形成\nC: 调节特定基因表达的转录因子, 即可使 HIV 在储存细胞中保持沉默\nD: 通过推广行为干预, 避免高危行为, 最好的治疗永远是预防\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_697",
        "problem": "果蝇的红眼 $\\left(X^{R}\\right)$ 对白眼 $\\left(X^{r}\\right)$ 为显性, 让红眼雄果蝇和白眼雌果蝇杂交, $F_{1}$ 中偶尔会出现极少数的不符合交叉遗传特点的例外子代, 结果如下表所示, 不考虑基因突变。下列说法正确的是（ ）\nP $\\quad$ 白眼 $+\\times$ 红眼 $ઊ$\n\n| $\\mathrm{F}_{1}$ | 正常 | 红眼O、白眼へ |\n| :---: | :---: | :---: |\n|  | 例外 | 红眼不育の (XO)、白眼可育早 (XXY) |\nA: $\\mathrm{F}_{1}$ 红眼雌果蝇既有纯合子也有杂合子\nB: Y 染色体的有无决定雌、雄果蝇的育性\nC: $F_{1}$ 例外出现的原因可能是母本减数分裂 I 或 II 异常\nD: $F_{1}$ 例外雌蝇与父本回交后代不会出现例外\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的红眼 $\\left(X^{R}\\right)$ 对白眼 $\\left(X^{r}\\right)$ 为显性, 让红眼雄果蝇和白眼雌果蝇杂交, $F_{1}$ 中偶尔会出现极少数的不符合交叉遗传特点的例外子代, 结果如下表所示, 不考虑基因突变。下列说法正确的是（ ）\nP $\\quad$ 白眼 $+\\times$ 红眼 $ઊ$\n\n| $\\mathrm{F}_{1}$ | 正常 | 红眼O、白眼へ |\n| :---: | :---: | :---: |\n|  | 例外 | 红眼不育の (XO)、白眼可育早 (XXY) |\n\nA: $\\mathrm{F}_{1}$ 红眼雌果蝇既有纯合子也有杂合子\nB: Y 染色体的有无决定雌、雄果蝇的育性\nC: $F_{1}$ 例外出现的原因可能是母本减数分裂 I 或 II 异常\nD: $F_{1}$ 例外雌蝇与父本回交后代不会出现例外\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_272",
        "problem": "Phylogenetic relationship among several species of the genus Nemesia (Scrophulariaceae), accompanied with geological time scale, is shown below. Predicted states for annual/perennial lifehistory have been mapped to the tree. Annual life-history is shown in black and the perennials are shown in blue (from Datson et al., 2008).\n[figure1]\nA: The phylogenetic tree suggests that Nemesia originated during the Miocene, but the majority of extant Nemesia species radiated during the Pliocene (Pl.) and\nB: The phylogenetic tree suggests that the most recent common ancestor of Nemesia had an annual life form.\nC: The phylogenetic tree does not show shift from annual to perennial habit.\nD: Diascia rigescens is the ancestral species of all other taxa in this phylogenetic tree.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPhylogenetic relationship among several species of the genus Nemesia (Scrophulariaceae), accompanied with geological time scale, is shown below. Predicted states for annual/perennial lifehistory have been mapped to the tree. Annual life-history is shown in black and the perennials are shown in blue (from Datson et al., 2008).\n[figure1]\n\nA: The phylogenetic tree suggests that Nemesia originated during the Miocene, but the majority of extant Nemesia species radiated during the Pliocene (Pl.) and\nB: The phylogenetic tree suggests that the most recent common ancestor of Nemesia had an annual life form.\nC: The phylogenetic tree does not show shift from annual to perennial habit.\nD: Diascia rigescens is the ancestral species of all other taxa in this phylogenetic tree.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-37.jpg?height=1392&width=1378&top_left_y=584&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1202",
        "problem": "The diagram shows the sequence of bases in a strand of DNA that codes for a polypeptide composed of ten amino acids. An additional base sequence coding for the start signal is on the left.\n\nT A C G G T C A A T C T G G T T C T G G T T C T T C T C A G C A A\n\nWhen the polypeptide for which this gene codes was hydrolysed, it yielded the amino acids shown in the table.\n\n| Amino <br> acid | Number of amino acid <br> residues per polypeptide |\n| :---: | :---: |\n| $w$ | 1 |\n| $x$ | 2 |\n| $y$ | 3 |\n| $z$ | 4 |\n\nThe correct sequence of amino acids in the polypeptide is\nA: $x y z x z y z z w y$\nB: $y z x y z z y z w x$\nC: $z x y z y z y y w z$\nD: $y x z y z y z z x w$\nE: $y x z y z y z z w x$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram shows the sequence of bases in a strand of DNA that codes for a polypeptide composed of ten amino acids. An additional base sequence coding for the start signal is on the left.\n\nT A C G G T C A A T C T G G T T C T G G T T C T T C T C A G C A A\n\nWhen the polypeptide for which this gene codes was hydrolysed, it yielded the amino acids shown in the table.\n\n| Amino <br> acid | Number of amino acid <br> residues per polypeptide |\n| :---: | :---: |\n| $w$ | 1 |\n| $x$ | 2 |\n| $y$ | 3 |\n| $z$ | 4 |\n\nThe correct sequence of amino acids in the polypeptide is\n\nA: $x y z x z y z z w y$\nB: $y z x y z z y z w x$\nC: $z x y z y z y y w z$\nD: $y x z y z y z z x w$\nE: $y x z y z y z z w x$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1308",
        "problem": "The graph shows how the live mass of a sheep changes as it develops.\n\n[figure1]\n\nWhich one of the following graphs $(A-E)$ correctly shows the rate of change of live mass of the sheep over the same period?\nA: ![]([figure2])\nB: ![]([figure3])\nC: ![]([figure4])\nD: ![]([figure5])\nE: ![]([figure6])\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows how the live mass of a sheep changes as it develops.\n\n[figure1]\n\nWhich one of the following graphs $(A-E)$ correctly shows the rate of change of live mass of the sheep over the same period?\n\nA: ![]([figure2])\nB: ![]([figure3])\nC: ![]([figure4])\nD: ![]([figure5])\nE: ![]([figure6])\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-11.jpg?height=429&width=574&top_left_y=385&top_left_x=747",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-11.jpg?height=269&width=446&top_left_y=979&top_left_x=128",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-11.jpg?height=269&width=462&top_left_y=982&top_left_x=634",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-11.jpg?height=265&width=459&top_left_y=981&top_left_x=1164",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-11.jpg?height=257&width=463&top_left_y=1322&top_left_x=405",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-11.jpg?height=257&width=460&top_left_y=1322&top_left_x=912"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_312",
        "problem": "孟买血型是由两对等位基因 $\\mathrm{I} / \\mathrm{i}$ (位于第 9 号染色体)和 $\\mathrm{H} / \\mathrm{h}$ (位于第 19 号染色体)相互作用产生的, 使 $\\mathrm{ABO}$ 血型的表型比例发生改变, 其机理如下图所示。下列有关叙述错\n误的是\n\n[图1]\nA: 两对基因的遗传遵循基因的自由组合定律\nB: H 基因表达的产物是 A、B 血型表现的基础\nC: 父母均为 $\\mathrm{AB}$ 型血时, 可能生出 $\\mathrm{O}$ 型血的后代\nD: $\\mathrm{O}$ 型血对应的基因型可能有 5 种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n孟买血型是由两对等位基因 $\\mathrm{I} / \\mathrm{i}$ (位于第 9 号染色体)和 $\\mathrm{H} / \\mathrm{h}$ (位于第 19 号染色体)相互作用产生的, 使 $\\mathrm{ABO}$ 血型的表型比例发生改变, 其机理如下图所示。下列有关叙述错\n误的是\n\n[图1]\n\nA: 两对基因的遗传遵循基因的自由组合定律\nB: H 基因表达的产物是 A、B 血型表现的基础\nC: 父母均为 $\\mathrm{AB}$ 型血时, 可能生出 $\\mathrm{O}$ 型血的后代\nD: $\\mathrm{O}$ 型血对应的基因型可能有 5 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-056.jpg?height=326&width=899&top_left_y=234&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_995",
        "problem": "The coefficient of relatedness between an uncle and his nephew is:\nA: 0.125\nB: 0.25\nC: 0.5\nD: 1.0\nE: cannot be determined\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe coefficient of relatedness between an uncle and his nephew is:\n\nA: 0.125\nB: 0.25\nC: 0.5\nD: 1.0\nE: cannot be determined\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_374",
        "problem": "果蝇灰身（B）和黑身（b）由一对常染色体上的等位基因控制。假定某果蝇种群有 10000 只果蝇, 其中黑身果蝇个体数量长期维持在 9\\%, 若再向该种群中引入 10000 只黑身果蝇, 在不考虑其他因素影响的前提下, 关于黑身果蝇引入后种群的叙述, 错误的是 ( )\nA: B 基因频率由 $70 \\%$ 降为 35\\%\nB: b 基因频率由 $30 \\%$ 升为 $75 \\%$\nC: 杂合子果蝇比例降低了 $50 \\%$\nD: 黑身果蝇比例增加了 $45.5 \\%$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇灰身（B）和黑身（b）由一对常染色体上的等位基因控制。假定某果蝇种群有 10000 只果蝇, 其中黑身果蝇个体数量长期维持在 9\\%, 若再向该种群中引入 10000 只黑身果蝇, 在不考虑其他因素影响的前提下, 关于黑身果蝇引入后种群的叙述, 错误的是 ( )\n\nA: B 基因频率由 $70 \\%$ 降为 35\\%\nB: b 基因频率由 $30 \\%$ 升为 $75 \\%$\nC: 杂合子果蝇比例降低了 $50 \\%$\nD: 黑身果蝇比例增加了 $45.5 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_252",
        "problem": "The lack of advanced molecular biology methods, such as DNA sequencing or site-directed mutagenesis, did not stop the pioneers of evolution from asking difficult questions concerning the fundamental aspects of biological systems. J. B. S. Haldane attempt to calculate the mutation rate in 1930s is an illuminating example.\n\nWith no direct genetic evidence, he focused on men living in London and hemophilia A, an X-linked recessive disorder. Assuming that men with this disorder do not reproduce, his calculations show that the mutation rate is three times the frequency of men with this disorder in London. His estimate of mutation per generation per locus is not that different from the more recent estimates for many genes.\nA: The purported relationship between mutation rate and frequency of that mutation is only limited to populations that have reached mutation selection equilibrium.\nB: For a recessive autosomal disorder that results in sterility, one would expect mutation rate to be equal to six times the frequency of that disorder in a population.\nC: The model assumes there is no drift.\nD: Linkage between a beneficial allele and hemophilia would have inflated Haldane's estimate of mutation rate for hemophilia.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe lack of advanced molecular biology methods, such as DNA sequencing or site-directed mutagenesis, did not stop the pioneers of evolution from asking difficult questions concerning the fundamental aspects of biological systems. J. B. S. Haldane attempt to calculate the mutation rate in 1930s is an illuminating example.\n\nWith no direct genetic evidence, he focused on men living in London and hemophilia A, an X-linked recessive disorder. Assuming that men with this disorder do not reproduce, his calculations show that the mutation rate is three times the frequency of men with this disorder in London. His estimate of mutation per generation per locus is not that different from the more recent estimates for many genes.\n\nA: The purported relationship between mutation rate and frequency of that mutation is only limited to populations that have reached mutation selection equilibrium.\nB: For a recessive autosomal disorder that results in sterility, one would expect mutation rate to be equal to six times the frequency of that disorder in a population.\nC: The model assumes there is no drift.\nD: Linkage between a beneficial allele and hemophilia would have inflated Haldane's estimate of mutation rate for hemophilia.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_596",
        "problem": "某育种专家在农田中发现一株大穗不抗病的小麦, 自花授粉后获得 160 粒种子, 这些种子发育成的小麦中有 30 株大穗抗病和若干株小穗抗病, 其余的都不抗病。若将这 30 试卷第 49 页，共 100 页\n株大穗抗病的小麦作为亲本自交, 在其 $F_{1}$ 中选择大穗抗病的再进行自交, 理论上 $F_{2}$ 中能稳定遗传的大穗抗病小麦占 $\\mathrm{F}_{2}$ 中所有大穗抗病小麦的\nA: $2 / 10$\nB: $7 / 10$\nC: $2 / 9$\nD: $7 / 9$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某育种专家在农田中发现一株大穗不抗病的小麦, 自花授粉后获得 160 粒种子, 这些种子发育成的小麦中有 30 株大穗抗病和若干株小穗抗病, 其余的都不抗病。若将这 30 试卷第 49 页，共 100 页\n株大穗抗病的小麦作为亲本自交, 在其 $F_{1}$ 中选择大穗抗病的再进行自交, 理论上 $F_{2}$ 中能稳定遗传的大穗抗病小麦占 $\\mathrm{F}_{2}$ 中所有大穗抗病小麦的\n\nA: $2 / 10$\nB: $7 / 10$\nC: $2 / 9$\nD: $7 / 9$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1218",
        "problem": "There were approximately 1733 tonnes of oil onboard the Rena when she ran aground. Approximately 350 tonnes were not recovered from the wreck and most of this washed ashore on Matakana Island and the coastline from Mt Maunganui to Maketu between 9 and 11 October 2011. Surveys examining the effects of the oil and debris washed onto the surf beaches focused on the northern tuatua (Paphies subtriangulata), as this is one of the most common species found burrowing in the sand on the open coast surf beaches that were most heavily fouled by oil from the Rena, and they are an important kai moana species.\n\nThe figure at right shows the level of total PAH in the tissue of tuatua from Papamoa and Omanu beaches from October 52011 to 30 June 2012. Before impact total PAH levels (background) were about $0.7 \\mu \\mathrm{g} / \\mathrm{kg}$ at Papamoa beach and $0.2 \\mu \\mathrm{g} / \\mathrm{kg}$ at Ōmanu Beach. These values were produced on a wet weight basis.\n\n[figure1]\n\nHeavily oiled\n\nModerately oiled\n\nLightly oiled\n\n[figure2]\n\nThe graph at left shows the total PAH\nlevels in tuatua in winter 2012 from beaches from Waihi - East Cape. They are colour-coded to represent the degree of oiling. No winter PAH data was available for Maketū Spit - Mid (shore level). Results obtained from Waihi and Ōhope beaches are considered background levels and an average between these levels (20.6 $\\mu \\mathrm{g} / \\mathrm{kg}$ ) is plotted as a dashed line. These values were produced on a dry weight basis.Considering the levels of total PAH over time, which statement is NOT correct.\nA: Total PAH levels in the tissue of tuatua from both Papamoa and Ōmanu beaches peaked on the $18^{\\text {th }}$ October.\nB: Total PAH levels in the tissue of tuatua returned to about pre-impact levels by 30 June 2012.\nC: Total PAH levels in the tissue of tuatua declined rapidly following the initial impact.\nD: Total PAH levels in the tissue of tuatua from Ōmanu beach showed a minor elevation in early December 2011.\nE: Minor elevations in the total PAH levels in the tissue of tuatua from Papamoa and Ōmanu beaches occurred on different dates.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nThere were approximately 1733 tonnes of oil onboard the Rena when she ran aground. Approximately 350 tonnes were not recovered from the wreck and most of this washed ashore on Matakana Island and the coastline from Mt Maunganui to Maketu between 9 and 11 October 2011. Surveys examining the effects of the oil and debris washed onto the surf beaches focused on the northern tuatua (Paphies subtriangulata), as this is one of the most common species found burrowing in the sand on the open coast surf beaches that were most heavily fouled by oil from the Rena, and they are an important kai moana species.\n\nThe figure at right shows the level of total PAH in the tissue of tuatua from Papamoa and Omanu beaches from October 52011 to 30 June 2012. Before impact total PAH levels (background) were about $0.7 \\mu \\mathrm{g} / \\mathrm{kg}$ at Papamoa beach and $0.2 \\mu \\mathrm{g} / \\mathrm{kg}$ at Ōmanu Beach. These values were produced on a wet weight basis.\n\n[figure1]\n\nHeavily oiled\n\nModerately oiled\n\nLightly oiled\n\n[figure2]\n\nThe graph at left shows the total PAH\nlevels in tuatua in winter 2012 from beaches from Waihi - East Cape. They are colour-coded to represent the degree of oiling. No winter PAH data was available for Maketū Spit - Mid (shore level). Results obtained from Waihi and Ōhope beaches are considered background levels and an average between these levels (20.6 $\\mu \\mathrm{g} / \\mathrm{kg}$ ) is plotted as a dashed line. These values were produced on a dry weight basis.\n\nproblem:\nConsidering the levels of total PAH over time, which statement is NOT correct.\n\nA: Total PAH levels in the tissue of tuatua from both Papamoa and Ōmanu beaches peaked on the $18^{\\text {th }}$ October.\nB: Total PAH levels in the tissue of tuatua returned to about pre-impact levels by 30 June 2012.\nC: Total PAH levels in the tissue of tuatua declined rapidly following the initial impact.\nD: Total PAH levels in the tissue of tuatua from Ōmanu beach showed a minor elevation in early December 2011.\nE: Minor elevations in the total PAH levels in the tissue of tuatua from Papamoa and Ōmanu beaches occurred on different dates.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-26.jpg?height=528&width=1105&top_left_y=196&top_left_x=772",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-26.jpg?height=708&width=1125&top_left_y=914&top_left_x=154"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_466",
        "problem": "唐氏综合征是一种严重的先天性智力发育不全疾病, 原因有 21 -三体型（有 3 条 21 号染色体)，和易位型 21 三体（如图 1) 等。易位型三体中的异常染色体 (14/21 易位染色体）是由 14 号染色体和 21 号染色体发生罗伯逊易位, 二者的长臂在着丝点处融合形成。该染色体携带者具有正常的表型，但在产生生殖细胞的过程中，染色体会在细胞中形成三联体, 如图 2。在进行减数分裂时, 三联体中任意两条染色体移向细胞一极,另一条染色体移向另外一极, 且各种情况概率相同。已知易位型 14 三体和单体类型在肧胎早期便不能存活，下列分析错误的是（）\n\n[图1]\n\n图1 易位型 21 三体\n\n[图2]\n\n图2 $14 / 21$ 易位染色体携带者染色体变化\nA: 易位型 21 三体导致的唐氏综合征患者体细胞中染色体数量正常\nB: 观察 14/21 易位染色体的携带者 MI 前期细胞, 可以发现 21 个四分体和 1 个三联体\nC: 14/21 易位染色体的携带者能产生 6 种不同类型的配子，但只有一种配子是完全正常的\nD: $14 / 21$ 易位染色体的携带者与正常人婚配, 后代中表型正常的概率为 $1 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n唐氏综合征是一种严重的先天性智力发育不全疾病, 原因有 21 -三体型（有 3 条 21 号染色体)，和易位型 21 三体（如图 1) 等。易位型三体中的异常染色体 (14/21 易位染色体）是由 14 号染色体和 21 号染色体发生罗伯逊易位, 二者的长臂在着丝点处融合形成。该染色体携带者具有正常的表型，但在产生生殖细胞的过程中，染色体会在细胞中形成三联体, 如图 2。在进行减数分裂时, 三联体中任意两条染色体移向细胞一极,另一条染色体移向另外一极, 且各种情况概率相同。已知易位型 14 三体和单体类型在肧胎早期便不能存活，下列分析错误的是（）\n\n[图1]\n\n图1 易位型 21 三体\n\n[图2]\n\n图2 $14 / 21$ 易位染色体携带者染色体变化\n\nA: 易位型 21 三体导致的唐氏综合征患者体细胞中染色体数量正常\nB: 观察 14/21 易位染色体的携带者 MI 前期细胞, 可以发现 21 个四分体和 1 个三联体\nC: 14/21 易位染色体的携带者能产生 6 种不同类型的配子，但只有一种配子是完全正常的\nD: $14 / 21$ 易位染色体的携带者与正常人婚配, 后代中表型正常的概率为 $1 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-86.jpg?height=302&width=303&top_left_y=2259&top_left_x=357",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-86.jpg?height=346&width=854&top_left_y=2237&top_left_x=721"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_64",
        "problem": "Animals possess mechanisms for maintaining their body temperature within permissible levels. For example, they show various responses to changes in room temperature. In addition, animals' body shapes are optimized to adapt to various climate changes, and their behaviors also regulate their body temperature.\nA: In each ordinary habitat, the body temperature of endotherms is always higher than that of ectotherms.\nB: In humans, the body temperature is elevated when the temperature of the hypothalamus is artificially increased.\nC: When a female Burmese python incubates eggs, her oxygen consumption in a cold room is less than that in warm room\nD: Ectotherms require less energy than endotherms for homeostasis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAnimals possess mechanisms for maintaining their body temperature within permissible levels. For example, they show various responses to changes in room temperature. In addition, animals' body shapes are optimized to adapt to various climate changes, and their behaviors also regulate their body temperature.\n\nA: In each ordinary habitat, the body temperature of endotherms is always higher than that of ectotherms.\nB: In humans, the body temperature is elevated when the temperature of the hypothalamus is artificially increased.\nC: When a female Burmese python incubates eggs, her oxygen consumption in a cold room is less than that in warm room\nD: Ectotherms require less energy than endotherms for homeostasis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_970",
        "problem": "What is the primary mechanism by which ADH exerts its effects on the kidney?\nA: Formation of water-impermeable junctions by the podocytes of nephron glomeruli\nB: Decreasing circulatory flux through renal afferent arterioles\nC: Raising the osmolarity of nephron medullae to $\\sim 1500 \\mathrm{mosm} / \\mathrm{L}$\nD: Increasing expression of aquaporins in the distal convoluted tubule\nE: Synthesis of additional cation pumps in the ascending Loop of Henle\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the primary mechanism by which ADH exerts its effects on the kidney?\n\nA: Formation of water-impermeable junctions by the podocytes of nephron glomeruli\nB: Decreasing circulatory flux through renal afferent arterioles\nC: Raising the osmolarity of nephron medullae to $\\sim 1500 \\mathrm{mosm} / \\mathrm{L}$\nD: Increasing expression of aquaporins in the distal convoluted tubule\nE: Synthesis of additional cation pumps in the ascending Loop of Henle\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_486",
        "problem": "果蝇的有眼与无眼、正常翅与裂翅分别由基因 $\\mathrm{D} / \\mathrm{d} 、 \\mathrm{~F} / \\mathrm{f}$ 控制, 已知这两对基因中只有一对位于 $\\mathrm{X}$ 染色体上, 且某一种基因型的个体存在胚胎致死现象。现用一对有眼裂翅雌雄果蝇杂交, $\\mathrm{F}_{1}$ 表型及数量如下表所示, 不考虑基因突变和染色体互换。下列说法错误的是（ ）\n\n|  | 有眼裂翅 | 有眼正常翅 | 无眼裂翅 | 无眼正常翅 |\n| :--- | :--- | :--- | :--- | :--- |\n| 雌蝇 (只) | 181 | 0 | 62 | 0 |\n| 雄蝇 (只) | 89 | 92 | 31 | 0 |\nA: 决定有眼与无眼、正常翅与裂翅的基因分别位于常染色体和 X 染色体上\nB: 亲本的基因型为 $D d X^{F} X^{f} 、 D d X^{F} Y$ ，基因型为 $d d X^{f} Y$ 的个体胚胎致死\nC: $F_{1}$ 中有眼个体随机交配, 后代成活个体中有眼正常翅雌蝇的占比为 $3 / 71$\nD: 用纯合无眼正常翅雌蝇和纯合无眼裂翅雄蝇杂交可验证胚胎致死的基因型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的有眼与无眼、正常翅与裂翅分别由基因 $\\mathrm{D} / \\mathrm{d} 、 \\mathrm{~F} / \\mathrm{f}$ 控制, 已知这两对基因中只有一对位于 $\\mathrm{X}$ 染色体上, 且某一种基因型的个体存在胚胎致死现象。现用一对有眼裂翅雌雄果蝇杂交, $\\mathrm{F}_{1}$ 表型及数量如下表所示, 不考虑基因突变和染色体互换。下列说法错误的是（ ）\n\n|  | 有眼裂翅 | 有眼正常翅 | 无眼裂翅 | 无眼正常翅 |\n| :--- | :--- | :--- | :--- | :--- |\n| 雌蝇 (只) | 181 | 0 | 62 | 0 |\n| 雄蝇 (只) | 89 | 92 | 31 | 0 |\n\nA: 决定有眼与无眼、正常翅与裂翅的基因分别位于常染色体和 X 染色体上\nB: 亲本的基因型为 $D d X^{F} X^{f} 、 D d X^{F} Y$ ，基因型为 $d d X^{f} Y$ 的个体胚胎致死\nC: $F_{1}$ 中有眼个体随机交配, 后代成活个体中有眼正常翅雌蝇的占比为 $3 / 71$\nD: 用纯合无眼正常翅雌蝇和纯合无眼裂翅雄蝇杂交可验证胚胎致死的基因型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_372",
        "problem": "某研究小组采用放射性同位素 ${ }^{14} \\mathrm{C}$ 进行了两组二倍体动物细胞学实验:\n\n实验一: 诱导 ${ }^{14} \\mathrm{C}$ 完全标记的细胞样本, 使其分别在只有 ${ }^{12} \\mathrm{C}$ 的培养基内进行有丝分裂和减数分裂, 实验期间收集到分裂中期的细胞样本甲和乙，统计样本放射性标记的染色体数和核 DNA 数如下表:\n\n| 样本 | 标记染色体数 | 标记 DNA <br> 数 |\n| :--- | :--- | :--- |\n| 甲 | 20 | 40 |\n| 乙 | 10 | 20 |\n\n实验二: 使用放射性同位素 ${ }^{14} \\mathrm{C}$ 分别标记尿嘧啶核苷酸和亮氨酸, 其后添加到两组细胞培养基中, 并对 ${ }^{14} \\mathrm{C}$ 在细胞中的分布进行跟踪测定, 实验过程中, 发现细胞对于放射性亮氨酸的吸收量远远高于同时期对放射性尿嘧啶核苷酸的吸收量。\n\n下列关于该实验的叙述错误的是 ( )\nA: 甲、乙细胞中染色体组的数目分别是 $2 、 1$\nB: 实验一的表格中所有的 DNA 分子都含 ${ }^{12} \\mathrm{C}$\nC: 甲细胞和乙细胞中均会发生非同源染色体的自由组合\nD: 实验二说明同时期蛋白质合成量多于 RNA 合成量\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某研究小组采用放射性同位素 ${ }^{14} \\mathrm{C}$ 进行了两组二倍体动物细胞学实验:\n\n实验一: 诱导 ${ }^{14} \\mathrm{C}$ 完全标记的细胞样本, 使其分别在只有 ${ }^{12} \\mathrm{C}$ 的培养基内进行有丝分裂和减数分裂, 实验期间收集到分裂中期的细胞样本甲和乙，统计样本放射性标记的染色体数和核 DNA 数如下表:\n\n| 样本 | 标记染色体数 | 标记 DNA <br> 数 |\n| :--- | :--- | :--- |\n| 甲 | 20 | 40 |\n| 乙 | 10 | 20 |\n\n实验二: 使用放射性同位素 ${ }^{14} \\mathrm{C}$ 分别标记尿嘧啶核苷酸和亮氨酸, 其后添加到两组细胞培养基中, 并对 ${ }^{14} \\mathrm{C}$ 在细胞中的分布进行跟踪测定, 实验过程中, 发现细胞对于放射性亮氨酸的吸收量远远高于同时期对放射性尿嘧啶核苷酸的吸收量。\n\n下列关于该实验的叙述错误的是 ( )\n\nA: 甲、乙细胞中染色体组的数目分别是 $2 、 1$\nB: 实验一的表格中所有的 DNA 分子都含 ${ }^{12} \\mathrm{C}$\nC: 甲细胞和乙细胞中均会发生非同源染色体的自由组合\nD: 实验二说明同时期蛋白质合成量多于 RNA 合成量\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_891",
        "problem": "苯丙酮尿症是由 PH 基因编码的苯丙氨酸羟化酶异常引起的一种遗传病。已知人群中染色体上 PH 基因两侧限制性内切酶 MspI酶切位点的分布存在两种形式（图 1)。一对夫妻婚后生育了一个患有苯丙酮尿症的孩子，(2)号个体再次怀孕（图 2)。为确定胎儿是否正常, 需要进行产前诊断, 提取该家庭所有成员的 DNA 经 MspI酶切后进行电泳分离, 并利用苂光标记的 PH 基因片段与酶切片段杂交，得到 DNA 条带分布情况如图\n\n3 . 下列叙述正确的是。\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\nA: (1)号个体 $23 \\mathrm{~Kb}$ 的 DNA 条带中一定含有正常 PH 基因\nB: (2)号个体 $23 \\mathrm{~Kb}$ 的 DNA 条带中一定含有正常 PH 基因\nC: 推测(4)号个体一定不是苯丙酮尿症患者\nD: (4)号个体为 $\\mathrm{PH}$ 基因杂合体的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n苯丙酮尿症是由 PH 基因编码的苯丙氨酸羟化酶异常引起的一种遗传病。已知人群中染色体上 PH 基因两侧限制性内切酶 MspI酶切位点的分布存在两种形式（图 1)。一对夫妻婚后生育了一个患有苯丙酮尿症的孩子，(2)号个体再次怀孕（图 2)。为确定胎儿是否正常, 需要进行产前诊断, 提取该家庭所有成员的 DNA 经 MspI酶切后进行电泳分离, 并利用苂光标记的 PH 基因片段与酶切片段杂交，得到 DNA 条带分布情况如图\n\n3 . 下列叙述正确的是。\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\n\nA: (1)号个体 $23 \\mathrm{~Kb}$ 的 DNA 条带中一定含有正常 PH 基因\nB: (2)号个体 $23 \\mathrm{~Kb}$ 的 DNA 条带中一定含有正常 PH 基因\nC: 推测(4)号个体一定不是苯丙酮尿症患者\nD: (4)号个体为 $\\mathrm{PH}$ 基因杂合体的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-092.jpg?height=257&width=351&top_left_y=2213&top_left_x=361",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-092.jpg?height=274&width=185&top_left_y=2213&top_left_x=730",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-092.jpg?height=271&width=329&top_left_y=2212&top_left_x=932"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_143",
        "problem": "Animal cap cells from the animal pole were removed from a Xenopus blastula embryo. These cells were then incubated in culture media containing different concentrations of activin. As seen in the table below, the cells differentiated into various tissues or cells depending on the concentration of activin.\n\n| [figure1] | medium | Concentration of <br> activin in medium | Tissues <br> or cells <br> differentiated |\n| :---: | :---: | :---: | :---: |\n|  |  | 0 (control) | epithelial cells |\n|  |  | $\\sim 0.1 \\mathrm{ng} / \\mathrm{mL}$ | blood cells |\n|  |  | $\\sim 1 \\mathrm{ng} / \\mathrm{mL}$ | muscles |\n|  |  | $\\sim 10 \\mathrm{ng} / \\mathrm{mL}$ | notochord |\n|  |  | $\\sim 100 \\mathrm{ng} / \\mathrm{mL}$ | heart |\n\nWhich of the following statement(s) regarding this experiment is/are correct?\n\nI. Ectodermal tissues are induced to differentiate into endodermal tissues according to the level of activin concentration.\n\nII. The fate of the animal cap cells was determined prior to the blastula stage.\n\nIII. Initially animal cap cells differentiate into epithelial tissue.\n\nIV. Cells from the vegetal pole are also able to differentiate into muscle or heart tissues if exposed to high activin concentrations.\nA: Only I\nB: Only III\nC: Only I and III\nD: Only II and IV\nE: II, III, and IV\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAnimal cap cells from the animal pole were removed from a Xenopus blastula embryo. These cells were then incubated in culture media containing different concentrations of activin. As seen in the table below, the cells differentiated into various tissues or cells depending on the concentration of activin.\n\n| [figure1] | medium | Concentration of <br> activin in medium | Tissues <br> or cells <br> differentiated |\n| :---: | :---: | :---: | :---: |\n|  |  | 0 (control) | epithelial cells |\n|  |  | $\\sim 0.1 \\mathrm{ng} / \\mathrm{mL}$ | blood cells |\n|  |  | $\\sim 1 \\mathrm{ng} / \\mathrm{mL}$ | muscles |\n|  |  | $\\sim 10 \\mathrm{ng} / \\mathrm{mL}$ | notochord |\n|  |  | $\\sim 100 \\mathrm{ng} / \\mathrm{mL}$ | heart |\n\nWhich of the following statement(s) regarding this experiment is/are correct?\n\nI. Ectodermal tissues are induced to differentiate into endodermal tissues according to the level of activin concentration.\n\nII. The fate of the animal cap cells was determined prior to the blastula stage.\n\nIII. Initially animal cap cells differentiate into epithelial tissue.\n\nIV. Cells from the vegetal pole are also able to differentiate into muscle or heart tissues if exposed to high activin concentrations.\n\nA: Only I\nB: Only III\nC: Only I and III\nD: Only II and IV\nE: II, III, and IV\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-22.jpg?height=725&width=234&top_left_y=672&top_left_x=454"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1426",
        "problem": "Energy can be stored in the body in different ways. Glucose and glycogen yield approximately $17 \\mathrm{~kJ}$ of energy per gram while lipids (fats) yield $37 \\mathrm{~kJ}$ per gram.\n\nIf the average person has stored $5 \\mathrm{~g}$ of free glucose in their blood, $480 \\mathrm{~g}$ total glycogen in their muscles and liver, $15 \\mathrm{~kg}$ of lipids in their adipose tissue and has an energy requirement of $8700 \\mathrm{~kJ}$ per day, approximately how many days could they theoretically survive under starvation conditions?\nA: 25 days\nB: 35 days\nC: 45 days\nD: 55 days\nE: 65 days\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEnergy can be stored in the body in different ways. Glucose and glycogen yield approximately $17 \\mathrm{~kJ}$ of energy per gram while lipids (fats) yield $37 \\mathrm{~kJ}$ per gram.\n\nIf the average person has stored $5 \\mathrm{~g}$ of free glucose in their blood, $480 \\mathrm{~g}$ total glycogen in their muscles and liver, $15 \\mathrm{~kg}$ of lipids in their adipose tissue and has an energy requirement of $8700 \\mathrm{~kJ}$ per day, approximately how many days could they theoretically survive under starvation conditions?\n\nA: 25 days\nB: 35 days\nC: 45 days\nD: 55 days\nE: 65 days\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_689",
        "problem": "$\\beta$-地中海贫血病人由于 $\\mathrm{H}$ 基因发生突变（如 $\\mathrm{h}_{1} 、 \\mathrm{~h}_{2}$ 等）, 导致 $\\beta$ 链的生成受到抑制,无法与 $\\alpha$ 链结合产生足量的血红蛋白 $\\mathrm{A}$ 。基因型为 $\\mathrm{h}_{1} \\mathrm{~h}_{1} 、 \\mathrm{~h}_{2} \\mathrm{~h}_{2} 、 \\mathrm{~h}_{1} \\mathrm{~h}_{2}$ 的个体均会表现为重型贫血, 而基因型为 $\\mathrm{Hh}_{1} 、 \\mathrm{Hh}_{2}$ 的个体病症轻微或无症状。研究发现, 很多重型地中海贫血病人由于携带 DNA 甲基转移酶 1 （DNMTI）的突变基因而造成 $\\gamma$ 链基因被激活, $\\gamma$ 链能替代缺失的 $\\beta$ 链与 $\\alpha$ 链形成血红蛋白 $\\mathrm{F}$, 从而减轻症状（人群中携带 DNMT1 突变基因的纯合个体所占比例为 $\\mathrm{p}$ )。下图 1 表示对患者家系进行基因测序后获得的结果 （相关基因位于常染色体上，且独立遗传)，图 2 是正常人和图 1 第II代中“某个体”相关蛋白的电泳图谱。下列说法错误的是（ ）\n\n[图1]\nA: 突变基因 $h_{1}$ 和 $h_{2}$ 的存在说明基因突变具有不定向性\nB: 图 2 中的“某个体”可能对应于图 1 中的 $\\mathrm{II}_{1} 、 \\mathrm{II}_{4} 、 \\mathrm{II}_{6}$\nC: $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$ 再生育一个与II2 基因型相同孩子的概率是 $3 / 8$\nD: 若 $\\mathrm{II}_{1}$ 与一个 $\\beta$-地中海贫血致病基因携带者婚配, 则后代表现为重型贫血的概率是 $3 / 8(1-\\sqrt{p})$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\beta$-地中海贫血病人由于 $\\mathrm{H}$ 基因发生突变（如 $\\mathrm{h}_{1} 、 \\mathrm{~h}_{2}$ 等）, 导致 $\\beta$ 链的生成受到抑制,无法与 $\\alpha$ 链结合产生足量的血红蛋白 $\\mathrm{A}$ 。基因型为 $\\mathrm{h}_{1} \\mathrm{~h}_{1} 、 \\mathrm{~h}_{2} \\mathrm{~h}_{2} 、 \\mathrm{~h}_{1} \\mathrm{~h}_{2}$ 的个体均会表现为重型贫血, 而基因型为 $\\mathrm{Hh}_{1} 、 \\mathrm{Hh}_{2}$ 的个体病症轻微或无症状。研究发现, 很多重型地中海贫血病人由于携带 DNA 甲基转移酶 1 （DNMTI）的突变基因而造成 $\\gamma$ 链基因被激活, $\\gamma$ 链能替代缺失的 $\\beta$ 链与 $\\alpha$ 链形成血红蛋白 $\\mathrm{F}$, 从而减轻症状（人群中携带 DNMT1 突变基因的纯合个体所占比例为 $\\mathrm{p}$ )。下图 1 表示对患者家系进行基因测序后获得的结果 （相关基因位于常染色体上，且独立遗传)，图 2 是正常人和图 1 第II代中“某个体”相关蛋白的电泳图谱。下列说法错误的是（ ）\n\n[图1]\n\nA: 突变基因 $h_{1}$ 和 $h_{2}$ 的存在说明基因突变具有不定向性\nB: 图 2 中的“某个体”可能对应于图 1 中的 $\\mathrm{II}_{1} 、 \\mathrm{II}_{4} 、 \\mathrm{II}_{6}$\nC: $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{2}$ 再生育一个与II2 基因型相同孩子的概率是 $3 / 8$\nD: 若 $\\mathrm{II}_{1}$ 与一个 $\\beta$-地中海贫血致病基因携带者婚配, 则后代表现为重型贫血的概率是 $3 / 8(1-\\sqrt{p})$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-25.jpg?height=1277&width=900&top_left_y=161&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1183",
        "problem": "## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAAThree factors influence the incidence of seabird ingestion of plastics: (1) foraging mode, (2) habitat use and (3) body size. Far-ranging species that feed opportunistically at the sea surface are most susceptible to plastic ingestion. Surface feeders have a greater rate of plastic ingestion. Diving birds also eat plastic but are not as susceptible as surface feeders. Oceanic species - which commonly range over vast areas in search of broadly distributed prey seem more prone to plastic ingestion that coastal species - which target dense aggregations of fish and zooplankton prey. Finally, because larger seabirds consume larger prey items, large-bodied species often ingest larger plastic fragments.\n\nSeabirds in New Zealand come from four orders:\n\n- Sphenisciformes - penguins\n- Procellariiformes - albatrosses, shearwaters and other petrels\n- Pelecaniformes - shags, gannets and their kin\n- Charadriiformes - terns, gulls and skuas.\n\nPenguins live only in the southern hemisphere. Of all the birds, penguins are the most accomplished divers, with some species capable of reaching depths of 100 metres or more. Their small wings or flippers, stiff oily plumage, dense bones and thick fat deposits are all adaptations to diving. They catch fish, crustaceans (such as krill) and squid by underwater pursuit.\n\nThe order Procellariiformes has about 124 species around the world. They range in size from tiny 35 -gram storm petrels to huge albatrosses weighing in at 9 kilograms, with a 3.5 metre wingspan. These birds find all their food at sea, and most species come to land only to breed. Petrels and shearwaters are adept divers - some shearwaters regularly dive to 60 metres. Storm petrels, prions and albatrosses obtain their food close to the water's surface.\n\nShags pursue their prey under water, using their feet to propel themselves whereas gannets sight fish while flying overhead and capture them by plunging into the water.\n\nMost members of the order Charadriiformes are not marine species. Gulls and some terns take a large proportion of their food ashore or from freshwater habitats. However, some tern species and all skuas are largely marine, and both of these groups have New Zealand representatives. Marine terns such as the white-fronted tern (Sterna striata) feed by dipping - hovering above the water then dropping to catch surface-shoaling fish. Skuas are well-known predators of eggs, chicks and small birds at seabird colonies but they do in fact take a large proportion of their food at sea, often by harassing smaller seabirds and forcing them to regurgitate their food. (information from Te Ara The encyclopedia of New Zealand).\n\nWhich of the following New Zealand seabirds would be at MOST risk of harm from marine debris?\nA: Little Blue penguins\nB: Shearwaters\nC: Storm petrels\nD: Australasian gannet\nE: Skua\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAA\n\nproblem:\nThree factors influence the incidence of seabird ingestion of plastics: (1) foraging mode, (2) habitat use and (3) body size. Far-ranging species that feed opportunistically at the sea surface are most susceptible to plastic ingestion. Surface feeders have a greater rate of plastic ingestion. Diving birds also eat plastic but are not as susceptible as surface feeders. Oceanic species - which commonly range over vast areas in search of broadly distributed prey seem more prone to plastic ingestion that coastal species - which target dense aggregations of fish and zooplankton prey. Finally, because larger seabirds consume larger prey items, large-bodied species often ingest larger plastic fragments.\n\nSeabirds in New Zealand come from four orders:\n\n- Sphenisciformes - penguins\n- Procellariiformes - albatrosses, shearwaters and other petrels\n- Pelecaniformes - shags, gannets and their kin\n- Charadriiformes - terns, gulls and skuas.\n\nPenguins live only in the southern hemisphere. Of all the birds, penguins are the most accomplished divers, with some species capable of reaching depths of 100 metres or more. Their small wings or flippers, stiff oily plumage, dense bones and thick fat deposits are all adaptations to diving. They catch fish, crustaceans (such as krill) and squid by underwater pursuit.\n\nThe order Procellariiformes has about 124 species around the world. They range in size from tiny 35 -gram storm petrels to huge albatrosses weighing in at 9 kilograms, with a 3.5 metre wingspan. These birds find all their food at sea, and most species come to land only to breed. Petrels and shearwaters are adept divers - some shearwaters regularly dive to 60 metres. Storm petrels, prions and albatrosses obtain their food close to the water's surface.\n\nShags pursue their prey under water, using their feet to propel themselves whereas gannets sight fish while flying overhead and capture them by plunging into the water.\n\nMost members of the order Charadriiformes are not marine species. Gulls and some terns take a large proportion of their food ashore or from freshwater habitats. However, some tern species and all skuas are largely marine, and both of these groups have New Zealand representatives. Marine terns such as the white-fronted tern (Sterna striata) feed by dipping - hovering above the water then dropping to catch surface-shoaling fish. Skuas are well-known predators of eggs, chicks and small birds at seabird colonies but they do in fact take a large proportion of their food at sea, often by harassing smaller seabirds and forcing them to regurgitate their food. (information from Te Ara The encyclopedia of New Zealand).\n\nWhich of the following New Zealand seabirds would be at MOST risk of harm from marine debris?\n\nA: Little Blue penguins\nB: Shearwaters\nC: Storm petrels\nD: Australasian gannet\nE: Skua\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=800&width=617&top_left_y=1299&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=817&width=1097&top_left_y=1299&top_left_x=845"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_993",
        "problem": "In a forest, a group consisting of all the maple trees would be considered a:\nA: Population\nB: Community\nC: Ecosystem\nD: Biome\nE: None or all of the above\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn a forest, a group consisting of all the maple trees would be considered a:\n\nA: Population\nB: Community\nC: Ecosystem\nD: Biome\nE: None or all of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1344",
        "problem": "The graphs show the effect of insulin on the glucose concentrations inside the cells of liver and of heart muscle when these two tissues were supplied with increasing concentrations of glucose.\n[figure1]\n\nThe best interpretation of these data is that:\nA: Liver cells are impermeable to glucose.\nB: Insulin increases the entry of glucose into heart and liver cells.\nC:  Insulin increases the entry of glucose into the heart tissue but not into liver cells.\nD:  Insulin is metabolised rapidly by liver cells.\nE:  Insulin is metabolised rapidly by heart cells.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graphs show the effect of insulin on the glucose concentrations inside the cells of liver and of heart muscle when these two tissues were supplied with increasing concentrations of glucose.\n[figure1]\n\nThe best interpretation of these data is that:\n\nA: Liver cells are impermeable to glucose.\nB: Insulin increases the entry of glucose into heart and liver cells.\nC:  Insulin increases the entry of glucose into the heart tissue but not into liver cells.\nD:  Insulin is metabolised rapidly by liver cells.\nE:  Insulin is metabolised rapidly by heart cells.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-19.jpg?height=442&width=1642&top_left_y=418&top_left_x=114"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1113",
        "problem": "Water potential in plant tissues is affected by three factors, namely dissolved solutes, matric forces acting on water (such as intermolecular attractions and binding of water to surfaces) and hydrostatic pressure exerted by cell walls. Thus total water potential $\\left(\\Psi_{\\mathrm{T}}\\right)$ in plants is expressed as:\n\n$\\Psi_{\\mathrm{T}}=\\Psi_{\\mathrm{s}}+\\Psi_{\\mathrm{m}}+\\Psi_{\\mathrm{p}}$ where $\\mathrm{T}=$ total, $\\mathrm{s}=$ solute, $\\mathrm{p}=$ pressure and $\\mathrm{m}=$ matric potential.\n\nWhich of the following is correct?\nA: In a rapidly transpiring spruce tree, water flows upwards, down the water potential gradient, from soil (-1.5 MPa) to terminal shoot (-0.04 MPa).\nB: Water potential in xylem of gutting plants whose hydathodes release water drops must be positive.\nC: Water potential of apoplast will have $\\Psi_{m}$ as the major contributing factor while that in a vacuole will have $\\Psi_{s}$ as a major contributing factor.\nD: If $\\Psi_{w}$ of vacuole is lower than $\\Psi_{w}$ of apoplast, then there will be a net efflux of water across cytoplasm resulting into a flaccid cell.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nWater potential in plant tissues is affected by three factors, namely dissolved solutes, matric forces acting on water (such as intermolecular attractions and binding of water to surfaces) and hydrostatic pressure exerted by cell walls. Thus total water potential $\\left(\\Psi_{\\mathrm{T}}\\right)$ in plants is expressed as:\n\n$\\Psi_{\\mathrm{T}}=\\Psi_{\\mathrm{s}}+\\Psi_{\\mathrm{m}}+\\Psi_{\\mathrm{p}}$ where $\\mathrm{T}=$ total, $\\mathrm{s}=$ solute, $\\mathrm{p}=$ pressure and $\\mathrm{m}=$ matric potential.\n\nWhich of the following is correct?\n\nA: In a rapidly transpiring spruce tree, water flows upwards, down the water potential gradient, from soil (-1.5 MPa) to terminal shoot (-0.04 MPa).\nB: Water potential in xylem of gutting plants whose hydathodes release water drops must be positive.\nC: Water potential of apoplast will have $\\Psi_{m}$ as the major contributing factor while that in a vacuole will have $\\Psi_{s}$ as a major contributing factor.\nD: If $\\Psi_{w}$ of vacuole is lower than $\\Psi_{w}$ of apoplast, then there will be a net efflux of water across cytoplasm resulting into a flaccid cell.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1235",
        "problem": "SAVING THE MAUI'S DOLPHINS - A STORY OF BIOLOGY, POLICY \\& CONSERVATION\n\n[figure1]\n\nWWF-New Zealand\n\nhttp://wwf.panda.org/wwf news/?206249/NZ-govt-fails-Mauisdolphins-on-global-stage\n\n[figure2]\n\nhttp://uww.doc.govt.nz/conservation/native-animals/marinemammals/dolphins/hectors-dolphin/docs-work/hectors-and-mauisdolphin-incident-databasel\n\nMaui's dolphin (Cephalorhynchus hectori maui) is one of the world's rarest dolphins and is found only on the west coast of the North Island of New Zealand (Resource Pack, Figure 1). It is a sub-species of Hector's dolphin (Cephalorhynchus hectori). The Maui's dolphin is protected by the West Coast North Island Marine Mammal Sanctuary (WCNIMMS) which restricts seabed mining activities, acoustic seismic survey work and commercial and recreational set netting. The boundaries of this sanctuary extend alongshore from Maunganui Bluff in Northland to Oakura Beach, Taranaki, in the south and from mean high water springs to the 12 nautical mile (nm) territorial sea limit. The total area of the sanctuary is approximately 1,200,086 hectares covering 2,164 km of coastline. (Resource Pack, Figure 1).\n\nMaui's dolphin is 'critically endangered' (IUCN Red List), with the population dropping from around 1000 individuals in 1970 to 111 in 2004 according to research by Assoc. Prof. Dr Liz Slooten and others from Otago University. Recent research by the Department of Conservation (DOC) suggests there are now even fewer Maui's dolphins remaining. Maui's dolphins are relatively short-lived (approximately 25 years), and are slow breeders. Females do not have their first calf until they are about seven or eight years old, and have a new calf only every two to four years. This means the species may be threatened by even occasional deaths caused by human activity. Fishing, particularly set netting, is the greatest known human threat to Maui's dolphins and thought to be responsible for about 75 per cent of reported deaths with a known cause. Other human threats include marine tourism, vessel traffic, mining, construction, coastal development pollution, sedimentation, oil spills, plastic bags, marine farming and climate change.\n\nIn recent weeks the need to develop effective management strategies for this species to prevent its extinction has hit the news. Liz Slooten presented the latest research to the International Whaling Commission in May this year, showing that the current protection measures are not sufficient to avoid the extinction of Maui's dolphin. WWF-New Zealand, Greenpeace, and Forest \\& Bird and international conservation groups including NABU and WDC in 2012 and 2013 are all campaigning to protect the Maui's dolphin and hundreds of angry protesters marched to the office of the Energy and Resources Minister Simon Bridges' office in Tauranga to protest the government allowing oil exploration within the West Coast North Island Marine Mammal Sanctuary in their annual tender process 'Block Offer 2014' (Resource Pack, Figure 3). This government decision came just two weeks after the International Whaling Committee (IWC) criticised New Zealand for not taking the necessary steps to save the Maui's dolphin. The IWC noted that a $350 \\mathrm{~km}^{2}$ set net restriction had been added to the WCNIMMS but commented that these measures fell significantly short of those required to reverse the Maui's dolphin decline as recommended by the IWC in 2012 and 2013. The IWC reiterated its extreme concern about the continued decline of such a small population \"as the humaninduced death of even one dolphin would increase the extinction risk for this subspecies\". In 2013 it strongly recommended that the NZ government should:\n\n\"take immediate management actions that will eliminate bycatch of Maui's dolphins. This includes full closures of any fisheries within the range of Maui's dolphins that are known to pose a risk of bycatch of small cetaceans (i.e. set net and trawl fisheries)\". Ensuring full protection of Maui's dolphins in all areas throughout their habitat, together with an ample buffer zone, would minimise the risk of bycatch and maximise the chances of population increase\". and \"commit to specific population increase targets and timelines\".\n\nBecause this recommendation was ignored by the NZ government, in 2014, the IWC recommended that the protected area should be extended south to Whanganui, offshore to 20 nautical miles and should include the harbours.\n\nThe resource pack contains maps of Maui's and Hector's dolphin sightings from 1970 - July 2013, Protection measures for Maui's dolphins on the West Coast North Island and the 'Block Offer 2014' for the Offshore Release Area: Taranaki Basin 14TAR-R1. It also has a table of all Maui's dolphin sightings from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti. Much of this area is included in the Block Offer 2014' Taranaki Basin 14TAR-R1 area. This table is modified from the Maui's dolphin sightings database:\n\nhttp://www.doc.govt.nz/conservation/native-animals/marine-mammals/dolphins/mauis-dolphin/docs-work/mauisdolphin-sightings/\n\nConservation Minister Nick Smith was questioned in parliament about the decision to open part of the sanctuary for exploration. \"The block offer is nowhere near where the Maui's live,\" Dr Smith said. \"There hasn't been a single observation of a Maui's dolphin, and the oil and gas industry hasn't been involved in a single Maui's dolphin incident in Taranaki over the past 40 years despite 23 wells being drilled\" (Source: http://www.3news.co.nz/Greenpeace-get23000-signatures-in-petition-to-sack-Bridges/tabid/423/articleID/349373/Default.aspx).Estimate the area of overlap between the West Coast North Island Marine Mammal Sanctuary and the 'Block Offer 2014' Taranaki Basin 14TAR-R1 using the information contained in Figures 2 and 3 in the resource Pack.\nA: $3000 \\mathrm{~km}^{2}$\nB: $3000 \\mathrm{~nm}^{2}$\nC: $1000 \\mathrm{~km}^{2}$\nD: $1000 \\mathrm{~nm}^{2}$\nE: $500 \\mathrm{~km}^{2}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nSAVING THE MAUI'S DOLPHINS - A STORY OF BIOLOGY, POLICY \\& CONSERVATION\n\n[figure1]\n\nWWF-New Zealand\n\nhttp://wwf.panda.org/wwf news/?206249/NZ-govt-fails-Mauisdolphins-on-global-stage\n\n[figure2]\n\nhttp://uww.doc.govt.nz/conservation/native-animals/marinemammals/dolphins/hectors-dolphin/docs-work/hectors-and-mauisdolphin-incident-databasel\n\nMaui's dolphin (Cephalorhynchus hectori maui) is one of the world's rarest dolphins and is found only on the west coast of the North Island of New Zealand (Resource Pack, Figure 1). It is a sub-species of Hector's dolphin (Cephalorhynchus hectori). The Maui's dolphin is protected by the West Coast North Island Marine Mammal Sanctuary (WCNIMMS) which restricts seabed mining activities, acoustic seismic survey work and commercial and recreational set netting. The boundaries of this sanctuary extend alongshore from Maunganui Bluff in Northland to Oakura Beach, Taranaki, in the south and from mean high water springs to the 12 nautical mile (nm) territorial sea limit. The total area of the sanctuary is approximately 1,200,086 hectares covering 2,164 km of coastline. (Resource Pack, Figure 1).\n\nMaui's dolphin is 'critically endangered' (IUCN Red List), with the population dropping from around 1000 individuals in 1970 to 111 in 2004 according to research by Assoc. Prof. Dr Liz Slooten and others from Otago University. Recent research by the Department of Conservation (DOC) suggests there are now even fewer Maui's dolphins remaining. Maui's dolphins are relatively short-lived (approximately 25 years), and are slow breeders. Females do not have their first calf until they are about seven or eight years old, and have a new calf only every two to four years. This means the species may be threatened by even occasional deaths caused by human activity. Fishing, particularly set netting, is the greatest known human threat to Maui's dolphins and thought to be responsible for about 75 per cent of reported deaths with a known cause. Other human threats include marine tourism, vessel traffic, mining, construction, coastal development pollution, sedimentation, oil spills, plastic bags, marine farming and climate change.\n\nIn recent weeks the need to develop effective management strategies for this species to prevent its extinction has hit the news. Liz Slooten presented the latest research to the International Whaling Commission in May this year, showing that the current protection measures are not sufficient to avoid the extinction of Maui's dolphin. WWF-New Zealand, Greenpeace, and Forest \\& Bird and international conservation groups including NABU and WDC in 2012 and 2013 are all campaigning to protect the Maui's dolphin and hundreds of angry protesters marched to the office of the Energy and Resources Minister Simon Bridges' office in Tauranga to protest the government allowing oil exploration within the West Coast North Island Marine Mammal Sanctuary in their annual tender process 'Block Offer 2014' (Resource Pack, Figure 3). This government decision came just two weeks after the International Whaling Committee (IWC) criticised New Zealand for not taking the necessary steps to save the Maui's dolphin. The IWC noted that a $350 \\mathrm{~km}^{2}$ set net restriction had been added to the WCNIMMS but commented that these measures fell significantly short of those required to reverse the Maui's dolphin decline as recommended by the IWC in 2012 and 2013. The IWC reiterated its extreme concern about the continued decline of such a small population \"as the humaninduced death of even one dolphin would increase the extinction risk for this subspecies\". In 2013 it strongly recommended that the NZ government should:\n\n\"take immediate management actions that will eliminate bycatch of Maui's dolphins. This includes full closures of any fisheries within the range of Maui's dolphins that are known to pose a risk of bycatch of small cetaceans (i.e. set net and trawl fisheries)\". Ensuring full protection of Maui's dolphins in all areas throughout their habitat, together with an ample buffer zone, would minimise the risk of bycatch and maximise the chances of population increase\". and \"commit to specific population increase targets and timelines\".\n\nBecause this recommendation was ignored by the NZ government, in 2014, the IWC recommended that the protected area should be extended south to Whanganui, offshore to 20 nautical miles and should include the harbours.\n\nThe resource pack contains maps of Maui's and Hector's dolphin sightings from 1970 - July 2013, Protection measures for Maui's dolphins on the West Coast North Island and the 'Block Offer 2014' for the Offshore Release Area: Taranaki Basin 14TAR-R1. It also has a table of all Maui's dolphin sightings from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti. Much of this area is included in the Block Offer 2014' Taranaki Basin 14TAR-R1 area. This table is modified from the Maui's dolphin sightings database:\n\nhttp://www.doc.govt.nz/conservation/native-animals/marine-mammals/dolphins/mauis-dolphin/docs-work/mauisdolphin-sightings/\n\nConservation Minister Nick Smith was questioned in parliament about the decision to open part of the sanctuary for exploration. \"The block offer is nowhere near where the Maui's live,\" Dr Smith said. \"There hasn't been a single observation of a Maui's dolphin, and the oil and gas industry hasn't been involved in a single Maui's dolphin incident in Taranaki over the past 40 years despite 23 wells being drilled\" (Source: http://www.3news.co.nz/Greenpeace-get23000-signatures-in-petition-to-sack-Bridges/tabid/423/articleID/349373/Default.aspx).\n\nproblem:\nEstimate the area of overlap between the West Coast North Island Marine Mammal Sanctuary and the 'Block Offer 2014' Taranaki Basin 14TAR-R1 using the information contained in Figures 2 and 3 in the resource Pack.\n\nA: $3000 \\mathrm{~km}^{2}$\nB: $3000 \\mathrm{~nm}^{2}$\nC: $1000 \\mathrm{~km}^{2}$\nD: $1000 \\mathrm{~nm}^{2}$\nE: $500 \\mathrm{~km}^{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-30.jpg?height=503&width=682&top_left_y=277&top_left_x=113",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-30.jpg?height=497&width=848&top_left_y=277&top_left_x=1049"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_9",
        "problem": "Sam is an imaginary single-celled organism. His body is covered with cilia which enable him to move. He has two nuclei: housekeeping genes are expressed in the macronucleus, while gene-expression tasks relevant to reproduction are undertaken by the micronucleus. Sam divides via a process called \"schizogony\", whereby nuclei undergo multiple fission and then the cell itself divides. Sam can eat food particles and even bacteria through his mouth, a characteristic that helps him to live freely as a saprophyte. Sam has a monomorphic shape throughout its life cycle and experience enlargement and division, consecutively. Sam has decided to abandon his free-living lifestyle and transform into a merry parasite (with limited harm to the host).\n\nJack is also a parasite and is more aggressive (damages its host) than Sam and his single-celled body is covered by an armour made out of protein. This protective layer is jagged and gives Jack a rugged look. Jack also has two nuclei. He has two flagella that allows him to move, attack, and penetrate. He is able to secrete enzymes that can degrade many animal tissues. immature individuals who belong to Jacks species lack traits mentioned above; they are small and proliferate rapidly. In addition, Immature individuals of Jack needs to find host to grow and cannot live independently.Assume that Sam and Jack chose a common host. Given the presence of these two parasites together\nA: Sam will probably win the competition against Jack.\nB: If a lethal bacterium spread through the population of the host, Sam's fitness will increase.\nC: Sam's fitness is negatively correlated with Jack's abundance.\nD: Sam's fitness is less affected by the life-history traits of the host than Jack's.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSam is an imaginary single-celled organism. His body is covered with cilia which enable him to move. He has two nuclei: housekeeping genes are expressed in the macronucleus, while gene-expression tasks relevant to reproduction are undertaken by the micronucleus. Sam divides via a process called \"schizogony\", whereby nuclei undergo multiple fission and then the cell itself divides. Sam can eat food particles and even bacteria through his mouth, a characteristic that helps him to live freely as a saprophyte. Sam has a monomorphic shape throughout its life cycle and experience enlargement and division, consecutively. Sam has decided to abandon his free-living lifestyle and transform into a merry parasite (with limited harm to the host).\n\nJack is also a parasite and is more aggressive (damages its host) than Sam and his single-celled body is covered by an armour made out of protein. This protective layer is jagged and gives Jack a rugged look. Jack also has two nuclei. He has two flagella that allows him to move, attack, and penetrate. He is able to secrete enzymes that can degrade many animal tissues. immature individuals who belong to Jacks species lack traits mentioned above; they are small and proliferate rapidly. In addition, Immature individuals of Jack needs to find host to grow and cannot live independently.Assume that Sam and Jack chose a common host. Given the presence of these two parasites together\n\nA: Sam will probably win the competition against Jack.\nB: If a lethal bacterium spread through the population of the host, Sam's fitness will increase.\nC: Sam's fitness is negatively correlated with Jack's abundance.\nD: Sam's fitness is less affected by the life-history traits of the host than Jack's.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_146",
        "problem": "Regional species diversity in ecosystems usually increases with primary productivity while local species diversity does not show any defined relation with primary productivity. It suggests that different processes have distinctive influence on species diversity at regional and local scale.\n\nFigure illustrates four hypothesized mechanisms and their influence on $\\alpha$-diversity (mean local diversity) and $\\beta$-diversity (diversity in species composition between local sites). (HNDD : Interspecific negative population density dependence. CNDD: Intraspecific negative population density dependence.).\n\n[figure1]\n\n## Local processes\n\n[figure2]\nA: Predators or pathogens specific to the dominant species will decrease $\\alpha$ diversity.\nB: Increasing CNDD has a homogenizing effect on community composition among sites.\nC: Regions with weak HNDD are expected to have higher degree of community differentiation.\nD: Environmental perturbation acts in a fashion similar to effect of increasing CNDD on alpha and beta diversity.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRegional species diversity in ecosystems usually increases with primary productivity while local species diversity does not show any defined relation with primary productivity. It suggests that different processes have distinctive influence on species diversity at regional and local scale.\n\nFigure illustrates four hypothesized mechanisms and their influence on $\\alpha$-diversity (mean local diversity) and $\\beta$-diversity (diversity in species composition between local sites). (HNDD : Interspecific negative population density dependence. CNDD: Intraspecific negative population density dependence.).\n\n[figure1]\n\n## Local processes\n\n[figure2]\n\nA: Predators or pathogens specific to the dominant species will decrease $\\alpha$ diversity.\nB: Increasing CNDD has a homogenizing effect on community composition among sites.\nC: Regions with weak HNDD are expected to have higher degree of community differentiation.\nD: Environmental perturbation acts in a fashion similar to effect of increasing CNDD on alpha and beta diversity.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-48.jpg?height=443&width=417&top_left_y=732&top_left_x=617",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-48.jpg?height=880&width=826&top_left_y=800&top_left_x=617"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_675",
        "problem": "某植物为二倍体雌雄同株同花植物, 自然状态下可以自花受粉或异花受粉。其花色受 $\\mathrm{A}$ （红色） $\\mathrm{A}^{\\mathrm{P}}$ (斑红色)、 $\\mathrm{A}^{\\mathrm{T}}$ （条红色）、a（白色） 4 个复等位基因控制, 4 个复等位基因的显隐性关系为 $\\mathrm{A}>\\mathrm{A}^{\\mathrm{P}}>\\mathrm{A}^{\\mathrm{T}}>\\mathrm{a}$ 。 $\\mathrm{A}^{\\mathrm{T}}$ 是一种“自私基因”，在产生配子时会导致同株一定比例的其他花粉死亡. 使其有更多的机会遗传下去。基因型为 $\\mathrm{A}^{\\mathrm{T}} \\mathrm{a}$ 的植株自交, $\\mathrm{F}_{1}$ 中条红色：白色=5：1。下列有关叙述错误的是（ ）\nA: 花色基因的遗传遵循孟德尔分离定律\nB: 两株花色不同的植株杂交，子代花色最多有 3 种\nC: 等比例的 $\\mathrm{AA}^{\\mathrm{P}}$ 与 $\\mathrm{A}^{\\mathrm{T}} \\mathrm{a}$ 植株随机交配, $\\mathrm{F}_{1}$ 中含“自私基因”的植株所占比例为 $13 / 28$\nD: 基因型为 $\\mathrm{Aa}$ 的植株自交, $\\mathrm{F}_{1}$ 条红色植株中能稳定遗传的占 $3 / 7$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某植物为二倍体雌雄同株同花植物, 自然状态下可以自花受粉或异花受粉。其花色受 $\\mathrm{A}$ （红色） $\\mathrm{A}^{\\mathrm{P}}$ (斑红色)、 $\\mathrm{A}^{\\mathrm{T}}$ （条红色）、a（白色） 4 个复等位基因控制, 4 个复等位基因的显隐性关系为 $\\mathrm{A}>\\mathrm{A}^{\\mathrm{P}}>\\mathrm{A}^{\\mathrm{T}}>\\mathrm{a}$ 。 $\\mathrm{A}^{\\mathrm{T}}$ 是一种“自私基因”，在产生配子时会导致同株一定比例的其他花粉死亡. 使其有更多的机会遗传下去。基因型为 $\\mathrm{A}^{\\mathrm{T}} \\mathrm{a}$ 的植株自交, $\\mathrm{F}_{1}$ 中条红色：白色=5：1。下列有关叙述错误的是（ ）\n\nA: 花色基因的遗传遵循孟德尔分离定律\nB: 两株花色不同的植株杂交，子代花色最多有 3 种\nC: 等比例的 $\\mathrm{AA}^{\\mathrm{P}}$ 与 $\\mathrm{A}^{\\mathrm{T}} \\mathrm{a}$ 植株随机交配, $\\mathrm{F}_{1}$ 中含“自私基因”的植株所占比例为 $13 / 28$\nD: 基因型为 $\\mathrm{Aa}$ 的植株自交, $\\mathrm{F}_{1}$ 条红色植株中能稳定遗传的占 $3 / 7$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1292",
        "problem": "The table below summarises the complications by age for measles cases in the United States from 1987-2000.\n\n|  |  | No. (\\%) of persons with complication, by age group |  |  |  |  |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: |\n| Complication | Overall $(67,032$ cases <br> with age information) | $(n=28$ years | $5-9$ years | $10-19$ years | $20-29$ years | $>30$ years |\n| $(n=28.730)$ | $(n=6492)$ | $(n=18,580)$ | $(n=9161)$ | $(n=4069)$ |  |  |\n| Any | $19,480(29.1)$ | $11,883(41.4)$ | $1173(18.1)$ | $2369(12.8)$ | $2656(29.0)$ | $1399(34.4)$ |\n| Death | $177(0.3)$ | $97(0.3)$ | $9(0.1)$ | $18(0.1)$ | $26(0.3)$ | $27(0.7)$ |\n| Diarrhea | $5482(8.2)$ | $3294(11.5)$ | $408(6.3)$ | $627(3.4)$ | $767(8.4)$ | $386(9.5)$ |\n| Encephalitis | $97(0.1)$ | $43(0.2)$ | $9(0.1)$ | $13(0.1)$ | $21(0.2)$ | $11(0.3)$ |\n| Hospitalization | $12,876(19.2)$ | $7470(26.0)$ | $612(9.4)$ | $1612(8.7)$ | $2075(22.7)$ | $1107(27.2)$ |\n| Otitis media | $4879(7.3)$ | $4009(14.0)$ | $305(4.7)$ | $338(1.8)$ | $157(1.7)$ | $70(1.7)$ |\n| Pneumonia | $3959(5.9)$ | $2480(8.6)$ | $183(2.8)$ | $363(2.0)$ | $554(6.1)$ | $379(9.3)$ |\n\nSource: Centers for Disease Control and Prevention.\n\nhttp://jid.oxfordjournals.org/content/189/Supplement_1/S4/F3.expansion.html\n\nBased on the data, which statement is correct?\nA: Measles is relatively more deadly in young people compared with older people\nB: Catching measles will give you pneumonia\nC: Most of the measles cases with complications reported were young people, below 19 years of age\nD: Hospitalization is rare with measles\nE: Encephalitis is a deadly complication of measles\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe table below summarises the complications by age for measles cases in the United States from 1987-2000.\n\n|  |  | No. (\\%) of persons with complication, by age group |  |  |  |  |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: |\n| Complication | Overall $(67,032$ cases <br> with age information) | $(n=28$ years | $5-9$ years | $10-19$ years | $20-29$ years | $>30$ years |\n| $(n=28.730)$ | $(n=6492)$ | $(n=18,580)$ | $(n=9161)$ | $(n=4069)$ |  |  |\n| Any | $19,480(29.1)$ | $11,883(41.4)$ | $1173(18.1)$ | $2369(12.8)$ | $2656(29.0)$ | $1399(34.4)$ |\n| Death | $177(0.3)$ | $97(0.3)$ | $9(0.1)$ | $18(0.1)$ | $26(0.3)$ | $27(0.7)$ |\n| Diarrhea | $5482(8.2)$ | $3294(11.5)$ | $408(6.3)$ | $627(3.4)$ | $767(8.4)$ | $386(9.5)$ |\n| Encephalitis | $97(0.1)$ | $43(0.2)$ | $9(0.1)$ | $13(0.1)$ | $21(0.2)$ | $11(0.3)$ |\n| Hospitalization | $12,876(19.2)$ | $7470(26.0)$ | $612(9.4)$ | $1612(8.7)$ | $2075(22.7)$ | $1107(27.2)$ |\n| Otitis media | $4879(7.3)$ | $4009(14.0)$ | $305(4.7)$ | $338(1.8)$ | $157(1.7)$ | $70(1.7)$ |\n| Pneumonia | $3959(5.9)$ | $2480(8.6)$ | $183(2.8)$ | $363(2.0)$ | $554(6.1)$ | $379(9.3)$ |\n\nSource: Centers for Disease Control and Prevention.\n\nhttp://jid.oxfordjournals.org/content/189/Supplement_1/S4/F3.expansion.html\n\nBased on the data, which statement is correct?\n\nA: Measles is relatively more deadly in young people compared with older people\nB: Catching measles will give you pneumonia\nC: Most of the measles cases with complications reported were young people, below 19 years of age\nD: Hospitalization is rare with measles\nE: Encephalitis is a deadly complication of measles\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_206",
        "problem": "Amphioxus (lancelet) is a marine slender, laterally compressed and translucent filter-feeder deutrostome that inhabits the sandy bottoms of coastal waters around the world. Sexes are separate, gonads are located in the atrial cavity and there are no ducts. The closed circulatory system of this fish-like animals, is complex for such a simple chordate. The flow pattern is remarkably similar to that of primitive vertebrates. The colourless blood is pumped forward in the ventral aorta by peristaltic-like contractions of the vessel wall only, then passes upward through branchial arteries (aortic arches) in the pharyngeal bars to paired dorsal aortas which join to become a single dorsal aorta. From here the blood is distributed to the body tissues by microcirculation and then is collected in veins, which return it to the ventral aorta.\n\n[figure1]\nA: The blood is circulated from anterior to posterior dorsally and from posterior to anterior ventrally.\nB: The heart consists of a sinuous venosus, a cardiac atrium, and a ventricle located ventrally and carries low oxygen blood.\nC: It differs from vertebrates in having numerous pharyngeal slits that are not open outside directly but acts as a trapping structure to collect small-suspended\nD: Sperms are released into the atrial cavity, and then pass out via genital pore and atriopore to the outside where fertilization occurs.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAmphioxus (lancelet) is a marine slender, laterally compressed and translucent filter-feeder deutrostome that inhabits the sandy bottoms of coastal waters around the world. Sexes are separate, gonads are located in the atrial cavity and there are no ducts. The closed circulatory system of this fish-like animals, is complex for such a simple chordate. The flow pattern is remarkably similar to that of primitive vertebrates. The colourless blood is pumped forward in the ventral aorta by peristaltic-like contractions of the vessel wall only, then passes upward through branchial arteries (aortic arches) in the pharyngeal bars to paired dorsal aortas which join to become a single dorsal aorta. From here the blood is distributed to the body tissues by microcirculation and then is collected in veins, which return it to the ventral aorta.\n\n[figure1]\n\nA: The blood is circulated from anterior to posterior dorsally and from posterior to anterior ventrally.\nB: The heart consists of a sinuous venosus, a cardiac atrium, and a ventricle located ventrally and carries low oxygen blood.\nC: It differs from vertebrates in having numerous pharyngeal slits that are not open outside directly but acts as a trapping structure to collect small-suspended\nD: Sperms are released into the atrial cavity, and then pass out via genital pore and atriopore to the outside where fertilization occurs.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-38.jpg?height=863&width=1582&top_left_y=802&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1394",
        "problem": "Which of the following may be pathogenic organisms?\n\nI. Bacteria\n\nII. Viruses\n\nIII. Prions\n\nIV. Fungi\n\nV. Protozoa\nA: I and II\nB: I, II and V\nC: I, IV and V\nD: I, II, IV and V\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following may be pathogenic organisms?\n\nI. Bacteria\n\nII. Viruses\n\nIII. Prions\n\nIV. Fungi\n\nV. Protozoa\n\nA: I and II\nB: I, II and V\nC: I, IV and V\nD: I, II, IV and V\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_810",
        "problem": "某团队因发现“剪接体的结构与分子机理”获生命科学奖。研究发现真核细胞中基因\n表达分三步进行, 如下图所示。剪接体主要由蛋白质和小分子的核 RNA 组成。下列叙述正确的是 ( )\n\n[图1]\nA: RNA 聚合酶和剪接体的彻底水解产物中均含有碱基 $\\mathrm{T}$ 和碱基 A\nB: 过程(3)中一个核糖体可结合多条 mRNA 链以提高蛋白质合成速率\nC: (2)过程剪接现象的发现是对传统基因表达过程的一个重要补充\nD: 若(2)过程位置出现差错, 最终编码的蛋白质结构不可能发生改变\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某团队因发现“剪接体的结构与分子机理”获生命科学奖。研究发现真核细胞中基因\n表达分三步进行, 如下图所示。剪接体主要由蛋白质和小分子的核 RNA 组成。下列叙述正确的是 ( )\n\n[图1]\n\nA: RNA 聚合酶和剪接体的彻底水解产物中均含有碱基 $\\mathrm{T}$ 和碱基 A\nB: 过程(3)中一个核糖体可结合多条 mRNA 链以提高蛋白质合成速率\nC: (2)过程剪接现象的发现是对传统基因表达过程的一个重要补充\nD: 若(2)过程位置出现差错, 最终编码的蛋白质结构不可能发生改变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-29.jpg?height=244&width=1422&top_left_y=352&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_631",
        "problem": "某昆虫的直翅与弯翅受一对等位基因控制, 现将一直翅昆虫与一弯翅昆虫杂交得到 $F_{1}$, 再让 $F_{1}$ 雌雄昆虫随机交配得到 $F_{2}$, 实验结果如表所示, 不考虑交换、突变以及性染色体的同源区段，下列相关叙述错误的是（）\n\n| $\\mathrm{P}$ | O:弯翅 | $\\sigma^{\\lambda}:$ 直翅 |\n| :--- | :--- | :--- |\n| $\\mathrm{F}_{1}$ | O $:$ 直翅 | $\\sigma^{\\lambda}:$ 直翅 |\n| $\\mathrm{F}_{2}$ | O $:$ 直翅:弯翅 $=1: 1$ | $\\delta^{\\lambda}:$ 直翅 |\n\n注: 子代中雌雄个体数相等\nA: 该对相对性状中, 直翅为显性性状\nB: 从 $F_{1}$ 到 $F_{2}$, 昆虫翅型的遗传遵循分离定律\nC: 控制该昆虫翅型的基因位于 $\\mathrm{X}$ 染色体上\nD: 子二代昆虫随机交配, 后代直翅:弯翅 $=13: 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某昆虫的直翅与弯翅受一对等位基因控制, 现将一直翅昆虫与一弯翅昆虫杂交得到 $F_{1}$, 再让 $F_{1}$ 雌雄昆虫随机交配得到 $F_{2}$, 实验结果如表所示, 不考虑交换、突变以及性染色体的同源区段，下列相关叙述错误的是（）\n\n| $\\mathrm{P}$ | O:弯翅 | $\\sigma^{\\lambda}:$ 直翅 |\n| :--- | :--- | :--- |\n| $\\mathrm{F}_{1}$ | O $:$ 直翅 | $\\sigma^{\\lambda}:$ 直翅 |\n| $\\mathrm{F}_{2}$ | O $:$ 直翅:弯翅 $=1: 1$ | $\\delta^{\\lambda}:$ 直翅 |\n\n注: 子代中雌雄个体数相等\n\nA: 该对相对性状中, 直翅为显性性状\nB: 从 $F_{1}$ 到 $F_{2}$, 昆虫翅型的遗传遵循分离定律\nC: 控制该昆虫翅型的基因位于 $\\mathrm{X}$ 染色体上\nD: 子二代昆虫随机交配, 后代直翅:弯翅 $=13: 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_488",
        "problem": "将抗冻基因 $\\mathrm{A}$ 导入西红柿细胞, 得到抗冻西红柿植株, 不含 $\\mathrm{A}$ 的染色体可看作 $\\mathrm{a}$ 。受 $\\mathrm{A}$ 基因的影响, 杂合子抗冻西红柿植株可正常产生雌配子, 但产生雄配子时, 一半含基因 $\\mathrm{a}$ 的配子死亡, 其他基因型植株产生配子时正常。现将某转基因西红柿杂合子植株自交得到 $F_{1}$, 再分别让 $F_{1}$ 自交和自由交配各自得到 $F_{2}$. 下列相关叙述错误的是 ( )\nA: 杂合子西红柿亲本产生的雌配子 $\\mathrm{A}: \\mathrm{a}=1: 1$\nB: $F_{1}$ 的基因型及比例为 $A A$ ：Aa： $a a=2: 3: 1$\nC: $F_{1}$ 自交得到的 $F_{2}$ 的性状分离比为 $3: 1$\nD: $F_{1}$ 进行随机受粉获得的 $F_{2}$ 中基因型为 aa 的植株所占比例为 $5 / 12$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将抗冻基因 $\\mathrm{A}$ 导入西红柿细胞, 得到抗冻西红柿植株, 不含 $\\mathrm{A}$ 的染色体可看作 $\\mathrm{a}$ 。受 $\\mathrm{A}$ 基因的影响, 杂合子抗冻西红柿植株可正常产生雌配子, 但产生雄配子时, 一半含基因 $\\mathrm{a}$ 的配子死亡, 其他基因型植株产生配子时正常。现将某转基因西红柿杂合子植株自交得到 $F_{1}$, 再分别让 $F_{1}$ 自交和自由交配各自得到 $F_{2}$. 下列相关叙述错误的是 ( )\n\nA: 杂合子西红柿亲本产生的雌配子 $\\mathrm{A}: \\mathrm{a}=1: 1$\nB: $F_{1}$ 的基因型及比例为 $A A$ ：Aa： $a a=2: 3: 1$\nC: $F_{1}$ 自交得到的 $F_{2}$ 的性状分离比为 $3: 1$\nD: $F_{1}$ 进行随机受粉获得的 $F_{2}$ 中基因型为 aa 的植株所占比例为 $5 / 12$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1397",
        "problem": "There are three types of muscle contraction, see diagram below. All these muscle contractions occur against a load. Concentric contraction occurs when a muscle shortens against a load. Eccentric contraction occurs when muscle lengthens against a load. Finally, isometric contraction occurs when muscle does not shorten or lengthen against a load.\n\n[figure1]\n\nCONCENTRIC (shortening)\n\n[figure2]\n\nECCENTRIC (lengthening)\n[figure3]\n\nISOMETRIC (no movement)\n\nWithin a muscle, there are two types of receptors: muscle spindles and golgi tendon organs, see diagram below. Muscle spindles are found within the belly of a muscle. These are activated when the muscle stretches. Golgi tendon organs are found in the tendon of a muscle. A tendon attaches muscle to bone. Golgi tendon organs are activated when the muscle contracts and shortens. This information is sent to the central nervous system via sensory neurons.\n\n[figure4]\n\nAfter drinking coffee, a person slowly lowers a mug down to the table. What happens to sensory neuron activity during this action?\nA: Muscle spindle nerve activity increase, Golgi tendon organ nerve activity decrease\nB: Muscle spindle nerve activity decrease, Golgi tendon organ nerve activity decrease\nC: Muscle spindle nerve activity increase, Golgi tendon organ nerve activity increase\nD: Muscle spindle nerve activity decrease, Golgi tendon organ nerve activity increase\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThere are three types of muscle contraction, see diagram below. All these muscle contractions occur against a load. Concentric contraction occurs when a muscle shortens against a load. Eccentric contraction occurs when muscle lengthens against a load. Finally, isometric contraction occurs when muscle does not shorten or lengthen against a load.\n\n[figure1]\n\nCONCENTRIC (shortening)\n\n[figure2]\n\nECCENTRIC (lengthening)\n[figure3]\n\nISOMETRIC (no movement)\n\nWithin a muscle, there are two types of receptors: muscle spindles and golgi tendon organs, see diagram below. Muscle spindles are found within the belly of a muscle. These are activated when the muscle stretches. Golgi tendon organs are found in the tendon of a muscle. A tendon attaches muscle to bone. Golgi tendon organs are activated when the muscle contracts and shortens. This information is sent to the central nervous system via sensory neurons.\n\n[figure4]\n\nAfter drinking coffee, a person slowly lowers a mug down to the table. What happens to sensory neuron activity during this action?\n\nA: Muscle spindle nerve activity increase, Golgi tendon organ nerve activity decrease\nB: Muscle spindle nerve activity decrease, Golgi tendon organ nerve activity decrease\nC: Muscle spindle nerve activity increase, Golgi tendon organ nerve activity increase\nD: Muscle spindle nerve activity decrease, Golgi tendon organ nerve activity increase\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-21.jpg?height=300&width=177&top_left_y=455&top_left_x=288",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-21.jpg?height=326&width=394&top_left_y=454&top_left_x=820",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-21.jpg?height=292&width=410&top_left_y=476&top_left_x=1369",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-21.jpg?height=985&width=1353&top_left_y=1501&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_589",
        "problem": "关于图甲、乙、丙的说法, 错误的是（ ）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n(8)\n\n(5)\n\n[图3]\n\n丙\nA: 图甲所示过程为转录, 其产物中三个相邻的碱基就是密码子\nB: 若图甲的(1)中 A 占 $23 \\% 、 U$ 占 $25 \\%$, 则对应 DNA 片段中 A 占 $24 \\%$\nC: 图乙所示过程相当于图丙的(9)过程, 所需原料是氨基酸\nD: 正常情况下, 图丙中在动、植物细胞中都不可能发生的是(7)(8)过程\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n关于图甲、乙、丙的说法, 错误的是（ ）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n(8)\n\n(5)\n\n[图3]\n\n丙\n\nA: 图甲所示过程为转录, 其产物中三个相邻的碱基就是密码子\nB: 若图甲的(1)中 A 占 $23 \\% 、 U$ 占 $25 \\%$, 则对应 DNA 片段中 A 占 $24 \\%$\nC: 图乙所示过程相当于图丙的(9)过程, 所需原料是氨基酸\nD: 正常情况下, 图丙中在动、植物细胞中都不可能发生的是(7)(8)过程\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-44.jpg?height=103&width=466&top_left_y=1967&top_left_x=475",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-44.jpg?height=300&width=348&top_left_y=1803&top_left_x=954",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-44.jpg?height=148&width=914&top_left_y=2179&top_left_x=408"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1009",
        "problem": "What provides the best evidence for the independent origin of microphylls and megaphylls?\nA: Microphylls have a single unbranched vein, while megaphylls have a branching vascular system\nB: Microphylls have a branching vascular system, while megaphylls have a single unbranched vein\nC: Microphylls are small, while megaphylls are larger\nD: Both are vascularized\nE: Both appear as lateral appendages\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat provides the best evidence for the independent origin of microphylls and megaphylls?\n\nA: Microphylls have a single unbranched vein, while megaphylls have a branching vascular system\nB: Microphylls have a branching vascular system, while megaphylls have a single unbranched vein\nC: Microphylls are small, while megaphylls are larger\nD: Both are vascularized\nE: Both appear as lateral appendages\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1300",
        "problem": "The diagrams show vertical sections of an ovule and of a pollen grain from the same species of flowering plant.\n\n[figure1]\n\nWhich one of the following combinations, A-E, represents the structure which, after fertilisation, becomes the endosperm?\nA: $\\mathrm{Z}+\\mathrm{Q}$\nB: $\\mathrm{Z}+\\mathrm{S}$\nC: $\\mathrm{Y}+\\mathrm{S}$\nD: $\\mathrm{P}+\\mathrm{Y}$\nE: $\\mathrm{Y}+\\mathrm{Q}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagrams show vertical sections of an ovule and of a pollen grain from the same species of flowering plant.\n\n[figure1]\n\nWhich one of the following combinations, A-E, represents the structure which, after fertilisation, becomes the endosperm?\n\nA: $\\mathrm{Z}+\\mathrm{Q}$\nB: $\\mathrm{Z}+\\mathrm{S}$\nC: $\\mathrm{Y}+\\mathrm{S}$\nD: $\\mathrm{P}+\\mathrm{Y}$\nE: $\\mathrm{Y}+\\mathrm{Q}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-11.jpg?height=485&width=786&top_left_y=842&top_left_x=667"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_988",
        "problem": "John, 45-years-old, ran his first marathon in Denver, Colorado, with a time of 3 hours and 43 minutes. Running hard at Mile-23, which statement is true?\nA: John's abdominal muscles increased his intra-abdominal pressure, which is forcing his abdominal viscera upward against his diaphragm. His intrapulmonary pressure at Mile-23 may have decreased by $80 \\mathrm{mmHg}$ since the beginning of the marathon\nB: John's abdominal muscles decreased his intra-abdominal pressure, which is forcing his abdominal viscera upward against his diaphragm. His intrapulmonary pressure at Mile-23 may have increased by $80 \\mathrm{mmHg}$ since the beginning of the marathon.\nC: John's abdominal muscles increased his intra-abdominal pressure, which is forcing his abdominal viscera upward against his diaphragm. His intrapulmonary pressure may have increased by $80 \\mathrm{mmHg}$ since the beginning of the marathon\nD: John's abdominal muscles decreased his intra-abdominal pressure, which is forcing his abdominal viscera upward against his diaphragm. His intrapulmonary pressure may have decreased by $80 \\mathrm{mmHg}$ since the beginning of the marathon\nE: John's abdominal muscles decreased his intra-abdominal pressure, which is forcing his abdominal viscera upward against his diaphragm. His intrapulmonary pressure stayed the same since the beginning of the marathon\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nJohn, 45-years-old, ran his first marathon in Denver, Colorado, with a time of 3 hours and 43 minutes. Running hard at Mile-23, which statement is true?\n\nA: John's abdominal muscles increased his intra-abdominal pressure, which is forcing his abdominal viscera upward against his diaphragm. His intrapulmonary pressure at Mile-23 may have decreased by $80 \\mathrm{mmHg}$ since the beginning of the marathon\nB: John's abdominal muscles decreased his intra-abdominal pressure, which is forcing his abdominal viscera upward against his diaphragm. His intrapulmonary pressure at Mile-23 may have increased by $80 \\mathrm{mmHg}$ since the beginning of the marathon.\nC: John's abdominal muscles increased his intra-abdominal pressure, which is forcing his abdominal viscera upward against his diaphragm. His intrapulmonary pressure may have increased by $80 \\mathrm{mmHg}$ since the beginning of the marathon\nD: John's abdominal muscles decreased his intra-abdominal pressure, which is forcing his abdominal viscera upward against his diaphragm. His intrapulmonary pressure may have decreased by $80 \\mathrm{mmHg}$ since the beginning of the marathon\nE: John's abdominal muscles decreased his intra-abdominal pressure, which is forcing his abdominal viscera upward against his diaphragm. His intrapulmonary pressure stayed the same since the beginning of the marathon\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1384",
        "problem": "The three recognised colours of Labradors are black, chocolate and golden. These coat colours result from differences in two genetic loci which have alleles B/b and E/e respectively.\n\nSupplied is a simplified flowchart representing pigment deposition in labradors. $\\mathrm{E}$ is the allele for functional MRC1 protein, while e is a mutated allele resulting in no MRC1 protein. B is the allele for functional TYRP1 enzyme, while $b$ is a mutated allele resulting in no TYRP1 enzyme activity. If there is no pigment deposition, the labrador is considered golden.\n\n[figure1]\n\nWhen a golden female Labrador was mated with a chocolate male, half of the puppies were chocolate and half were golden. What is the genotype of this golden female?\nA: BBee\nB: Bbee\nC: bbee\nD: bbEe\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe three recognised colours of Labradors are black, chocolate and golden. These coat colours result from differences in two genetic loci which have alleles B/b and E/e respectively.\n\nSupplied is a simplified flowchart representing pigment deposition in labradors. $\\mathrm{E}$ is the allele for functional MRC1 protein, while e is a mutated allele resulting in no MRC1 protein. B is the allele for functional TYRP1 enzyme, while $b$ is a mutated allele resulting in no TYRP1 enzyme activity. If there is no pigment deposition, the labrador is considered golden.\n\n[figure1]\n\nWhen a golden female Labrador was mated with a chocolate male, half of the puppies were chocolate and half were golden. What is the genotype of this golden female?\n\nA: BBee\nB: Bbee\nC: bbee\nD: bbEe\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-19.jpg?height=780&width=926&top_left_y=1312&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_848",
        "problem": "鹦武鸟的性别决定为 ZW 型, 其毛色由两对等位基因决定, 其中一对位于性染色体上,决定机制如图。让一只绿色雌性鹦武鸟（甲）先后与乙、丙杂交，结果如表所示。有关分析正确的是\n\n## 基因 A（Z染色体上）\n\n[图1]\n白色物质 $\\longrightarrow$ 酶B 黄色物质}\n\n## 基因 B（常染色体上）\n\n| 组别 | 亲代 | 子代 |  |\n| :--- | :--- | :--- | :--- |\n|  |  | 雌 | 雄 |\n| 第一组 | 甲 $\\times$ 乙 | 黄色 $\\frac{3}{8}$, 白 | 绿色 $\\frac{3}{8}$, 蓝 |\n\n\n|  |  | 色 $\\frac{1}{8}$ | 色 $\\frac{1}{8}$ |\n| :--- | :--- | :--- | :--- |\n| 第二组 | 甲×丙 | 绿色 $\\frac{1}{4}$, 蓝 | 绿色 $\\frac{1}{4}$, 蓝 |\n|  |  | 色 $\\frac{1}{4}$ | 色 $\\frac{1}{4}$ |\nA: 鹦鞄乙、鹦武丙的基因型分别是 $B b Z^{A} Z^{A}$ 和 $b b Z^{a} Z^{a}$\nB: 在自然条件下, 白色雄鹦鹉的比例大于白色雌鹦鹉\nC: 第一组子代绿色雄鹦武鸟和第二组子代绿色雌鹦鹉杂交, 后代出现绿色鹦武的概率为 $5 / 8$\nD: 欲通过两纯合鹦㘱鸟杂交得到雄性全为绿色、雌性全为黄色的子代, 则亲代基因组合有两种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n鹦武鸟的性别决定为 ZW 型, 其毛色由两对等位基因决定, 其中一对位于性染色体上,决定机制如图。让一只绿色雌性鹦武鸟（甲）先后与乙、丙杂交，结果如表所示。有关分析正确的是\n\n## 基因 A（Z染色体上）\n\n[图1]\n白色物质 $\\longrightarrow$ 酶B 黄色物质}\n\n## 基因 B（常染色体上）\n\n| 组别 | 亲代 | 子代 |  |\n| :--- | :--- | :--- | :--- |\n|  |  | 雌 | 雄 |\n| 第一组 | 甲 $\\times$ 乙 | 黄色 $\\frac{3}{8}$, 白 | 绿色 $\\frac{3}{8}$, 蓝 |\n\n\n|  |  | 色 $\\frac{1}{8}$ | 色 $\\frac{1}{8}$ |\n| :--- | :--- | :--- | :--- |\n| 第二组 | 甲×丙 | 绿色 $\\frac{1}{4}$, 蓝 | 绿色 $\\frac{1}{4}$, 蓝 |\n|  |  | 色 $\\frac{1}{4}$ | 色 $\\frac{1}{4}$ |\n\nA: 鹦鞄乙、鹦武丙的基因型分别是 $B b Z^{A} Z^{A}$ 和 $b b Z^{a} Z^{a}$\nB: 在自然条件下, 白色雄鹦鹉的比例大于白色雌鹦鹉\nC: 第一组子代绿色雄鹦武鸟和第二组子代绿色雌鹦鹉杂交, 后代出现绿色鹦武的概率为 $5 / 8$\nD: 欲通过两纯合鹦㘱鸟杂交得到雄性全为绿色、雌性全为黄色的子代, 则亲代基因组合有两种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-021.jpg?height=144&width=998&top_left_y=1892&top_left_x=354"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_787",
        "problem": "某二倍体高等动物 $(2 n=4)$ 的基因型为 $\\mathrm{AaBb}$, 现将其一个精原细胞的 DNA 都用放射性同位素 ${ }^{32} \\mathrm{P}$ 标记并放入含 ${ }^{31} \\mathrm{P}$ 的培养基中培养并分裂, 其中某个子细胞染色体及其基因位置如图所示，且形成该细胞的过程中只发生了一次遗传物质的异常变化。下列叙述错误的是（ ）\n\n[图1]\nA: 若只有 1 条染色体有放射性, 说明形成该细胞过程中发生了基因突变\nB: 若只有 2 条染色体有放射性, 说明该精原细胞在减数分裂前只进行了一次有丝分裂\nC: 若只有 3 条染色体有放射性, 说明形成该细胞过程中发生了染色体互换\nD: 若 4 条染色体均有放射性, 说明该精原细胞经过一次有丝分裂后进行减数分裂。\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某二倍体高等动物 $(2 n=4)$ 的基因型为 $\\mathrm{AaBb}$, 现将其一个精原细胞的 DNA 都用放射性同位素 ${ }^{32} \\mathrm{P}$ 标记并放入含 ${ }^{31} \\mathrm{P}$ 的培养基中培养并分裂, 其中某个子细胞染色体及其基因位置如图所示，且形成该细胞的过程中只发生了一次遗传物质的异常变化。下列叙述错误的是（ ）\n\n[图1]\n\nA: 若只有 1 条染色体有放射性, 说明形成该细胞过程中发生了基因突变\nB: 若只有 2 条染色体有放射性, 说明该精原细胞在减数分裂前只进行了一次有丝分裂\nC: 若只有 3 条染色体有放射性, 说明形成该细胞过程中发生了染色体互换\nD: 若 4 条染色体均有放射性, 说明该精原细胞经过一次有丝分裂后进行减数分裂。\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-076.jpg?height=429&width=303&top_left_y=2267&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1305",
        "problem": "## RATITE EVOLUTION\n\nRatites are named for their flat \"raft-like\" sternum that lacks a keel and thus cannot provide sufficient anchorage for flight muscles to effectively generate the power required to fly. Cassowary (Australia and Papua New Guinea), Emu (Australia), Rhea (South America), Ostrich (Africa), Kiwi (New Zealand), and extinct Moa (New Zealand) are all examples of ground dwelling, flightless ratites. Given their large, flightless nature, biologists have proposed that they have all descended from a common ancestor present in ancient Gondwana before it broke apart. The exact timing and evolutionary origin of the Moa and Kiwi in New Zealand have been greatly debated in the scientific literature.\n\n[figure1]\nfrom Antarctica and India\n\n- 130 mya South America breaks free of Africa\n- 80 mya Zealandia breaks free\n- 60 mya New Zealand separates from Australia\n- 40 mya Australia separates from Antarctica\n- 30 mya South America breaks from West Antarctica (Antarctica freezes over)\n\n[figure2]\n\nFuture Zealandia\n\n## Flying Cousins\n\nRatites are one of only two groups of birds belonging to the \"old jaw\" Paleognaths originating in Gondwana, the other comprises the 47 living Tinamou species of Central and South America. Tinamou species are generally ground dwelling, though they do have wings that allow for limited flight.\n\nIt is generally accepted that loss of flight in birds is due to the development of successful foraging behaviours and diminished predation on eggs and nests.\n\n## Fossil Evidence\n\nExamination of the oldest known Moa fossils, dating from 19 million years ago, determined they had thickened leg bones and no wing structures at all. At least two species of flightless Moa were present in New Zealand at this time.\n\nThe oldest Tinamou fossils in South America are 10 million years old.\n\nThe oldest Kiwi fossil dates back 1 million years. Though diminished in size, kiwis have wings and flight feathers. Kiwis are the only known bird to have nostrils at the end of their beaks and the only ratite with two functioning ovaries.Proportional to their body size, Kiwi have the largest egg of all the ratites. Which of the following inferences about Kiwi is LEAST likely to be true?\nA: Kiwi have evolved from a much larger species.\nB: Having two ovaries allows them to produce larger eggs.\nC: Kiwi have been flightless for a very long time.\nD: Larger eggs allow for better survival of Kiwi chicks.\nE: Larger eggs evolved after flightlessness.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## RATITE EVOLUTION\n\nRatites are named for their flat \"raft-like\" sternum that lacks a keel and thus cannot provide sufficient anchorage for flight muscles to effectively generate the power required to fly. Cassowary (Australia and Papua New Guinea), Emu (Australia), Rhea (South America), Ostrich (Africa), Kiwi (New Zealand), and extinct Moa (New Zealand) are all examples of ground dwelling, flightless ratites. Given their large, flightless nature, biologists have proposed that they have all descended from a common ancestor present in ancient Gondwana before it broke apart. The exact timing and evolutionary origin of the Moa and Kiwi in New Zealand have been greatly debated in the scientific literature.\n\n[figure1]\nfrom Antarctica and India\n\n- 130 mya South America breaks free of Africa\n- 80 mya Zealandia breaks free\n- 60 mya New Zealand separates from Australia\n- 40 mya Australia separates from Antarctica\n- 30 mya South America breaks from West Antarctica (Antarctica freezes over)\n\n[figure2]\n\nFuture Zealandia\n\n## Flying Cousins\n\nRatites are one of only two groups of birds belonging to the \"old jaw\" Paleognaths originating in Gondwana, the other comprises the 47 living Tinamou species of Central and South America. Tinamou species are generally ground dwelling, though they do have wings that allow for limited flight.\n\nIt is generally accepted that loss of flight in birds is due to the development of successful foraging behaviours and diminished predation on eggs and nests.\n\n## Fossil Evidence\n\nExamination of the oldest known Moa fossils, dating from 19 million years ago, determined they had thickened leg bones and no wing structures at all. At least two species of flightless Moa were present in New Zealand at this time.\n\nThe oldest Tinamou fossils in South America are 10 million years old.\n\nThe oldest Kiwi fossil dates back 1 million years. Though diminished in size, kiwis have wings and flight feathers. Kiwis are the only known bird to have nostrils at the end of their beaks and the only ratite with two functioning ovaries.\n\nproblem:\nProportional to their body size, Kiwi have the largest egg of all the ratites. Which of the following inferences about Kiwi is LEAST likely to be true?\n\nA: Kiwi have evolved from a much larger species.\nB: Having two ovaries allows them to produce larger eggs.\nC: Kiwi have been flightless for a very long time.\nD: Larger eggs allow for better survival of Kiwi chicks.\nE: Larger eggs evolved after flightlessness.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-24.jpg?height=711&width=585&top_left_y=689&top_left_x=135",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-24.jpg?height=708&width=1259&top_left_y=688&top_left_x=747"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1391",
        "problem": "The absorption spectrum of a photosynthetic pigment and the spectrum of visible light are given below.\n\nO.D.\n[figure1]\n\nWhat is the most likely colour of this photosynthetic pigment?\nA: Green\nB: Red\nC: Violet\nD: Yellow\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe absorption spectrum of a photosynthetic pigment and the spectrum of visible light are given below.\n\nO.D.\n[figure1]\n\nWhat is the most likely colour of this photosynthetic pigment?\n\nA: Green\nB: Red\nC: Violet\nD: Yellow\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-30.jpg?height=1356&width=1376&top_left_y=454&top_left_x=266"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1116",
        "problem": "The total food given to a Chinese hamster under experimental conditions was equivalent to $270 \\mathrm{kcal} / \\mathrm{animal} / \\mathrm{day}$. As per the \"material-balance calculations\", $37 \\%$ of the total consumption is lost as feces. How many calories are assimilated per day in\nconditions of hypothermia that accounts for a further $67 \\%$ decrement in diet consumption? (Write your answer upto one decimal place.)",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe total food given to a Chinese hamster under experimental conditions was equivalent to $270 \\mathrm{kcal} / \\mathrm{animal} / \\mathrm{day}$. As per the \"material-balance calculations\", $37 \\%$ of the total consumption is lost as feces. How many calories are assimilated per day in\nconditions of hypothermia that accounts for a further $67 \\%$ decrement in diet consumption? (Write your answer upto one decimal place.)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of kcal, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "kcal"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_475",
        "problem": "研究发现, 某种植株（性别决定方式为 XY 型）存在性反转现象, 基因 a 纯合时可使雌株发生性反转成为雄株、而雄株不会发生性反转; 皆因 A 无此现象。让纯合雄株与杂合雌株杂交, $F_{1}$ 中雌株:雄株=1:3。不考虑突变, 下列分析错误的是 ( )\nA: 若基因 $A 、 a$ 位于常染色体上, 则 $F_{1}$ 雄株中的纯合子占 $1 / 3$\nB: $F_{1}$ 中发生性反转产生的雄株占 $F_{1}$ 雄株的比例为 $1 / 3$, 且无法判断 $A / a$ 基因的位 置\nC: $F_{1}$ 雄株的基因型有 3 种或 2 种\nD: $F_{1}$ 中雌株的基因型与亲本雌株的基因型均相同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究发现, 某种植株（性别决定方式为 XY 型）存在性反转现象, 基因 a 纯合时可使雌株发生性反转成为雄株、而雄株不会发生性反转; 皆因 A 无此现象。让纯合雄株与杂合雌株杂交, $F_{1}$ 中雌株:雄株=1:3。不考虑突变, 下列分析错误的是 ( )\n\nA: 若基因 $A 、 a$ 位于常染色体上, 则 $F_{1}$ 雄株中的纯合子占 $1 / 3$\nB: $F_{1}$ 中发生性反转产生的雄株占 $F_{1}$ 雄株的比例为 $1 / 3$, 且无法判断 $A / a$ 基因的位 置\nC: $F_{1}$ 雄株的基因型有 3 种或 2 种\nD: $F_{1}$ 中雌株的基因型与亲本雌株的基因型均相同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1554",
        "problem": "Francis Crick stated the central dogma of biology: * There is a one-directional flow of information from the DNA, through RNA, to proteins. Proteins then build and control bodies.\n\nDo these observations follow from the central dogma (showing it is true) or are they exceptions to it (showing it is false)?\n\nWhich of these observations follow from the central dogma (showing it is often true)?\nA: Evolution by inheritance of acquired characteristics (proposed by Lamarck).\nB: Retroviruses like HIV have reverse transcriptase enzymes.\nC: When a mutant allele is correlated with a disease, we are fairly certain the allele causes disease.\nD: Prions are misfolded proteins which cause more proteins to misfold, and can spread between animals.\nE: Women who suffered starvation as children are more likely to have obese children.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFrancis Crick stated the central dogma of biology: * There is a one-directional flow of information from the DNA, through RNA, to proteins. Proteins then build and control bodies.\n\nDo these observations follow from the central dogma (showing it is true) or are they exceptions to it (showing it is false)?\n\nWhich of these observations follow from the central dogma (showing it is often true)?\n\nA: Evolution by inheritance of acquired characteristics (proposed by Lamarck).\nB: Retroviruses like HIV have reverse transcriptase enzymes.\nC: When a mutant allele is correlated with a disease, we are fairly certain the allele causes disease.\nD: Prions are misfolded proteins which cause more proteins to misfold, and can spread between animals.\nE: Women who suffered starvation as children are more likely to have obese children.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_587",
        "problem": "噬菌体的基因组比较小, 但又必须要编码一些维持其生命和复制所必需的基因, 在选择的压力下, 形成了重叠基因。重叠基因是指两个或两个以上的基因共有一段 DNA 序列的不同可读框，编码不同的蛋白质。重叠基因有多种重叠方式，例如，大基因内包含小基因(如图); 前后两个基因首尾重叠; 几个基因的重叠等。近年来，在果蝇和人中也存在重叠基因, 例如人类神经纤维病 I型基因内含子中含有 3 个小基因(由互补链编码)。\n\n[图1]\nA: 这是一种充分利用碱基资源的机制\nB: 基因的重叠可能对基因表达具有调控作用\nC: 重叠基因的转录方向可能不是一致的 D\nD: 同一个基因的编码区中可能存在多个起始密码子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n噬菌体的基因组比较小, 但又必须要编码一些维持其生命和复制所必需的基因, 在选择的压力下, 形成了重叠基因。重叠基因是指两个或两个以上的基因共有一段 DNA 序列的不同可读框，编码不同的蛋白质。重叠基因有多种重叠方式，例如，大基因内包含小基因(如图); 前后两个基因首尾重叠; 几个基因的重叠等。近年来，在果蝇和人中也存在重叠基因, 例如人类神经纤维病 I型基因内含子中含有 3 个小基因(由互补链编码)。\n\n[图1]\n\nA: 这是一种充分利用碱基资源的机制\nB: 基因的重叠可能对基因表达具有调控作用\nC: 重叠基因的转录方向可能不是一致的 D\nD: 同一个基因的编码区中可能存在多个起始密码子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-067.jpg?height=437&width=1042&top_left_y=227&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_226",
        "problem": "While snapdragon normally has bilateral flowers, flowers of its mutant defective in gene $\\boldsymbol{G}$ lose bilateral symmetry and have radial symmetry, thereby indicating that gene $\\boldsymbol{G}$ confers bilateral symmetry to the flower.\n\nIn the inflorescence of wild-type sunflower, the outer region has ligulate florets, whereas the inner region has tubular florets (Figure 1). Variants $\\boldsymbol{x}, \\boldsymbol{y}$, and $\\boldsymbol{z}$ of sunflower have DNA insertions in gene $\\boldsymbol{G}$, a sunflower orthologue of the gene $\\boldsymbol{G}$ from snapdragon (Figure 2). As a result of these insertions, variant $\\boldsymbol{x}$ has only ligulate florets over the entire inflorescence, and variants $y$ and $z$ have only tubular florets over the entire inflorescence.\n\n[figure1]\n\nLigulate\n\n[figure2]\n\nTubular\n\nFigure 1 Ligulate and tubular florets of sunflower\n\n## Gene $\\boldsymbol{G}$\n\nIntron\n\n[figure3]\n\nFigure 2 DNA insertions in gene $\\boldsymbol{G}^{\\boldsymbol{\\prime}}$ in the sunflower variants\n\nVariant $\\boldsymbol{y}$ has two DNA insertions, while variants $\\boldsymbol{x}$ and $\\boldsymbol{z}$ have one insertion.\nA: In the wild-type sunflower, gene $\\boldsymbol{G}$, is not expressed in the florets that form early during inflorescence development, but is expressed in the florets that form later.\nB: In variant $\\boldsymbol{x}$, expression of gene $\\boldsymbol{G}$ ' is decreased due to the DNA insertion.\nC: Variant $\\boldsymbol{y}$ is a loss-of-function mutant of gene $\\boldsymbol{G}$.\nD: Variant $\\boldsymbol{y}$ is more closely related to variant $\\boldsymbol{x}$ than to variant $\\boldsymbol{z}$ in the lineage of sunflower variants.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nWhile snapdragon normally has bilateral flowers, flowers of its mutant defective in gene $\\boldsymbol{G}$ lose bilateral symmetry and have radial symmetry, thereby indicating that gene $\\boldsymbol{G}$ confers bilateral symmetry to the flower.\n\nIn the inflorescence of wild-type sunflower, the outer region has ligulate florets, whereas the inner region has tubular florets (Figure 1). Variants $\\boldsymbol{x}, \\boldsymbol{y}$, and $\\boldsymbol{z}$ of sunflower have DNA insertions in gene $\\boldsymbol{G}$, a sunflower orthologue of the gene $\\boldsymbol{G}$ from snapdragon (Figure 2). As a result of these insertions, variant $\\boldsymbol{x}$ has only ligulate florets over the entire inflorescence, and variants $y$ and $z$ have only tubular florets over the entire inflorescence.\n\n[figure1]\n\nLigulate\n\n[figure2]\n\nTubular\n\nFigure 1 Ligulate and tubular florets of sunflower\n\n## Gene $\\boldsymbol{G}$\n\nIntron\n\n[figure3]\n\nFigure 2 DNA insertions in gene $\\boldsymbol{G}^{\\boldsymbol{\\prime}}$ in the sunflower variants\n\nVariant $\\boldsymbol{y}$ has two DNA insertions, while variants $\\boldsymbol{x}$ and $\\boldsymbol{z}$ have one insertion.\n\nA: In the wild-type sunflower, gene $\\boldsymbol{G}$, is not expressed in the florets that form early during inflorescence development, but is expressed in the florets that form later.\nB: In variant $\\boldsymbol{x}$, expression of gene $\\boldsymbol{G}$ ' is decreased due to the DNA insertion.\nC: Variant $\\boldsymbol{y}$ is a loss-of-function mutant of gene $\\boldsymbol{G}$.\nD: Variant $\\boldsymbol{y}$ is more closely related to variant $\\boldsymbol{x}$ than to variant $\\boldsymbol{z}$ in the lineage of sunflower variants.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-46.jpg?height=409&width=300&top_left_y=932&top_left_x=775",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-46.jpg?height=228&width=94&top_left_y=1114&top_left_x=1072",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-46.jpg?height=354&width=1263&top_left_y=1625&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_88",
        "problem": "According to the evolutionary concepts, symbiosis is a general characteristic of living organisms and can help to understand complex features and phenotypes. Gut microbiota and host evolve in symbiosis and developed an integrative circuitry essential for their survival. This complex relationship resulted in a personal evolutionary adapted ecosystem. Recently developed genomics and metagenomics approaches helped to discover the link between the gut microbiota alteration with the host health and disease.\n\nGagnière and his colleagues (2016) highlighted the unexpected role of human gut microbiota in colorectal cancer. More recently in a study conducted by Flemer and his colleagues (2018) it has been emphasized that the heterogeneity of colorectal cancer might be related to the types of gut microbiota that either predispose or offer resistance to the disease.\n\n[figure1]\n\nBoxplots show the relative abundances of three bacterial groups from individuals with colorectal cancer (CRC), individuals with polyps (the intermediate stage between healthy and cancerous tissue which does not always turning to CRC) and healthy control (HC). An asterisk represents the significant $p$ value which represented as ${ }^{* *} p$ value $<0.001$; ${ }^{* *} p$ value $<0.01 ;{ }^{*} p$ value $<0.05 ; p$ value $>0.05$.\nA: The microbiota compositional differences between patients with CRC and the control are secondary to the onset of cancer.\nB: Changes in the abundance of the bacteria group 2 compared to the control are restricted to polyp.\nC: Any bacterial group that populates the gut is in association with an increased risk of CRC.\nD: The polyp-associated microbiota can be used as an indicator for individuals with higher risk of developing CRC.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAccording to the evolutionary concepts, symbiosis is a general characteristic of living organisms and can help to understand complex features and phenotypes. Gut microbiota and host evolve in symbiosis and developed an integrative circuitry essential for their survival. This complex relationship resulted in a personal evolutionary adapted ecosystem. Recently developed genomics and metagenomics approaches helped to discover the link between the gut microbiota alteration with the host health and disease.\n\nGagnière and his colleagues (2016) highlighted the unexpected role of human gut microbiota in colorectal cancer. More recently in a study conducted by Flemer and his colleagues (2018) it has been emphasized that the heterogeneity of colorectal cancer might be related to the types of gut microbiota that either predispose or offer resistance to the disease.\n\n[figure1]\n\nBoxplots show the relative abundances of three bacterial groups from individuals with colorectal cancer (CRC), individuals with polyps (the intermediate stage between healthy and cancerous tissue which does not always turning to CRC) and healthy control (HC). An asterisk represents the significant $p$ value which represented as ${ }^{* *} p$ value $<0.001$; ${ }^{* *} p$ value $<0.01 ;{ }^{*} p$ value $<0.05 ; p$ value $>0.05$.\n\nA: The microbiota compositional differences between patients with CRC and the control are secondary to the onset of cancer.\nB: Changes in the abundance of the bacteria group 2 compared to the control are restricted to polyp.\nC: Any bacterial group that populates the gut is in association with an increased risk of CRC.\nD: The polyp-associated microbiota can be used as an indicator for individuals with higher risk of developing CRC.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-45.jpg?height=840&width=1242&top_left_y=868&top_left_x=407"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_921",
        "problem": "纯合亲本白眼长翅和红眼残翅果蝇进行杂交, 结果如图。 $\\mathrm{F}_{2}$ 中每种表型都有雌、雄个体。根据杂交结果，下列推测错误的是（）\n\n\\section*{$P \\quad$ 白眼长翅(우) $X$ 红眼残翅( $\\left.0^{*}\\right)$ <br> $F_{1} \\quad$ 红眼长翅(우) 白眼长翅 $\\left(O^{*}\\right)$ $\\mathrm{F}_{1}$ 雌、雄交配\n\n[图1]\nA: 根据 $\\mathrm{F}_{1}$ 结果可推测长翅对残翅为显性\nB: 控制红眼和白眼的基因位于 X 染色体上\nC: $F_{2}$ 中白眼长翅果蝇有 2 种基因型\nD: $F_{2}$ 中红眼残翅果蝇相互交配, 雄性子代中红眼残翅: 白眼残翅 $=1: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n纯合亲本白眼长翅和红眼残翅果蝇进行杂交, 结果如图。 $\\mathrm{F}_{2}$ 中每种表型都有雌、雄个体。根据杂交结果，下列推测错误的是（）\n\n\\section*{$P \\quad$ 白眼长翅(우) $X$ 红眼残翅( $\\left.0^{*}\\right)$ <br> $F_{1} \\quad$ 红眼长翅(우) 白眼长翅 $\\left(O^{*}\\right)$ $\\mathrm{F}_{1}$ 雌、雄交配\n\n[图1]\n\nA: 根据 $\\mathrm{F}_{1}$ 结果可推测长翅对残翅为显性\nB: 控制红眼和白眼的基因位于 X 染色体上\nC: $F_{2}$ 中白眼长翅果蝇有 2 种基因型\nD: $F_{2}$ 中红眼残翅果蝇相互交配, 雄性子代中红眼残翅: 白眼残翅 $=1: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-94.jpg?height=146&width=1197&top_left_y=1663&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_259",
        "problem": "Pulse- chase experiments performed in cells are experiments in which cells are first exposed to labelled precursors of specific molecules for a short period of time (the pulse), unincorporated labelled precursors are then washed away, and presence of label in molecules of interest are followed through time (the chase). In an experiment designed to study gene expression, cells were exposed to labelled UTP during the pulse, and results of the chase part of the experiment are summarized in the figure below:\n\n[figure1]\nA: Based on the data presented, most ( $\\sim 90 \\%$ of mass) of RNA that is synthesized in the nucleus is degraded in the nucleus without ever entering the cytoplasm.\nB: Based on data presented, the complexity of RNA (which refers to number of different sequences) in the nucleus is higher than in the cytoplasm.\nC: Assuming that introns constitute $60 \\%$ of primary transcripts, splicing can account for the observed difference in amount of label in the nucleus at start of\nD: It is expected that chases of longer than 8 hours would have ultimately shown a much higher amount of label in the cytoplasm than seen at 8 hours.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPulse- chase experiments performed in cells are experiments in which cells are first exposed to labelled precursors of specific molecules for a short period of time (the pulse), unincorporated labelled precursors are then washed away, and presence of label in molecules of interest are followed through time (the chase). In an experiment designed to study gene expression, cells were exposed to labelled UTP during the pulse, and results of the chase part of the experiment are summarized in the figure below:\n\n[figure1]\n\nA: Based on the data presented, most ( $\\sim 90 \\%$ of mass) of RNA that is synthesized in the nucleus is degraded in the nucleus without ever entering the cytoplasm.\nB: Based on data presented, the complexity of RNA (which refers to number of different sequences) in the nucleus is higher than in the cytoplasm.\nC: Assuming that introns constitute $60 \\%$ of primary transcripts, splicing can account for the observed difference in amount of label in the nucleus at start of\nD: It is expected that chases of longer than 8 hours would have ultimately shown a much higher amount of label in the cytoplasm than seen at 8 hours.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-06.jpg?height=693&width=868&top_left_y=670&top_left_x=594"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1104",
        "problem": "Birds can sustain high levels of activity much longer than mammals even at high altitudes. Which of the following features found in birds helps in this?\nA: More efficient expansion and contraction of air sacs as compared to mammals.\nB: Size of lungs larger than that of a mammal of comparable size.\nC: Air sacs allowing unidirectional air flow with minimum dead space.\nD: Circulatory system with more number of erythrocytes per unit volume of blood as compared to mammals.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBirds can sustain high levels of activity much longer than mammals even at high altitudes. Which of the following features found in birds helps in this?\n\nA: More efficient expansion and contraction of air sacs as compared to mammals.\nB: Size of lungs larger than that of a mammal of comparable size.\nC: Air sacs allowing unidirectional air flow with minimum dead space.\nD: Circulatory system with more number of erythrocytes per unit volume of blood as compared to mammals.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_780",
        "problem": "将全部 DNA 分子双链经 ${ }^{32} \\mathrm{P}$ 标记的雄性动物细胞（染色体数为 $2 \\mathrm{~N}$ ) 置于不含 ${ }^{32} \\mathrm{P}$ 的培养基中培养。经过连续 3 次细胞分裂后产生 8 个子细胞，检测子细胞中的情况。下列推断正确的是（ ）\nA: 若只进行有丝分裂, 则含 ${ }^{32} \\mathrm{P}$ 染色体的子细胞比例一定为 $1 / 2$\nB: 若进行一次有丝分裂再进行一次减数分裂, 则含 ${ }^{32} \\mathrm{P}$ 染色体的子细胞比例至少占 $1 / 2$\nC: 若子细胞中的染色体都含 ${ }^{32} \\mathrm{P}$, 则一定进行有丝分裂\nD: 若子细胞中的染色体都不含 ${ }^{32} \\mathrm{P}$, 则一定进行减数分裂\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将全部 DNA 分子双链经 ${ }^{32} \\mathrm{P}$ 标记的雄性动物细胞（染色体数为 $2 \\mathrm{~N}$ ) 置于不含 ${ }^{32} \\mathrm{P}$ 的培养基中培养。经过连续 3 次细胞分裂后产生 8 个子细胞，检测子细胞中的情况。下列推断正确的是（ ）\n\nA: 若只进行有丝分裂, 则含 ${ }^{32} \\mathrm{P}$ 染色体的子细胞比例一定为 $1 / 2$\nB: 若进行一次有丝分裂再进行一次减数分裂, 则含 ${ }^{32} \\mathrm{P}$ 染色体的子细胞比例至少占 $1 / 2$\nC: 若子细胞中的染色体都含 ${ }^{32} \\mathrm{P}$, 则一定进行有丝分裂\nD: 若子细胞中的染色体都不含 ${ }^{32} \\mathrm{P}$, 则一定进行减数分裂\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1080",
        "problem": "Elaborate pathway of human blood clotting is shown. A pathway very similar to that indicated by the box $\\mathrm{X}$ is found in lobster.\n\n[figure1]\n\nWhich of the following is correct?\nA: Multistep pathway found in humans would have advantage as it would greatly amplify the original signal.\nB: Clotting of blood is an essential requirement only for a closed circulatory system as the blood flows through vessels \\& capillaries.\nC: Multistep blood clotting mechanism is likely to evolve in high volume high pressure circulatory systems than a low volume low pressure systems.\nD: The hypothesis that blood clotting evolved in vertebrate lineage is supported by the presence of fibrinogen-like protein in early vertebrates that do not possess clotting pathway.\nE: Lesser the steps required for clotting, longer will it take to produce substantial clot.\nF: If the clotting factors show extensive homology, the hypothesis that they are formed by gene duplication is supported.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nElaborate pathway of human blood clotting is shown. A pathway very similar to that indicated by the box $\\mathrm{X}$ is found in lobster.\n\n[figure1]\n\nWhich of the following is correct?\n\nA: Multistep pathway found in humans would have advantage as it would greatly amplify the original signal.\nB: Clotting of blood is an essential requirement only for a closed circulatory system as the blood flows through vessels \\& capillaries.\nC: Multistep blood clotting mechanism is likely to evolve in high volume high pressure circulatory systems than a low volume low pressure systems.\nD: The hypothesis that blood clotting evolved in vertebrate lineage is supported by the presence of fibrinogen-like protein in early vertebrates that do not possess clotting pathway.\nE: Lesser the steps required for clotting, longer will it take to produce substantial clot.\nF: If the clotting factors show extensive homology, the hypothesis that they are formed by gene duplication is supported.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E, F].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-21.jpg?height=998&width=1203&top_left_y=257&top_left_x=575"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1547",
        "problem": "The 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nHere are some observations:\n\n1. The average speed with which allele frequency changes\n2. Some alleles takeover very large populations even when they have impacts too small to measure in laboratories\n3. The probability an allele will takeover a population\n4. The speed an allele's frequency changes in any one population at any one time\n5. The frequency at which an allele is initially introduced alters the probability it will takeover a population\n6. More fit alleles sometimes go extinct before less fit alleles\n\n\nEvolution is often thought to occur slowly, although we have seen this is not always true. When will 'slow and steady' evolution over thousands of generations dominate?\nA: Evolving adaptations with slight benefit in very large populations.\nB: Evolving adaptations to resist pathogens.\nC: Evolving adaptations when populations are regularly separated for long periods of time.\nD: Evolving adaptations when there is only one selective pressure on a trait.\nE: Evolving non-adaptive traits due to genetic drift.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nHere are some observations:\n\n1. The average speed with which allele frequency changes\n2. Some alleles takeover very large populations even when they have impacts too small to measure in laboratories\n3. The probability an allele will takeover a population\n4. The speed an allele's frequency changes in any one population at any one time\n5. The frequency at which an allele is initially introduced alters the probability it will takeover a population\n6. More fit alleles sometimes go extinct before less fit alleles\n\n\nEvolution is often thought to occur slowly, although we have seen this is not always true. When will 'slow and steady' evolution over thousands of generations dominate?\n\nA: Evolving adaptations with slight benefit in very large populations.\nB: Evolving adaptations to resist pathogens.\nC: Evolving adaptations when populations are regularly separated for long periods of time.\nD: Evolving adaptations when there is only one selective pressure on a trait.\nE: Evolving non-adaptive traits due to genetic drift.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=970&width=704&top_left_y=778&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=966&width=706&top_left_y=778&top_left_x=1118"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_952",
        "problem": "某雌雄异体二倍体动物 $(2 n=48)$, 雌性个体体细胞有两条相同的性染色体 $\\mathrm{X}$, 雄性个体体细胞仅有一条性染色体 $\\mathrm{X}$ 。下列关于基因型为 $\\mathrm{AaX} \\mathrm{X}^{\\mathrm{B}}$ 的雄性个体进行有丝分裂和减数分裂过程的叙述, 错误的是 ( )\nA: 处于有丝分裂后期的精原细胞有 46 对同源染色体、 2 条性染色体\nB: 处于减数第一次分裂后期的初级精母细胞有 2 种基因型\nC: 次级精母细胞中含有的性染色体数可为 0 条、 1 条或 2 条\nD: 该雄性个体减数分裂可能产生基因型为 $\\mathrm{A} 、 \\mathrm{a} 、 \\mathrm{AX}^{\\mathrm{B}} 、 \\mathrm{aX}^{\\mathrm{B}}$ 的 4 种精子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某雌雄异体二倍体动物 $(2 n=48)$, 雌性个体体细胞有两条相同的性染色体 $\\mathrm{X}$, 雄性个体体细胞仅有一条性染色体 $\\mathrm{X}$ 。下列关于基因型为 $\\mathrm{AaX} \\mathrm{X}^{\\mathrm{B}}$ 的雄性个体进行有丝分裂和减数分裂过程的叙述, 错误的是 ( )\n\nA: 处于有丝分裂后期的精原细胞有 46 对同源染色体、 2 条性染色体\nB: 处于减数第一次分裂后期的初级精母细胞有 2 种基因型\nC: 次级精母细胞中含有的性染色体数可为 0 条、 1 条或 2 条\nD: 该雄性个体减数分裂可能产生基因型为 $\\mathrm{A} 、 \\mathrm{a} 、 \\mathrm{AX}^{\\mathrm{B}} 、 \\mathrm{aX}^{\\mathrm{B}}$ 的 4 种精子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_51",
        "problem": "The following figures indicate changes in phosphate concentration as the filtrate passes through regions $a$ and $b$, according to the increase in plasma phosphate concentration.\n[figure1]\n\nUsing this information, choose the most appropriate graph that depicts the changes in the renal reabsorption rate of phosphate ions according to the increase of its concentration in the plasma.\nA: [figure2]\nB: [figure3]\nC: [figure4]\nD: [figure5]\nE: [figure6]\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following figures indicate changes in phosphate concentration as the filtrate passes through regions $a$ and $b$, according to the increase in plasma phosphate concentration.\n[figure1]\n\nUsing this information, choose the most appropriate graph that depicts the changes in the renal reabsorption rate of phosphate ions according to the increase of its concentration in the plasma.\n\nA: [figure2]\nB: [figure3]\nC: [figure4]\nD: [figure5]\nE: [figure6]\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-26.jpg?height=678&width=1450&top_left_y=481&top_left_x=310",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-26.jpg?height=340&width=494&top_left_y=1606&top_left_x=330",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-26.jpg?height=337&width=491&top_left_y=1605&top_left_x=882",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-26.jpg?height=371&width=488&top_left_y=1571&top_left_x=1435",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-26.jpg?height=343&width=497&top_left_y=2010&top_left_x=328",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-26.jpg?height=343&width=491&top_left_y=2010&top_left_x=882"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_136",
        "problem": "For the growth of plants, the supply of nutrient inorganic ions is essential. A certain crop was grown in two different soils (X, Y). The concentrations of nutrients (potassium ions and chloride ions) in each type of soil are shown in the table. The estimated cytosolic concentrations of each ion in the root epidermal cells of this crop are also shown. When the membrane potential of the epidermal cell is $-150 \\mathrm{mV}$, how is each ion transported into the cell?\n\nIon movement is determined by electrical and concentration gradients. The membrane potential which would counterbalance the concentration gradient is given by the Nernst equilibrium potential equation:\n\n$E=-\\frac{60}{\\mathbf{z}} \\log \\frac{\\mathbf{C}_{\\mathbf{i}}}{\\mathbf{C}_{\\mathbf{o}}}(\\mathrm{mV})$\n\n$E$ : the Nernst equilibrium potential\n\n$\\mathbf{Z}$ : the charge of the ion, e.g. $\\mathbf{z}$ for $\\mathrm{Ca}^{2+}=+2$\n\n$\\mathbf{C}_{\\mathbf{i}}$ : the molar concentration of the ion in the cytosol\n\n$\\mathbf{C}_{0}$ : the extracellular molar concentration (here, the concentration in the soil) of that ion.\n\nThe direction of ion transport is determined by comparing the Nernst equilibrium potential with the membrane potential of the cell. Here, transport against the electrochemical gradient of each ion is called \"active transport\" and transport according to the electrochemical gradient of each ion is called \"passive transport.\"\n\n|  | Soil X | Soil Y | The estimated cytosolic <br> concentrations of each ion in the <br> root epidermal cells |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{K}^{+}$ | $1 \\mathrm{mM}$ | $0.01 \\mathrm{mM}$ | $100 \\mathrm{mM}$ |\n| $\\mathrm{Cl}^{-}$ | $0.5 \\mathrm{mM}$ | $5 \\mathrm{mM}$ | $5 \\mathrm{mM}$ |\nA: In soil X, potassium ions are absorbed by the active transport system.\nB: In soil Y, potassium ions are absorbed by the active transport system.\nC: In soil X, chloride ions are absorbed by the passive transport system. .....\nD: In soil Y, chloride ions are absorbed by the passive transport system. .....\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor the growth of plants, the supply of nutrient inorganic ions is essential. A certain crop was grown in two different soils (X, Y). The concentrations of nutrients (potassium ions and chloride ions) in each type of soil are shown in the table. The estimated cytosolic concentrations of each ion in the root epidermal cells of this crop are also shown. When the membrane potential of the epidermal cell is $-150 \\mathrm{mV}$, how is each ion transported into the cell?\n\nIon movement is determined by electrical and concentration gradients. The membrane potential which would counterbalance the concentration gradient is given by the Nernst equilibrium potential equation:\n\n$E=-\\frac{60}{\\mathbf{z}} \\log \\frac{\\mathbf{C}_{\\mathbf{i}}}{\\mathbf{C}_{\\mathbf{o}}}(\\mathrm{mV})$\n\n$E$ : the Nernst equilibrium potential\n\n$\\mathbf{Z}$ : the charge of the ion, e.g. $\\mathbf{z}$ for $\\mathrm{Ca}^{2+}=+2$\n\n$\\mathbf{C}_{\\mathbf{i}}$ : the molar concentration of the ion in the cytosol\n\n$\\mathbf{C}_{0}$ : the extracellular molar concentration (here, the concentration in the soil) of that ion.\n\nThe direction of ion transport is determined by comparing the Nernst equilibrium potential with the membrane potential of the cell. Here, transport against the electrochemical gradient of each ion is called \"active transport\" and transport according to the electrochemical gradient of each ion is called \"passive transport.\"\n\n|  | Soil X | Soil Y | The estimated cytosolic <br> concentrations of each ion in the <br> root epidermal cells |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{K}^{+}$ | $1 \\mathrm{mM}$ | $0.01 \\mathrm{mM}$ | $100 \\mathrm{mM}$ |\n| $\\mathrm{Cl}^{-}$ | $0.5 \\mathrm{mM}$ | $5 \\mathrm{mM}$ | $5 \\mathrm{mM}$ |\n\nA: In soil X, potassium ions are absorbed by the active transport system.\nB: In soil Y, potassium ions are absorbed by the active transport system.\nC: In soil X, chloride ions are absorbed by the passive transport system. .....\nD: In soil Y, chloride ions are absorbed by the passive transport system. .....\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_348",
        "problem": "果蝇的长翅和残翅受一对等位基因（A、a）控制, 灰体和黄体受另外一对等位基因 (B、b) 控制。某兴趣小组用一只长翅灰体雌果蝇和一只长翅灰体雄果蝇为亲本进行杂交, 子代表型及比例为长翅灰体䧳:残翅灰体䧳: 长翅灰体雄: 残翅灰体雄: 长翅黄体雄:残翅黄体雄 $=6: 2: 3: 1: 3: 1$, 不考虑 X、Y 染色体的同源区段。下列有关说法正确的是 ( )\nA: 控制果蝇翅型性状与体色性状的基因的遗传不遵循基因自由组合定律\nB: 两个亲本果蝇一定是杂合子，子代残翅灰体雌蝇可能是纯合子\nC: 子代残翅灰体雌果蝇与长翅黄体雄果蝇杂交, 子二代残翅黄体雄蝇的比例为 $1 / 12$\nD: 子代所有雌雄个体随机交配, 子二代中 $\\mathrm{A}$ 的基因频率与 B 的基因频率会发生变化\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的长翅和残翅受一对等位基因（A、a）控制, 灰体和黄体受另外一对等位基因 (B、b) 控制。某兴趣小组用一只长翅灰体雌果蝇和一只长翅灰体雄果蝇为亲本进行杂交, 子代表型及比例为长翅灰体䧳:残翅灰体䧳: 长翅灰体雄: 残翅灰体雄: 长翅黄体雄:残翅黄体雄 $=6: 2: 3: 1: 3: 1$, 不考虑 X、Y 染色体的同源区段。下列有关说法正确的是 ( )\n\nA: 控制果蝇翅型性状与体色性状的基因的遗传不遵循基因自由组合定律\nB: 两个亲本果蝇一定是杂合子，子代残翅灰体雌蝇可能是纯合子\nC: 子代残翅灰体雌果蝇与长翅黄体雄果蝇杂交, 子二代残翅黄体雄蝇的比例为 $1 / 12$\nD: 子代所有雌雄个体随机交配, 子二代中 $\\mathrm{A}$ 的基因频率与 B 的基因频率会发生变化\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_299",
        "problem": "$P o a$ and Stipa are two perennial grasses in the steppes. The former is highly preferred by domestic and wild herbivores, while the latter is relatively unpalatable. A student grew Poa plants in different distances to Stipa plants, with or without root barrier, in different grazing levels, and then recorded the growth of Poa plants.\n[figure1]\n\nFigure Q. 46\n\nEffects of distance (close and far) and barrier (without barrier (-B) and with the barrier present (+B)) to root competition manipulations on Poa female (left panels) and male plants (right panels) total biomass, at each grazing intensity level. \"*\" and “**\": Statistically significant differences; n.s.: non significant.\nA: Distance to the less palatable Stipa neighbours significantly affects the Poa biomass of females and males at the ungrazed site.\nB: Plants at the moderate grazing site, whether male or female, generally perform better near Stipa tussocks than far from them, demonstrating a positive effect of Stipa canopies on both genders.\nC: There is a strong below-ground competition between Poa females and Stipa neighbours at the ungrazed site.\nD: Population sex ratio drift between female and male bias may be influenced by domestic grazing intensity.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\n$P o a$ and Stipa are two perennial grasses in the steppes. The former is highly preferred by domestic and wild herbivores, while the latter is relatively unpalatable. A student grew Poa plants in different distances to Stipa plants, with or without root barrier, in different grazing levels, and then recorded the growth of Poa plants.\n[figure1]\n\nFigure Q. 46\n\nEffects of distance (close and far) and barrier (without barrier (-B) and with the barrier present (+B)) to root competition manipulations on Poa female (left panels) and male plants (right panels) total biomass, at each grazing intensity level. \"*\" and “**\": Statistically significant differences; n.s.: non significant.\n\nA: Distance to the less palatable Stipa neighbours significantly affects the Poa biomass of females and males at the ungrazed site.\nB: Plants at the moderate grazing site, whether male or female, generally perform better near Stipa tussocks than far from them, demonstrating a positive effect of Stipa canopies on both genders.\nC: There is a strong below-ground competition between Poa females and Stipa neighbours at the ungrazed site.\nD: Population sex ratio drift between female and male bias may be influenced by domestic grazing intensity.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-100.jpg?height=1516&width=1134&top_left_y=593&top_left_x=494"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_347",
        "problem": "果蝇的眼色有红眼和白眼 (由 $\\mathrm{A} 、 \\mathrm{a}$ 控制), 体色有灰身和黑身 (由 $\\mathrm{B} 、 \\mathrm{~b}$ 控制)。\n\n某科研小组用一对表型均为红眼灰身的雌、雄果蝇进行杂交实验, 在特定的实验环境下培养子代, 某种基因型的合子完全致死, 结果如下表所示。下列说法不正确的是（）\n\n## P 红眼灰身雌果蝇 $\\times$ 红眼灰身雄果蝇\n\n| $F_{1}$ | 红眼灰身 | 红眼黑身 | 白眼灰身 | 白眼黑身 |\n| :---: | :---: | :---: | :---: | :---: |\n| 雄性 | 301 | 99 | 298 | 103 |\n| 雌性 | 501 | 203 | 0 | 0 |\nA: 控制眼色和体色的基因分别位于 $\\mathrm{X}$ 染色体和常染色体上\nB: 在特定的实验环境下培养, $F_{1}$ 雌性中的红眼灰身存在致死现象\nC: 在特定的实验环境下培养, $F_{1}$ 红眼灰身雌性均为杂合子\nD: 选择 $F_{1}$ 中的红眼黑身雌性与白眼灰身雄性杂交, 子代白眼黑身雄性所占比例为 $1 / 24$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的眼色有红眼和白眼 (由 $\\mathrm{A} 、 \\mathrm{a}$ 控制), 体色有灰身和黑身 (由 $\\mathrm{B} 、 \\mathrm{~b}$ 控制)。\n\n某科研小组用一对表型均为红眼灰身的雌、雄果蝇进行杂交实验, 在特定的实验环境下培养子代, 某种基因型的合子完全致死, 结果如下表所示。下列说法不正确的是（）\n\n## P 红眼灰身雌果蝇 $\\times$ 红眼灰身雄果蝇\n\n| $F_{1}$ | 红眼灰身 | 红眼黑身 | 白眼灰身 | 白眼黑身 |\n| :---: | :---: | :---: | :---: | :---: |\n| 雄性 | 301 | 99 | 298 | 103 |\n| 雌性 | 501 | 203 | 0 | 0 |\n\nA: 控制眼色和体色的基因分别位于 $\\mathrm{X}$ 染色体和常染色体上\nB: 在特定的实验环境下培养, $F_{1}$ 雌性中的红眼灰身存在致死现象\nC: 在特定的实验环境下培养, $F_{1}$ 红眼灰身雌性均为杂合子\nD: 选择 $F_{1}$ 中的红眼黑身雌性与白眼灰身雄性杂交, 子代白眼黑身雄性所占比例为 $1 / 24$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_197",
        "problem": "Some researchers studied the changes in the level of saliva cortisol and 2-AG (2arachidonoylglycerol) concentration in blood in two groups of people with motion sickness and without motion sickness (no sickness) during parabolic flight maneuvers (PFs). During PFs, the saliva cortisol levels and 2-AG blood concentrations were measured from samples taken in-flight before start of the parabolic maneuvers (T0), after 10 parabolas (T1), 20 parabolas (T2), and 30 parabolas (T3), termination of PFs (T4) and $24 \\mathrm{~h}$ later (T5). The results are shown in Figure Q. 78.\n\n[figure1]\n\nFigure Q.78A.\n\n[figure2]\n\nFigure Q.78B.\nA: A 2-AG inhibitor can be used to reduce motion sickness.\nB: In motion sickness group, the blood glucose level at T4 was higher than at T1.\nC: At T2, the blood ACTH (adrenocorticotropic hormone) level in the no sickness group was higher than that in the motion sickness group.\nD: In motion sickness group, the blood CRH (corticotropin-releasing hormone) level at $\\mathrm{T} 5$ was lower than that at $\\mathrm{T} 2$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nSome researchers studied the changes in the level of saliva cortisol and 2-AG (2arachidonoylglycerol) concentration in blood in two groups of people with motion sickness and without motion sickness (no sickness) during parabolic flight maneuvers (PFs). During PFs, the saliva cortisol levels and 2-AG blood concentrations were measured from samples taken in-flight before start of the parabolic maneuvers (T0), after 10 parabolas (T1), 20 parabolas (T2), and 30 parabolas (T3), termination of PFs (T4) and $24 \\mathrm{~h}$ later (T5). The results are shown in Figure Q. 78.\n\n[figure1]\n\nFigure Q.78A.\n\n[figure2]\n\nFigure Q.78B.\n\nA: A 2-AG inhibitor can be used to reduce motion sickness.\nB: In motion sickness group, the blood glucose level at T4 was higher than at T1.\nC: At T2, the blood ACTH (adrenocorticotropic hormone) level in the no sickness group was higher than that in the motion sickness group.\nD: In motion sickness group, the blood CRH (corticotropin-releasing hormone) level at $\\mathrm{T} 5$ was lower than that at $\\mathrm{T} 2$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-066.jpg?height=567&width=827&top_left_y=984&top_left_x=199",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-066.jpg?height=570&width=800&top_left_y=977&top_left_x=1070"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1297",
        "problem": "The Census of Marine Life also examined the global threats to marine biodiversity and these are summarized in the table below.\n\n|  | Overfishing | Habitat loss | Pollution | Alien species | Temperature | Hypoxia | Acidification | Total | Median |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Mediterranean | 5 | 5 | 4 | 5 | 5 | 2 | 1 | 27 | 5.0 |\n| Gulf of Mexico | 5 | 5 | 5 | 4 | 2 | 3 | 1 | 25 | 4.0 |\n| China | 5 | 5 | 5 | 2 | 2 | 3 | 1 | 23 | 3.0 |\n| Baltic | 4 | 3 | 4 | 3 | 4 | 3 | 1 | 22 | 3.0 |\n| Caribbean | 4 | 4 | 4 | 4 | 2 | 2 | 2 | 22 | 4.0 |\n| USA Southeast | 4 | 4 | 3 | 3 | 3 | 2 | 3 | 22 | 3.0 |\n| Brazil and Tropical West Atlantic | 4 | 4 | 3 | 3 | 3 | 3 | 2 | 22 | 3.0 |\n| Humboldt Current and Patagonian Shelf | 4 | 3 | 3 | 3 | 2 | 4 | 2 | 21 | 3.0 |\n| North Indian Ocean | 3 | 4 | 4 | 3 | 3 | 2 | 2 | 21 | 3.0 |\n| Tropical East Pacific | 3 | 3 | 3 | 3 | 3 | 3 | 2 | 20 | 3.0 |\n| South Africa | 3 | 2 | 4 | 4 | 2 | 4 | 1 | 20 | 3.0 |\n| New Zealand | 4 | 3 | 2 | 4 | 2 | 1 | 3 | 19 | 3.0 |\n| Atlantic Europe | 4 | 2 | 4 | 2 | 4 | 1 | 2 | 19 | 2.0 |\n| USA Northeast | 4 | 3 | 3 | 2 | 3 | 2 | 1 | 18 | 3.0 |\n| Japan | 3 | 3 | 3 | 2 | 3 | 1 | 2 | 17 | 3.0 |\n| Canada (all) | 2 | 4 | 2 | 2 | 5 | 0 | 1 | 16 | 2.0 |\n| Australia | 3 | 3 | 2 | 3 | 2 | 0 | 1 | 14 | 2.0 |\n| Antarctica | 2 | 2 | 2 | 0 | 1 | 0 | 2 | 9 | 2.0 |\n| Total | 66 | 62 | 60 | 52 | 51 | 36 | 30 |  |  |\n| Median | 4.0 | 3.0 | 3.0 | 3.0 | 3.0 | 2.0 | 2.0 |  |  |\n\nEach threat was scored from 1 to 5 (minimum to maximum) across a comparative scale among different regions. Some regions (e.g., Australia) reported only known threats rather than predicted threats. Table is sorted by reported greatest threats and areas with greatest impacts. Median values of each threat and for each region are also reported. oi:10.1371/journal.pone.0012110.t005The region whose marine biodiversity is under the most threat from temperature changes such as global warming is?\nA: Mediterranean only\nB: Mediterranean and Canada\nC: Mediterranean and Baltic\nD: Canada and Baltic\nE: Canada only\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nThe Census of Marine Life also examined the global threats to marine biodiversity and these are summarized in the table below.\n\n|  | Overfishing | Habitat loss | Pollution | Alien species | Temperature | Hypoxia | Acidification | Total | Median |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Mediterranean | 5 | 5 | 4 | 5 | 5 | 2 | 1 | 27 | 5.0 |\n| Gulf of Mexico | 5 | 5 | 5 | 4 | 2 | 3 | 1 | 25 | 4.0 |\n| China | 5 | 5 | 5 | 2 | 2 | 3 | 1 | 23 | 3.0 |\n| Baltic | 4 | 3 | 4 | 3 | 4 | 3 | 1 | 22 | 3.0 |\n| Caribbean | 4 | 4 | 4 | 4 | 2 | 2 | 2 | 22 | 4.0 |\n| USA Southeast | 4 | 4 | 3 | 3 | 3 | 2 | 3 | 22 | 3.0 |\n| Brazil and Tropical West Atlantic | 4 | 4 | 3 | 3 | 3 | 3 | 2 | 22 | 3.0 |\n| Humboldt Current and Patagonian Shelf | 4 | 3 | 3 | 3 | 2 | 4 | 2 | 21 | 3.0 |\n| North Indian Ocean | 3 | 4 | 4 | 3 | 3 | 2 | 2 | 21 | 3.0 |\n| Tropical East Pacific | 3 | 3 | 3 | 3 | 3 | 3 | 2 | 20 | 3.0 |\n| South Africa | 3 | 2 | 4 | 4 | 2 | 4 | 1 | 20 | 3.0 |\n| New Zealand | 4 | 3 | 2 | 4 | 2 | 1 | 3 | 19 | 3.0 |\n| Atlantic Europe | 4 | 2 | 4 | 2 | 4 | 1 | 2 | 19 | 2.0 |\n| USA Northeast | 4 | 3 | 3 | 2 | 3 | 2 | 1 | 18 | 3.0 |\n| Japan | 3 | 3 | 3 | 2 | 3 | 1 | 2 | 17 | 3.0 |\n| Canada (all) | 2 | 4 | 2 | 2 | 5 | 0 | 1 | 16 | 2.0 |\n| Australia | 3 | 3 | 2 | 3 | 2 | 0 | 1 | 14 | 2.0 |\n| Antarctica | 2 | 2 | 2 | 0 | 1 | 0 | 2 | 9 | 2.0 |\n| Total | 66 | 62 | 60 | 52 | 51 | 36 | 30 |  |  |\n| Median | 4.0 | 3.0 | 3.0 | 3.0 | 3.0 | 2.0 | 2.0 |  |  |\n\nEach threat was scored from 1 to 5 (minimum to maximum) across a comparative scale among different regions. Some regions (e.g., Australia) reported only known threats rather than predicted threats. Table is sorted by reported greatest threats and areas with greatest impacts. Median values of each threat and for each region are also reported. oi:10.1371/journal.pone.0012110.t005\n\nproblem:\nThe region whose marine biodiversity is under the most threat from temperature changes such as global warming is?\n\nA: Mediterranean only\nB: Mediterranean and Canada\nC: Mediterranean and Baltic\nD: Canada and Baltic\nE: Canada only\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_815",
        "problem": "已知一家四口人关于某单基因遗传病相关基因的电泳结果如图所示, 其中 $1 、 2$ 为亲代, 3、4 为子代。不考虑基因突变等情况, 下列叙述错误的是（）\n\n[图1]\nA: 若 $1 、 2$ 均正常, 则 4 一定是男孩且患病\nB: 若 $1 、 2$ 均患病，则再生一个男孩患病的概率为 $1 / 4$\nC: 若 3、 4 均正常, 则该病不可能为常染色体隐性遗传病\nD: 若 $3 、 4$ 均患病，则 $1 、 2$ 再生一个患病女孩的概率一定为 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知一家四口人关于某单基因遗传病相关基因的电泳结果如图所示, 其中 $1 、 2$ 为亲代, 3、4 为子代。不考虑基因突变等情况, 下列叙述错误的是（）\n\n[图1]\n\nA: 若 $1 、 2$ 均正常, 则 4 一定是男孩且患病\nB: 若 $1 、 2$ 均患病，则再生一个男孩患病的概率为 $1 / 4$\nC: 若 3、 4 均正常, 则该病不可能为常染色体隐性遗传病\nD: 若 $3 、 4$ 均患病，则 $1 、 2$ 再生一个患病女孩的概率一定为 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-065.jpg?height=285&width=517&top_left_y=577&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_629",
        "problem": "果蝇的直毛和分叉毛受基因 $\\mathrm{D} / \\mathrm{d}$ 的控制。让两只表现型不同的果蝇杂交, $\\mathrm{F}_{1}$ 个体均表现为直毛, 再让 $F_{1}$ 雌雄果蝇随机交配, $F_{2}$ 中的雌果蝇均表现为直毛, 雄果蝇中直毛:分叉毛=1：1. 不考虑 X、Y 染色体的同源区段, 下列分析正确的是（ ）\nA: 基因 $\\mathrm{D} / \\mathrm{d}$ 位于常染色体上\nB: 亲代雌果蝇和 $F_{1}$ 雌果蝇的基因型相同\nC: $F_{2}$ 中分叉毛果蝇所占比例为 $1 / 4$\nD: $F_{2}$ 直毛雌果蝇均为纯合子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的直毛和分叉毛受基因 $\\mathrm{D} / \\mathrm{d}$ 的控制。让两只表现型不同的果蝇杂交, $\\mathrm{F}_{1}$ 个体均表现为直毛, 再让 $F_{1}$ 雌雄果蝇随机交配, $F_{2}$ 中的雌果蝇均表现为直毛, 雄果蝇中直毛:分叉毛=1：1. 不考虑 X、Y 染色体的同源区段, 下列分析正确的是（ ）\n\nA: 基因 $\\mathrm{D} / \\mathrm{d}$ 位于常染色体上\nB: 亲代雌果蝇和 $F_{1}$ 雌果蝇的基因型相同\nC: $F_{2}$ 中分叉毛果蝇所占比例为 $1 / 4$\nD: $F_{2}$ 直毛雌果蝇均为纯合子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_111",
        "problem": "Nasopharyngeal carcinoma is closely associated with the Epstein-Barr virus (EBV) infection. In a recent prospective study involving more than 20,000 participants, detection of EBV DNA in plasma was shown to be useful in screening for nasopharyngeal carcinoma. However, $5.4 \\%$ of participants without nasopharyngeal carcinoma had detectable EBV DNA in plasma at recruitment. Figure below shows the effect of ambient temperature on test results.\n\n- Sensitivity refers to the test's ability to correctly detect affected patients who do have the condition. It is calculated as: number of affected patients with positive tests/total number of affected individuals studied.\n- Specificity relates to the test's ability to correctly identified those without disease. It is calculated as: number of healthy individuals with negative tests/total number of healthy individuals studied.\n\n[figure1]\nA: Based on Figure above, performing the test in warmer places will increase the specificity.\nB: When a test becomes more sensitive, the specificity of the test will increase.\nC: A $100 \\%$ specific test will be $100 \\%$ sensitive.\nD: To rule out a disease, it is better to use a test with high sensitivity instead of a test with high specificity.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nNasopharyngeal carcinoma is closely associated with the Epstein-Barr virus (EBV) infection. In a recent prospective study involving more than 20,000 participants, detection of EBV DNA in plasma was shown to be useful in screening for nasopharyngeal carcinoma. However, $5.4 \\%$ of participants without nasopharyngeal carcinoma had detectable EBV DNA in plasma at recruitment. Figure below shows the effect of ambient temperature on test results.\n\n- Sensitivity refers to the test's ability to correctly detect affected patients who do have the condition. It is calculated as: number of affected patients with positive tests/total number of affected individuals studied.\n- Specificity relates to the test's ability to correctly identified those without disease. It is calculated as: number of healthy individuals with negative tests/total number of healthy individuals studied.\n\n[figure1]\n\nA: Based on Figure above, performing the test in warmer places will increase the specificity.\nB: When a test becomes more sensitive, the specificity of the test will increase.\nC: A $100 \\%$ specific test will be $100 \\%$ sensitive.\nD: To rule out a disease, it is better to use a test with high sensitivity instead of a test with high specificity.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-24.jpg?height=862&width=1145&top_left_y=911&top_left_x=455"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_516",
        "problem": "现有两瓶世代连续的果蝇, 甲瓶中的个体全为灰身, 乙瓶中个体既有灰身也有黑身个体。让乙瓶中的全部灰身个体与异性黑身果蝇交配，若后代都不出现性状分离则可以认定（ ）\nA: 甲为乙的亲本, 乙中灰身果蝇为杂合体\nB: 甲为乙的亲本, 乙中灰身果蝇为纯合体\nC: 乙为甲的亲本，乙中灰身果蝇为杂合体\nD: 乙为甲的亲本，乙中灰身果蝇为纯合体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n现有两瓶世代连续的果蝇, 甲瓶中的个体全为灰身, 乙瓶中个体既有灰身也有黑身个体。让乙瓶中的全部灰身个体与异性黑身果蝇交配，若后代都不出现性状分离则可以认定（ ）\n\nA: 甲为乙的亲本, 乙中灰身果蝇为杂合体\nB: 甲为乙的亲本, 乙中灰身果蝇为纯合体\nC: 乙为甲的亲本，乙中灰身果蝇为杂合体\nD: 乙为甲的亲本，乙中灰身果蝇为纯合体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_997",
        "problem": "A spectrophotometer is used to measure concentrations of proteins, DNA and many other chemical compounds. Assume the input light intensity at a particular wavelength $\\lambda$ is $I_{\\omega}$ and the intensity of the light transmitted through the second cuvette is $I_{1}$. If the concentration is doubled which of the following statements are TRUE.\n\n[figure1]\nA: The percent transmission of the first sample will be two times greater than the second sample\nB: The absorbance, $\\mathbf{A}_{\\lambda}$, of the second sample will be twice the absorbance of the first sample\nC: The increase in concentration will shift the wavelength of maximum absorbance to a longer wavelength\nD: The percent transmission of the second sample will be twice the first sample\nE: There is insufficient information to determine the correct answer\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA spectrophotometer is used to measure concentrations of proteins, DNA and many other chemical compounds. Assume the input light intensity at a particular wavelength $\\lambda$ is $I_{\\omega}$ and the intensity of the light transmitted through the second cuvette is $I_{1}$. If the concentration is doubled which of the following statements are TRUE.\n\n[figure1]\n\nA: The percent transmission of the first sample will be two times greater than the second sample\nB: The absorbance, $\\mathbf{A}_{\\lambda}$, of the second sample will be twice the absorbance of the first sample\nC: The increase in concentration will shift the wavelength of maximum absorbance to a longer wavelength\nD: The percent transmission of the second sample will be twice the first sample\nE: There is insufficient information to determine the correct answer\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_c11579df664240d6158ag-02.jpg?height=294&width=638&top_left_y=622&top_left_x=556"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_568",
        "problem": "利用农杆菌转化法, 将含有基因修饰系统的 T--DNA 插入到水稻细胞 $\\mathrm{M}$ 的某条染色体上, 在该修饰系统的作用下, 一个 DNA 分子单链上的一个 $\\mathrm{C}$ 脱去氨基变为 $\\mathrm{U}$, 脱氨基过程在细胞 $\\mathrm{M}$ 中只发生一次。将细胞 $\\mathrm{M}$ 培育成植株 $\\mathrm{N}$ 。下列说法错误的\n是 ( )\nA: $\\mathrm{N}$ 的每一个细胞中都含有 T--DNA\nB: $\\mathrm{N}$ 自交, 子一代中含 T--DNA 的植株占 $3 / 4$\nC: $\\mathrm{M}$ 经 $\\mathrm{n}\\left(\\mathrm{n} \\geq 1\\right.$ ) 次有丝分裂后, 脱氨基位点为 $\\mathrm{A}-\\mathrm{U}$ 的细胞占 $1 / 2^{\\mathrm{n}}$\nD: M 经 2 次有丝分裂后, 含 T--DNA 且脱氨基位点为 A-T 的细胞占 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n利用农杆菌转化法, 将含有基因修饰系统的 T--DNA 插入到水稻细胞 $\\mathrm{M}$ 的某条染色体上, 在该修饰系统的作用下, 一个 DNA 分子单链上的一个 $\\mathrm{C}$ 脱去氨基变为 $\\mathrm{U}$, 脱氨基过程在细胞 $\\mathrm{M}$ 中只发生一次。将细胞 $\\mathrm{M}$ 培育成植株 $\\mathrm{N}$ 。下列说法错误的\n是 ( )\n\nA: $\\mathrm{N}$ 的每一个细胞中都含有 T--DNA\nB: $\\mathrm{N}$ 自交, 子一代中含 T--DNA 的植株占 $3 / 4$\nC: $\\mathrm{M}$ 经 $\\mathrm{n}\\left(\\mathrm{n} \\geq 1\\right.$ ) 次有丝分裂后, 脱氨基位点为 $\\mathrm{A}-\\mathrm{U}$ 的细胞占 $1 / 2^{\\mathrm{n}}$\nD: M 经 2 次有丝分裂后, 含 T--DNA 且脱氨基位点为 A-T 的细胞占 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1041",
        "problem": "What is the concentration of hydrogen ions at $\\mathrm{pH}$ 5.7?\nA: $7.0 \\times 10^{-5} \\mathrm{M} \\mathrm{H}^{+}$\nB: $2.0 \\times 10^{-6} \\mathrm{M} \\mathrm{H}^{+}$\nC: $1.0 \\times 10^{-5} \\mathrm{M} \\mathrm{H}^{+}$\nD: $1.0 \\times 10^{-6} \\mathrm{M} \\mathrm{H}^{+}$\nE: $3.0 \\times 10^{-5} \\mathrm{M} \\mathrm{H}^{+}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the concentration of hydrogen ions at $\\mathrm{pH}$ 5.7?\n\nA: $7.0 \\times 10^{-5} \\mathrm{M} \\mathrm{H}^{+}$\nB: $2.0 \\times 10^{-6} \\mathrm{M} \\mathrm{H}^{+}$\nC: $1.0 \\times 10^{-5} \\mathrm{M} \\mathrm{H}^{+}$\nD: $1.0 \\times 10^{-6} \\mathrm{M} \\mathrm{H}^{+}$\nE: $3.0 \\times 10^{-5} \\mathrm{M} \\mathrm{H}^{+}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_154",
        "problem": "The table below shows the results of experimental tests on skin graft rejection between two different mouse strains. (Strains [A] and [B] are genetically identical except for the MHC loci.)\n\n| Exp. | Skin donor <br> mouse | Skin recipient <br> mouse | Skin rejection |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  |  |  | $6 \\sim 8 \\mathrm{~d}$ | $10 \\sim 13 \\mathrm{~d}$ |\n| I | $[\\mathrm{A}]$ | $[\\mathrm{A}]$ | did not occur | did not occur |\n| II | $[\\mathrm{A}]$ | $[\\mathrm{B}]$ | did not occur | occurred weakly |\n| III | $[\\mathrm{A}]$ | mouse which had <br> previously received <br> strain [A] skin | occurred strongly |  |\n| IV | $[\\mathrm{A}]$ | mouse which has <br> received lymphocytes from <br> a strain-[A]-skin-grafted <br> $[B]$ mouse | occurred strongly |  |\n\nWhich of the following explanations for the results is not correct?\nA: Graft rejection is considered to be the result of immune responses.\nB: MHC genes are mainly responsible for the graft rejection.\nC: If strain [B] skin is grafted onto mouse [A], the result would be the same as the result of Exp. II.\nD: If strain $[A]$ skin is grafted onto an offspring from a mating between $[A]$ and $[B]$ mice (e.g. F1, $[\\mathrm{A}] \\mathrm{x}[\\mathrm{B}]$, the result would be the same as that of Exp. III.\nE: The result observed in Exp. III is due to the formation of memory cells in the previously-exposed [B] mouse against strain [A] MHC antigens.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe table below shows the results of experimental tests on skin graft rejection between two different mouse strains. (Strains [A] and [B] are genetically identical except for the MHC loci.)\n\n| Exp. | Skin donor <br> mouse | Skin recipient <br> mouse | Skin rejection |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  |  |  | $6 \\sim 8 \\mathrm{~d}$ | $10 \\sim 13 \\mathrm{~d}$ |\n| I | $[\\mathrm{A}]$ | $[\\mathrm{A}]$ | did not occur | did not occur |\n| II | $[\\mathrm{A}]$ | $[\\mathrm{B}]$ | did not occur | occurred weakly |\n| III | $[\\mathrm{A}]$ | mouse which had <br> previously received <br> strain [A] skin | occurred strongly |  |\n| IV | $[\\mathrm{A}]$ | mouse which has <br> received lymphocytes from <br> a strain-[A]-skin-grafted <br> $[B]$ mouse | occurred strongly |  |\n\nWhich of the following explanations for the results is not correct?\n\nA: Graft rejection is considered to be the result of immune responses.\nB: MHC genes are mainly responsible for the graft rejection.\nC: If strain [B] skin is grafted onto mouse [A], the result would be the same as the result of Exp. II.\nD: If strain $[A]$ skin is grafted onto an offspring from a mating between $[A]$ and $[B]$ mice (e.g. F1, $[\\mathrm{A}] \\mathrm{x}[\\mathrm{B}]$, the result would be the same as that of Exp. III.\nE: The result observed in Exp. III is due to the formation of memory cells in the previously-exposed [B] mouse against strain [A] MHC antigens.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1010",
        "problem": "Bacteria including cyanobacteria accumulate a glycogen-like polysaccharide for storing glucose. Which of the following can reasonably explain the evolution of storage polysaccharides?\n\nThe common ancestor of plants and animals could synthesize:\nA: Both amylopectin and glycogen, but plants have lost the ability of glycogen synthesis during evolution\nB: Both amylopectin and glycogen, but animals have lost the ability of amylopectin synthesis\nC: Amylopectin but not glycogen, and animals have acquired the ability of glycogen synthesis\nD: Glycogen but not amylopectin, and plants have acquired the ability of amylopectin synthesis\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBacteria including cyanobacteria accumulate a glycogen-like polysaccharide for storing glucose. Which of the following can reasonably explain the evolution of storage polysaccharides?\n\nThe common ancestor of plants and animals could synthesize:\n\nA: Both amylopectin and glycogen, but plants have lost the ability of glycogen synthesis during evolution\nB: Both amylopectin and glycogen, but animals have lost the ability of amylopectin synthesis\nC: Amylopectin but not glycogen, and animals have acquired the ability of glycogen synthesis\nD: Glycogen but not amylopectin, and plants have acquired the ability of amylopectin synthesis\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1161",
        "problem": "A homozygous tall maize plant is fertilised by a dwarf maize plant. From which one of the following tissues of the resultant grain will the allele for dwarfness be absent?\nA: cotyledon\nB: endosperm\nC: fruit wall\nD: coleoptile\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA homozygous tall maize plant is fertilised by a dwarf maize plant. From which one of the following tissues of the resultant grain will the allele for dwarfness be absent?\n\nA: cotyledon\nB: endosperm\nC: fruit wall\nD: coleoptile\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_607",
        "problem": "某二倍体 $(2 n=14)$ 植物的红花和白花是一对相对性状, 该性状同时受多对独立遗传的等位基因控制，每对等位基因中至少有一个显性基因时才开红花。利用甲、乙、丙三种纯合品系进行了如下杂交实验。\n\n实验一：甲 $\\times$ 乙 $\\rightarrow F_{1}$ (红花) $\\rightarrow F_{2}$ 红花：白花 $=2709: 3689$\n\n实验二：甲 $\\times$ 丙 $\\rightarrow F_{1}($ 红花 $) \\rightarrow F_{2}$ 红花: 白花 $=907: 699$\n\n实验三: 乙 $\\times$ 丙 $\\rightarrow F_{1}($ 白花 $) \\rightarrow F_{2}$ 白花\n\n有关说法错误的是（ ）\nA: 控制该相对性状的基因数量至少为 3 对, 最多是 7 对\nB: 这三个品系中至少有一种是红花纯合子\nC: 上述杂交组合中 $F_{2}$ 白花纯合子比例最低是实验三\nD: 实验一的 $\\mathrm{F}_{2}$ 白花植株中自交后代不发生性状分离的比例为 $7 / 37$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某二倍体 $(2 n=14)$ 植物的红花和白花是一对相对性状, 该性状同时受多对独立遗传的等位基因控制，每对等位基因中至少有一个显性基因时才开红花。利用甲、乙、丙三种纯合品系进行了如下杂交实验。\n\n实验一：甲 $\\times$ 乙 $\\rightarrow F_{1}$ (红花) $\\rightarrow F_{2}$ 红花：白花 $=2709: 3689$\n\n实验二：甲 $\\times$ 丙 $\\rightarrow F_{1}($ 红花 $) \\rightarrow F_{2}$ 红花: 白花 $=907: 699$\n\n实验三: 乙 $\\times$ 丙 $\\rightarrow F_{1}($ 白花 $) \\rightarrow F_{2}$ 白花\n\n有关说法错误的是（ ）\n\nA: 控制该相对性状的基因数量至少为 3 对, 最多是 7 对\nB: 这三个品系中至少有一种是红花纯合子\nC: 上述杂交组合中 $F_{2}$ 白花纯合子比例最低是实验三\nD: 实验一的 $\\mathrm{F}_{2}$ 白花植株中自交后代不发生性状分离的比例为 $7 / 37$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_270",
        "problem": "The phagocytic activity of macrophages against a certain pathogenic bacteria is described below.\n\n[figure1]\nWhich of the following conclusions best explains the results above?\nA: Non-specific immunity enhances acquired immunity.\nB: Humoral immunity enhances acquired immunity.\nC: Humoral immunity enhances non-specific immunity.\nD: Cell-mediated immunity enhances humoral immunity.\nE: Cell-mediated immunity enhances non-specific immunity.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe phagocytic activity of macrophages against a certain pathogenic bacteria is described below.\n\n[figure1]\nWhich of the following conclusions best explains the results above?\n\nA: Non-specific immunity enhances acquired immunity.\nB: Humoral immunity enhances acquired immunity.\nC: Humoral immunity enhances non-specific immunity.\nD: Cell-mediated immunity enhances humoral immunity.\nE: Cell-mediated immunity enhances non-specific immunity.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://i.postimg.cc/GpcYz8PG/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1022",
        "problem": "Listed are three statements about lipids.\n1. Lipids are made up of polymers of fatty acids\n2. Lipids are hydrophobic\n3. Lipids that are made up of fatty acids with a high degree of saturation are more likely to be solids at room temperature\n\nChoose the most accurate statement from below:\nA: Statements 1 and 2 are true\nB: Statements 1 and 3 are true\nC: Statements 2 and 3 are true\nD: Only statement 2 is true\nE: All three statements are true\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nListed are three statements about lipids.\n1. Lipids are made up of polymers of fatty acids\n2. Lipids are hydrophobic\n3. Lipids that are made up of fatty acids with a high degree of saturation are more likely to be solids at room temperature\n\nChoose the most accurate statement from below:\n\nA: Statements 1 and 2 are true\nB: Statements 1 and 3 are true\nC: Statements 2 and 3 are true\nD: Only statement 2 is true\nE: All three statements are true\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_286",
        "problem": "Soil salinity $(\\mathrm{NaCl})$ affects the growth of plants. As the increase of osmotic pressure induced by soil salinity reduces the ability of plants to take up water and minerals, soil salinity elicits osmotic stress. Additionally, because cytosolic $\\mathrm{Na}^{+}$interferes with the activities of metabolic enzymes, soil salinity also elicits ionic stress. Thus, $\\mathrm{NaCl}$ elicits two primary effects on plant cells, which both trigger a signaling cascade that start with the elevation of the intracellular $\\mathrm{Ca}^{2+}$ concentration $\\left(\\left[\\mathrm{Ca}^{2+}\\right] \\mathrm{i}\\right)$. In contrast, sorbitol, a sugar alcohol often used as an osmoticum, elicits only osmotic stress because sorbitol is non-ionic. $x$ and $y$ are mutants of Arabidopsis isolated as defective in $\\mathrm{NaCl}$-induced increases in $\\left[\\mathrm{Ca}^{2+}\\right]$ i. Figure 1 illustrated below shows the dosedependent $\\left[\\mathrm{Ca}^{2+}\\right]$ increases induced by $\\mathrm{NaCl}$ or sorbitol in the seedlings of the wild type (WT) and mutants $x$ and $y$.\n\n[figure1]\n\nSorbitol (mM)\n\n[figure2]\n\n$\\mathrm{NaCl}(\\mathrm{mM})$\n\nFigure 1\nA: Mutant $x$ is defective in sensing osmotic stress.\nB: Mutant $y$ can sense ionic stress.\nC: The dose-dependent $\\left[\\mathrm{Ca}^{2+}\\right]$ increases induced by $\\mathrm{NaCl}$ of the $x y$ double mutant are expected to be equivalent to those of the $x$ single mutant. .....\nD: The dose-dependent $\\left[\\mathrm{Ca}^{2+}\\right] \\mathrm{i}$ increases induced by sorbitol of the $x y$ double mutant are expected to be equivalent to those of the $y$ single mutant.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nSoil salinity $(\\mathrm{NaCl})$ affects the growth of plants. As the increase of osmotic pressure induced by soil salinity reduces the ability of plants to take up water and minerals, soil salinity elicits osmotic stress. Additionally, because cytosolic $\\mathrm{Na}^{+}$interferes with the activities of metabolic enzymes, soil salinity also elicits ionic stress. Thus, $\\mathrm{NaCl}$ elicits two primary effects on plant cells, which both trigger a signaling cascade that start with the elevation of the intracellular $\\mathrm{Ca}^{2+}$ concentration $\\left(\\left[\\mathrm{Ca}^{2+}\\right] \\mathrm{i}\\right)$. In contrast, sorbitol, a sugar alcohol often used as an osmoticum, elicits only osmotic stress because sorbitol is non-ionic. $x$ and $y$ are mutants of Arabidopsis isolated as defective in $\\mathrm{NaCl}$-induced increases in $\\left[\\mathrm{Ca}^{2+}\\right]$ i. Figure 1 illustrated below shows the dosedependent $\\left[\\mathrm{Ca}^{2+}\\right]$ increases induced by $\\mathrm{NaCl}$ or sorbitol in the seedlings of the wild type (WT) and mutants $x$ and $y$.\n\n[figure1]\n\nSorbitol (mM)\n\n[figure2]\n\n$\\mathrm{NaCl}(\\mathrm{mM})$\n\nFigure 1\n\nA: Mutant $x$ is defective in sensing osmotic stress.\nB: Mutant $y$ can sense ionic stress.\nC: The dose-dependent $\\left[\\mathrm{Ca}^{2+}\\right]$ increases induced by $\\mathrm{NaCl}$ of the $x y$ double mutant are expected to be equivalent to those of the $x$ single mutant. .....\nD: The dose-dependent $\\left[\\mathrm{Ca}^{2+}\\right] \\mathrm{i}$ increases induced by sorbitol of the $x y$ double mutant are expected to be equivalent to those of the $y$ single mutant.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-48.jpg?height=665&width=759&top_left_y=1141&top_left_x=203",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-48.jpg?height=748&width=760&top_left_y=1125&top_left_x=1028"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_653",
        "problem": "某卯原细胞的基因组成为 $\\mathrm{Ee}$, 其减数分裂 I 时可能发生交叉互换, 减数分裂 II 时,姐妹染色单体可分别将自身两端粘在一起, 着丝粒分开后, 2 个环状染色体互锁在一起,如图所示。 2 个环状染色体随机交换一部分染色体片段后分开, 分别进入 2 个子细胞,交换的部分大小可不相等，位置随机。不考虑其他突变和基因被破坏的情况，关于该卵原细胞所形成子细胞的基因组成，下列叙述错误的是（）\n\n[图1]\n\n## 染色体\nA: 卵细胞基因组成最多有 6 种可能\nB: 若卵细胞为 $\\mathrm{Ee}$, 则第二极体可能为 $\\mathrm{EE}$ 或 ee\nC: 若卵细胞为 $\\mathrm{E}$ 且第一极体不含 $\\mathrm{E}$, 则第二极体最多有 4 种可能\nD: 若卵细胞不含 $\\mathrm{E} 、 \\mathrm{e}$ 且一个第二极体为 $\\mathrm{E}$, 则第一极体最多有 2 种可能\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某卯原细胞的基因组成为 $\\mathrm{Ee}$, 其减数分裂 I 时可能发生交叉互换, 减数分裂 II 时,姐妹染色单体可分别将自身两端粘在一起, 着丝粒分开后, 2 个环状染色体互锁在一起,如图所示。 2 个环状染色体随机交换一部分染色体片段后分开, 分别进入 2 个子细胞,交换的部分大小可不相等，位置随机。不考虑其他突变和基因被破坏的情况，关于该卵原细胞所形成子细胞的基因组成，下列叙述错误的是（）\n\n[图1]\n\n## 染色体\n\nA: 卵细胞基因组成最多有 6 种可能\nB: 若卵细胞为 $\\mathrm{Ee}$, 则第二极体可能为 $\\mathrm{EE}$ 或 ee\nC: 若卵细胞为 $\\mathrm{E}$ 且第一极体不含 $\\mathrm{E}$, 则第二极体最多有 4 种可能\nD: 若卵细胞不含 $\\mathrm{E} 、 \\mathrm{e}$ 且一个第二极体为 $\\mathrm{E}$, 则第一极体最多有 2 种可能\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-058.jpg?height=297&width=691&top_left_y=160&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_26",
        "problem": "A plant is normally red-flowered. Plant breeders obtained three genetically different pure mutated lines of white-flowered plants (designated as a, b and c). They performed crosses and observed progeny phenotypes as follows:\n\n| Cross | Parents | Progeny |\n| :---: | :---: | :---: |\n| 1 | Line $\\mathrm{a} \\times$ line $\\mathrm{b}$ | $\\mathrm{F}_{1}$ all white |\n| 2 | Line $\\mathrm{a} \\times$ line $\\mathrm{c}$ | $\\mathrm{F}_{1}$ all red |\n| 3 | Line $\\mathrm{b} \\times$ line $\\mathrm{c}$ | $\\mathrm{F}_{1}$ all white |\n| 4 | Red $\\mathrm{F}_{1}$ from cross $2 \\times$ line $\\mathrm{a}$ | $1 / 4$ red $: 3 / 4$ white |\n| 5 | Red $\\mathrm{F}_{1}$ from cross $2 \\times$ line $\\mathrm{b}$ | $1 / 8$ red $: 7 / 8$ white |\n| 6 | Red $\\mathrm{F}_{1}$ from cross $2 \\times$ line $\\mathrm{c}$ | $1 / 2$ red $: 1 / 2$ white |\nA: Line (a) is homozygous for one mutated gene.\nB: Line (b) shares two mutated homozygous genes with line $c$.\nC: Line (c) shares one mutated homozygous gene with line a.\nD: Line (b) has three mutated homozygous genes.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA plant is normally red-flowered. Plant breeders obtained three genetically different pure mutated lines of white-flowered plants (designated as a, b and c). They performed crosses and observed progeny phenotypes as follows:\n\n| Cross | Parents | Progeny |\n| :---: | :---: | :---: |\n| 1 | Line $\\mathrm{a} \\times$ line $\\mathrm{b}$ | $\\mathrm{F}_{1}$ all white |\n| 2 | Line $\\mathrm{a} \\times$ line $\\mathrm{c}$ | $\\mathrm{F}_{1}$ all red |\n| 3 | Line $\\mathrm{b} \\times$ line $\\mathrm{c}$ | $\\mathrm{F}_{1}$ all white |\n| 4 | Red $\\mathrm{F}_{1}$ from cross $2 \\times$ line $\\mathrm{a}$ | $1 / 4$ red $: 3 / 4$ white |\n| 5 | Red $\\mathrm{F}_{1}$ from cross $2 \\times$ line $\\mathrm{b}$ | $1 / 8$ red $: 7 / 8$ white |\n| 6 | Red $\\mathrm{F}_{1}$ from cross $2 \\times$ line $\\mathrm{c}$ | $1 / 2$ red $: 1 / 2$ white |\n\nA: Line (a) is homozygous for one mutated gene.\nB: Line (b) shares two mutated homozygous genes with line $c$.\nC: Line (c) shares one mutated homozygous gene with line a.\nD: Line (b) has three mutated homozygous genes.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_775",
        "problem": "免疫球蛋白（ $\\mathrm{Ig}$ ）由两条重链（ $\\mathrm{H}$ 链）和两条轻链（ $\\mathrm{L}$ 链）组成。 $\\mathrm{H}$ 链基因由 $\\mathrm{V}$ 、\n\n$\\mathrm{D} 、 \\mathrm{~J} 、 \\mathrm{C}$ 四组基因片段组成， $\\mathrm{L}$ 链基因由 $\\mathrm{V} 、 \\mathrm{~J} 、 \\mathrm{C}$ 三组基因片段组成，只有经 $\\mathrm{V} 、$\n\n（D、） $\\mathrm{J}$ 基因片段重排连接在一起, 才具有转录功能, 其中两种重排举例如图。下列叙述错误的是（ ）\n\n## $-\\mathrm{V}_{1}-\\mathrm{V}_{2}-1 /-\\mathrm{V}_{\\mathrm{n}}-1 /-\\mathrm{D}_{1}-\\mathrm{D}_{2}-1 /-\\mathrm{D}_{\\mathrm{n}}-1 /-\\mathrm{J}_{1}-\\mathrm{J}_{2}-\\mathrm{J}_{3}-1 /-\\mathrm{J}_{\\mathrm{n}}-1 /-\\mathrm{C}-$\n\n[图1]\nA: 重排过程发生了基因突变, 重排提高了突变率, 增加了多样性\nB: 第二次重排时上游的 $\\mathrm{V}_{1}$ 片段可取代第一次重排后的 $\\mathrm{V}_{2}$ 片段\nC: 未重排的各基因片段间被插入了序列，不能作为独立单位表达\nD: Ig 基因重排的机制决定了 $\\operatorname{Ig}$ 的多样性\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n免疫球蛋白（ $\\mathrm{Ig}$ ）由两条重链（ $\\mathrm{H}$ 链）和两条轻链（ $\\mathrm{L}$ 链）组成。 $\\mathrm{H}$ 链基因由 $\\mathrm{V}$ 、\n\n$\\mathrm{D} 、 \\mathrm{~J} 、 \\mathrm{C}$ 四组基因片段组成， $\\mathrm{L}$ 链基因由 $\\mathrm{V} 、 \\mathrm{~J} 、 \\mathrm{C}$ 三组基因片段组成，只有经 $\\mathrm{V} 、$\n\n（D、） $\\mathrm{J}$ 基因片段重排连接在一起, 才具有转录功能, 其中两种重排举例如图。下列叙述错误的是（ ）\n\n## $-\\mathrm{V}_{1}-\\mathrm{V}_{2}-1 /-\\mathrm{V}_{\\mathrm{n}}-1 /-\\mathrm{D}_{1}-\\mathrm{D}_{2}-1 /-\\mathrm{D}_{\\mathrm{n}}-1 /-\\mathrm{J}_{1}-\\mathrm{J}_{2}-\\mathrm{J}_{3}-1 /-\\mathrm{J}_{\\mathrm{n}}-1 /-\\mathrm{C}-$\n\n[图1]\n\nA: 重排过程发生了基因突变, 重排提高了突变率, 增加了多样性\nB: 第二次重排时上游的 $\\mathrm{V}_{1}$ 片段可取代第一次重排后的 $\\mathrm{V}_{2}$ 片段\nC: 未重排的各基因片段间被插入了序列，不能作为独立单位表达\nD: Ig 基因重排的机制决定了 $\\operatorname{Ig}$ 的多样性\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-082.jpg?height=314&width=876&top_left_y=1722&top_left_x=710"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_762",
        "problem": "果蝇 $(2 n=8)$ 卵原细胞减数分裂存在“逆反”减数分裂现象, 下图是“逆反减数分裂过程 (MI、MII分别表示减数第一次分裂和减数第二次分裂), 相关叙述错误的是 ( )\n\n[图1]\nA: “逆反”减数分裂的 MI发生非姐妹染色单体的互换和姐妹染色单体分离\nB: “逆反”减数分裂的 MII发生同源染色体的分离和非同源染色体自由组合\nC: “逆反”减数分裂的 MII过程中核 DNA 数目和染色体数目都发生减半\nD: MII过程重组染色体高概率的进入卵细胞有利于提高子代的变异率\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇 $(2 n=8)$ 卵原细胞减数分裂存在“逆反”减数分裂现象, 下图是“逆反减数分裂过程 (MI、MII分别表示减数第一次分裂和减数第二次分裂), 相关叙述错误的是 ( )\n\n[图1]\n\nA: “逆反”减数分裂的 MI发生非姐妹染色单体的互换和姐妹染色单体分离\nB: “逆反”减数分裂的 MII发生同源染色体的分离和非同源染色体自由组合\nC: “逆反”减数分裂的 MII过程中核 DNA 数目和染色体数目都发生减半\nD: MII过程重组染色体高概率的进入卵细胞有利于提高子代的变异率\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-62.jpg?height=614&width=1151&top_left_y=704&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_115",
        "problem": "Immature lymphocyte B cells differentiate in an area of the peripheral lymph organ called the embryo center. Myeloma cells are tumor cells that produce one type of mature immunoglobulin. mRNAs for full-length or only 3 'half of the immunoglobulin light chain gene were purified from a myeloma cell and were radioisotope-labeled. Genomic DNA fragments obtained from either the embryo center or myeloma cells were digested with a restriction enzyme and size fractionated by agarose electrophoresis. These DNA were hybridized with radiolabeled mRNA, and radiation was measured after the removal of unhybridized mRNA (the experimental flow is shown in Figure 1). The results are shown in Figure 2.\n\n[figure1]\n\nFigure 1\n\n|  | DNA | mRNA |\n| :--- | :---: | :---: |\n| - | myeloma | Full length |\n| ---- | myeloma | 3'end half |\n| - | embryo | Full length |\n| ---- | embryo | 3'end half |\n\n[figure2]\n\nFigure 2\nA: The immunoglobulin light chain gene contained in the embryo center cells is shorter than that in the myeloma cells.\nB: The running distance depends on the length of DNA hybridized with mRNA.\nC: The nucleotide sequence of DNA region hybridized with 3'-end mRNA is different between the myelomaderived DNA and the embryo center-derived DNA.\nD: The full-length immunoglobulin light chain mRNA isolated from myeloma cells contains sequences from two different parts of the DNA genome of the embryo center cells.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nImmature lymphocyte B cells differentiate in an area of the peripheral lymph organ called the embryo center. Myeloma cells are tumor cells that produce one type of mature immunoglobulin. mRNAs for full-length or only 3 'half of the immunoglobulin light chain gene were purified from a myeloma cell and were radioisotope-labeled. Genomic DNA fragments obtained from either the embryo center or myeloma cells were digested with a restriction enzyme and size fractionated by agarose electrophoresis. These DNA were hybridized with radiolabeled mRNA, and radiation was measured after the removal of unhybridized mRNA (the experimental flow is shown in Figure 1). The results are shown in Figure 2.\n\n[figure1]\n\nFigure 1\n\n|  | DNA | mRNA |\n| :--- | :---: | :---: |\n| - | myeloma | Full length |\n| ---- | myeloma | 3'end half |\n| - | embryo | Full length |\n| ---- | embryo | 3'end half |\n\n[figure2]\n\nFigure 2\n\nA: The immunoglobulin light chain gene contained in the embryo center cells is shorter than that in the myeloma cells.\nB: The running distance depends on the length of DNA hybridized with mRNA.\nC: The nucleotide sequence of DNA region hybridized with 3'-end mRNA is different between the myelomaderived DNA and the embryo center-derived DNA.\nD: The full-length immunoglobulin light chain mRNA isolated from myeloma cells contains sequences from two different parts of the DNA genome of the embryo center cells.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-38.jpg?height=711&width=1628&top_left_y=1038&top_left_x=197",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-38.jpg?height=685&width=757&top_left_y=1873&top_left_x=1049"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_228",
        "problem": "The direction of shell coiling in the snail Limnaea peregra is either dextral or sinistral. Coiling direction is determined by a pair of autosomal alleles. The allele for dextral $\\left(S^{+}\\right)$is dominant over the allele for sinistral (s). Experimental results of two reciprocal monohybrid crosses are shown below.\n\n[figure1]\n\nWhat is the genetic phenomenon that explains the inheritance pattern for coiling direction?\nA: Cytoplasmic inheritance.\nB: Epistasis.\nC: Genetic imprinting.\nD: Maternal effect.\nE: Sex-limited inheritance.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe direction of shell coiling in the snail Limnaea peregra is either dextral or sinistral. Coiling direction is determined by a pair of autosomal alleles. The allele for dextral $\\left(S^{+}\\right)$is dominant over the allele for sinistral (s). Experimental results of two reciprocal monohybrid crosses are shown below.\n\n[figure1]\n\nWhat is the genetic phenomenon that explains the inheritance pattern for coiling direction?\n\nA: Cytoplasmic inheritance.\nB: Epistasis.\nC: Genetic imprinting.\nD: Maternal effect.\nE: Sex-limited inheritance.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-48.jpg?height=848&width=1647&top_left_y=567&top_left_x=293"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_12",
        "problem": "Animal personality is defined as individual differences in behaviour that are consistent across time and ecological context. One of the important drivers of personality difference is factors that alter the value of a decision. It was hypothesized that altering the feeding regimes of some spider species affect some aspects of their personality as you can see in graphs below.\n[figure1]\n\nEstimates of repeatability of boldness and aggressiveness response of three spider species under different feeding regimes. Error bars represent $95 \\%$ confidence intervals for our repeatability estimates (After Lichtenstein, et al. 2016).\nA: This study failed to demonstrate association between boldness and aggressiveness regardless of feeding regimes.\nB: Repeatability of aggressiveness response in L. hesperus is independent of feeding regimes.\nC: Behavioural variation of each individual always decreases when spiders suffered prolonged food restrictions.\nD: L. hesperus shows different repeatability in behavioural response than two other species regarding to different feeding regime, thus we can assume different species have different behaviour over time and contexts.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAnimal personality is defined as individual differences in behaviour that are consistent across time and ecological context. One of the important drivers of personality difference is factors that alter the value of a decision. It was hypothesized that altering the feeding regimes of some spider species affect some aspects of their personality as you can see in graphs below.\n[figure1]\n\nEstimates of repeatability of boldness and aggressiveness response of three spider species under different feeding regimes. Error bars represent $95 \\%$ confidence intervals for our repeatability estimates (After Lichtenstein, et al. 2016).\n\nA: This study failed to demonstrate association between boldness and aggressiveness regardless of feeding regimes.\nB: Repeatability of aggressiveness response in L. hesperus is independent of feeding regimes.\nC: Behavioural variation of each individual always decreases when spiders suffered prolonged food restrictions.\nD: L. hesperus shows different repeatability in behavioural response than two other species regarding to different feeding regime, thus we can assume different species have different behaviour over time and contexts.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-60.jpg?height=1568&width=1034&top_left_y=640&top_left_x=517"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_23",
        "problem": "Arsenic (As) in the soil has become an environmental concern worldwide because it is difficult to remediate and can adversely impact human health. The fern, Athyrium yokoscense is well known as a $\\mathrm{Cd}$ hyperaccumulator as well as a $\\mathrm{Cu}, \\mathrm{Pb}$ and $\\mathrm{Zn}$ tolerant plant. However, no information is available on As accumulation by A. yokoscense, although it often grows in soils containing high levels of several heavy metals and As. To understand As accumulation in A. yokoscense, a student conducted an experiment in which young ferns collected from a mining area were grown in media containing Asspiked paddy soils or mine soil in a greenhouse for 21 weeks. Before transplanting fern biomass was $0.26 \\pm 0.08 \\mathrm{~g} \\mathrm{plant}^{-1} \\mathrm{DW}$ and As concentrations of young and old fronds were $7.8 \\pm 0.3$ and $57.7 \\pm 2.2 \\mathrm{mg} \\mathrm{kg}^{-1}$, respectively.\n\n[figure1]\n\nFig.Q16- 1. Dry biomass of $A$. yokoscense after 21 weeks of cultivation in a greenhouse\n\n[figure2]\n\nFig.Q16-2. Arsenic concentration in different parts of $A$. yokoscense cultivated in Asspiked soils and mine soils. a,b,c letter above each bar indicate the significant difference of the same plant parts\nA: Moderate As levels in soils might promote the growth of ferns.\nB: Concentration of As in root grown in arsenite-spiked media is lower than those in arsenate treatment, resulting in the increase of total biomass.\nC: Arsenic concentration increases from young to old fronds and is correlated to As levels in the soil\nD: The transfer of As from root to frond of $A$. yokoscense in mine soil is similar with those in arsenate-spiked soil.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nArsenic (As) in the soil has become an environmental concern worldwide because it is difficult to remediate and can adversely impact human health. The fern, Athyrium yokoscense is well known as a $\\mathrm{Cd}$ hyperaccumulator as well as a $\\mathrm{Cu}, \\mathrm{Pb}$ and $\\mathrm{Zn}$ tolerant plant. However, no information is available on As accumulation by A. yokoscense, although it often grows in soils containing high levels of several heavy metals and As. To understand As accumulation in A. yokoscense, a student conducted an experiment in which young ferns collected from a mining area were grown in media containing Asspiked paddy soils or mine soil in a greenhouse for 21 weeks. Before transplanting fern biomass was $0.26 \\pm 0.08 \\mathrm{~g} \\mathrm{plant}^{-1} \\mathrm{DW}$ and As concentrations of young and old fronds were $7.8 \\pm 0.3$ and $57.7 \\pm 2.2 \\mathrm{mg} \\mathrm{kg}^{-1}$, respectively.\n\n[figure1]\n\nFig.Q16- 1. Dry biomass of $A$. yokoscense after 21 weeks of cultivation in a greenhouse\n\n[figure2]\n\nFig.Q16-2. Arsenic concentration in different parts of $A$. yokoscense cultivated in Asspiked soils and mine soils. a,b,c letter above each bar indicate the significant difference of the same plant parts\n\nA: Moderate As levels in soils might promote the growth of ferns.\nB: Concentration of As in root grown in arsenite-spiked media is lower than those in arsenate treatment, resulting in the increase of total biomass.\nC: Arsenic concentration increases from young to old fronds and is correlated to As levels in the soil\nD: The transfer of As from root to frond of $A$. yokoscense in mine soil is similar with those in arsenate-spiked soil.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-037.jpg?height=997&width=1559&top_left_y=1409&top_left_x=294",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-038.jpg?height=993&width=1610&top_left_y=234&top_left_x=242"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_678",
        "problem": "马蛔虫的体细胞中只含有两对同源染色体。研究小组将马蛔虫的一个 置于含放射性同位素 ${ }^{32} \\mathrm{P}$ 的培养基中进行一次有丝分裂, 再将其转移至普通培养基中进行减数分裂,获得 8 个精细胞。这 8 个精细胞中不含放射性同位素 ${ }^{32} \\mathrm{P}$ 标记的细胞数目不可能是 $(\\quad)$ (不考虑染色体变异和互换)\nA: 0 个\nB: 1 个\nC: 4 个\nD: 5 个\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n马蛔虫的体细胞中只含有两对同源染色体。研究小组将马蛔虫的一个 置于含放射性同位素 ${ }^{32} \\mathrm{P}$ 的培养基中进行一次有丝分裂, 再将其转移至普通培养基中进行减数分裂,获得 8 个精细胞。这 8 个精细胞中不含放射性同位素 ${ }^{32} \\mathrm{P}$ 标记的细胞数目不可能是 $(\\quad)$ (不考虑染色体变异和互换)\n\nA: 0 个\nB: 1 个\nC: 4 个\nD: 5 个\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_212",
        "problem": "In 1876, Van Beneden, found a group of microscopic ciliated wormlike organisms that inhabited the renal sacs of cephalopods, mainly octopuses and cuttlefishes and named them as Dicyemids.\n\nBiologists have been fascinated because of their highly-simplified body organization and complex life cycles of asexual (under normal condition producing vermiform larva in host kidney) and sexual (under crowded condition producing infusoriform larva excreted in host urine). These ciliated animals are composed of approximately 20-30 (or 40) cells arranged in two layers, and they lack coeloms, circulatory systems, and other differentiated tissues and their embryos employ spiral cleavage. Due to their simple body plans, it can be considered as intermediates between the Protozoa and the Metazoa.\n[figure1]\n\nThe phylogenetic position of dicyemids suggested by previous phylogenetic studies remains controversial.\nA: Based on the above explanations, infusiform larva is a mean of dispersal for these animals.\nB: Based on the above molecular data, these animals are missing link between Metazoa and protozoa.\nC: Based on the data, these described animals faced a regression evolution during its evolutionary history.\nD: Based on the given data, it is more likely that Dicyemida is more closely related to Mollusca than to Ecdysozoa.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn 1876, Van Beneden, found a group of microscopic ciliated wormlike organisms that inhabited the renal sacs of cephalopods, mainly octopuses and cuttlefishes and named them as Dicyemids.\n\nBiologists have been fascinated because of their highly-simplified body organization and complex life cycles of asexual (under normal condition producing vermiform larva in host kidney) and sexual (under crowded condition producing infusoriform larva excreted in host urine). These ciliated animals are composed of approximately 20-30 (or 40) cells arranged in two layers, and they lack coeloms, circulatory systems, and other differentiated tissues and their embryos employ spiral cleavage. Due to their simple body plans, it can be considered as intermediates between the Protozoa and the Metazoa.\n[figure1]\n\nThe phylogenetic position of dicyemids suggested by previous phylogenetic studies remains controversial.\n\nA: Based on the above explanations, infusiform larva is a mean of dispersal for these animals.\nB: Based on the above molecular data, these animals are missing link between Metazoa and protozoa.\nC: Based on the data, these described animals faced a regression evolution during its evolutionary history.\nD: Based on the given data, it is more likely that Dicyemida is more closely related to Mollusca than to Ecdysozoa.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-29.jpg?height=396&width=1536&top_left_y=758&top_left_x=266"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1063",
        "problem": "The shaker (sh) gene in Drosophila, when mutated, shows a typical behavior of continuous leg shaking. When the action potential of axons of such a shaker mutant of Drosophila was studied, it showed the following graph.\n\n[figure1]\n\nThe graph indicates that the defect lies in the:\nA: functioning of activation gate of voltage-gated $\\mathrm{Na}^{+}$channel.\nB: functioning of inactivation gate of voltage-gated $\\mathrm{Na}^{+}$channel.\nC: functioning of voltage-gated $\\mathrm{K}^{+}$channel.\nD: strength of the electrical stimulus.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe shaker (sh) gene in Drosophila, when mutated, shows a typical behavior of continuous leg shaking. When the action potential of axons of such a shaker mutant of Drosophila was studied, it showed the following graph.\n\n[figure1]\n\nThe graph indicates that the defect lies in the:\n\nA: functioning of activation gate of voltage-gated $\\mathrm{Na}^{+}$channel.\nB: functioning of inactivation gate of voltage-gated $\\mathrm{Na}^{+}$channel.\nC: functioning of voltage-gated $\\mathrm{K}^{+}$channel.\nD: strength of the electrical stimulus.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-03.jpg?height=624&width=1046&top_left_y=130&top_left_x=472"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1102",
        "problem": "Diagrams $A-D$ show the external or internal structures of leaves from different groups of plants.\n\n[figure1]\n\nA\n\n[figure2]\n\nB\n\n[figure3]\n\nC\n\n[figure4]\n\n$\\mathrm{D}$\n\nMatch these diagrams with the plants described by statements I - IV.\n\nI. The sexual reproductive structures are formed on an independent non-dominant plant body.\n\nII. Plants in this group have leaves with a haploid number of chromosomes.\n\nIII. Plants in this group are seed-bearing and amongst the earliest evolved extant trees.\n\nIV. The mega gametophytes develop within enclosed ovules.\nA: I-B, II-A, III-D, IV-C\nB: I-B, II-D, III- C, IV-A\nC: I-C, II-A, III-B, IV-D\nD: I-D, II-B, III-A, IV-C\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDiagrams $A-D$ show the external or internal structures of leaves from different groups of plants.\n\n[figure1]\n\nA\n\n[figure2]\n\nB\n\n[figure3]\n\nC\n\n[figure4]\n\n$\\mathrm{D}$\n\nMatch these diagrams with the plants described by statements I - IV.\n\nI. The sexual reproductive structures are formed on an independent non-dominant plant body.\n\nII. Plants in this group have leaves with a haploid number of chromosomes.\n\nIII. Plants in this group are seed-bearing and amongst the earliest evolved extant trees.\n\nIV. The mega gametophytes develop within enclosed ovules.\n\nA: I-B, II-A, III-D, IV-C\nB: I-B, II-D, III- C, IV-A\nC: I-C, II-A, III-B, IV-D\nD: I-D, II-B, III-A, IV-C\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-03.jpg?height=309&width=832&top_left_y=1252&top_left_x=297",
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-03.jpg?height=179&width=691&top_left_y=1320&top_left_x=1194",
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-03.jpg?height=529&width=889&top_left_y=1638&top_left_x=282",
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-03.jpg?height=521&width=678&top_left_y=1645&top_left_x=1190"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1346",
        "problem": "Radioactive phosphate $\\left({ }^{32} \\mathrm{PO}_{4}\\right)$ was applied to a plot of natural grassland. During the next 35 days, samples of four species of arthropod were investigated for radioactive content. The graph shows the relative amounts of radioactivity found in the four species, $\\mathrm{W}, \\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$.\n\n[figure1]\n\nWhich one of the rows $\\mathbf{A}-\\mathrm{E}$ represents the organisms in the graph?\nA: $\\mathrm{W}$: decomposer, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: herbivore, $\\mathbf{Z}$: carnivore\nB: $\\mathrm{W}$: herbivore, $\\mathbf{X}$: carnivore, $\\mathbf{Y}$: decomposer, $\\mathbf{Z}$: herbivore\nC: $\\mathrm{W}$: carnivore, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: decomposer, $\\mathbf{Z}$: herbivore\nD: $\\mathrm{W}$: grasses, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: carnivore, $\\mathbf{Z}$: decomposer\nE: $\\mathrm{W}$: herbivore, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: carnivore, $\\mathbf{Z}$: decomposer\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRadioactive phosphate $\\left({ }^{32} \\mathrm{PO}_{4}\\right)$ was applied to a plot of natural grassland. During the next 35 days, samples of four species of arthropod were investigated for radioactive content. The graph shows the relative amounts of radioactivity found in the four species, $\\mathrm{W}, \\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$.\n\n[figure1]\n\nWhich one of the rows $\\mathbf{A}-\\mathrm{E}$ represents the organisms in the graph?\n\nA: $\\mathrm{W}$: decomposer, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: herbivore, $\\mathbf{Z}$: carnivore\nB: $\\mathrm{W}$: herbivore, $\\mathbf{X}$: carnivore, $\\mathbf{Y}$: decomposer, $\\mathbf{Z}$: herbivore\nC: $\\mathrm{W}$: carnivore, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: decomposer, $\\mathbf{Z}$: herbivore\nD: $\\mathrm{W}$: grasses, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: carnivore, $\\mathbf{Z}$: decomposer\nE: $\\mathrm{W}$: herbivore, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: carnivore, $\\mathbf{Z}$: decomposer\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-05.jpg?height=388&width=998&top_left_y=348&top_left_x=561"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_913",
        "problem": "科研人员用一种甜瓜 $(2 n)$ 的纯合亲本进行杂交得到 $F_{1}, F_{1}$ 经自交得到 $F_{2}$, 结果如下表。已知 $\\mathrm{A} 、 \\mathrm{E}$ 基因同在一条染色体上， $\\mathrm{a} 、 \\mathrm{e}$ 基因同在另一条染色体上，当 $\\mathrm{E}$ 和 $\\mathrm{F}$同时存在时果皮才表现出有覆纹性状。不考虑互换、染色体变异、基因突变等情况。下列叙述错误的是（ ）\n\n| 性状 | 控制基因及其所在染色 <br> 体 | 母本 | 父本 | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 果皮底 <br> 色 | $\\mathrm{A} / \\mathrm{a}, 4$ 号染色体 | 黄绿 <br> 色 | 黄色 | 黄绿 <br> 色 | 黄绿色:黄色 3:1 |\n| 果肉颜 <br> 色 | $\\mathrm{B} / \\mathrm{b}, 9$ 号染色体 | 白色 | 橘红 <br> 色 | 橘红 <br> 色 | 橘红色:白色 $\\approx 3: 1$ |\n| 果皮覆 <br> 纹 | $\\mathrm{E} / \\mathrm{e}, 4$ 号染色体 F/f, 2 <br> 号染色体 | 无覆 <br> 纹 | 无覆 <br> 纹 | 有覆 <br> 纹 | 有覆纹: 无覆纹 $\\approx 9: 7$ |\nA: 果肉颜色的显性性状是橘红色\nB: $\\mathrm{F}_{1}$ 个体所产配子类型有 8 种\nC: $F_{2}$ 中黄色无覆纹果皮橘红色果肉的植株中杂合子所占比例是 $5 / 8$\nD: $F_{2}$ 中黄绿色有覆纹果皮、黄绿色无覆纹果皮、黄色无覆纹果皮的植株数量比是 9:3:4\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n科研人员用一种甜瓜 $(2 n)$ 的纯合亲本进行杂交得到 $F_{1}, F_{1}$ 经自交得到 $F_{2}$, 结果如下表。已知 $\\mathrm{A} 、 \\mathrm{E}$ 基因同在一条染色体上， $\\mathrm{a} 、 \\mathrm{e}$ 基因同在另一条染色体上，当 $\\mathrm{E}$ 和 $\\mathrm{F}$同时存在时果皮才表现出有覆纹性状。不考虑互换、染色体变异、基因突变等情况。下列叙述错误的是（ ）\n\n| 性状 | 控制基因及其所在染色 <br> 体 | 母本 | 父本 | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 果皮底 <br> 色 | $\\mathrm{A} / \\mathrm{a}, 4$ 号染色体 | 黄绿 <br> 色 | 黄色 | 黄绿 <br> 色 | 黄绿色:黄色 3:1 |\n| 果肉颜 <br> 色 | $\\mathrm{B} / \\mathrm{b}, 9$ 号染色体 | 白色 | 橘红 <br> 色 | 橘红 <br> 色 | 橘红色:白色 $\\approx 3: 1$ |\n| 果皮覆 <br> 纹 | $\\mathrm{E} / \\mathrm{e}, 4$ 号染色体 F/f, 2 <br> 号染色体 | 无覆 <br> 纹 | 无覆 <br> 纹 | 有覆 <br> 纹 | 有覆纹: 无覆纹 $\\approx 9: 7$ |\n\nA: 果肉颜色的显性性状是橘红色\nB: $\\mathrm{F}_{1}$ 个体所产配子类型有 8 种\nC: $F_{2}$ 中黄色无覆纹果皮橘红色果肉的植株中杂合子所占比例是 $5 / 8$\nD: $F_{2}$ 中黄绿色有覆纹果皮、黄绿色无覆纹果皮、黄色无覆纹果皮的植株数量比是 9:3:4\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1215",
        "problem": "In the following pedigree of a human family carrying the sex-linked allele for red-green colour-blindness, colourblind individuals are shown by shaded symbols.\n\n[figure1]\n\nAssuming no mutations, from which grandparent, A, B, C or D did X inherit his colour blindness?\nA: A\nB: B\nC: C\nD: D\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the following pedigree of a human family carrying the sex-linked allele for red-green colour-blindness, colourblind individuals are shown by shaded symbols.\n\n[figure1]\n\nAssuming no mutations, from which grandparent, A, B, C or D did X inherit his colour blindness?\n\nA: A\nB: B\nC: C\nD: D\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-18.jpg?height=368&width=593&top_left_y=301&top_left_x=758"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1223",
        "problem": "There were approximately 1733 tonnes of oil onboard the Rena when she ran aground. Approximately 350 tonnes were not recovered from the wreck and most of this washed ashore on Matakana Island and the coastline from Mt Maunganui to Maketu between 9 and 11 October 2011. Surveys examining the effects of the oil and debris washed onto the surf beaches focused on the northern tuatua (Paphies subtriangulata), as this is one of the most common species found burrowing in the sand on the open coast surf beaches that were most heavily fouled by oil from the Rena, and they are an important kai moana species.\n\nThe figure at right shows the level of total PAH in the tissue of tuatua from Papamoa and Omanu beaches from October 52011 to 30 June 2012. Before impact total PAH levels (background) were about $0.7 \\mu \\mathrm{g} / \\mathrm{kg}$ at Papamoa beach and $0.2 \\mu \\mathrm{g} / \\mathrm{kg}$ at Ōmanu Beach. These values were produced on a wet weight basis.\n\n[figure1]\n\nHeavily oiled\n\nModerately oiled\n\nLightly oiled\n\n[figure2]\n\nThe graph at left shows the total PAH\nlevels in tuatua in winter 2012 from beaches from Waihi - East Cape. They are colour-coded to represent the degree of oiling. No winter PAH data was available for Maketū Spit - Mid (shore level). Results obtained from Waihi and Ōhope beaches are considered background levels and an average between these levels (20.6 $\\mu \\mathrm{g} / \\mathrm{kg}$ ) is plotted as a dashed line. These values were produced on a dry weight basis.Total PAH levels in tuatua in winter 2012 were elevated above background levels at:\nA: Matakana Pipeline only.\nB: Papamoa Domain and Papamoa Taylors reserve only.\nC: Matakana Pipeline, Papamoa Domain and Papamoa Taylors reserve only.\nD: Matakana Pipeline, Papamoa Domain, Papamoa Taylors reserve and Maketu Spit only.\nE: Papamoa Domain, Papamoa Taylors reserve and Maketu Spit only.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nThere were approximately 1733 tonnes of oil onboard the Rena when she ran aground. Approximately 350 tonnes were not recovered from the wreck and most of this washed ashore on Matakana Island and the coastline from Mt Maunganui to Maketu between 9 and 11 October 2011. Surveys examining the effects of the oil and debris washed onto the surf beaches focused on the northern tuatua (Paphies subtriangulata), as this is one of the most common species found burrowing in the sand on the open coast surf beaches that were most heavily fouled by oil from the Rena, and they are an important kai moana species.\n\nThe figure at right shows the level of total PAH in the tissue of tuatua from Papamoa and Omanu beaches from October 52011 to 30 June 2012. Before impact total PAH levels (background) were about $0.7 \\mu \\mathrm{g} / \\mathrm{kg}$ at Papamoa beach and $0.2 \\mu \\mathrm{g} / \\mathrm{kg}$ at Ōmanu Beach. These values were produced on a wet weight basis.\n\n[figure1]\n\nHeavily oiled\n\nModerately oiled\n\nLightly oiled\n\n[figure2]\n\nThe graph at left shows the total PAH\nlevels in tuatua in winter 2012 from beaches from Waihi - East Cape. They are colour-coded to represent the degree of oiling. No winter PAH data was available for Maketū Spit - Mid (shore level). Results obtained from Waihi and Ōhope beaches are considered background levels and an average between these levels (20.6 $\\mu \\mathrm{g} / \\mathrm{kg}$ ) is plotted as a dashed line. These values were produced on a dry weight basis.\n\nproblem:\nTotal PAH levels in tuatua in winter 2012 were elevated above background levels at:\n\nA: Matakana Pipeline only.\nB: Papamoa Domain and Papamoa Taylors reserve only.\nC: Matakana Pipeline, Papamoa Domain and Papamoa Taylors reserve only.\nD: Matakana Pipeline, Papamoa Domain, Papamoa Taylors reserve and Maketu Spit only.\nE: Papamoa Domain, Papamoa Taylors reserve and Maketu Spit only.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-26.jpg?height=528&width=1105&top_left_y=196&top_left_x=772",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-26.jpg?height=708&width=1125&top_left_y=914&top_left_x=154"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1338",
        "problem": "Evolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).Tinamous and the moa last shared a common ancestor?\nA: $68.1 \\mathrm{Ma}$\nB: $65.3 \\mathrm{Ma}$\nC: $58.1 \\mathrm{Ma}$\nD: $41.7 \\mathrm{Ma}$\nE: $7.7 \\mathrm{Ma}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nEvolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).\n\nproblem:\nTinamous and the moa last shared a common ancestor?\n\nA: $68.1 \\mathrm{Ma}$\nB: $65.3 \\mathrm{Ma}$\nC: $58.1 \\mathrm{Ma}$\nD: $41.7 \\mathrm{Ma}$\nE: $7.7 \\mathrm{Ma}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-28.jpg?height=1868&width=1627&top_left_y=748&top_left_x=206"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1299",
        "problem": "In the Hill reaction, methylene blue may be reduced by illuminated isolated chloroplasts. The donor of electrons is\nA: Water\nB: Chlorophyll\nC: Oxygen\nD: ATP\nE: NADP\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the Hill reaction, methylene blue may be reduced by illuminated isolated chloroplasts. The donor of electrons is\n\nA: Water\nB: Chlorophyll\nC: Oxygen\nD: ATP\nE: NADP\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_167",
        "problem": "Jack has isolated five different polypeptides containing five amino acids (named A, B, C, D, E). He determines the molecular weight and the sequence of each polypeptide. The data, which he obtained is shown on the table below.\n\n| Polypeptide | Amino Acids Sequence (in form of <br> the tube containing it before) | Mass (Da) |\n| :---: | :---: | :---: |\n| $\\mathbf{1}$ | BCDACCDEDCB | 966 |\n| $\\mathbf{2}$ | ABBCAEEDECB | 1099 |\n| $\\mathbf{3}$ | BACDAEAEECA | 1357 |\n| $\\mathbf{4}$ | CACADBACAEB | 1279 |\n| $\\mathbf{5}$ | EDDCABBCCEE | 1014 |\n\nThe mass of individual amino acids are shown in the table below.\n\n| Amino Acids | Mass (Da) | Amino Acids | Mass (Da) |\n| :---: | :---: | :---: | :---: |\n| Alanine | 89 | Leucine | 131 |\n| Arginine | 174 | Lysine | 146 |\n| Asparagine | 132 | Methionine | 149 |\n| Aspartic Acid | 133 | Phenylalanine | 165 |\n| Cysteine | 121 | Proline | 115 |\n| Glutamic Acid | 147 | Serine | 105 |\n| Glutamine | 146 | Threonine | 119 |\n| Glycine | 75 | Tryptophan | 204 |\n| Histidine | 155 | Tyrosine | 181 |\n| Isoleucine | 131 | Valine | 117 |\n\nHint : in polymerization reaction, to form a peptide, when two different ends of amino acids are joined, a water molecule (mass : $18 \\mathrm{Da}$ ) is released.\nA: Amino acid named $\\mathrm{C}$ is serine\nB: Amino acid named $\\mathrm{A}$ is tyrosine\nC: Amino acid named $\\mathrm{E}$ is cysteine\nD: Amino acid named B is glycine\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nJack has isolated five different polypeptides containing five amino acids (named A, B, C, D, E). He determines the molecular weight and the sequence of each polypeptide. The data, which he obtained is shown on the table below.\n\n| Polypeptide | Amino Acids Sequence (in form of <br> the tube containing it before) | Mass (Da) |\n| :---: | :---: | :---: |\n| $\\mathbf{1}$ | BCDACCDEDCB | 966 |\n| $\\mathbf{2}$ | ABBCAEEDECB | 1099 |\n| $\\mathbf{3}$ | BACDAEAEECA | 1357 |\n| $\\mathbf{4}$ | CACADBACAEB | 1279 |\n| $\\mathbf{5}$ | EDDCABBCCEE | 1014 |\n\nThe mass of individual amino acids are shown in the table below.\n\n| Amino Acids | Mass (Da) | Amino Acids | Mass (Da) |\n| :---: | :---: | :---: | :---: |\n| Alanine | 89 | Leucine | 131 |\n| Arginine | 174 | Lysine | 146 |\n| Asparagine | 132 | Methionine | 149 |\n| Aspartic Acid | 133 | Phenylalanine | 165 |\n| Cysteine | 121 | Proline | 115 |\n| Glutamic Acid | 147 | Serine | 105 |\n| Glutamine | 146 | Threonine | 119 |\n| Glycine | 75 | Tryptophan | 204 |\n| Histidine | 155 | Tyrosine | 181 |\n| Isoleucine | 131 | Valine | 117 |\n\nHint : in polymerization reaction, to form a peptide, when two different ends of amino acids are joined, a water molecule (mass : $18 \\mathrm{Da}$ ) is released.\n\nA: Amino acid named $\\mathrm{C}$ is serine\nB: Amino acid named $\\mathrm{A}$ is tyrosine\nC: Amino acid named $\\mathrm{E}$ is cysteine\nD: Amino acid named B is glycine\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1103",
        "problem": "Some parasitoid flies respond to acoustic stimuli from crickets. Once the host is located, these flies lay their larvae on these insects. The maggots burrow into the bodies of crickets killing them. The graph depicts the tuning curves of parasitoid fly \\& cricket. In the accompanying figure, \" $c$ \" indicates frequency versus intensity spectrum of cricket song.\n\n[figure1]\n\nWhich of the following is correct?\nA: The parasitoid species would exhibit the call of very similar frequency distribution as that of cricket.\nB: The tympanic membrane \\& associated receptors that respond to frequencies lower than 4 $\\&$ greater than 8 will be selected in the course of evolution.\nC: The frequency of best sensitivity lies in the range of $5-8 \\mathrm{KHz}$ for a parasitoid fly.\nD: Mechanical filtering for narrow band hearing is more conspicuous in fly \" $X$ \" than in fly \" $Y$ \".\nE: \" $X$ \" is likely to represent the response of a female predator and \"Y\" of a male prey insect.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nSome parasitoid flies respond to acoustic stimuli from crickets. Once the host is located, these flies lay their larvae on these insects. The maggots burrow into the bodies of crickets killing them. The graph depicts the tuning curves of parasitoid fly \\& cricket. In the accompanying figure, \" $c$ \" indicates frequency versus intensity spectrum of cricket song.\n\n[figure1]\n\nWhich of the following is correct?\n\nA: The parasitoid species would exhibit the call of very similar frequency distribution as that of cricket.\nB: The tympanic membrane \\& associated receptors that respond to frequencies lower than 4 $\\&$ greater than 8 will be selected in the course of evolution.\nC: The frequency of best sensitivity lies in the range of $5-8 \\mathrm{KHz}$ for a parasitoid fly.\nD: Mechanical filtering for narrow band hearing is more conspicuous in fly \" $X$ \" than in fly \" $Y$ \".\nE: \" $X$ \" is likely to represent the response of a female predator and \"Y\" of a male prey insect.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-27.jpg?height=686&width=1000&top_left_y=1592&top_left_x=557"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_343",
        "problem": "图为细胞缺乏氨基酸时, 负载 tRNA 和空载 tRNA 调控基因表达的相关过程。下列叙述错误的是（ ）\n\n[图1]\nA: 过程(2)中多个核糖体同时合成一条多肽链可以加快翻译的速度\nB: 空载 tRNA 增多将导致相应 $\\mathrm{mRNA}$ 减少，从而避免细胞物质和能量的浪费\nC: 图示若干核糖体中, 核糖体 a 距起始密码子最远\nD: 当细胞中缺乏氨基酸时, 图中空载 tRNA 通过两条途径抑制基因表达\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图为细胞缺乏氨基酸时, 负载 tRNA 和空载 tRNA 调控基因表达的相关过程。下列叙述错误的是（ ）\n\n[图1]\n\nA: 过程(2)中多个核糖体同时合成一条多肽链可以加快翻译的速度\nB: 空载 tRNA 增多将导致相应 $\\mathrm{mRNA}$ 减少，从而避免细胞物质和能量的浪费\nC: 图示若干核糖体中, 核糖体 a 距起始密码子最远\nD: 当细胞中缺乏氨基酸时, 图中空载 tRNA 通过两条途径抑制基因表达\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-41.jpg?height=568&width=774&top_left_y=2035&top_left_x=367"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_397",
        "problem": "红眼雌果蝇与白眼雄果蝇交配, 子代雌雄果蝇都表现红眼, 这些雌雄果蝇交配产生的后代中, 红眼雄果蝇占 $1 / 4$, 白眼雄果蝇占 $1 / 4$, 红眼雌果蝇占 $1 / 2$ 。下列叙述错误的是 $(\\quad)$\nA: 红眼对白眼是显性\nB: 眼色遗传符合分离规律\nC: 眼色和性别表现自由组合\nD: 红眼和白眼基因位于 X 染色体上\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n红眼雌果蝇与白眼雄果蝇交配, 子代雌雄果蝇都表现红眼, 这些雌雄果蝇交配产生的后代中, 红眼雄果蝇占 $1 / 4$, 白眼雄果蝇占 $1 / 4$, 红眼雌果蝇占 $1 / 2$ 。下列叙述错误的是 $(\\quad)$\n\nA: 红眼对白眼是显性\nB: 眼色遗传符合分离规律\nC: 眼色和性别表现自由组合\nD: 红眼和白眼基因位于 X 染色体上\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_448",
        "problem": "图 1 为某单基因遗传病的系谱图。科研人员对 II 个体的该病相关基因用某种限制酶处理, 并进行电泳分析, 得到如图 2 结果。若不考虑致病基因位于 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体同源区\n\n[图1]\n\n图1\n\n[图2]\n\n患病女性\nA: 致病基因位于常染色体上\nB: 致病基因内部存在一个该种限制酶的酶切位点\nC: 图 2 中表示该致病基因的是条带(2)\nD: $\\mathrm{III}_{2}$ 和 $\\mathrm{III}_{4}$ 所生正常女儿携带致病基因的概率是 $3 / 5$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 1 为某单基因遗传病的系谱图。科研人员对 II 个体的该病相关基因用某种限制酶处理, 并进行电泳分析, 得到如图 2 结果。若不考虑致病基因位于 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体同源区\n\n[图1]\n\n图1\n\n[图2]\n\n患病女性\n\nA: 致病基因位于常染色体上\nB: 致病基因内部存在一个该种限制酶的酶切位点\nC: 图 2 中表示该致病基因的是条带(2)\nD: $\\mathrm{III}_{2}$ 和 $\\mathrm{III}_{4}$ 所生正常女儿携带致病基因的概率是 $3 / 5$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-035.jpg?height=248&width=462&top_left_y=270&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-035.jpg?height=246&width=218&top_left_y=271&top_left_x=859"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_256",
        "problem": "Students cultured plants, including four grass species (A. capillaris, $A$. odoratum, $F$. rubra, and $H$. lanatus) and four herbs (C. jacea, $L$. vulgare, P. lanceolata, and R. acetosa) without legumes, in different blocks with treatments of monoculture and mixtures of two, four or all eight species. They then measured different parameters as functions of plant species richness, as shown in the figure below. Values are shown in $\\log _{2}$ scale.\n\n[figure1]\n\n(d)\n\n[figure2]\n\n[figure3]\n\n[figure4]\n\n(e)\n\n[figure5]\n\n[figure6]\n\nFigure Q. 96\nA: In monoculture, both above ground and standing root biomass are lower than those in the mixtures of two, four or eight species.\nB: Plant species richness promotes soil $\\mathrm{C}$ stocks mainly through enhanced plant productivity, despite accelerated soil organic $\\mathrm{C}$ decomposition.\nC: Greater soil $\\mathrm{N}$ stocks at higher species richness is mainly attributed to increased $\\mathrm{N}$ retention, rather than $\\mathrm{N}$ input, with enhanced plant productivity.\nD: More diverse ecosystems can increase the potential for $\\mathrm{C}$ sequestration in terrestrial ecosystems.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nStudents cultured plants, including four grass species (A. capillaris, $A$. odoratum, $F$. rubra, and $H$. lanatus) and four herbs (C. jacea, $L$. vulgare, P. lanceolata, and R. acetosa) without legumes, in different blocks with treatments of monoculture and mixtures of two, four or all eight species. They then measured different parameters as functions of plant species richness, as shown in the figure below. Values are shown in $\\log _{2}$ scale.\n\n[figure1]\n\n(d)\n\n[figure2]\n\n[figure3]\n\n[figure4]\n\n(e)\n\n[figure5]\n\n[figure6]\n\nFigure Q. 96\n\nA: In monoculture, both above ground and standing root biomass are lower than those in the mixtures of two, four or eight species.\nB: Plant species richness promotes soil $\\mathrm{C}$ stocks mainly through enhanced plant productivity, despite accelerated soil organic $\\mathrm{C}$ decomposition.\nC: Greater soil $\\mathrm{N}$ stocks at higher species richness is mainly attributed to increased $\\mathrm{N}$ retention, rather than $\\mathrm{N}$ input, with enhanced plant productivity.\nD: More diverse ecosystems can increase the potential for $\\mathrm{C}$ sequestration in terrestrial ecosystems.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-101.jpg?height=433&width=525&top_left_y=679&top_left_x=251",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-101.jpg?height=463&width=510&top_left_y=1172&top_left_x=264",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-101.jpg?height=434&width=507&top_left_y=664&top_left_x=819",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-101.jpg?height=431&width=494&top_left_y=663&top_left_x=1379",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-101.jpg?height=468&width=510&top_left_y=1161&top_left_x=812",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-101.jpg?height=499&width=528&top_left_y=1120&top_left_x=1368"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_552",
        "problem": "噬菌体展示技术 (如下图) 可将某些蛋白质呈递至噬菌体表面, 便于对目标蛋白进行笁选、鉴定。以下对该技术的分析错误的是（）\n\n[图1]\n\n## 建立噬菌体展示库\nA: 聚合酶\nB: 可利用抗原-抗体杂交技术篮选目标蛋白\nC: 可用细菌细胞作宿主培养噬菌体\nD: 该技术可用于获得与抗原亲和力更强的抗体及其基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n噬菌体展示技术 (如下图) 可将某些蛋白质呈递至噬菌体表面, 便于对目标蛋白进行笁选、鉴定。以下对该技术的分析错误的是（）\n\n[图1]\n\n## 建立噬菌体展示库\n\nA: 聚合酶\nB: 可利用抗原-抗体杂交技术篮选目标蛋白\nC: 可用细菌细胞作宿主培养噬菌体\nD: 该技术可用于获得与抗原亲和力更强的抗体及其基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-07.jpg?height=445&width=1399&top_left_y=163&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_485",
        "problem": "如图是某同学调查的某家族遗传病系谱图, 与该病相关的基因为 $\\mathrm{G} 、 \\mathrm{~g}$ 。下列有关说法错误的是 ( )\n\n[图1]\nA: 该病为常染色体显性遗传病\nB: $\\mathrm{I}_{2} 、 \\mathrm{II}_{4} 、 \\mathrm{II}_{5}$ 个体的基因型均为 $\\mathrm{Gg}$\nC: $\\mathrm{III}_{7} 、 \\mathrm{III}_{9}$ 个体的基因型均与 $\\mathrm{I}_{1}$ 相同\nD: $\\mathrm{III}_{7} 、 \\mathrm{III}_{8}$ 婚配后生了一男孩，其正常的概率为 $1 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图是某同学调查的某家族遗传病系谱图, 与该病相关的基因为 $\\mathrm{G} 、 \\mathrm{~g}$ 。下列有关说法错误的是 ( )\n\n[图1]\n\nA: 该病为常染色体显性遗传病\nB: $\\mathrm{I}_{2} 、 \\mathrm{II}_{4} 、 \\mathrm{II}_{5}$ 个体的基因型均为 $\\mathrm{Gg}$\nC: $\\mathrm{III}_{7} 、 \\mathrm{III}_{9}$ 个体的基因型均与 $\\mathrm{I}_{1}$ 相同\nD: $\\mathrm{III}_{7} 、 \\mathrm{III}_{8}$ 婚配后生了一男孩，其正常的概率为 $1 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-16.jpg?height=397&width=946&top_left_y=815&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1275",
        "problem": "Tuberculosis (TB) kills more than two million people annually and is a disease studied by scientists around the world.\n\n[figure1]\n\nhttp://www.plosone.org/article/info\\%3Adoi\\%2F10.1371\\%2Fjournal.p one. 0091024\nThe diagram on the left shows the chronology of a TB outbreak in New Zealand. \"Each square represent a subject at the time of the TB diagnosis. Broken lines represent known close direct contact with the initial index case \" $X$ \" during $X$ 's period of infectiousness. Solid lines show assumed connections between a case of presumed reactivation (C1 to $\\mathrm{C} 2$ ) and a case of potential child-parent transmission (A to $\\mathrm{T}$ )\".\n\nWhat is the best conclusion to be made from these data?\nA: Tuberculosis is widespread in New Zealand.\nB: Tuberculosis can stay dormant in a human before symptoms are shown.\nC: Tuberculosis outbreaks occur in geographical patterns.\nD: Patient $\\mathrm{X}$ and Patient $\\mathrm{A}$ are responsible for all of the tuberculosis infections in this study.\nE: None of these conclusions are valid.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTuberculosis (TB) kills more than two million people annually and is a disease studied by scientists around the world.\n\n[figure1]\n\nhttp://www.plosone.org/article/info\\%3Adoi\\%2F10.1371\\%2Fjournal.p one. 0091024\nThe diagram on the left shows the chronology of a TB outbreak in New Zealand. \"Each square represent a subject at the time of the TB diagnosis. Broken lines represent known close direct contact with the initial index case \" $X$ \" during $X$ 's period of infectiousness. Solid lines show assumed connections between a case of presumed reactivation (C1 to $\\mathrm{C} 2$ ) and a case of potential child-parent transmission (A to $\\mathrm{T}$ )\".\n\nWhat is the best conclusion to be made from these data?\n\nA: Tuberculosis is widespread in New Zealand.\nB: Tuberculosis can stay dormant in a human before symptoms are shown.\nC: Tuberculosis outbreaks occur in geographical patterns.\nD: Patient $\\mathrm{X}$ and Patient $\\mathrm{A}$ are responsible for all of the tuberculosis infections in this study.\nE: None of these conclusions are valid.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-19.jpg?height=653&width=883&top_left_y=1421&top_left_x=124"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1200",
        "problem": "## MARINE BIODIVERSITY - THE CENSUS OF MARINE LIFE\n\nThe Census of Marine Life (2000-2010) is the largest global research programme on marine biodiversity. Its findings have recently been integrated in an article authored by Mark Costello who is based at Leigh Marine Laboratory, University of Auckland (PLoS ONE 5(8): e12110). The study found that many habitats were poorly sampled and that there are major gaps in our knowledge of marine organisms worldwide that limit our ability to understand species of economic and ecological importance.\n\nThe table below gives the number of endemic plants, invertebrates, and vertebrates reported for specific geographic regions. Endemic species are those found only in one specific geographical area.\n\n|  |  |  |  |  |  |  |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| NRIC region | Plants | Invertebrates | Fish | Other vertebrates | Total | Number of species | $\\%$ endemics |\n| Antarctica | - | - | - | - | 3,700 | 8,200 | 45 |\n| Australia | - | 7987 | 1298 | - | 9,286 | 32,889 | 28 |\n| Baltic | 1 | 0 | 0 | 0 | 1 | 5,865 | 2 |\n| Caribbean | - | 868 | 704 | 1 | 1,573 | 12,046 | 13 |\n| China | 142 | 1387 | 70 | 2 | 1,601 | 22,365 | 7 |\n| Japan | - | 1508 | 364 | 0 | 1,872 | 32,777 | 6 |\n| Mediterranean | 171 | 844 | 80 | 3 | 1,098 | 16,845 | 7 |\n| New Zealand | 225 | 6014 | 278 | 43 | 6,560 | 12,780 | 51 |\n| South Africa | - | 3269 | 280 | - | 3,549 | 12,715 | 28 |\n| Total | 538 | 21,639 | 3,074 | 49 | 25,300 | 150,617 | 17 |\n\ndoi:10.1371/journal.pone.0012110.t004\n\nNote: - means no specific data is available.Which area has the greatest proportion of endemic invertebrates?\nA: Antarctica\nB: Australia\nC: Japan\nD: New Zealand\nE: South Africa\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## MARINE BIODIVERSITY - THE CENSUS OF MARINE LIFE\n\nThe Census of Marine Life (2000-2010) is the largest global research programme on marine biodiversity. Its findings have recently been integrated in an article authored by Mark Costello who is based at Leigh Marine Laboratory, University of Auckland (PLoS ONE 5(8): e12110). The study found that many habitats were poorly sampled and that there are major gaps in our knowledge of marine organisms worldwide that limit our ability to understand species of economic and ecological importance.\n\nThe table below gives the number of endemic plants, invertebrates, and vertebrates reported for specific geographic regions. Endemic species are those found only in one specific geographical area.\n\n|  |  |  |  |  |  |  |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| NRIC region | Plants | Invertebrates | Fish | Other vertebrates | Total | Number of species | $\\%$ endemics |\n| Antarctica | - | - | - | - | 3,700 | 8,200 | 45 |\n| Australia | - | 7987 | 1298 | - | 9,286 | 32,889 | 28 |\n| Baltic | 1 | 0 | 0 | 0 | 1 | 5,865 | 2 |\n| Caribbean | - | 868 | 704 | 1 | 1,573 | 12,046 | 13 |\n| China | 142 | 1387 | 70 | 2 | 1,601 | 22,365 | 7 |\n| Japan | - | 1508 | 364 | 0 | 1,872 | 32,777 | 6 |\n| Mediterranean | 171 | 844 | 80 | 3 | 1,098 | 16,845 | 7 |\n| New Zealand | 225 | 6014 | 278 | 43 | 6,560 | 12,780 | 51 |\n| South Africa | - | 3269 | 280 | - | 3,549 | 12,715 | 28 |\n| Total | 538 | 21,639 | 3,074 | 49 | 25,300 | 150,617 | 17 |\n\ndoi:10.1371/journal.pone.0012110.t004\n\nNote: - means no specific data is available.\n\nproblem:\nWhich area has the greatest proportion of endemic invertebrates?\n\nA: Antarctica\nB: Australia\nC: Japan\nD: New Zealand\nE: South Africa\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_974",
        "problem": "The flippers of dolphins are exposed to cold water away from the main mass of the body and employ countercurrent heat exchange. Which of the following statements is FALSE?\nA: Each artery in the flipper is surrounded by several veins, allowing efficient heat exchange.\nB: As long as there is a difference in temperatures, heat transfer occurs from the warmer vessel to the cooler one.\nC: The coolest blood is found at the tip of the dolphin flipper.\nD: Dolphins are the only mammals known to employ countercurrent heat exchange.\nE: Once the arterial blood has been cooled past a certain point, it will no longer transfer heat to the veins.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe flippers of dolphins are exposed to cold water away from the main mass of the body and employ countercurrent heat exchange. Which of the following statements is FALSE?\n\nA: Each artery in the flipper is surrounded by several veins, allowing efficient heat exchange.\nB: As long as there is a difference in temperatures, heat transfer occurs from the warmer vessel to the cooler one.\nC: The coolest blood is found at the tip of the dolphin flipper.\nD: Dolphins are the only mammals known to employ countercurrent heat exchange.\nE: Once the arterial blood has been cooled past a certain point, it will no longer transfer heat to the veins.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_964",
        "problem": "The completion of meiosis in males produces four spermatids, each containing\nA: 23 chromosomes\nB: 23 pairs of chromosomes\nC: Diploid number of chromosomes\nD: Haploid number of chromosomes\nE: None of the above\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe completion of meiosis in males produces four spermatids, each containing\n\nA: 23 chromosomes\nB: 23 pairs of chromosomes\nC: Diploid number of chromosomes\nD: Haploid number of chromosomes\nE: None of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1538",
        "problem": "The RECOVERY trial run by the NHS is the most successful trial to identify proven treatments for COVID-19, and has disproven several popular candidates.\n\nWhat things can a randomised control trial like RECOVERY NOT conclude?\nA: What causes a disease.\nB: Whether a drug causes the improvement or is just correlated with it.\nC: Whether a drug is better than placebos.\nD: Whether a drug has side-effects or they are just symptoms of diseases.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe RECOVERY trial run by the NHS is the most successful trial to identify proven treatments for COVID-19, and has disproven several popular candidates.\n\nWhat things can a randomised control trial like RECOVERY NOT conclude?\n\nA: What causes a disease.\nB: Whether a drug causes the improvement or is just correlated with it.\nC: Whether a drug is better than placebos.\nD: Whether a drug has side-effects or they are just symptoms of diseases.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_878",
        "problem": "食物长于无名指为长食指, 反之为短食指, 该相对性状由常染色体上一对等位基因控制( $\\mathrm{T}^{\\mathrm{s}}$ 表示短食指基因, $\\mathrm{T}^{\\mathrm{L}}$ 为长食指基因).此等位基因表达受性激素影响, $\\mathrm{T}^{\\mathrm{S}}$ 在男性为显性, $\\mathrm{T}^{\\mathrm{L}}$ 在女性为显性。若一对夫妇均为短食指,所生孩子中既有长食指又有短食指，则该夫妇再生一个孩子是长食指的概率为（ )\nA: $\\frac{1}{4}$\nB: $\\frac{i}{3}$\nC: $\\frac{1}{2}$\nD: $\\frac{3}{4}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n食物长于无名指为长食指, 反之为短食指, 该相对性状由常染色体上一对等位基因控制( $\\mathrm{T}^{\\mathrm{s}}$ 表示短食指基因, $\\mathrm{T}^{\\mathrm{L}}$ 为长食指基因).此等位基因表达受性激素影响, $\\mathrm{T}^{\\mathrm{S}}$ 在男性为显性, $\\mathrm{T}^{\\mathrm{L}}$ 在女性为显性。若一对夫妇均为短食指,所生孩子中既有长食指又有短食指，则该夫妇再生一个孩子是长食指的概率为（ )\n\nA: $\\frac{1}{4}$\nB: $\\frac{i}{3}$\nC: $\\frac{1}{2}$\nD: $\\frac{3}{4}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1409",
        "problem": "On the left is corn yield, on the right is seal reproduction. Interpret these and answer the questions below:\n[figure1]\n\nIn the seal figure, the positive line may suggest:\nA: more food available as the seals age\nB: as the population density increases, the seals have less food\nC: as the population density decreases, seal age at reproduction increases.\nD: less food available as the seals age\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOn the left is corn yield, on the right is seal reproduction. Interpret these and answer the questions below:\n[figure1]\n\nIn the seal figure, the positive line may suggest:\n\nA: more food available as the seals age\nB: as the population density increases, the seals have less food\nC: as the population density decreases, seal age at reproduction increases.\nD: less food available as the seals age\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-06.jpg?height=520&width=1200&top_left_y=1599&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1545",
        "problem": "The adult cells of baobab trees have 168 chromosomes, compared to 46 in humans. These contain four copies of the genome, compared to two copies in humans.\n\n[figure1]\n\nHow many chromosomes does a baobab gamete contain?\nA: 168\nB: 84\nC: 63\nD: 42\nE: 21\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe adult cells of baobab trees have 168 chromosomes, compared to 46 in humans. These contain four copies of the genome, compared to two copies in humans.\n\n[figure1]\n\nHow many chromosomes does a baobab gamete contain?\n\nA: 168\nB: 84\nC: 63\nD: 42\nE: 21\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-49.jpg?height=1228&width=925&top_left_y=474&top_left_x=240"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_869",
        "problem": "某家系的遗传系谱图及部分个体基因型如图所示，A1、A2、A3 是位于 $\\mathrm{X}$ 染色体上的等位基因。下列推断正确的是( )\n\n[图1]\nA: II-2 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 2}$ 的概率是 $1 / 4$\nB: III-1 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{Y}$ 的概率是 $1 / 4$\nC: III- 2 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 2}$ 的概率是 $1 / 8$\nD: IV- 1 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 1}$ 概率是 $1 / 8$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某家系的遗传系谱图及部分个体基因型如图所示，A1、A2、A3 是位于 $\\mathrm{X}$ 染色体上的等位基因。下列推断正确的是( )\n\n[图1]\n\nA: II-2 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 2}$ 的概率是 $1 / 4$\nB: III-1 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{Y}$ 的概率是 $1 / 4$\nC: III- 2 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 2}$ 的概率是 $1 / 8$\nD: IV- 1 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 1}$ 概率是 $1 / 8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-09.jpg?height=371&width=439&top_left_y=1071&top_left_x=203"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_796",
        "problem": "下图(1)染色体着丝粒两端的染色体臂称为长臂和短臂, 若只在长臂或短臂内部发生倒位为臂内倒位; 若长臂和短臂各发生一次断裂, 断片倒转 180 度发生的倒位为臂间倒位。在发生倒位的杂合体中, 若倒位环内的非姐妹染色单体间发生了一次单交换, 其过程如图(2) $\\rightarrow$ (3) $\\rightarrow$ (4)所示。相关叙述错误的是（ )\n[图1]\n\n[图2]\n\n(3)\n\n[图3]\n\n(4)\nA: 图(2)倒位环的形成是由于染色体片段 DE 发生了臂内倒位\nB: 图(2)倒位环内的非姐妹染色单体间交换发生于减数第一次分裂\nC: (2) $\\rightarrow$ (3) (4)中, 交换的产物都带有缺失或重复, 不能形成有功能的配子\nD: 正常同源染色体的非姐妹染色单体间一次单交换后也可产生 4 种配子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图(1)染色体着丝粒两端的染色体臂称为长臂和短臂, 若只在长臂或短臂内部发生倒位为臂内倒位; 若长臂和短臂各发生一次断裂, 断片倒转 180 度发生的倒位为臂间倒位。在发生倒位的杂合体中, 若倒位环内的非姐妹染色单体间发生了一次单交换, 其过程如图(2) $\\rightarrow$ (3) $\\rightarrow$ (4)所示。相关叙述错误的是（ )\n[图1]\n\n[图2]\n\n(3)\n\n[图3]\n\n(4)\n\nA: 图(2)倒位环的形成是由于染色体片段 DE 发生了臂内倒位\nB: 图(2)倒位环内的非姐妹染色单体间交换发生于减数第一次分裂\nC: (2) $\\rightarrow$ (3) (4)中, 交换的产物都带有缺失或重复, 不能形成有功能的配子\nD: 正常同源染色体的非姐妹染色单体间一次单交换后也可产生 4 种配子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-79.jpg?height=202&width=694&top_left_y=590&top_left_x=333",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-79.jpg?height=265&width=345&top_left_y=490&top_left_x=1027",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-79.jpg?height=258&width=380&top_left_y=522&top_left_x=1406"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1362",
        "problem": "Like many mammals, cat species use the $X Y$ sex-determination system where males have the sex chromosomes XY and females have the sex chromosomes XX. In a specific breed of cats called tortoiseshell cats, the major gene for fur colour is located on the Xchromosome. There are two fur colour alleles for this gene: one for orange fur and one for black fur.\n\nIn a specific litter of kittens, there are three kittens born with the following fur colours: pure orange fur, pure black fur, and one with both orange and black fur.\n\n[figure1]\n\nAn orange fur female tortoiseshell cat was crossed with a black fur male tortoiseshell cat and had a litter of six kittens. Which of the following offspring colours are most likely?\nA: 3 black fur male kittens and 3 mixed fur (orange and black fur) female kittens\nB: 4 orange fur male kittens and 2 mixed fur female kittens\nC: 2 black fur kittens, 2 orange fur kittens, 2 mixed fur kittens\nD: 1 mixed fur male kitten, 5 orange female kittens\nE: All kittens will be of a mixed fur pattern\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nLike many mammals, cat species use the $X Y$ sex-determination system where males have the sex chromosomes XY and females have the sex chromosomes XX. In a specific breed of cats called tortoiseshell cats, the major gene for fur colour is located on the Xchromosome. There are two fur colour alleles for this gene: one for orange fur and one for black fur.\n\nIn a specific litter of kittens, there are three kittens born with the following fur colours: pure orange fur, pure black fur, and one with both orange and black fur.\n\n[figure1]\n\nAn orange fur female tortoiseshell cat was crossed with a black fur male tortoiseshell cat and had a litter of six kittens. Which of the following offspring colours are most likely?\n\nA: 3 black fur male kittens and 3 mixed fur (orange and black fur) female kittens\nB: 4 orange fur male kittens and 2 mixed fur female kittens\nC: 2 black fur kittens, 2 orange fur kittens, 2 mixed fur kittens\nD: 1 mixed fur male kitten, 5 orange female kittens\nE: All kittens will be of a mixed fur pattern\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-31.jpg?height=325&width=468&top_left_y=674&top_left_x=794"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_186",
        "problem": "Speciation rates are variable in different lineages of organisms. Some lineages have many species; others have only a few.\nA: The larger the number of species in a lineage, the larger number of opportunities for new species to form.\nB: Animal-pollinated plant families have, on average, more species than closely related families pollinated by wind.\nC: Animals with complex behavior are likely to form new species at a low rate.\nD: Oscillations of climates may increase the speciation rate.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nSpeciation rates are variable in different lineages of organisms. Some lineages have many species; others have only a few.\n\nA: The larger the number of species in a lineage, the larger number of opportunities for new species to form.\nB: Animal-pollinated plant families have, on average, more species than closely related families pollinated by wind.\nC: Animals with complex behavior are likely to form new species at a low rate.\nD: Oscillations of climates may increase the speciation rate.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1278",
        "problem": "A suspension of microscopic green algae was divided into two equal samples. Each was given the SAME total amount of light energy. Sample I was exposed to continuous light. Sample II was exposed to bursts of light for 5 - 10 seconds duration, followed by dark periods. Photosynthesis took place in both samples, but more occurred in sample I. From this evidence we may conclude that\nA: More photosynthesis occurs in the dark than in the light.\nB: Some part of the photosynthetic process can occur in darkness.\nC: Photosynthesis requires darkness as well as light.\nD: Photosynthesis is a very rapid process.\nE: Photosynthesis involves enzymes as well as light.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA suspension of microscopic green algae was divided into two equal samples. Each was given the SAME total amount of light energy. Sample I was exposed to continuous light. Sample II was exposed to bursts of light for 5 - 10 seconds duration, followed by dark periods. Photosynthesis took place in both samples, but more occurred in sample I. From this evidence we may conclude that\n\nA: More photosynthesis occurs in the dark than in the light.\nB: Some part of the photosynthetic process can occur in darkness.\nC: Photosynthesis requires darkness as well as light.\nD: Photosynthesis is a very rapid process.\nE: Photosynthesis involves enzymes as well as light.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_837",
        "problem": "家鸡的正常啄和交叉啄分别由 $\\mathrm{Z}$ 染色体上的 $\\mathrm{E}$ 和 e 基因控制, 其中某种基因型雌性致死。现在有一对家鸡杂交, 子一代中 $O^{\\lambda}: Q=2: 1$ 。下列分析合理的是（）\nA: 若该家鸡种群中存在基因型为 $\\mathrm{ZZ}$ 的个体，则推测致死基因型为 $\\mathrm{Z}^{\\mathrm{E}} \\mathrm{W}$\nB: 若子一代雄性个体中有杂合子，则推测致死基因型为 $Z^{e} W$\nC: 若子一代雄性个体均为纯合子，则推测致死基因型为 $Z^{E} W$\nD: 若子一代中出现正常喙的概率为 $1 / 3$, 则推测致死基因型为 $Z^{E} W$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n家鸡的正常啄和交叉啄分别由 $\\mathrm{Z}$ 染色体上的 $\\mathrm{E}$ 和 e 基因控制, 其中某种基因型雌性致死。现在有一对家鸡杂交, 子一代中 $O^{\\lambda}: Q=2: 1$ 。下列分析合理的是（）\n\nA: 若该家鸡种群中存在基因型为 $\\mathrm{ZZ}$ 的个体，则推测致死基因型为 $\\mathrm{Z}^{\\mathrm{E}} \\mathrm{W}$\nB: 若子一代雄性个体中有杂合子，则推测致死基因型为 $Z^{e} W$\nC: 若子一代雄性个体均为纯合子，则推测致死基因型为 $Z^{E} W$\nD: 若子一代中出现正常喙的概率为 $1 / 3$, 则推测致死基因型为 $Z^{E} W$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_17",
        "problem": "Figure Q. 21 shows how lung ventilation is affected by physical activity. As the intensity of exercise increases, humans respond to the increased need for gas exchange in two ways: increase in ventilation rate and increase in tidal volume. The experimental data for a runner on a treadmill are shown in Figure Q.21.\n\n[figure1]\n\nFigure Q. 21\nA: At the onset of physical exercise, the rate of ventilation increases before the depth of ventilation increases.\nB: In intense physical exercise ( $>15 \\mathrm{~km} / \\mathrm{h}$ treadmill speed), the increase in the ventilatory minute volume is mainly caused by increased ventilation rate.\nC: At the treadmill speed of $15 \\mathrm{~km} / \\mathrm{h}$, the ventilatory volume per minute is approximately $120 \\mathrm{dm}^{3}$.\nD: In an adult human, a tidal volume of $200 \\mathrm{~cm}^{3}$ and ventilation rate of 30 breaths per minute can provide equally effective gas exchange as a tidal volume of $600 \\mathrm{~cm}^{3}$ and a ventilation rate of 10 breaths per minute.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFigure Q. 21 shows how lung ventilation is affected by physical activity. As the intensity of exercise increases, humans respond to the increased need for gas exchange in two ways: increase in ventilation rate and increase in tidal volume. The experimental data for a runner on a treadmill are shown in Figure Q.21.\n\n[figure1]\n\nFigure Q. 21\n\nA: At the onset of physical exercise, the rate of ventilation increases before the depth of ventilation increases.\nB: In intense physical exercise ( $>15 \\mathrm{~km} / \\mathrm{h}$ treadmill speed), the increase in the ventilatory minute volume is mainly caused by increased ventilation rate.\nC: At the treadmill speed of $15 \\mathrm{~km} / \\mathrm{h}$, the ventilatory volume per minute is approximately $120 \\mathrm{dm}^{3}$.\nD: In an adult human, a tidal volume of $200 \\mathrm{~cm}^{3}$ and ventilation rate of 30 breaths per minute can provide equally effective gas exchange as a tidal volume of $600 \\mathrm{~cm}^{3}$ and a ventilation rate of 10 breaths per minute.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-048.jpg?height=980&width=1469&top_left_y=729&top_left_x=291",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-049.jpg?height=819&width=1192&top_left_y=1296&top_left_x=431"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1263",
        "problem": "[figure1]\n\nThe diagram shows the daily energy budget of a small rodent weighing $23 \\mathrm{~g}$ living in captivity during the summer.\n\nThe information in the diagram suggests that:\nA: Less than $1 \\%$ of the food eaten is used for new growth.\nB: In captivity the vole spends most of its time in its nest.\nC: The rate of heat loss is lower when the vole is active.\nD: The vole's metabolism is adapted to warm weather conditions.\nE: The vole converts food into flesh more efficiently when in its nest.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nThe diagram shows the daily energy budget of a small rodent weighing $23 \\mathrm{~g}$ living in captivity during the summer.\n\nThe information in the diagram suggests that:\n\nA: Less than $1 \\%$ of the food eaten is used for new growth.\nB: In captivity the vole spends most of its time in its nest.\nC: The rate of heat loss is lower when the vole is active.\nD: The vole's metabolism is adapted to warm weather conditions.\nE: The vole converts food into flesh more efficiently when in its nest.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-13.jpg?height=542&width=1056&top_left_y=1531&top_left_x=503"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_821",
        "problem": "很多因素都会造成 DNA 损伤, 机体也能够对损伤 DNA 进行修复, 当 DNA 损伤较大时，损伤部位难以直接被修复，可先进行复制再修复，修复过程如图所示。下列叙述试卷第 90 页，共 93 页\n错误的是\n\n复制：损伤部位不能做损伤部位\n\n模板，子链对应\n\n[图1]\nA: DNA 修复是以未损伤的 DNA 互补链为模板, 利用脱氧核苷酸为原料进行的\nB: 过程(2)(3)中, 母链缺口的产生需要在限制酶的作用下切开磷酸二酯键\nC: 修补过程, 需要解旋酶、限制酶和 DNA 聚合酶\nD: 某受损伤的 DNA 进行 10 次复制并按图示进行损伤修复, 则有损伤的 DNA 分子将占 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n很多因素都会造成 DNA 损伤, 机体也能够对损伤 DNA 进行修复, 当 DNA 损伤较大时，损伤部位难以直接被修复，可先进行复制再修复，修复过程如图所示。下列叙述试卷第 90 页，共 93 页\n错误的是\n\n复制：损伤部位不能做损伤部位\n\n模板，子链对应\n\n[图1]\n\nA: DNA 修复是以未损伤的 DNA 互补链为模板, 利用脱氧核苷酸为原料进行的\nB: 过程(2)(3)中, 母链缺口的产生需要在限制酶的作用下切开磷酸二酯键\nC: 修补过程, 需要解旋酶、限制酶和 DNA 聚合酶\nD: 某受损伤的 DNA 进行 10 次复制并按图示进行损伤修复, 则有损伤的 DNA 分子将占 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-91.jpg?height=1025&width=691&top_left_y=290&top_left_x=791"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1495",
        "problem": "Last year, scientists were stunned when captive sturgeon fish and paddlefish mated and produced offspring. These fish species diverged about 200 million years ago at the beginning of the dinosaur era.\n\n[figure1]\n\nWhich factors usually do NOT stop different species breeding?\nA: Live in different places.\nB: Not sexually attracted to each other.\nC: Different numbers of chromosomes.\nD: Different genetic triplet codes.\nE: Embryo does not develop.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nLast year, scientists were stunned when captive sturgeon fish and paddlefish mated and produced offspring. These fish species diverged about 200 million years ago at the beginning of the dinosaur era.\n\n[figure1]\n\nWhich factors usually do NOT stop different species breeding?\n\nA: Live in different places.\nB: Not sexually attracted to each other.\nC: Different numbers of chromosomes.\nD: Different genetic triplet codes.\nE: Embryo does not develop.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-20.jpg?height=742&width=1485&top_left_y=477&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_851",
        "problem": "林某（女）在新生儿体检中被诊断为 C6PD（葡萄糖-6-磷酸脱氢酶）缺乏, C6PD 基因 (记作 $\\mathrm{G} / \\mathrm{g}$ ) 位于 X 染色体上, 对其家系的进一步检查的结果如下表。正常的 C6PD 基因可以表达有活性的 C6PD。雌性哺乳动物在胚胎发育早期，体细胞中的 X 染色体会有一条随机失活（部分基因不表达）。下列叙述正确的是（）\n\n| 家庭成 <br> 员 | G6PD 活 <br> 性 | G6PD 基因测序 | 染色体组成 |\n| :---: | :---: | :---: | :---: |\n| 父亲 | 0.54 | 第 1388 位碱基突变 $(\\mathrm{G} \\rightarrow \\mathrm{A})$; 纯 <br> 合 | $44+X Y$ |\n| 母亲 | 2.31 | 无突变 | $44+X X$ |\n| 林某 | 0.62 | 第 1388 位碱基突变 $(\\mathrm{G} \\rightarrow \\mathrm{A})$; 杂 <br> 合 | $44+X \\quad(67 \\%) 、 44+X X \\quad(33 \\%)$ |\nA: 能确定林某父亲的基因型是 $X^{\\mathrm{G}} Y$ 、母亲的基因型为 $X^{g} X^{g}$\nB: 林某部分细胞染色体组成异常是因为母亲形成的配子中 X 染色体丢失\nC: 林某丢失的染色体中, 携带正常 G6PD 基因的 X 染色体占比较高\nD: 胚胎发育早期携带有突变 G6PD 基因的 X 染色体优先失活\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n林某（女）在新生儿体检中被诊断为 C6PD（葡萄糖-6-磷酸脱氢酶）缺乏, C6PD 基因 (记作 $\\mathrm{G} / \\mathrm{g}$ ) 位于 X 染色体上, 对其家系的进一步检查的结果如下表。正常的 C6PD 基因可以表达有活性的 C6PD。雌性哺乳动物在胚胎发育早期，体细胞中的 X 染色体会有一条随机失活（部分基因不表达）。下列叙述正确的是（）\n\n| 家庭成 <br> 员 | G6PD 活 <br> 性 | G6PD 基因测序 | 染色体组成 |\n| :---: | :---: | :---: | :---: |\n| 父亲 | 0.54 | 第 1388 位碱基突变 $(\\mathrm{G} \\rightarrow \\mathrm{A})$; 纯 <br> 合 | $44+X Y$ |\n| 母亲 | 2.31 | 无突变 | $44+X X$ |\n| 林某 | 0.62 | 第 1388 位碱基突变 $(\\mathrm{G} \\rightarrow \\mathrm{A})$; 杂 <br> 合 | $44+X \\quad(67 \\%) 、 44+X X \\quad(33 \\%)$ |\n\nA: 能确定林某父亲的基因型是 $X^{\\mathrm{G}} Y$ 、母亲的基因型为 $X^{g} X^{g}$\nB: 林某部分细胞染色体组成异常是因为母亲形成的配子中 X 染色体丢失\nC: 林某丢失的染色体中, 携带正常 G6PD 基因的 X 染色体占比较高\nD: 胚胎发育早期携带有突变 G6PD 基因的 X 染色体优先失活\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1460",
        "problem": "Anna is studying a population of seabirds which roost on one particular island off the coast of Tasmania. She hypothesises that seabirds with longer beaks lay heavier eggs.\n\nAnna recorded the beak length and egg weight of a sample of 40 birds. Her results and the population of seabirds are plotted below\n\n[figure1]\n\nWhich of the following is true?\nA: The difference between Anna's sample and the population could be attributed to random error\nB: There is a clear relationship between egg weight and beak length in the population\nC: Anna's sample overestimates the average egg weight of the population\nD: Increasing the sample size would not improve Anna's study\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAnna is studying a population of seabirds which roost on one particular island off the coast of Tasmania. She hypothesises that seabirds with longer beaks lay heavier eggs.\n\nAnna recorded the beak length and egg weight of a sample of 40 birds. Her results and the population of seabirds are plotted below\n\n[figure1]\n\nWhich of the following is true?\n\nA: The difference between Anna's sample and the population could be attributed to random error\nB: There is a clear relationship between egg weight and beak length in the population\nC: Anna's sample overestimates the average egg weight of the population\nD: Increasing the sample size would not improve Anna's study\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-05.jpg?height=791&width=1314&top_left_y=1752&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1307",
        "problem": "## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAACalculate the length of the piece of white plastic second from the top in Bolus 2. Record this value on your answer sheet.",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAA\n\nproblem:\nCalculate the length of the piece of white plastic second from the top in Bolus 2. Record this value on your answer sheet.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of cm, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=800&width=617&top_left_y=1299&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=817&width=1097&top_left_y=1299&top_left_x=845"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "cm"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_288",
        "problem": "The plant genus Xcontains 6 species ( $X$. messa, $X$. obnoxia, $X$. beatifica, $X$. confusa, $X$. foetida, $X$. nerda). All of them share many character states that distinguish this genus from all closely related genera. The species differ from each other, however, as described below. All species in the sister genus, $\\mathrm{Y}$, are vining plants with palmately compound, alternate leaves, sweet-smelling flowers with pink, free petals and 10 stamens, and drupes.\n\n$X$. messa: Plant upright, stems glabrous; leaves opposite, palmately compound; petals purple, free; stamens 5; flowers sweet-smelling; fruit a berry\n\n$X$. obnoxia: Plant vining, stems glabrous; leaves opposite, simple; petals red, free; stamens 10 ;\n\nflowers with a rotten meat smell; fruit a berry\n\n$X$. beatifica: Plant upright, stems glabrous; leaves opposite, palmately compound; petals pink, connate; stamens 5; flowers sweet-smelling; fruit a berry\n\n$X$. confusa: Plant upright, stems glabrous; leaves opposite, simple; petals purple, free; stamens\n\n5; flowers sweet-smelling; fruit a berry\n\nX. foetida: Plant vining, stems hispid; leaves opposite, palmately compound; petals red, free; stamens 10; flowers with a rotten meat smell; fruit a berry\n\n$X$. nerda: Plant upright, stems glabrous; leaves opposite, palmately compound; petals pink, connate; stamens 4; flowers sweet-smelling; fruit a berry\n\nThe data matrix, in which characters are coded according to whether character states are plesiomorphic $(0)$ or apomorphic $(1,2)$ was presented in the following table. The most parsimonious cladogram was constructed from the data matrix.\n\n|  | Habit | Stem <br> hairs | Leaf <br> position | Leaf <br> structure | Petal <br> colour | Petal <br> fusion | Stamen <br> number | Flower <br> odour | Fruit <br> type |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| X. messa | 1 | 0 | 1 | 0 | 2 | 0 | 1 | 0 | 1 |\n| X. obnoxia | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |\n| X. beatifica | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |\n| X. confusa | 1 | 0 | 1 | 1 | 2 | 0 | 1 | 0 | 1 |\n| X. foetida | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 |\n| X. nerda | 1 | 0 | 1 | 0 | 0 | 1 | 2 | 0 | 1 |\n| Outgroup (Y) | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |\n\n[figure1]\nA: Fruit as a berry and leaves opposite support the monophyly of the genus X.\nB: X. obnoxia, $X$. beatifica and $X$. nerda form a monophyletic group.\nC: Hispid stems and 4 stamens are autapomorphic (unique) for $X$. foetida and $X$. nerda, respectively.\nD: Simple leaves in X. obnoxia and X. confuse appear to have evolved independently.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe plant genus Xcontains 6 species ( $X$. messa, $X$. obnoxia, $X$. beatifica, $X$. confusa, $X$. foetida, $X$. nerda). All of them share many character states that distinguish this genus from all closely related genera. The species differ from each other, however, as described below. All species in the sister genus, $\\mathrm{Y}$, are vining plants with palmately compound, alternate leaves, sweet-smelling flowers with pink, free petals and 10 stamens, and drupes.\n\n$X$. messa: Plant upright, stems glabrous; leaves opposite, palmately compound; petals purple, free; stamens 5; flowers sweet-smelling; fruit a berry\n\n$X$. obnoxia: Plant vining, stems glabrous; leaves opposite, simple; petals red, free; stamens 10 ;\n\nflowers with a rotten meat smell; fruit a berry\n\n$X$. beatifica: Plant upright, stems glabrous; leaves opposite, palmately compound; petals pink, connate; stamens 5; flowers sweet-smelling; fruit a berry\n\n$X$. confusa: Plant upright, stems glabrous; leaves opposite, simple; petals purple, free; stamens\n\n5; flowers sweet-smelling; fruit a berry\n\nX. foetida: Plant vining, stems hispid; leaves opposite, palmately compound; petals red, free; stamens 10; flowers with a rotten meat smell; fruit a berry\n\n$X$. nerda: Plant upright, stems glabrous; leaves opposite, palmately compound; petals pink, connate; stamens 4; flowers sweet-smelling; fruit a berry\n\nThe data matrix, in which characters are coded according to whether character states are plesiomorphic $(0)$ or apomorphic $(1,2)$ was presented in the following table. The most parsimonious cladogram was constructed from the data matrix.\n\n|  | Habit | Stem <br> hairs | Leaf <br> position | Leaf <br> structure | Petal <br> colour | Petal <br> fusion | Stamen <br> number | Flower <br> odour | Fruit <br> type |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| X. messa | 1 | 0 | 1 | 0 | 2 | 0 | 1 | 0 | 1 |\n| X. obnoxia | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |\n| X. beatifica | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |\n| X. confusa | 1 | 0 | 1 | 1 | 2 | 0 | 1 | 0 | 1 |\n| X. foetida | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 |\n| X. nerda | 1 | 0 | 1 | 0 | 0 | 1 | 2 | 0 | 1 |\n| Outgroup (Y) | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |\n\n[figure1]\n\nA: Fruit as a berry and leaves opposite support the monophyly of the genus X.\nB: X. obnoxia, $X$. beatifica and $X$. nerda form a monophyletic group.\nC: Hispid stems and 4 stamens are autapomorphic (unique) for $X$. foetida and $X$. nerda, respectively.\nD: Simple leaves in X. obnoxia and X. confuse appear to have evolved independently.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-44.jpg?height=668&width=1082&top_left_y=1896&top_left_x=493"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_18",
        "problem": "Compounds $\\mathrm{X}$ and $\\mathrm{Y}$ are precursors in the pathway of $\\mathrm{Z}$ synthesis. $\\mathrm{Z}$ is essential for growth. Wild type (WT) Neurospora are prototrophs, meaning that they can grow on minimal medium (MM).\n\nGrowth Experiment: Four Z- neurospora mutants were isolated, and results of experiments with these mutants are presented below ( $+=$ growth, $-=$ no growth).\n\n| Cell Type | MM | MM + Y | MM + X | MM + Z | Accumulated compound <br> during growth in MM |\n| :--- | :---: | :---: | :---: | :---: | :---: |\n| WT | + | + | + | + |  |\n| Mutant 1 | - | - | + | + | $Y$ |\n| Mutant 2 | - | - | - | + | $X$ |\n| Mutant 3 | - | - | - | + | $X$ |\n| Mutant 4 | - | - | - | + | $Y$ |\n\nComplementation Tests: Complementation tests were performed by testing for growth of heterokaryons in minimal medium. Results were as follows ( $+=$ growth, $-=$ no growth).\n\nMutant 1 Mutant 2 Mutant 3 Mutant 4\n\n| Mutant 1 | - |  |  |\n| :--- | :--- | :--- | :--- |\n| Mutant 2 | + | - |  |\n| Mutant 3 | + | - | - |\n| Mutant 4 | - | - | - |\n\nMating Experiment: Mutant strains were mated and \\% prototrophs among spores of each mating were determined. Results were as follows.\n\n| Mating | \\% prototroph spores |\n| :--- | :---: |\n| Mutant $1 \\times$ Mutant 2 | $25 \\%$ |\n| Mutant $1 \\times$ Mutant 3 | $25 \\%$ |\n| Mutant $1 \\times$ Mutant 4 | $0 \\%$ (many spores counted) |\n| Mutant $2 \\times$ Mutant 3 | $0.004 \\%$ |\n| Mutant $2 \\times$ Mutant 4 | $0.001 \\%$ |\n| Mutant 3 x Mutant 4 | $0.001 \\%$ |\nA: The mutations in the mutated strains were distributed in two genes\nB: Considering all the mutant strains, at least five different sites of the Neurospora genome were mutated in one or more strains.\nC: At least two of the mutant strains were double mutants (i.e. $>1$ site was mutated in each of these strains).\nD: The mutation site of Mutant 3 is positioned between the mutation site in Mutant 2 and the mutation site in Mutant 4.\nE: It is expected that $25 \\%$ of spores resulting from mating between Mutant 2 and wild type strain will be prototrophs.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCompounds $\\mathrm{X}$ and $\\mathrm{Y}$ are precursors in the pathway of $\\mathrm{Z}$ synthesis. $\\mathrm{Z}$ is essential for growth. Wild type (WT) Neurospora are prototrophs, meaning that they can grow on minimal medium (MM).\n\nGrowth Experiment: Four Z- neurospora mutants were isolated, and results of experiments with these mutants are presented below ( $+=$ growth, $-=$ no growth).\n\n| Cell Type | MM | MM + Y | MM + X | MM + Z | Accumulated compound <br> during growth in MM |\n| :--- | :---: | :---: | :---: | :---: | :---: |\n| WT | + | + | + | + |  |\n| Mutant 1 | - | - | + | + | $Y$ |\n| Mutant 2 | - | - | - | + | $X$ |\n| Mutant 3 | - | - | - | + | $X$ |\n| Mutant 4 | - | - | - | + | $Y$ |\n\nComplementation Tests: Complementation tests were performed by testing for growth of heterokaryons in minimal medium. Results were as follows ( $+=$ growth, $-=$ no growth).\n\nMutant 1 Mutant 2 Mutant 3 Mutant 4\n\n| Mutant 1 | - |  |  |\n| :--- | :--- | :--- | :--- |\n| Mutant 2 | + | - |  |\n| Mutant 3 | + | - | - |\n| Mutant 4 | - | - | - |\n\nMating Experiment: Mutant strains were mated and \\% prototrophs among spores of each mating were determined. Results were as follows.\n\n| Mating | \\% prototroph spores |\n| :--- | :---: |\n| Mutant $1 \\times$ Mutant 2 | $25 \\%$ |\n| Mutant $1 \\times$ Mutant 3 | $25 \\%$ |\n| Mutant $1 \\times$ Mutant 4 | $0 \\%$ (many spores counted) |\n| Mutant $2 \\times$ Mutant 3 | $0.004 \\%$ |\n| Mutant $2 \\times$ Mutant 4 | $0.001 \\%$ |\n| Mutant 3 x Mutant 4 | $0.001 \\%$ |\n\nA: The mutations in the mutated strains were distributed in two genes\nB: Considering all the mutant strains, at least five different sites of the Neurospora genome were mutated in one or more strains.\nC: At least two of the mutant strains were double mutants (i.e. $>1$ site was mutated in each of these strains).\nD: The mutation site of Mutant 3 is positioned between the mutation site in Mutant 2 and the mutation site in Mutant 4.\nE: It is expected that $25 \\%$ of spores resulting from mating between Mutant 2 and wild type strain will be prototrophs.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_711",
        "problem": "如图表示某哺乳动物一个正在分裂的性原细胞中, 两对同源染色体及染色体上的基因种类、位置示意图。下列叙述错误的是（ ）\n\n[图1]\nA: 图示细胞一定是精原细胞, 2 为 $\\mathrm{Y}$ 染色体\nB: 2、4 染色体上可能存在相同或等位基因\nC: 1 染色体上出现 $\\mathrm{a}$ 基因的原因是基因突变或互换\nD: 该细胞的一个子细胞基因组成为 $\\mathrm{AaB}$ ，则其他三个子细胞基因组成为 $\\mathrm{B} 、 \\mathrm{~A} 、 \\mathrm{~A}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图表示某哺乳动物一个正在分裂的性原细胞中, 两对同源染色体及染色体上的基因种类、位置示意图。下列叙述错误的是（ ）\n\n[图1]\n\nA: 图示细胞一定是精原细胞, 2 为 $\\mathrm{Y}$ 染色体\nB: 2、4 染色体上可能存在相同或等位基因\nC: 1 染色体上出现 $\\mathrm{a}$ 基因的原因是基因突变或互换\nD: 该细胞的一个子细胞基因组成为 $\\mathrm{AaB}$ ，则其他三个子细胞基因组成为 $\\mathrm{B} 、 \\mathrm{~A} 、 \\mathrm{~A}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-55.jpg?height=468&width=485&top_left_y=651&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_770",
        "problem": "血友病是一种伴 $\\mathrm{X}$ 隐性遗传病, 经研究发现其致病基因 $\\mathrm{d}$ 有两种突变形式, 记作 $\\mathrm{dA}$\n\n与 $\\mathrm{dB}$ 。如图表示某血友病家族系谱图, $\\mathrm{II}_{1}$ 同时还患有克氏综合征 (性染色体组成为 XXY)。不考虑新的突变和染色体片段交换, 下列分析正确的是（）\n\n[图1]\n\nII\n\n[图2]\n\nB\n\n[图3]\n\n$\\square$ 正常男性、女姓\n\n患伴X染色体隐性遗传病男性、女性\n\n$\\mathrm{Ad}$ 基因突变形式是 $\\mathrm{dA}$ $B d$ 基因突变形式是 $d B$ $\\mathrm{AB} \\mathrm{d}$ 基因突变形式兼有 $\\mathrm{d} A$ 和 $\\mathrm{d} B$\nA: $\\mathrm{II}_{1}$ 性染色体异常, 可能是因为 $\\mathrm{I}_{1}$ 个体减数分裂 $\\mathrm{I}$ 时同源染色体不分离\nB: $\\mathrm{II}_{2}$ 与正常女性婚配, 所生子女患有该伴 X 染色体隐性遗传病的概率是 $1 / 2$\nC: $\\mathrm{I}_{1}$ 与 $\\mathrm{I}_{2}$ 婚配所生子女中, 女儿一定带有致病基因, 儿子一定不携带致病基因\nD: $\\mathrm{II}_{4}$ 与基因型与 $\\mathrm{II}_{2}$ 相同的个体结婚, 所生子女患血友病的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n血友病是一种伴 $\\mathrm{X}$ 隐性遗传病, 经研究发现其致病基因 $\\mathrm{d}$ 有两种突变形式, 记作 $\\mathrm{dA}$\n\n与 $\\mathrm{dB}$ 。如图表示某血友病家族系谱图, $\\mathrm{II}_{1}$ 同时还患有克氏综合征 (性染色体组成为 XXY)。不考虑新的突变和染色体片段交换, 下列分析正确的是（）\n\n[图1]\n\nII\n\n[图2]\n\nB\n\n[图3]\n\n$\\square$ 正常男性、女姓\n\n患伴X染色体隐性遗传病男性、女性\n\n$\\mathrm{Ad}$ 基因突变形式是 $\\mathrm{dA}$ $B d$ 基因突变形式是 $d B$ $\\mathrm{AB} \\mathrm{d}$ 基因突变形式兼有 $\\mathrm{d} A$ 和 $\\mathrm{d} B$\n\nA: $\\mathrm{II}_{1}$ 性染色体异常, 可能是因为 $\\mathrm{I}_{1}$ 个体减数分裂 $\\mathrm{I}$ 时同源染色体不分离\nB: $\\mathrm{II}_{2}$ 与正常女性婚配, 所生子女患有该伴 X 染色体隐性遗传病的概率是 $1 / 2$\nC: $\\mathrm{I}_{1}$ 与 $\\mathrm{I}_{2}$ 婚配所生子女中, 女儿一定带有致病基因, 儿子一定不携带致病基因\nD: $\\mathrm{II}_{4}$ 与基因型与 $\\mathrm{II}_{2}$ 相同的个体结婚, 所生子女患血友病的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-091.jpg?height=63&width=43&top_left_y=1939&top_left_x=391",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-091.jpg?height=140&width=237&top_left_y=1929&top_left_x=595",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-091.jpg?height=191&width=488&top_left_y=2112&top_left_x=470"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_34",
        "problem": "Which of following statements is/are correct?\n\nWith increasing atmospheric $\\mathrm{CO}_{2}$,\n\nI. inorganic nutrients in soil will be increasingly limiting for plant growth.\n\nII. C4 plants will grow increasingly better than C3 plants in environments where water is limiting.\n\nIII. the increased $\\mathrm{C}: \\mathrm{N}$ ratio in litter will cause an increase in decomposition rate by soil microorganisms.\nA: Only I\nB: Only II\nC: Only III\nD: Only I and II\nE: Only II and III\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of following statements is/are correct?\n\nWith increasing atmospheric $\\mathrm{CO}_{2}$,\n\nI. inorganic nutrients in soil will be increasingly limiting for plant growth.\n\nII. C4 plants will grow increasingly better than C3 plants in environments where water is limiting.\n\nIII. the increased $\\mathrm{C}: \\mathrm{N}$ ratio in litter will cause an increase in decomposition rate by soil microorganisms.\n\nA: Only I\nB: Only II\nC: Only III\nD: Only I and II\nE: Only II and III\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1190",
        "problem": "Understanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.The native plant species consumed most frequently by the orange-fronted parakeet on Maud Island was?\nA: Sycamore\nB: Makomako\nC: Tree lucerne\nD: Manuka\nE: Mahoe\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nUnderstanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.\n\nproblem:\nThe native plant species consumed most frequently by the orange-fronted parakeet on Maud Island was?\n\nA: Sycamore\nB: Makomako\nC: Tree lucerne\nD: Manuka\nE: Mahoe\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_356",
        "problem": "鸡的性别决定方式为 ZW 型，控制其芦花羽与非芦花羽的基因仅位于 $\\mathrm{Z}$ 染色体上,芦花羽对非芦花羽为显性。某种鸡的毛腿与光腿、单冠与复冠各由一对基因控制, 且两对基因在染色体上的位置关系未知（不考虑 $\\mathrm{Z} 、 \\mathrm{~W}$ 同源区段）。现有纯合芦花羽光腿单冠雄鸡与非芦花羽毛腿复冠雌鸡杂交, $F_{1}$ 相互杂交, $F_{2}$ 中腿毛与鸡冠的表型及比例为毛腿单冠: 毛腿复冠: 光腿单冠: 光腿复冠 $=9: 3: 3: 1$ 。 $F_{2}$ 表型中不可能出现 ( )\nA: 光腿全为雌性\nB: 非芦花全为雌性\nC: 复冠全为雌性\nD: 复冠全为非芦花羽\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n鸡的性别决定方式为 ZW 型，控制其芦花羽与非芦花羽的基因仅位于 $\\mathrm{Z}$ 染色体上,芦花羽对非芦花羽为显性。某种鸡的毛腿与光腿、单冠与复冠各由一对基因控制, 且两对基因在染色体上的位置关系未知（不考虑 $\\mathrm{Z} 、 \\mathrm{~W}$ 同源区段）。现有纯合芦花羽光腿单冠雄鸡与非芦花羽毛腿复冠雌鸡杂交, $F_{1}$ 相互杂交, $F_{2}$ 中腿毛与鸡冠的表型及比例为毛腿单冠: 毛腿复冠: 光腿单冠: 光腿复冠 $=9: 3: 3: 1$ 。 $F_{2}$ 表型中不可能出现 ( )\n\nA: 光腿全为雌性\nB: 非芦花全为雌性\nC: 复冠全为雌性\nD: 复冠全为非芦花羽\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_280",
        "problem": "Organismal senescence is the aging of the whole organism. It is a widespread phenomenon in nature but the exact etiology is still largely unclear. An evolutionary theory was proposed by George C.\n\nWilliams which involves the following assumptions:\n\n1. Soma (referring to all somatic cells in the body) is essential for reproductive success, but it does not pass on to the offspring.\n2. Alleles are affected by Natural selection.\n3. Pleiotropic genes can have opposite effects on fitness at different ages or somatic environments.\n4. The probability of reproduction decreases with increasing age.\n\nWilliams suggested the following example: Perhaps a gene was selected for because it codes for calcium deposition in bones, which promotes juvenile survival, but the same gene promotes calcium deposition in arteries.\nA: Lower adult death rates can be associated with lower rates of senescence, according to the theory.\nB: Where there is a sex difference, the sex with higher rate of decrease in fecundity after maturation (but with equal death rate to the other sex) should undergo the more rapid senescence, according to the theory.\nC: The frequencies of an allele in a gene that behaves similar to the aforementioned calcium deposition gene in a cohort will be smallest at early age and greatest at late age.\nD: This model views senescence as an adaptation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nOrganismal senescence is the aging of the whole organism. It is a widespread phenomenon in nature but the exact etiology is still largely unclear. An evolutionary theory was proposed by George C.\n\nWilliams which involves the following assumptions:\n\n1. Soma (referring to all somatic cells in the body) is essential for reproductive success, but it does not pass on to the offspring.\n2. Alleles are affected by Natural selection.\n3. Pleiotropic genes can have opposite effects on fitness at different ages or somatic environments.\n4. The probability of reproduction decreases with increasing age.\n\nWilliams suggested the following example: Perhaps a gene was selected for because it codes for calcium deposition in bones, which promotes juvenile survival, but the same gene promotes calcium deposition in arteries.\n\nA: Lower adult death rates can be associated with lower rates of senescence, according to the theory.\nB: Where there is a sex difference, the sex with higher rate of decrease in fecundity after maturation (but with equal death rate to the other sex) should undergo the more rapid senescence, according to the theory.\nC: The frequencies of an allele in a gene that behaves similar to the aforementioned calcium deposition gene in a cohort will be smallest at early age and greatest at late age.\nD: This model views senescence as an adaptation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1163",
        "problem": "The scatter graphs show the relation between the number of mayfly nymphs of two species, $X$ and $Y$, and the flow rate of the water in which they live. The correlation coefficients for species $X$ is 0.34 and for species $Y$ is 0.87 .\n\n\n[figure1]\n\nFlow rate in metres per second\n\nThe correlation coefficient for species $\\mathrm{X}$ is lower than that for species $Y$ because\nA: The scattering of the points is greater.\nB: The mean flow rate is less.\nC: There are fewer points in its scatter graph.\nD: The slope of the line is less.\nE: The number of mayfly nymphs per square metre is less.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe scatter graphs show the relation between the number of mayfly nymphs of two species, $X$ and $Y$, and the flow rate of the water in which they live. The correlation coefficients for species $X$ is 0.34 and for species $Y$ is 0.87 .\n\n\n[figure1]\n\nFlow rate in metres per second\n\nThe correlation coefficient for species $\\mathrm{X}$ is lower than that for species $Y$ because\n\nA: The scattering of the points is greater.\nB: The mean flow rate is less.\nC: There are fewer points in its scatter graph.\nD: The slope of the line is less.\nE: The number of mayfly nymphs per square metre is less.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-10.jpg?height=488&width=808&top_left_y=1389&top_left_x=1092"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1415",
        "problem": "[figure1]\n\nA prion is a rare type of misfolded protein that can trigger other correctly folded proteins of the same type to fold incorrectly.\n\nWhich of the following is LEAST likely to be an origin for prion disease in humans?\nA: Unusually rapid cellular proliferation\nB: Eating contaminated vegetables\nC: Spontaneous mutation\nD: Environmental factors leading to abnormal cellular conditions\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nA prion is a rare type of misfolded protein that can trigger other correctly folded proteins of the same type to fold incorrectly.\n\nWhich of the following is LEAST likely to be an origin for prion disease in humans?\n\nA: Unusually rapid cellular proliferation\nB: Eating contaminated vegetables\nC: Spontaneous mutation\nD: Environmental factors leading to abnormal cellular conditions\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-36.jpg?height=560&width=1394&top_left_y=388&top_left_x=248"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_379",
        "problem": "某动物 $(2 n=38)$ 细胞内发生了两条染色体融合成一条染色体的情况, 过程如图所示 (图中 $\\mathrm{A} 、 \\mathrm{~B}$ 表示染色体片段)，融合后的染色体称为重接染色体，脱离的小残片最终会丢失。下列相关叙述正确的是（ ）\n\n[图1]\n\n13号染色体\n\n[图2]\n\n17 号染色体\n\n[图3]\n\n重接染色体\n\n[图4]\n\n丢失的残片\nA: 图示过程中只发生了染色体数目变异\nB: 该动物细胞中染色体组数可能是 $1 、 2 、 3$ 或 4\nC: 该变异在有丝分裂和减数分裂过程中均可发生\nD: 该动物形成的生殖细胞中染色体数量一定减少\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某动物 $(2 n=38)$ 细胞内发生了两条染色体融合成一条染色体的情况, 过程如图所示 (图中 $\\mathrm{A} 、 \\mathrm{~B}$ 表示染色体片段)，融合后的染色体称为重接染色体，脱离的小残片最终会丢失。下列相关叙述正确的是（ ）\n\n[图1]\n\n13号染色体\n\n[图2]\n\n17 号染色体\n\n[图3]\n\n重接染色体\n\n[图4]\n\n丢失的残片\n\nA: 图示过程中只发生了染色体数目变异\nB: 该动物细胞中染色体组数可能是 $1 、 2 、 3$ 或 4\nC: 该变异在有丝分裂和减数分裂过程中均可发生\nD: 该动物形成的生殖细胞中染色体数量一定减少\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-15.jpg?height=343&width=100&top_left_y=585&top_left_x=407",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-15.jpg?height=339&width=306&top_left_y=590&top_left_x=541",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-15.jpg?height=297&width=208&top_left_y=617&top_left_x=844",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-15.jpg?height=77&width=120&top_left_y=818&top_left_x=1139"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_282",
        "problem": "Some flowering plants are aquatic and have adapted to live in aquatic environments. They are divided into 3 types, depending on how much of the plant is normally positioned inside or outside of the water. These groups include emergent, floating, and submerged plants. The organs of submerged aquatic plants grow completely under water.\nA: They do not have xylem as they can absorb water from all of their surfaces.\nB: Given that some species do have stomata, their opening and closing of the stomata is not expected to be driven by circadian rhythm.\nC: These plants are heterophyllous, with thin and dark green leaves.\nD: These plants have a well-developed supporting mechanical tissues with thick walls.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSome flowering plants are aquatic and have adapted to live in aquatic environments. They are divided into 3 types, depending on how much of the plant is normally positioned inside or outside of the water. These groups include emergent, floating, and submerged plants. The organs of submerged aquatic plants grow completely under water.\n\nA: They do not have xylem as they can absorb water from all of their surfaces.\nB: Given that some species do have stomata, their opening and closing of the stomata is not expected to be driven by circadian rhythm.\nC: These plants are heterophyllous, with thin and dark green leaves.\nD: These plants have a well-developed supporting mechanical tissues with thick walls.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_396",
        "problem": "如图所示为甲、乙两种遗传病的家族系谱图（II-1 不携带任何致病基因）。下列叙述错误的是（ ）\n\nI\n\nII\n\n[图1]\n\n正常女性正常男性甲病女性甲病男性 $\\mathbb{N}$ 乙病男性两种病男性\nA: 甲病为常染色体上的显性遗传, 乙病为 X 染色体上的隐性遗传\nB: 若III-1 与III-5 结婚, 生患病男孩的概率为 $1 / 4$\nC: 若III-1 与III-4 结婚, 生两病兼患孩子的概率为 $1 / 12$\nD: 若III- 1 与III-4 结婚, 后代中只患甲病的概率为 $7 / 12$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图所示为甲、乙两种遗传病的家族系谱图（II-1 不携带任何致病基因）。下列叙述错误的是（ ）\n\nI\n\nII\n\n[图1]\n\n正常女性正常男性甲病女性甲病男性 $\\mathbb{N}$ 乙病男性两种病男性\n\nA: 甲病为常染色体上的显性遗传, 乙病为 X 染色体上的隐性遗传\nB: 若III-1 与III-5 结婚, 生患病男孩的概率为 $1 / 4$\nC: 若III-1 与III-4 结婚, 生两病兼患孩子的概率为 $1 / 12$\nD: 若III- 1 与III-4 结婚, 后代中只患甲病的概率为 $7 / 12$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-076.jpg?height=605&width=806&top_left_y=500&top_left_x=408"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_870",
        "problem": "果蝇的灰体 (B) 与黑体 (b), 红褐色眼 (A) 与鲜红色眼 (a), 长刚毛 (H) 与钩状刚毛（h）是常染色体上基因控制的三对相对性状。用三对基因均杂合的雌蝇进行了测交实验，据表分析正确的是（ ）\n\n| 组 <br> 别 | 杂交亲本 | 子代性状及数量 |\n| :--- | :--- | :--- |\n| 一 | 杂合雌蝇 $\\times$ 黑体鲜红 <br> 色眼雄蝇 | 黑体鲜红色眼 253、黑体红褐色眼 256、灰体鲜红色眼 <br> $238 、$ 灰体红褐色眼 253 |\n| 二 | 杂合雌蝇 $\\times$ 钩状刚毛 <br> 鲜红色眼雄蝇 | 钩状刚毛鲜红色眼 236、长刚毛鲜红色眼 255、钩状刚毛 <br> 红褐色眼 250、长刚毛红褐色眼 259 |\n| 三 | 杂合雌蝇 $\\times$ 黑体钩状 <br> 刚毛雄蝇 | 黑体钩状刚毛 25、黑体长刚毛 484、灰体钩状刚毛 461、 <br> 灰体长刚毛 30 |\nA: 组别三中亲本杂合雌蝇产生的配子中 BH 多于 Bh\nB: 控制体色和刚毛的两对等位基因在遗传时遵循自由组合定律\nC: 将组别一子代中“黑体红褐色眼”与“灰体鲜红色眼”的个体杂交可以验证自由组\nD: 将组别二子代中鲜红色眼的果蝇淘汰, 让其余果蝇自由交配, $\\mathrm{F}_{2}$ 中钧状刚毛果蝇占比为 $9 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的灰体 (B) 与黑体 (b), 红褐色眼 (A) 与鲜红色眼 (a), 长刚毛 (H) 与钩状刚毛（h）是常染色体上基因控制的三对相对性状。用三对基因均杂合的雌蝇进行了测交实验，据表分析正确的是（ ）\n\n| 组 <br> 别 | 杂交亲本 | 子代性状及数量 |\n| :--- | :--- | :--- |\n| 一 | 杂合雌蝇 $\\times$ 黑体鲜红 <br> 色眼雄蝇 | 黑体鲜红色眼 253、黑体红褐色眼 256、灰体鲜红色眼 <br> $238 、$ 灰体红褐色眼 253 |\n| 二 | 杂合雌蝇 $\\times$ 钩状刚毛 <br> 鲜红色眼雄蝇 | 钩状刚毛鲜红色眼 236、长刚毛鲜红色眼 255、钩状刚毛 <br> 红褐色眼 250、长刚毛红褐色眼 259 |\n| 三 | 杂合雌蝇 $\\times$ 黑体钩状 <br> 刚毛雄蝇 | 黑体钩状刚毛 25、黑体长刚毛 484、灰体钩状刚毛 461、 <br> 灰体长刚毛 30 |\n\nA: 组别三中亲本杂合雌蝇产生的配子中 BH 多于 Bh\nB: 控制体色和刚毛的两对等位基因在遗传时遵循自由组合定律\nC: 将组别一子代中“黑体红褐色眼”与“灰体鲜红色眼”的个体杂交可以验证自由组\nD: 将组别二子代中鲜红色眼的果蝇淘汰, 让其余果蝇自由交配, $\\mathrm{F}_{2}$ 中钧状刚毛果蝇占比为 $9 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_894",
        "problem": "人体细胞中的染色体有时会出现结构和行为的异常。如图, 某男性精原细胞中的一\n\n条 3 号染色体和一条 7 号染色体连在一起, 成为一条染色体, 减数分裂时染色体的相应片段将正常联会形成特殊的结构, 在减数第一次分裂后期, 该结构中的任意两条染色体移向一极,剩下的一条移向另一极,减数第二次分裂正常。下列相关叙述错误的是 ( )\n\n[图1]\nA: 该男性细胞中染色体的形态有 25 种\nB: 该男性细胞中染色体数最多为 90 条\nC: 该男性细胞经减数分裂形成正常精子的概率是 $1 / 2$\nD: 该男性细胞中等位基因 A、a 在减数第一次分裂时不一定发生分离\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人体细胞中的染色体有时会出现结构和行为的异常。如图, 某男性精原细胞中的一\n\n条 3 号染色体和一条 7 号染色体连在一起, 成为一条染色体, 减数分裂时染色体的相应片段将正常联会形成特殊的结构, 在减数第一次分裂后期, 该结构中的任意两条染色体移向一极,剩下的一条移向另一极,减数第二次分裂正常。下列相关叙述错误的是 ( )\n\n[图1]\n\nA: 该男性细胞中染色体的形态有 25 种\nB: 该男性细胞中染色体数最多为 90 条\nC: 该男性细胞经减数分裂形成正常精子的概率是 $1 / 2$\nD: 该男性细胞中等位基因 A、a 在减数第一次分裂时不一定发生分离\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-002.jpg?height=359&width=514&top_left_y=166&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1217",
        "problem": "In Guinea pigs, black hair (B) is dominant over white (b), rough coat texture (R) is dominant over smooth ( $r$ ), and short hair (S) is dominant over long hair (s). These genes are found on separate chromosomes. A homozygous black, rough, short-haired Guinea pig was crossed with a white, smooth, long-haired one. The offspring of this cross were mated and 200 offspring where produced. How many of these F2 offspring would you expect to be black with a rough long haired coat?\nA: 9\nB: 14\nC: 28\nD: 42\nE: 84\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn Guinea pigs, black hair (B) is dominant over white (b), rough coat texture (R) is dominant over smooth ( $r$ ), and short hair (S) is dominant over long hair (s). These genes are found on separate chromosomes. A homozygous black, rough, short-haired Guinea pig was crossed with a white, smooth, long-haired one. The offspring of this cross were mated and 200 offspring where produced. How many of these F2 offspring would you expect to be black with a rough long haired coat?\n\nA: 9\nB: 14\nC: 28\nD: 42\nE: 84\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_325",
        "problem": "某种昆虫的红眼 (E)对白眼 (e)为显性，该对基因位于 $\\mathrm{X}$ 染色体上。用染色体组成正常的白眼雌虫和红眼雄虫杂交, 子代雌、雄虫中均有红眼和白眼个体。研究发现出现异\n常子代的原因是母本产生了染色体数目异常的卵细胞。已知该昆虫的 X 染色体数量等于染色体组数时为雌性, 性染色体组成为 $X Y$ 或 $X O(O$ 表示无 $X$ 染色体 $)$ 时为雄性, 受精卵中 X 染色体多于 2 条或不含 X 染色体时, 受精卵无法正常发育。亲本产生的受精卵总数用 $m$ 表示, 其中染色体数目异常的受精卵数用 $n$ 表示。下列说法正确的是（）\nA: 母本产生的异常卵细胞中性染色体组成为 X 或 $\\mathrm{O}$\nB: 子代中性染色体组成为 $O Y 、 X X Y$ 的受精卵无法正常发育\nC: 子代中红眼雄虫和白眼雌虫的性染色体组成分别为 $X^{E} O$ 和 $X^{e} X^{e}$\nD: 子代中性染色体组成为 $X Y$ 的昆虫个体数目的占比是 $(m-n) /(2 m-n)$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种昆虫的红眼 (E)对白眼 (e)为显性，该对基因位于 $\\mathrm{X}$ 染色体上。用染色体组成正常的白眼雌虫和红眼雄虫杂交, 子代雌、雄虫中均有红眼和白眼个体。研究发现出现异\n常子代的原因是母本产生了染色体数目异常的卵细胞。已知该昆虫的 X 染色体数量等于染色体组数时为雌性, 性染色体组成为 $X Y$ 或 $X O(O$ 表示无 $X$ 染色体 $)$ 时为雄性, 受精卵中 X 染色体多于 2 条或不含 X 染色体时, 受精卵无法正常发育。亲本产生的受精卵总数用 $m$ 表示, 其中染色体数目异常的受精卵数用 $n$ 表示。下列说法正确的是（）\n\nA: 母本产生的异常卵细胞中性染色体组成为 X 或 $\\mathrm{O}$\nB: 子代中性染色体组成为 $O Y 、 X X Y$ 的受精卵无法正常发育\nC: 子代中红眼雄虫和白眼雌虫的性染色体组成分别为 $X^{E} O$ 和 $X^{e} X^{e}$\nD: 子代中性染色体组成为 $X Y$ 的昆虫个体数目的占比是 $(m-n) /(2 m-n)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_853",
        "problem": "人类对遗传物质本质的探索经历了漫长的过程, 下列有关叙述正确的是 ( )\nA: 孟德尔发现遗传因子并证实了其传递规律和化学本质\nB: 噬菌体侵染细菌实验比肺炎双球菌体外转化实验更具说服力\nC: 沃森和克里克提出在 DNA 双螺旋结构中嘧啶数不等于嘌呤数\nD: 烟草花叶病毒感染烟草实验说明所有病毒的遗传物质是 RNA\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人类对遗传物质本质的探索经历了漫长的过程, 下列有关叙述正确的是 ( )\n\nA: 孟德尔发现遗传因子并证实了其传递规律和化学本质\nB: 噬菌体侵染细菌实验比肺炎双球菌体外转化实验更具说服力\nC: 沃森和克里克提出在 DNA 双螺旋结构中嘧啶数不等于嘌呤数\nD: 烟草花叶病毒感染烟草实验说明所有病毒的遗传物质是 RNA\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_985",
        "problem": "Rank the following biological molecules in order of how readily they diffuse across the plasma membrane from the most diffusible to the least diffusible.\n\nI. $\\mathrm{C O}_{2}$\n\nII. $\\mathrm{Cl}^{-}$\n\nIII. Sucrose\n\nIV. Glycerol\nA: I, III, IV, II\nB: II, IV, III, I\nC: III,II, IV, I\nD: I, IV, III, II\nE: II, I, III, IV\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRank the following biological molecules in order of how readily they diffuse across the plasma membrane from the most diffusible to the least diffusible.\n\nI. $\\mathrm{C O}_{2}$\n\nII. $\\mathrm{Cl}^{-}$\n\nIII. Sucrose\n\nIV. Glycerol\n\nA: I, III, IV, II\nB: II, IV, III, I\nC: III,II, IV, I\nD: I, IV, III, II\nE: II, I, III, IV\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_313",
        "problem": "为探究 kdr 基因突变对杀虫剂抗性的影响, 使用溴氰菊酯和氯菊酯两种杀虫剂处理白纹伊蚊, 结果如下图。相关叙述正确的是（ ）\n\n[图1]\nA: 杀虫剂的使用使 kdr 基因发生突变\nB: kdr 基因突变型个体对氯菊酯更敏感\nC: 实验结果说明突变型对杀虫剂的抵抗效果更好\nD: 轮换使用两种杀虫剂可延缓白纹伊蚊产生抗性\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n为探究 kdr 基因突变对杀虫剂抗性的影响, 使用溴氰菊酯和氯菊酯两种杀虫剂处理白纹伊蚊, 结果如下图。相关叙述正确的是（ ）\n\n[图1]\n\nA: 杀虫剂的使用使 kdr 基因发生突变\nB: kdr 基因突变型个体对氯菊酯更敏感\nC: 实验结果说明突变型对杀虫剂的抵抗效果更好\nD: 轮换使用两种杀虫剂可延缓白纹伊蚊产生抗性\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-66.jpg?height=728&width=874&top_left_y=167&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_924",
        "problem": "某家系患有甲、乙两种单基因遗传病，其基因分别用 $\\mathrm{A} 、 \\mathrm{a} 、 \\mathrm{~B} 、 \\mathrm{~b}$ 表示。甲病是伴性遗传病, 1 号个体的双亲只有父亲患甲病, 下图是家系部分系谱图和核酸分子杂交诊断结果示意图。在不考虑家系内发生新的基因突变和染色体变异的情况下，据图判断下列叙述错误的是（ ）\n[图1]\nA: 甲病是伴 X 染色体显性遗传病\nB: 根据杂交诊断图中 1、2、3 可以判定乙病遗传方式\nC: 根据 4 号个体可以判定 1 号个体的基因型为 $\\mathrm{X}^{\\mathrm{Ab}} \\mathrm{X}^{\\mathrm{aB}}$\nD: 6 号个体可能只患甲病或者正常\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某家系患有甲、乙两种单基因遗传病，其基因分别用 $\\mathrm{A} 、 \\mathrm{a} 、 \\mathrm{~B} 、 \\mathrm{~b}$ 表示。甲病是伴性遗传病, 1 号个体的双亲只有父亲患甲病, 下图是家系部分系谱图和核酸分子杂交诊断结果示意图。在不考虑家系内发生新的基因突变和染色体变异的情况下，据图判断下列叙述错误的是（ ）\n[图1]\n\nA: 甲病是伴 X 染色体显性遗传病\nB: 根据杂交诊断图中 1、2、3 可以判定乙病遗传方式\nC: 根据 4 号个体可以判定 1 号个体的基因型为 $\\mathrm{X}^{\\mathrm{Ab}} \\mathrm{X}^{\\mathrm{aB}}$\nD: 6 号个体可能只患甲病或者正常\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-75.jpg?height=416&width=1422&top_left_y=1136&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_706",
        "problem": "果蝇 $(2 \\mathrm{~N}=8)$ 的精巢中，处于分裂期的 $\\mathrm{a} 、 \\mathrm{~b} 、 \\mathrm{c}$ 三个细胞中的染色体组数、四分体个数和染色单体数如表所示 (不考虑基因突变和染色体变异)。下列相关叙述错误的是\n\n| 细胞 | 染色体组数 | 四分体个数 | 染色单体数 |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{a}$ | 2 | 0 | 16 |\n| $\\mathrm{b}$ | 2 | 4 | 16 |\n| $\\mathrm{c}$ | 2 | 0 | 0 |\nA: c 细胞可能含有 0 条、 1 条或 2 条 Y 染色体\nB: b 细胞可能发生了非姐妹染色单体间片段的互换导致基因重组\nC: $a 、 b 、 c$ 细胞中的染色体数相同\nD: a 细胞可能进行有丝分裂, 也可能进行减数分裂\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇 $(2 \\mathrm{~N}=8)$ 的精巢中，处于分裂期的 $\\mathrm{a} 、 \\mathrm{~b} 、 \\mathrm{c}$ 三个细胞中的染色体组数、四分体个数和染色单体数如表所示 (不考虑基因突变和染色体变异)。下列相关叙述错误的是\n\n| 细胞 | 染色体组数 | 四分体个数 | 染色单体数 |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{a}$ | 2 | 0 | 16 |\n| $\\mathrm{b}$ | 2 | 4 | 16 |\n| $\\mathrm{c}$ | 2 | 0 | 0 |\n\nA: c 细胞可能含有 0 条、 1 条或 2 条 Y 染色体\nB: b 细胞可能发生了非姐妹染色单体间片段的互换导致基因重组\nC: $a 、 b 、 c$ 细胞中的染色体数相同\nD: a 细胞可能进行有丝分裂, 也可能进行减数分裂\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1407",
        "problem": "Review the following map and answer the 5 subsequent questions.\n\n[figure1]\n\nEcologists have been studying the small archipelago shown in the map above. They have been comparing the ecology of the mainland with those on the nearby islands and have come up with a few discoveries. The area has not been known to have been inhabited by humans, so it remains very natural. The mainland contains a low mountain range approximately $50 \\mathrm{~km}$ inland from the coast (highest peak approximately $400 \\mathrm{~m}$ above sea level) and the prevailing winds are westerlies.\n\nWhich Island is expected to contain the greatest species diversity?\nA: 1\nB: 5\nC: 3\nD: 2\nE: 6\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nReview the following map and answer the 5 subsequent questions.\n\n[figure1]\n\nEcologists have been studying the small archipelago shown in the map above. They have been comparing the ecology of the mainland with those on the nearby islands and have come up with a few discoveries. The area has not been known to have been inhabited by humans, so it remains very natural. The mainland contains a low mountain range approximately $50 \\mathrm{~km}$ inland from the coast (highest peak approximately $400 \\mathrm{~m}$ above sea level) and the prevailing winds are westerlies.\n\nWhich Island is expected to contain the greatest species diversity?\n\nA: 1\nB: 5\nC: 3\nD: 2\nE: 6\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-39.jpg?height=925&width=1479&top_left_y=451&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1502",
        "problem": "The NDMAR protein is a membrane channel found on neurones. It binds glutamate then opens to allow ions into the cell. Several drugs (NO, ethanol, ketamine) inhibit NDMAR and can cause powerful psychological dissociation. The effect of several inhibitors is shown.\n\n$$\n\\text { Ion entry }\n$$\n\n$$\n\\text { into neurone }\n$$\n[figure1]\n\nWhich binds inside the channel of open NDMAR and prevents ions flowing?\nA: A\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: All of them\nE: None of them\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe NDMAR protein is a membrane channel found on neurones. It binds glutamate then opens to allow ions into the cell. Several drugs (NO, ethanol, ketamine) inhibit NDMAR and can cause powerful psychological dissociation. The effect of several inhibitors is shown.\n\n$$\n\\text { Ion entry }\n$$\n\n$$\n\\text { into neurone }\n$$\n[figure1]\n\nWhich binds inside the channel of open NDMAR and prevents ions flowing?\n\nA: A\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: All of them\nE: None of them\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-10.jpg?height=960&width=1476&top_left_y=591&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_654",
        "problem": "葛莱弗德氏综合征是一种性染色体数目异常的疾病。如图为甲、乙两种单基因遗传病的家系图, 其中一种病为伴性遗传。已知甲病正常人群中携带者占 $1 / 3$, 且含甲病基因的精子成活率为 $1 / 2$, 雌配子正常。 $\\mathrm{I}_{1}$ 所生女儿一定携带乙病致病基因, 家系中无基因突变发生。下列叙述正确的是（ ）\n\n[图1]\nA: $\\mathrm{III}_{12}$ 的性染色体来自 $\\mathrm{I}_{2}$ 和 $\\mathrm{I}_{3}$ 的概率相同\nB: 仅考虑甲病, $\\mathrm{III}_{11}$ 与 $\\mathrm{III}_{12}$ 基因型相同的概率为 $13 / 25$\nC: 若 $\\mathrm{II}_{5}$ 和 $\\mathrm{II}_{6}$ 生一个葛莱弗德氏综合征的乙病小孩, 可能 $\\mathrm{II}_{6}$ 在减数第一次分裂出现异常，减数第二次分裂过程正常\nD: $\\mathrm{III}_{10}$ 和人群中正常男性结婚, 生育一个两病兼患男孩的概率为 $1 / 96$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n葛莱弗德氏综合征是一种性染色体数目异常的疾病。如图为甲、乙两种单基因遗传病的家系图, 其中一种病为伴性遗传。已知甲病正常人群中携带者占 $1 / 3$, 且含甲病基因的精子成活率为 $1 / 2$, 雌配子正常。 $\\mathrm{I}_{1}$ 所生女儿一定携带乙病致病基因, 家系中无基因突变发生。下列叙述正确的是（ ）\n\n[图1]\n\nA: $\\mathrm{III}_{12}$ 的性染色体来自 $\\mathrm{I}_{2}$ 和 $\\mathrm{I}_{3}$ 的概率相同\nB: 仅考虑甲病, $\\mathrm{III}_{11}$ 与 $\\mathrm{III}_{12}$ 基因型相同的概率为 $13 / 25$\nC: 若 $\\mathrm{II}_{5}$ 和 $\\mathrm{II}_{6}$ 生一个葛莱弗德氏综合征的乙病小孩, 可能 $\\mathrm{II}_{6}$ 在减数第一次分裂出现异常，减数第二次分裂过程正常\nD: $\\mathrm{III}_{10}$ 和人群中正常男性结婚, 生育一个两病兼患男孩的概率为 $1 / 96$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-05.jpg?height=434&width=1168&top_left_y=1459&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-06.jpg?height=46&width=1380&top_left_y=742&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1108",
        "problem": "The famous experiment by Avery, MacLeod and McCarty showed that neither heat killed S-type nor live R-type pneumococci could kill mice (experiment 1) but simultaneous injection of both killed the mice just as efficiently as the live S-type (experiment 2). S-type pneumococci are known to possess a capsule structure outermost to the cell wall. Structurally, capsule is a linear homopolymer of sialic acid residues. Apart from the capsule, there are other virulence factors present in these bacteria such as adhesion proteins that recognize a wide range of molecular motifs and provide targeting of the bacteria to specific tissue surfaces in the host.\n\nWhich of the following is correct?\nA: One can hypothesise that sialic acid residues are likely to be poorly immunogenic if the same homopolymer is a structural component of the host.\nB: If the capsular deficient mutants are used in place of S-type bacteria and adhesion deficient mutants in place of R-type in the Avery, MacLeod and McCarty experiment, mice would be killed.\nC: If genomic DNA from S-type is extracted and injected sub-cutaneously into the mice, most of the mice would be killed.\nD: In the experiments performed by Avery, MacLeod and McCarty, all types of live R strains recovered from dead mice would also be virulent.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe famous experiment by Avery, MacLeod and McCarty showed that neither heat killed S-type nor live R-type pneumococci could kill mice (experiment 1) but simultaneous injection of both killed the mice just as efficiently as the live S-type (experiment 2). S-type pneumococci are known to possess a capsule structure outermost to the cell wall. Structurally, capsule is a linear homopolymer of sialic acid residues. Apart from the capsule, there are other virulence factors present in these bacteria such as adhesion proteins that recognize a wide range of molecular motifs and provide targeting of the bacteria to specific tissue surfaces in the host.\n\nWhich of the following is correct?\n\nA: One can hypothesise that sialic acid residues are likely to be poorly immunogenic if the same homopolymer is a structural component of the host.\nB: If the capsular deficient mutants are used in place of S-type bacteria and adhesion deficient mutants in place of R-type in the Avery, MacLeod and McCarty experiment, mice would be killed.\nC: If genomic DNA from S-type is extracted and injected sub-cutaneously into the mice, most of the mice would be killed.\nD: In the experiments performed by Avery, MacLeod and McCarty, all types of live R strains recovered from dead mice would also be virulent.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_645",
        "problem": "下图为甲、乙两种单基因遗传病的遗传家系图, 其中一种遗传病为伴性遗传, 相关基因均不位于 Y 染色体上，人群中乙病的发病率为 $1 / 256$ 。下列叙述错误的是（ ）\n\n[图1]\nA: 甲病为伴 X 染色体显性遗传病，乙病为常染色体隐性遗传病\nB: $\\mathrm{I}_{4} 、 \\mathrm{II}_{5}$ 和 $\\mathrm{III}_{6}$ 与乙病有关的基因型均相同\nC: $\\mathrm{II}_{2}$ 和 $\\mathrm{II}_{3}$ 生育男孩可避免子代患甲病\nD: $\\mathrm{III}_{1}$ 与某正常男性结婚, 所生正常孩子的概率为 $95 / 192$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为甲、乙两种单基因遗传病的遗传家系图, 其中一种遗传病为伴性遗传, 相关基因均不位于 Y 染色体上，人群中乙病的发病率为 $1 / 256$ 。下列叙述错误的是（ ）\n\n[图1]\n\nA: 甲病为伴 X 染色体显性遗传病，乙病为常染色体隐性遗传病\nB: $\\mathrm{I}_{4} 、 \\mathrm{II}_{5}$ 和 $\\mathrm{III}_{6}$ 与乙病有关的基因型均相同\nC: $\\mathrm{II}_{2}$ 和 $\\mathrm{II}_{3}$ 生育男孩可避免子代患甲病\nD: $\\mathrm{III}_{1}$ 与某正常男性结婚, 所生正常孩子的概率为 $95 / 192$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-70.jpg?height=348&width=934&top_left_y=1468&top_left_x=367"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_809",
        "problem": "在引进优良家畜（如牛、羊）时一般引进雄性个体, 让其与本地的同种雌性个体杂\n\n交，考虑到经济与遗传方面的因素，一般以引进 4 头雄性家畜为宜，具体做法是：让一头引种雄畜与一头本地种雌性个体杂交得 $F_{1}, F_{1}$ 与第二头引种雄畜相交得 $F_{2}, F_{2}$ 与第三头引种雄畜相交得 $\\mathrm{F}_{3}, \\mathrm{~F}_{3}$ 与第四头引种雄畜相交得 $\\mathrm{F}_{4}$, 试问 $\\mathrm{F}_{4}$ 个体中, 细胞核内的遗传物质属本地种的约有多少（ ）\nA: $\\frac{1}{8}$\nB: $\\frac{1}{16}$\nC: $\\frac{15}{16}$\nD: $\\frac{1}{2}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在引进优良家畜（如牛、羊）时一般引进雄性个体, 让其与本地的同种雌性个体杂\n\n交，考虑到经济与遗传方面的因素，一般以引进 4 头雄性家畜为宜，具体做法是：让一头引种雄畜与一头本地种雌性个体杂交得 $F_{1}, F_{1}$ 与第二头引种雄畜相交得 $F_{2}, F_{2}$ 与第三头引种雄畜相交得 $\\mathrm{F}_{3}, \\mathrm{~F}_{3}$ 与第四头引种雄畜相交得 $\\mathrm{F}_{4}$, 试问 $\\mathrm{F}_{4}$ 个体中, 细胞核内的遗传物质属本地种的约有多少（ ）\n\nA: $\\frac{1}{8}$\nB: $\\frac{1}{16}$\nC: $\\frac{15}{16}$\nD: $\\frac{1}{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1453",
        "problem": "Auxin (IAA) is a hormone produced primarily in shoot apical meristems and young leaves. Root apical meristems also produce some auxin, although the root depends on the shoot for much of its auxin. Amongst its many functions in plant growth, auxin plays an important role in gravitropism.\n\nFigure A shows the anatomy of a root. Figure B shows the relationship between auxin concentration and growth of roots and shoots.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\nWhat concentration/s of auxin leads to a percentage stimulation of shoot growth closest to $50 \\%$ ?\nA: $10^{-1}$ parts per million\nB: $10^{-1}$ parts per million and 10 parts per million\nC: $10^{-2}$ parts per million and 10 parts per million\nD: $2 \\times 10^{-2}$ parts per million and 30 parts per million\nE: $6 \\times 10^{-2}$ parts per million and 70 parts per million\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAuxin (IAA) is a hormone produced primarily in shoot apical meristems and young leaves. Root apical meristems also produce some auxin, although the root depends on the shoot for much of its auxin. Amongst its many functions in plant growth, auxin plays an important role in gravitropism.\n\nFigure A shows the anatomy of a root. Figure B shows the relationship between auxin concentration and growth of roots and shoots.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\nWhat concentration/s of auxin leads to a percentage stimulation of shoot growth closest to $50 \\%$ ?\n\nA: $10^{-1}$ parts per million\nB: $10^{-1}$ parts per million and 10 parts per million\nC: $10^{-2}$ parts per million and 10 parts per million\nD: $2 \\times 10^{-2}$ parts per million and 30 parts per million\nE: $6 \\times 10^{-2}$ parts per million and 70 parts per million\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-17.jpg?height=731&width=599&top_left_y=700&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-17.jpg?height=671&width=683&top_left_y=767&top_left_x=1052"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_421",
        "problem": "如图 1 为甲、乙两种单基因遗传病的遗传系谱图, 已知甲病相关基因经限制酶 $S m a \\mathrm{I}$酶切后的电泳图如图 2 所示, 且在男性中甲病的患病概率为 $1 / 7$; 乙病性状的遗传与性别无关，II-4 不带有乙病致病基因。下列分析错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n注:不考虑 X、Y染色体的同源区段。\n\n注：不考虑 X、Y 染色体的同源区段。\nA: 甲病为伴 X 染色体隐性遗传病, 乙病为常染色体显性遗传病\nB: 限制酶 SmaI 不能识别切割甲病致病基因\nC: II-5 与II-6 生了一个男孩, 正常的概率为 $1 / 6$\nD: III- 3 携带甲病致病基因的概率是 $1 / 8$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图 1 为甲、乙两种单基因遗传病的遗传系谱图, 已知甲病相关基因经限制酶 $S m a \\mathrm{I}$酶切后的电泳图如图 2 所示, 且在男性中甲病的患病概率为 $1 / 7$; 乙病性状的遗传与性别无关，II-4 不带有乙病致病基因。下列分析错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n注:不考虑 X、Y染色体的同源区段。\n\n注：不考虑 X、Y 染色体的同源区段。\n\nA: 甲病为伴 X 染色体隐性遗传病, 乙病为常染色体显性遗传病\nB: 限制酶 SmaI 不能识别切割甲病致病基因\nC: II-5 与II-6 生了一个男孩, 正常的概率为 $1 / 6$\nD: III- 3 携带甲病致病基因的概率是 $1 / 8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-01.jpg?height=312&width=643&top_left_y=752&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-01.jpg?height=255&width=411&top_left_y=792&top_left_x=1368"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_940",
        "problem": "人类有一种隐性遗传病, 其致病基因 $\\mathrm{a}$ 是由基因 $\\mathrm{A}$ 编码序列缺失部分碱基产生的。\n\n图 1 表示与该遗传病有关的某家系成员的关系, 未标明患病个体。通过对该家系成员部分个体进行基因检测, 得到的条带检测结果如图 2 所示。不考虑其他变异, 下列分析错误的是 ( )\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 基因 A 发生的变异属于基因突变\nB: 控制该隐性遗传病的致病基因 a 位于常染色体上\nC: 若 5 号和 6 号再生了一个患病孩子，则该患病孩子一定为男孩\nD: 若 7 号和一个正常男性结婚, 则生下患病孩子的概率为 $1 / 8$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人类有一种隐性遗传病, 其致病基因 $\\mathrm{a}$ 是由基因 $\\mathrm{A}$ 编码序列缺失部分碱基产生的。\n\n图 1 表示与该遗传病有关的某家系成员的关系, 未标明患病个体。通过对该家系成员部分个体进行基因检测, 得到的条带检测结果如图 2 所示。不考虑其他变异, 下列分析错误的是 ( )\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 基因 A 发生的变异属于基因突变\nB: 控制该隐性遗传病的致病基因 a 位于常染色体上\nC: 若 5 号和 6 号再生了一个患病孩子，则该患病孩子一定为男孩\nD: 若 7 号和一个正常男性结婚, 则生下患病孩子的概率为 $1 / 8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-40.jpg?height=346&width=577&top_left_y=2374&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-40.jpg?height=335&width=832&top_left_y=2374&top_left_x=886"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1100",
        "problem": "In Angiosperms, double fertilization provides the stimulus to the formation of seeds. Growth (measured as change in volume) of ovule, embryo and endosperm in Pisum is plotted in the graph.\n\n[figure1]\n\nI, II and III respectively represent:\nA: endosperm, ovule and embryo.\nB: ovule, endosperm and embryo.\nC: embryo, ovule and endosperm.\nD: endosperm, embryo and ovule.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn Angiosperms, double fertilization provides the stimulus to the formation of seeds. Growth (measured as change in volume) of ovule, embryo and endosperm in Pisum is plotted in the graph.\n\n[figure1]\n\nI, II and III respectively represent:\n\nA: endosperm, ovule and embryo.\nB: ovule, endosperm and embryo.\nC: embryo, ovule and endosperm.\nD: endosperm, embryo and ovule.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-05.jpg?height=748&width=813&top_left_y=754&top_left_x=970"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_903",
        "problem": "某雌雄同株植物花色的紫色和红色分别由基因 A、a 控制，大花和小花分别由基因 $\\mathrm{B} 、 \\mathrm{~b}$ 控制, 且两对基因位于一对同源染色体上。用紫色大花植株（ $\\mathrm{AaBb}$ ）作亲本进行连续自交, $\\mathrm{F}_{\\mathrm{n}}$ 中杂合紫色大花个体的比例为 $1 / 2^{\\mathrm{n}}$ 。仅考虑上述基因且不考虑基因突变及染色体互换，下列叙述正确的是（）\nA: 组成基因 $\\mathrm{A}$ 和基因 $\\mathrm{B}$ 的碱基种类不同\nB: 该亲本植株的基因 $\\mathrm{A}$ 和基因 $\\mathrm{B}$ 位于同一条染色体上\nC: 每一代的子代中均会出现 4 种表型\nD: 子代中不同基因型的纯合个体所占的比例相同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某雌雄同株植物花色的紫色和红色分别由基因 A、a 控制，大花和小花分别由基因 $\\mathrm{B} 、 \\mathrm{~b}$ 控制, 且两对基因位于一对同源染色体上。用紫色大花植株（ $\\mathrm{AaBb}$ ）作亲本进行连续自交, $\\mathrm{F}_{\\mathrm{n}}$ 中杂合紫色大花个体的比例为 $1 / 2^{\\mathrm{n}}$ 。仅考虑上述基因且不考虑基因突变及染色体互换，下列叙述正确的是（）\n\nA: 组成基因 $\\mathrm{A}$ 和基因 $\\mathrm{B}$ 的碱基种类不同\nB: 该亲本植株的基因 $\\mathrm{A}$ 和基因 $\\mathrm{B}$ 位于同一条染色体上\nC: 每一代的子代中均会出现 4 种表型\nD: 子代中不同基因型的纯合个体所占的比例相同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_360",
        "problem": "某同学用标有 $\\mathrm{B}$ 和 $\\mathrm{b}$ 的卡片建立某符合遗传平衡定律的人群中红绿色盲的遗传模型,向甲（男）、乙（女）两个信封放入若干 $\\mathrm{B}$ 和 $\\mathrm{b}$ 的卡片，随机从每个信封中取出一张卡片放在一起并记录, 记录后将卡片放回原信封内, 重复多次。已知男性群体的红绿色盲率接近 $7 \\%$ ，下列叙述错误的是（）\nA: 甲、乙信封中的 B 与 $\\mathrm{b}$ 卡片数之比约为 $93: 7$\nB: 甲信封中还应装入与 $\\mathrm{B} 、 \\mathrm{~b}$ 卡片数之和相等的空白卡片\nC: 乙信封模拟的女性群体中红绿色盲占比约为 $0.5 \\%$\nD: 完善模型后, 重复多次实验, 可推出人群中携带者的概率约为 $13 \\%$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某同学用标有 $\\mathrm{B}$ 和 $\\mathrm{b}$ 的卡片建立某符合遗传平衡定律的人群中红绿色盲的遗传模型,向甲（男）、乙（女）两个信封放入若干 $\\mathrm{B}$ 和 $\\mathrm{b}$ 的卡片，随机从每个信封中取出一张卡片放在一起并记录, 记录后将卡片放回原信封内, 重复多次。已知男性群体的红绿色盲率接近 $7 \\%$ ，下列叙述错误的是（）\n\nA: 甲、乙信封中的 B 与 $\\mathrm{b}$ 卡片数之比约为 $93: 7$\nB: 甲信封中还应装入与 $\\mathrm{B} 、 \\mathrm{~b}$ 卡片数之和相等的空白卡片\nC: 乙信封模拟的女性群体中红绿色盲占比约为 $0.5 \\%$\nD: 完善模型后, 重复多次实验, 可推出人群中携带者的概率约为 $13 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1114",
        "problem": "The diploid chromosome number in the leaf cell of Pisum is 14 . The diagram given below depicts the ovule structure immediately after fertilization. (Note: PEN refers to primary endosperm nucleus)\n\n[figure1]\n\nStructures 1 - 4 and the number of chromosomes present in them respectively are:\nA: nucellus:7; integument:14; PEN:21; egg cell:7.\nB: nucellus:14; integument:14; PEN:21; zygote:14.\nC: embryo sac:7; integument:14; PEN:14; zygote:14.\nD: nucellus:14; integument:7; PEN:14; zygote:14.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diploid chromosome number in the leaf cell of Pisum is 14 . The diagram given below depicts the ovule structure immediately after fertilization. (Note: PEN refers to primary endosperm nucleus)\n\n[figure1]\n\nStructures 1 - 4 and the number of chromosomes present in them respectively are:\n\nA: nucellus:7; integument:14; PEN:21; egg cell:7.\nB: nucellus:14; integument:14; PEN:21; zygote:14.\nC: embryo sac:7; integument:14; PEN:14; zygote:14.\nD: nucellus:14; integument:7; PEN:14; zygote:14.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-05.jpg?height=776&width=650&top_left_y=1691&top_left_x=735"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1194",
        "problem": "The diagram on the left illustrates the different body planes anatomists and radiologists use when talking about structures within the human body.\n[figure1]\n\nWhich plane would the picture on the RIGHT represent?\nA:  Sagittal plane.\nB: Coronal plane.\nC: Transverse plane.\nD: Oblique plane.\nE: None of the above.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram on the left illustrates the different body planes anatomists and radiologists use when talking about structures within the human body.\n[figure1]\n\nWhich plane would the picture on the RIGHT represent?\n\nA:  Sagittal plane.\nB: Coronal plane.\nC: Transverse plane.\nD: Oblique plane.\nE: None of the above.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-06.jpg?height=1020&width=1438&top_left_y=362&top_left_x=134"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1173",
        "problem": "## AVOIDING FREEZING IN ANTARCTICA\n\nThe research group of Associate Professor Clive Evans at the University of Auckland has an active Antarctic Programme investigating how fish avoid freezing in the frigid waters of the Southern Ocean surrounding Antarctica.\n\nSeawater temperature hovers close to its freezing point of around $-1.93^{\\circ} \\mathrm{C}$ throughout the year but most marine fish freeze at about $-0.7^{\\circ} \\mathrm{C}$ so they cannot survive in the icy Antarctic waters. Notothenioid fishes (icefish) thrive in this freezing environment because they are able to produce antifreeze glycoproteins (AFGPs). AFGPs bind to and inhibit the growth of minute ice crystals that occasionally enter the fish, thus preventing their body fluids from freezing. This key evolutionary innovation allowed the icefish to colonize the frigid waters of the Southern Ocean some 5-15 million years ago.\n\n[figure1]\n\nIcefish risk freezing of the intestinal tract by swallowing ice in ingested seawater or food. This suggests that AFGPs should be present in the stomach and intestinal fluids to decrease the risk of freezing initiated by ingested ice. In the diagram below, the activity of antifreeze in these fluids is examined by measuring the difference between melting and freezing points in degrees Celsius, a measure of thermal hysteresis (TH). TH is represented by the gray box.\n\n[figure2]\n\nCheng C C et al. PNAS 2006;103:10491-10496The species of fish is written in italics on the left hand side of the diagram. Which species shows the greatest TH of the intestinal fluid?\nA: D. mawsoni\nB: P. borchgrevinki\nC: T. bernacchii\nD: G. acuticeps\nE: L. dearborni\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## AVOIDING FREEZING IN ANTARCTICA\n\nThe research group of Associate Professor Clive Evans at the University of Auckland has an active Antarctic Programme investigating how fish avoid freezing in the frigid waters of the Southern Ocean surrounding Antarctica.\n\nSeawater temperature hovers close to its freezing point of around $-1.93^{\\circ} \\mathrm{C}$ throughout the year but most marine fish freeze at about $-0.7^{\\circ} \\mathrm{C}$ so they cannot survive in the icy Antarctic waters. Notothenioid fishes (icefish) thrive in this freezing environment because they are able to produce antifreeze glycoproteins (AFGPs). AFGPs bind to and inhibit the growth of minute ice crystals that occasionally enter the fish, thus preventing their body fluids from freezing. This key evolutionary innovation allowed the icefish to colonize the frigid waters of the Southern Ocean some 5-15 million years ago.\n\n[figure1]\n\nIcefish risk freezing of the intestinal tract by swallowing ice in ingested seawater or food. This suggests that AFGPs should be present in the stomach and intestinal fluids to decrease the risk of freezing initiated by ingested ice. In the diagram below, the activity of antifreeze in these fluids is examined by measuring the difference between melting and freezing points in degrees Celsius, a measure of thermal hysteresis (TH). TH is represented by the gray box.\n\n[figure2]\n\nCheng C C et al. PNAS 2006;103:10491-10496\n\nproblem:\nThe species of fish is written in italics on the left hand side of the diagram. Which species shows the greatest TH of the intestinal fluid?\n\nA: D. mawsoni\nB: P. borchgrevinki\nC: T. bernacchii\nD: G. acuticeps\nE: L. dearborni\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-30.jpg?height=554&width=813&top_left_y=440&top_left_x=1084",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-30.jpg?height=1011&width=868&top_left_y=1413&top_left_x=114"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_647",
        "problem": "为研究四倍体亚洲百合 (48 条染色体) 花粉生活力较低的原因, 研究人员对四倍体亚洲百合的花粉母细胞进行观察, 发现在减数分裂过程中存在滞后染色体 (图 1)、细胞分裂不同步（图 2) 及两者同时存在（图 3) 等异常现象, 染色体发生交错纠缠或断裂、倒位等变异导致配子不育。下列相关分析错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\nA: 染色体滞后、细胞分裂不同步以及染色体结构变异均可在显微镜下观察到\nB: 图 1 染色体滞后发生在减数分裂II后期, 可能形成四个染色体数目异常的配子\nC: 图 2 细胞的分裂不同步现象可能与纺锤体异常有关, 导致形成三个具核细胞\nD: 四倍体亚洲百合细胞有丝分裂和减数分裂都可能发生同源染色体间的互换\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n为研究四倍体亚洲百合 (48 条染色体) 花粉生活力较低的原因, 研究人员对四倍体亚洲百合的花粉母细胞进行观察, 发现在减数分裂过程中存在滞后染色体 (图 1)、细胞分裂不同步（图 2) 及两者同时存在（图 3) 等异常现象, 染色体发生交错纠缠或断裂、倒位等变异导致配子不育。下列相关分析错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\n\nA: 染色体滞后、细胞分裂不同步以及染色体结构变异均可在显微镜下观察到\nB: 图 1 染色体滞后发生在减数分裂II后期, 可能形成四个染色体数目异常的配子\nC: 图 2 细胞的分裂不同步现象可能与纺锤体异常有关, 导致形成三个具核细胞\nD: 四倍体亚洲百合细胞有丝分裂和减数分裂都可能发生同源染色体间的互换\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-01.jpg?height=226&width=303&top_left_y=818&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-01.jpg?height=234&width=294&top_left_y=820&top_left_x=664",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-01.jpg?height=234&width=297&top_left_y=820&top_left_x=1002"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1129",
        "problem": "Eosin is an acidic stain that is widely used for staining cytoplasm of eukaryotic cells. Which of the following is responsible for this specificity?\nA: Ability of eosin to bind with water molecules in cytoplasm.\nB: Ability of eosin to bind with amino acids in cytoplasm.\nC: Eosin can cross the plasma membrane but cannot cross the nuclear membrane and hence accumulates in cytoplasm.\nD: Ability of eosin to bind with carbohydrate moieties in the cytoplasm.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEosin is an acidic stain that is widely used for staining cytoplasm of eukaryotic cells. Which of the following is responsible for this specificity?\n\nA: Ability of eosin to bind with water molecules in cytoplasm.\nB: Ability of eosin to bind with amino acids in cytoplasm.\nC: Eosin can cross the plasma membrane but cannot cross the nuclear membrane and hence accumulates in cytoplasm.\nD: Ability of eosin to bind with carbohydrate moieties in the cytoplasm.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1255",
        "problem": "## DEAR ENEMIES \\& NASTY NEIGHBORS\n\n\"Dear Enemy, I curse you, and hope that something slightly unpleasant happens to you, like an onion falling on your head.\"\n\n## Blackadder I, The Archbishop\n\nMany species show territorial behaviour (actively defending an area containing resources) allowing the species to monopolise resources such as food or mates. Territoriality is costly in terms of energy and time so some species reduce these costs by being less aggressive towards their neighbours than towards unfamiliar strangers, the so called \"dear enemy\" hypothesis. The converse are \"nasty neighbours\" in which species are more, not less aggressive towards their neighbours.\n\nDr Brunton's research group, at the Institute of Natural Sciences at Massey University, has been studying the behaviour of the New Zealand bellbird,\n\n[figure1]\n(Anthornis melanura) on Tiritiri Matangi Island. Both sexes sing prolifically and are known to use vocalisations to recognise individuals. Male and female bellbirds counter-sing in response to the vocalisations of their neighbours in a territorial behaviour that may lead to chasing of an individual bird intruding on a territory.\n\nThe researchers used speakers to play the song of neighbouring females or stranger females at different breeding stages. They recorded the responses of male and female bellbirds, including counter-singing and whether the birds approached the speaker. An asterisk indicates significant differences between the response to neighbours and strangers. Due to small sample sizes, the responses to the neighbours versus strangers were not tested during the incubation stage. Control 1 was silence, control 2 played back the\n\n[figure2]\nsong of a different species.\n[figure3]\n\nb) Counter-singing\n[figure4]\n\nUsing the information and the graphs above answer Questions 54 - 58 on the following page.The average percentage of female bellbirds responding to playback of neighboring female song by counter-singing over the entire breeding season was?\nA: $90 \\%$\nB: $82 \\%$\nC: $77 \\%$\nD: $62 \\%$\nE: Unable to be determined\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## DEAR ENEMIES \\& NASTY NEIGHBORS\n\n\"Dear Enemy, I curse you, and hope that something slightly unpleasant happens to you, like an onion falling on your head.\"\n\n## Blackadder I, The Archbishop\n\nMany species show territorial behaviour (actively defending an area containing resources) allowing the species to monopolise resources such as food or mates. Territoriality is costly in terms of energy and time so some species reduce these costs by being less aggressive towards their neighbours than towards unfamiliar strangers, the so called \"dear enemy\" hypothesis. The converse are \"nasty neighbours\" in which species are more, not less aggressive towards their neighbours.\n\nDr Brunton's research group, at the Institute of Natural Sciences at Massey University, has been studying the behaviour of the New Zealand bellbird,\n\n[figure1]\n(Anthornis melanura) on Tiritiri Matangi Island. Both sexes sing prolifically and are known to use vocalisations to recognise individuals. Male and female bellbirds counter-sing in response to the vocalisations of their neighbours in a territorial behaviour that may lead to chasing of an individual bird intruding on a territory.\n\nThe researchers used speakers to play the song of neighbouring females or stranger females at different breeding stages. They recorded the responses of male and female bellbirds, including counter-singing and whether the birds approached the speaker. An asterisk indicates significant differences between the response to neighbours and strangers. Due to small sample sizes, the responses to the neighbours versus strangers were not tested during the incubation stage. Control 1 was silence, control 2 played back the\n\n[figure2]\nsong of a different species.\n[figure3]\n\nb) Counter-singing\n[figure4]\n\nUsing the information and the graphs above answer Questions 54 - 58 on the following page.\n\nproblem:\nThe average percentage of female bellbirds responding to playback of neighboring female song by counter-singing over the entire breeding season was?\n\nA: $90 \\%$\nB: $82 \\%$\nC: $77 \\%$\nD: $62 \\%$\nE: Unable to be determined\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=357&width=571&top_left_y=524&top_left_x=1368",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=485&width=311&top_left_y=931&top_left_x=1495",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=486&width=1734&top_left_y=1493&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=534&width=1744&top_left_y=2042&top_left_x=108"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_408",
        "problem": "色盲有红色盲、绿色盲和蓝色盲，分别由 $\\mathrm{LW} 、 \\mathrm{MW}$ 和 SW 基因突变引起。调查发现, 红色盲和绿色盲不能由爷爷传给孙子。下图是三种色盲的遗传图谱, 下列说法错误的是 ( )\n\n[图1]\nA: 三种色盲的遗传方式均为隐性遗传\nB: 红色盲和绿色盲在男性人群中的发病率高\nC: II-5 和II-6 生出色觉正常孩子的概率为 $3 / 8$\nD: III- 1 与II- 2 基因型相同的概率为 $2 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n色盲有红色盲、绿色盲和蓝色盲，分别由 $\\mathrm{LW} 、 \\mathrm{MW}$ 和 SW 基因突变引起。调查发现, 红色盲和绿色盲不能由爷爷传给孙子。下图是三种色盲的遗传图谱, 下列说法错误的是 ( )\n\n[图1]\n\nA: 三种色盲的遗传方式均为隐性遗传\nB: 红色盲和绿色盲在男性人群中的发病率高\nC: II-5 和II-6 生出色觉正常孩子的概率为 $3 / 8$\nD: III- 1 与II- 2 基因型相同的概率为 $2 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-53.jpg?height=331&width=1345&top_left_y=180&top_left_x=367"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_160",
        "problem": "A man had lost approximately $700 \\mathrm{~mL}$ of blood in a severe injury of a major artery in a motorcycle accident. At the time of accident, his blood pressure was $90 / 50 \\mathrm{mmHg}$. Several physiological changes should be expected in response to hemorrhage.\nA: Oxygen affinity of hemoglobin was increased getting more oxygen for cellular respiration.\nB: Total peripheral resistance was increased.\nC: Hyperpolarization occurred in the cells of the sinoatrial node.\nD: Vasoconstriction occurred in the brain and in the coronary arteries.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA man had lost approximately $700 \\mathrm{~mL}$ of blood in a severe injury of a major artery in a motorcycle accident. At the time of accident, his blood pressure was $90 / 50 \\mathrm{mmHg}$. Several physiological changes should be expected in response to hemorrhage.\n\nA: Oxygen affinity of hemoglobin was increased getting more oxygen for cellular respiration.\nB: Total peripheral resistance was increased.\nC: Hyperpolarization occurred in the cells of the sinoatrial node.\nD: Vasoconstriction occurred in the brain and in the coronary arteries.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_341",
        "problem": "如图为某家族的遗传系谱图, 已知 $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{3}$ 均患有白化病。假设白化病在人群中的发病率约为 $1 / 10000$ 。下列相关说法正确的是 ( )\n\n[图1]\nA: $\\mathrm{II}_{2}$ 与 $\\mathrm{II}_{4}$ 基因型相同的概率为 $1 / 4$\nB: $\\mathrm{II}_{1}$ 与 $\\mathrm{II}_{2}$ 基因型相同的概率为 $1 / 2$\nC: 若 $\\mathrm{IV}_{1}$ 与 $\\mathrm{IV}_{2}$ 结婚, 所生后代患白化病的概率小于 $1 / 10000$\nD: 若 $\\mathrm{II}_{1}$ 为携带者, 则 $\\mathrm{III}_{2}$ 含有来自 $\\mathrm{II}_{2}$ 的致病基因的概率为 $1 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图为某家族的遗传系谱图, 已知 $\\mathrm{I}_{1}$ 和 $\\mathrm{I}_{3}$ 均患有白化病。假设白化病在人群中的发病率约为 $1 / 10000$ 。下列相关说法正确的是 ( )\n\n[图1]\n\nA: $\\mathrm{II}_{2}$ 与 $\\mathrm{II}_{4}$ 基因型相同的概率为 $1 / 4$\nB: $\\mathrm{II}_{1}$ 与 $\\mathrm{II}_{2}$ 基因型相同的概率为 $1 / 2$\nC: 若 $\\mathrm{IV}_{1}$ 与 $\\mathrm{IV}_{2}$ 结婚, 所生后代患白化病的概率小于 $1 / 10000$\nD: 若 $\\mathrm{II}_{1}$ 为携带者, 则 $\\mathrm{III}_{2}$ 含有来自 $\\mathrm{II}_{2}$ 的致病基因的概率为 $1 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-02.jpg?height=323&width=776&top_left_y=318&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-02.jpg?height=54&width=1039&top_left_y=2360&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_610",
        "problem": "下图是某家系甲、乙、丙三种单基因遗传病的系谱图, 其基因分别用 $A 、 a 、 B 、 b$和 D、 $d$ 表示。甲病是伴性遗传病, II-7 不携带乙病的致病基因。III-15 为丙病基因携带者的概率是 $1 / 100$, 且为乙病致病基因的杂合子。在不考虑家系内发生新的基因突变的情况下, 下列说法正确的是（ ）\n[图1]\nA: 甲、乙病的遗传方式都是伴 X 染色体隐性遗传\nB: II-6 个体的细胞中甲乙病致病基因一定不在同一条染色体上\nC: III-13 患两种遗传病的原因是II-6 的次级卵母细胞发生了变异\nD: III-15 和 III-16 结婚, 所生的子女只患一种病的概率是 301/1200, 患丙病的女孩的概率是 $1 / 1200$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图是某家系甲、乙、丙三种单基因遗传病的系谱图, 其基因分别用 $A 、 a 、 B 、 b$和 D、 $d$ 表示。甲病是伴性遗传病, II-7 不携带乙病的致病基因。III-15 为丙病基因携带者的概率是 $1 / 100$, 且为乙病致病基因的杂合子。在不考虑家系内发生新的基因突变的情况下, 下列说法正确的是（ ）\n[图1]\n\nA: 甲、乙病的遗传方式都是伴 X 染色体隐性遗传\nB: II-6 个体的细胞中甲乙病致病基因一定不在同一条染色体上\nC: III-13 患两种遗传病的原因是II-6 的次级卵母细胞发生了变异\nD: III-15 和 III-16 结婚, 所生的子女只患一种病的概率是 301/1200, 患丙病的女孩的概率是 $1 / 1200$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-002.jpg?height=396&width=1072&top_left_y=586&top_left_x=387"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_417",
        "problem": "下列病毒的构成组合中错误的是( )\n(1)DNA\n(2)RNA\n(3)蛋白质\n(4)磷脂\nA: (2)(3)(4)\nB: (1)(2)(3)\nC: (1)(3)\nD: (2)(3)\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列病毒的构成组合中错误的是( )\n(1)DNA\n(2)RNA\n(3)蛋白质\n(4)磷脂\n\nA: (2)(3)(4)\nB: (1)(2)(3)\nC: (1)(3)\nD: (2)(3)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1319",
        "problem": "The light reaction of photosynthesis can be made to occur experimentally in the absence of\nA: water\nB: carbon dioxide\nC: chlorophyll\nD: ADP\nE: NADP\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe light reaction of photosynthesis can be made to occur experimentally in the absence of\n\nA: water\nB: carbon dioxide\nC: chlorophyll\nD: ADP\nE: NADP\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1053",
        "problem": "Thermal denaturation of three double stranded DNA samples is represented in the figure.\n\n[figure1]\n\nwhich of the following is correct?\nA: It is likely that GC content of sample ' $A$ ' is highest among all.\nB: If all three DNA samples are polyG:polyC homopolymers, then ' $A$ ' is likely to be the longest of the three.\nC: If $A, B, C$ represent the same DNA sample then the melting curves obtained (from $A \\rightarrow C$ ) could be due to increasing concentration of salt in the solutions.\nD: The graph indicates that double helices have lower molecular absorptivity for UV light as compared to denatured single strands.\nE: The pattern of thermal denaturation shown is characteristic and unique for each DNA molecule and will not change if external conditions such as $\\mathrm{pH}$ or rate of cooling are changed.\nF: If phosphate and pentose sugar moieties are cleaved from DNA molecule ' $A$ ', it will show absorbance close to 1.0 .\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThermal denaturation of three double stranded DNA samples is represented in the figure.\n\n[figure1]\n\nwhich of the following is correct?\n\nA: It is likely that GC content of sample ' $A$ ' is highest among all.\nB: If all three DNA samples are polyG:polyC homopolymers, then ' $A$ ' is likely to be the longest of the three.\nC: If $A, B, C$ represent the same DNA sample then the melting curves obtained (from $A \\rightarrow C$ ) could be due to increasing concentration of salt in the solutions.\nD: The graph indicates that double helices have lower molecular absorptivity for UV light as compared to denatured single strands.\nE: The pattern of thermal denaturation shown is characteristic and unique for each DNA molecule and will not change if external conditions such as $\\mathrm{pH}$ or rate of cooling are changed.\nF: If phosphate and pentose sugar moieties are cleaved from DNA molecule ' $A$ ', it will show absorbance close to 1.0 .\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E, F].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-16.jpg?height=826&width=1027&top_left_y=91&top_left_x=511"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_483",
        "problem": "三体细胞（染色体比正常体细胞多 1 条）在有丝分裂时, 三条染色体中的一条随机丢失，可产生染色体数目正常的体细胞，这种现象称为“三体自救”。某家庭中的男性为正常个体，女性患有进行性肌营养不良（伴 X 染色体隐性遗传病），这对夫妇生下一个染色体正常的患病女孩。下列说法错误的是（ ）\nA: 若发育为该女孩的受精卵发生“三体自救”，则丢失的染色体不可能来自父方\nB: 该女孩患病的原因可能是来自父方 X 染色体上的正常基因突变成致病基因\nC: 该患病女孩的两条 X 染色体上的进行性肌营养不良致病基因可能均来自母方\nD: 该女孩患病的原因可能是母亲在产生卵细胞的过程中, 减数分裂II后期发生异常\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n三体细胞（染色体比正常体细胞多 1 条）在有丝分裂时, 三条染色体中的一条随机丢失，可产生染色体数目正常的体细胞，这种现象称为“三体自救”。某家庭中的男性为正常个体，女性患有进行性肌营养不良（伴 X 染色体隐性遗传病），这对夫妇生下一个染色体正常的患病女孩。下列说法错误的是（ ）\n\nA: 若发育为该女孩的受精卵发生“三体自救”，则丢失的染色体不可能来自父方\nB: 该女孩患病的原因可能是来自父方 X 染色体上的正常基因突变成致病基因\nC: 该患病女孩的两条 X 染色体上的进行性肌营养不良致病基因可能均来自母方\nD: 该女孩患病的原因可能是母亲在产生卵细胞的过程中, 减数分裂II后期发生异常\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_268",
        "problem": "Polarity, charge and molecular weight of molecules can affect their rate of passive diffusion through membranes. Amino acids and drugs like aspirin differ in both efficiency and location of absorption. In the figure below the chemical structure the $\\mathrm{pKa}$ values of aspirin and arginine are represented.\n[figure1]\nA: Aspirin diffuses through membranes mainly in the stomach because more aspirin molecules are in deprotonated form at $\\mathrm{pH}$ of about 1.6 in the stomach.\nB: Arginine diffuses less efficiently than aspirin because of its higher molecular weight.\nC: Optimal pH range for arginine absorption by passive diffusion is between 2.18 and 9.04 .\nD: The proton pump inhibitor, omeprazole, blocks the entry of aspirin into the blood in the initial few minutes after oral administration.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPolarity, charge and molecular weight of molecules can affect their rate of passive diffusion through membranes. Amino acids and drugs like aspirin differ in both efficiency and location of absorption. In the figure below the chemical structure the $\\mathrm{pKa}$ values of aspirin and arginine are represented.\n[figure1]\n\nA: Aspirin diffuses through membranes mainly in the stomach because more aspirin molecules are in deprotonated form at $\\mathrm{pH}$ of about 1.6 in the stomach.\nB: Arginine diffuses less efficiently than aspirin because of its higher molecular weight.\nC: Optimal pH range for arginine absorption by passive diffusion is between 2.18 and 9.04 .\nD: The proton pump inhibitor, omeprazole, blocks the entry of aspirin into the blood in the initial few minutes after oral administration.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://i.postimg.cc/vH6MYVW8/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_133",
        "problem": "In a Xenopus embryo, the dorsal-ventral axis is determined through cortical rotation after fertilization. On the dorsal side of an embryo, the Spemann-Mangold organizer is necessary to determine the body plan of the embryo. When the organizer formation is inhibited, a head defect occurs in embryos. On the other hand, the head is enlarged when the organizer region expands.\n\n$\\beta$-catenin ( $\\beta$-cat) and GSK3 $\\beta$ are involved in organizer formation. The table below shows the results of phenotype of tadpoles microinjected with $\\beta$-cat, GSK3 $\\beta$, an DN $\\beta$-cat ( $\\beta$-catenin inhibition factor), and DN GSK3 $\\beta$ (GSK3 $\\beta$ inhibition factor) into the dorsal or ventral side of the embryo.\n\n| mRNA | Dorsal injection | Ventral injection |\n| :--- | :--- | :--- |\n| $\\beta$-cat | Large head | Secondary head formation |\n| GSK3 $\\beta$ | Head defect | No effect |\n| $\\beta$-cat + GSK3 $\\beta$ | No effect | No effect |\n| DN $\\beta$-cat | Head defect | No effect |\n| DN GSK3 $\\beta$ | No effect | Secondary head formation |\nA: This experiment shows that GSK3 $\\beta$ inhibited organizer formation.\nB: This experiment shows that GSK3 $\\beta$ inhibits $\\beta$-cat activity.\nC: This experiment shows that $\\beta$-cat is not expressed in ventral region.\nD: This experiment shows that GSK3 $\\beta$ works downstream of $\\beta$-cat.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn a Xenopus embryo, the dorsal-ventral axis is determined through cortical rotation after fertilization. On the dorsal side of an embryo, the Spemann-Mangold organizer is necessary to determine the body plan of the embryo. When the organizer formation is inhibited, a head defect occurs in embryos. On the other hand, the head is enlarged when the organizer region expands.\n\n$\\beta$-catenin ( $\\beta$-cat) and GSK3 $\\beta$ are involved in organizer formation. The table below shows the results of phenotype of tadpoles microinjected with $\\beta$-cat, GSK3 $\\beta$, an DN $\\beta$-cat ( $\\beta$-catenin inhibition factor), and DN GSK3 $\\beta$ (GSK3 $\\beta$ inhibition factor) into the dorsal or ventral side of the embryo.\n\n| mRNA | Dorsal injection | Ventral injection |\n| :--- | :--- | :--- |\n| $\\beta$-cat | Large head | Secondary head formation |\n| GSK3 $\\beta$ | Head defect | No effect |\n| $\\beta$-cat + GSK3 $\\beta$ | No effect | No effect |\n| DN $\\beta$-cat | Head defect | No effect |\n| DN GSK3 $\\beta$ | No effect | Secondary head formation |\n\nA: This experiment shows that GSK3 $\\beta$ inhibited organizer formation.\nB: This experiment shows that GSK3 $\\beta$ inhibits $\\beta$-cat activity.\nC: This experiment shows that $\\beta$-cat is not expressed in ventral region.\nD: This experiment shows that GSK3 $\\beta$ works downstream of $\\beta$-cat.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1046",
        "problem": "Males of a small tropical bird species spend 6 months of the year jumping around in a small area on the forest floor exposing their bright yellow beard during landing. Males don't contribute anything other than sperm to the production of their offspring. Which of the following statements best explains why males display themselves in such an obvious manner that also would make them an easy prey?\nA: Males that display more frequently and move faster obtain more matings, as females use displaying capabilities as indicators of male quality on which they base their mate choice.\nB: Males that display more frequently and faster obtain less matings, as females prefer males that perform such displays as little and slow as possible so they don't get eaten by predators.\nC: Males that perform the courtship behavior more frequently and faster obtain more matings as this is direct evidence of good male genetic quality important for healthy offspring.\nD: Males perform such noisy and startling behaviors as there are very few predators in the tropics.\nE: Males that display more frequently and move faster are selected for because displays improve their chance of survival.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMales of a small tropical bird species spend 6 months of the year jumping around in a small area on the forest floor exposing their bright yellow beard during landing. Males don't contribute anything other than sperm to the production of their offspring. Which of the following statements best explains why males display themselves in such an obvious manner that also would make them an easy prey?\n\nA: Males that display more frequently and move faster obtain more matings, as females use displaying capabilities as indicators of male quality on which they base their mate choice.\nB: Males that display more frequently and faster obtain less matings, as females prefer males that perform such displays as little and slow as possible so they don't get eaten by predators.\nC: Males that perform the courtship behavior more frequently and faster obtain more matings as this is direct evidence of good male genetic quality important for healthy offspring.\nD: Males perform such noisy and startling behaviors as there are very few predators in the tropics.\nE: Males that display more frequently and move faster are selected for because displays improve their chance of survival.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_189",
        "problem": "The citric acid cycle is central to metabolism, for the supply energy and various key compounds. In citric acid cycle, the enzyme aconitase catalyzes the reversible conversion between citrate and isocitrate. In this reaction, $\\mathrm{OH}$ group at $\\mathrm{C} 3$ and $\\mathrm{H}$ group at $\\mathrm{C} 4$ of citrate are removed as water, thereafter a water molecule is added back in a reverse manner to generate isocitrate (Figure 1). However, OH group is never added at C2.\n\n[figure1]\n\nFigure 1\nA: Citrate has enantiomers. 1\nB: Isocitrate has enantiomers. 2\nC: Two $-\\mathrm{CH}_{2} \\mathrm{COO}^{-}$groups are stereochemically equivalent when citrate is free in solution.\nD: Two $-\\mathrm{CH}_{2} \\mathrm{COO}^{-}$groups are stereochemically equivalent when citrate is bound to aconitase.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe citric acid cycle is central to metabolism, for the supply energy and various key compounds. In citric acid cycle, the enzyme aconitase catalyzes the reversible conversion between citrate and isocitrate. In this reaction, $\\mathrm{OH}$ group at $\\mathrm{C} 3$ and $\\mathrm{H}$ group at $\\mathrm{C} 4$ of citrate are removed as water, thereafter a water molecule is added back in a reverse manner to generate isocitrate (Figure 1). However, OH group is never added at C2.\n\n[figure1]\n\nFigure 1\n\nA: Citrate has enantiomers. 1\nB: Isocitrate has enantiomers. 2\nC: Two $-\\mathrm{CH}_{2} \\mathrm{COO}^{-}$groups are stereochemically equivalent when citrate is free in solution.\nD: Two $-\\mathrm{CH}_{2} \\mathrm{COO}^{-}$groups are stereochemically equivalent when citrate is bound to aconitase.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-05.jpg?height=660&width=1422&top_left_y=795&top_left_x=223"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1489",
        "problem": "A species of green flower is a popular ingredient in Chinese medicine, and so it is picked in the wild by professional harvesters. In recent years, many individuals have appeared with textured grey leaves and flowers.\n\n[figure1]\n\nWhich are correct descriptions of what is occurring in this species?\nA: Evolution by sexual selection\nB: Evolution by natural selection\nC: Evolution by artificial selection\nD: Evolution by genetic drift\nE: Disease\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA species of green flower is a popular ingredient in Chinese medicine, and so it is picked in the wild by professional harvesters. In recent years, many individuals have appeared with textured grey leaves and flowers.\n\n[figure1]\n\nWhich are correct descriptions of what is occurring in this species?\n\nA: Evolution by sexual selection\nB: Evolution by natural selection\nC: Evolution by artificial selection\nD: Evolution by genetic drift\nE: Disease\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-05.jpg?height=1031&width=1702&top_left_y=478&top_left_x=240"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1011",
        "problem": "Stressful stimuli cause a number of physiological reactions. Which of the following is NOT a correctly matched statement about stress-related hormones?\nA: Epinephrine - Stimulate glucose production from glycogen\nB: Mineralocorticoids - Increase blood pressure and volume\nC: Glucocorticoids - Reduce immune system activity\nD: Norepinephrine - Increase breathing rate\nE: ACTH - Stimulate adrenal medulla to secrete hormones\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nStressful stimuli cause a number of physiological reactions. Which of the following is NOT a correctly matched statement about stress-related hormones?\n\nA: Epinephrine - Stimulate glucose production from glycogen\nB: Mineralocorticoids - Increase blood pressure and volume\nC: Glucocorticoids - Reduce immune system activity\nD: Norepinephrine - Increase breathing rate\nE: ACTH - Stimulate adrenal medulla to secrete hormones\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_990",
        "problem": "The presence of which of the following amino acids in the channel region of aquaporins contributes to the electrostatic selectivity to allow water, but not other molecules, to pass through?\nA: Valine\nB: Tryptophan\nC: Asparagine\nD: Methionine\nE: Leucine\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe presence of which of the following amino acids in the channel region of aquaporins contributes to the electrostatic selectivity to allow water, but not other molecules, to pass through?\n\nA: Valine\nB: Tryptophan\nC: Asparagine\nD: Methionine\nE: Leucine\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1175",
        "problem": "A suspension of microscopic green algae was divided into two equal samples. Each was given the same total amount of light energy. Sample I was exposed to continuous light. Sample II was exposed to light flashes of $10^{-5}$ seconds duration followed by dark periods. Photosynthesis took place in both samples, but more occurred in Sample II. From this evidence we may conclude that\nWhich of the following sections of the graph shows a region during early/mid luteal phase?\nA: More photosynthesis occurs in the dark than in the light.\nB: Some part of the photosynthetic process can occur in darkness.\nC: Photosynthesis requires darkness as well as light.\nD: Photosynthesis is a very rapid process.\nE: Photosynthesis involves enzymes as well as light.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA suspension of microscopic green algae was divided into two equal samples. Each was given the same total amount of light energy. Sample I was exposed to continuous light. Sample II was exposed to light flashes of $10^{-5}$ seconds duration followed by dark periods. Photosynthesis took place in both samples, but more occurred in Sample II. From this evidence we may conclude that\nWhich of the following sections of the graph shows a region during early/mid luteal phase?\n\nA: More photosynthesis occurs in the dark than in the light.\nB: Some part of the photosynthetic process can occur in darkness.\nC: Photosynthesis requires darkness as well as light.\nD: Photosynthesis is a very rapid process.\nE: Photosynthesis involves enzymes as well as light.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1089",
        "problem": "Two alleles are found in Calico cats for coat color and are located on the $X$ chromosome. One allele is responsible for black color while the other for orange color. Which of the following is correct about the phenotype of Calico cats?\nA: All male cats will have the same coat color.\nB: Female cats can have mosaic pattern.\nC: All females will show intermediate coat color.\nD: All female cats born to mothers with black coat color will have black coat color.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTwo alleles are found in Calico cats for coat color and are located on the $X$ chromosome. One allele is responsible for black color while the other for orange color. Which of the following is correct about the phenotype of Calico cats?\n\nA: All male cats will have the same coat color.\nB: Female cats can have mosaic pattern.\nC: All females will show intermediate coat color.\nD: All female cats born to mothers with black coat color will have black coat color.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_868",
        "problem": "将某动物（染色体数为 $2 \\mathrm{n}$ ) 的精原细胞中的核 DNA 用 ${ }^{32} \\mathrm{P}$ 充分标记, 随后转入能让精原细胞按照特定分裂要求的培养基（无 ${ }^{32} \\mathrm{P}$ ）中继续培养，连续两次正常分裂。下列叙述错误的是 ( )\nA: 第二次有丝分裂后期一个细胞中含有 ${ }^{32} \\mathrm{P}$ 标记的染色体有 $n$ 条\nB: 减数分裂 $\\mathrm{I}$ 后期的一个细胞中含有 ${ }^{32} \\mathrm{P}$ 标记的染色体有 $2 n$ 条\nC: 两次有丝分裂后能产生含有 ${ }^{32} \\mathrm{P}$ 标记的细胞至少有 2 个, 最多有 4 个\nD: 经减数分裂产生的每一个精细胞中含有 ${ }^{32} \\mathrm{P}$ 标记的染色体有 $n$ 条\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将某动物（染色体数为 $2 \\mathrm{n}$ ) 的精原细胞中的核 DNA 用 ${ }^{32} \\mathrm{P}$ 充分标记, 随后转入能让精原细胞按照特定分裂要求的培养基（无 ${ }^{32} \\mathrm{P}$ ）中继续培养，连续两次正常分裂。下列叙述错误的是 ( )\n\nA: 第二次有丝分裂后期一个细胞中含有 ${ }^{32} \\mathrm{P}$ 标记的染色体有 $n$ 条\nB: 减数分裂 $\\mathrm{I}$ 后期的一个细胞中含有 ${ }^{32} \\mathrm{P}$ 标记的染色体有 $2 n$ 条\nC: 两次有丝分裂后能产生含有 ${ }^{32} \\mathrm{P}$ 标记的细胞至少有 2 个, 最多有 4 个\nD: 经减数分裂产生的每一个精细胞中含有 ${ }^{32} \\mathrm{P}$ 标记的染色体有 $n$ 条\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_735",
        "problem": "杜氏肌营养不良（DMD）是由单基因突变引起的伴 X 隐性遗传病，男性中发病率约为 $1 / 4000$. 甲、乙家系中两患者的外祖父均表现正常, 家系乙II- 2 还患有红绿色盲。两家系部分成员 DMD 基因测序结果（显示部分序列，其他未显示序列均正常）如图。下列叙述错误的是（）\n\n正常女性\n\n$\\square$ 正常男性\n\n[图1]\n\n家系甲基因正常序列-GACTCAAACAACTT-\nI-2个体基因系列-GACTCAAACAACTT-\nII-1个体基因系列-GACTCAAAACAACT-\n\n[图2]\n基因正常序列 -GGGAIGAACA-\nI-2个体基因系列-GGGATGAACA-\nII-2个体基因系列-GGGATGAACA-\nA: 家系乙II-2 遗传其母亲的 DMD 致病基因\nB: 若家系乙I-1 和I-2 再生育一个儿子, 儿子患两种病的概率比患一种病的概率低\nC: 不考虑其他突变, 家系甲II- 2 和家系乙II- 1 婚后生出患 DMD 儿子的概率为 $1 / 8$\nD: 人群中女性 DMD 患者频率远低于男性，女性中携带者的频率约为 $1 / 4000$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n杜氏肌营养不良（DMD）是由单基因突变引起的伴 X 隐性遗传病，男性中发病率约为 $1 / 4000$. 甲、乙家系中两患者的外祖父均表现正常, 家系乙II- 2 还患有红绿色盲。两家系部分成员 DMD 基因测序结果（显示部分序列，其他未显示序列均正常）如图。下列叙述错误的是（）\n\n正常女性\n\n$\\square$ 正常男性\n\n[图1]\n\n家系甲基因正常序列-GACTCAAACAACTT-\nI-2个体基因系列-GACTCAAACAACTT-\nII-1个体基因系列-GACTCAAAACAACT-\n\n[图2]\n基因正常序列 -GGGAIGAACA-\nI-2个体基因系列-GGGATGAACA-\nII-2个体基因系列-GGGATGAACA-\n\nA: 家系乙II-2 遗传其母亲的 DMD 致病基因\nB: 若家系乙I-1 和I-2 再生育一个儿子, 儿子患两种病的概率比患一种病的概率低\nC: 不考虑其他突变, 家系甲II- 2 和家系乙II- 1 婚后生出患 DMD 儿子的概率为 $1 / 8$\nD: 人群中女性 DMD 患者频率远低于男性，女性中携带者的频率约为 $1 / 4000$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-071.jpg?height=157&width=205&top_left_y=1481&top_left_x=617",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-071.jpg?height=194&width=215&top_left_y=1673&top_left_x=612"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1071",
        "problem": "When these bacteria were grown for many generations at $37^{\\circ} \\mathrm{C} \\&$ then plated at $42^{\\circ} \\mathrm{C}$, growth was observed. Among the observed colonies, very few could utilize lactose as a source of carbon but unable to utilize galactose. Which of the following can best explain these results?\nA: Proliferative multiplication of some plasmid molecules at $42^{\\circ} \\mathrm{C}$.\nB: Integration of plasmid into chromosomal DNA.\nC: Reversal of $l a c^{-}$mutation to yield $\\mathrm{lac}^{+}$gene triggered by higher temperature.\nD: Insertion of plasmid DNA within gal gene of bacterium.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen these bacteria were grown for many generations at $37^{\\circ} \\mathrm{C} \\&$ then plated at $42^{\\circ} \\mathrm{C}$, growth was observed. Among the observed colonies, very few could utilize lactose as a source of carbon but unable to utilize galactose. Which of the following can best explain these results?\n\nA: Proliferative multiplication of some plasmid molecules at $42^{\\circ} \\mathrm{C}$.\nB: Integration of plasmid into chromosomal DNA.\nC: Reversal of $l a c^{-}$mutation to yield $\\mathrm{lac}^{+}$gene triggered by higher temperature.\nD: Insertion of plasmid DNA within gal gene of bacterium.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_433",
        "problem": "天使综合征（AS）患者发育迟缓、智力低下。研究发现，人类 15 号染色体上的 UA 和 UB 基因互为等位基因, UA 基因控制神经细胞的正常生理活动, UB 基因是天使综合征的致病基因, 在神经细胞内, 只有来自母本 15 号染色体上的 UA 或 UB 基因能得\n是 ( )\n\n[图1]\nA: II-3 和II-4 再生一个健康女孩的概率为 $1 / 4$\nB: 神经细胞与其他体细胞中 UA 基因表达有差异, 体现了细胞的分化\nC: III-1 个体的基因型可能为 UBUB 或 UAUB,其致病基因来自 I-2\nD: 若基因型为 UAUB 的正常男性与一不携带致病基因的正常女性婚配, 后代患病概率为 0\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n天使综合征（AS）患者发育迟缓、智力低下。研究发现，人类 15 号染色体上的 UA 和 UB 基因互为等位基因, UA 基因控制神经细胞的正常生理活动, UB 基因是天使综合征的致病基因, 在神经细胞内, 只有来自母本 15 号染色体上的 UA 或 UB 基因能得\n是 ( )\n\n[图1]\n\nA: II-3 和II-4 再生一个健康女孩的概率为 $1 / 4$\nB: 神经细胞与其他体细胞中 UA 基因表达有差异, 体现了细胞的分化\nC: III-1 个体的基因型可能为 UBUB 或 UAUB,其致病基因来自 I-2\nD: 若基因型为 UAUB 的正常男性与一不携带致病基因的正常女性婚配, 后代患病概率为 0\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-36.jpg?height=488&width=1174&top_left_y=316&top_left_x=378"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1026",
        "problem": "There are $n+1$ alleles at a particular locus on an autosome. The frequency of one allele is $1 / 2$ and the frequencies of the other alleles are $1 / 2 n$. Under the assumption of Hardy-Weinberg equilibrium, what is the total frequency of heterozygotes?\nA: $n-1 / 2 n$\nB: $2 n-1 / 3 n$\nC: $3 n-1 / 4 n$\nD: $4 n-1 / 5 n$\nE: $5 n-1 / 6 n$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThere are $n+1$ alleles at a particular locus on an autosome. The frequency of one allele is $1 / 2$ and the frequencies of the other alleles are $1 / 2 n$. Under the assumption of Hardy-Weinberg equilibrium, what is the total frequency of heterozygotes?\n\nA: $n-1 / 2 n$\nB: $2 n-1 / 3 n$\nC: $3 n-1 / 4 n$\nD: $4 n-1 / 5 n$\nE: $5 n-1 / 6 n$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1493",
        "problem": "Glucagon is secreted from pancreatic cells and signals via receptors on the cells of target tissues. The figure shows the amount of receptor mRNA in the same number of cells from different rat tissues.\n\n[figure1]\n\nWhich of these is true?\nA: Glucagon expression is often low when an animal first wakes up from sleeping.\nB: Glucagon stimulates the liver to produce glycogen.\nC: Glucagon reduces glucose uptake into muscles.\nD: The liver responds to lower levels of glucagon than other organs.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGlucagon is secreted from pancreatic cells and signals via receptors on the cells of target tissues. The figure shows the amount of receptor mRNA in the same number of cells from different rat tissues.\n\n[figure1]\n\nWhich of these is true?\n\nA: Glucagon expression is often low when an animal first wakes up from sleeping.\nB: Glucagon stimulates the liver to produce glycogen.\nC: Glucagon reduces glucose uptake into muscles.\nD: The liver responds to lower levels of glucagon than other organs.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-12.jpg?height=974&width=1390&top_left_y=484&top_left_x=230"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_434",
        "problem": "一对表现型正常的夫妻进行了生育遗传咨询和染色体检查, 结果发现妻子的染色体出现了异常。如图是该个体染色体异常的形成过程示意图 (其余染色体正常)。下列分析错误的是 ( )\n\n[图1]\nA: 妻子的体细胞染色体数目是 45 条\nB: 异常染色体的形成原因是染色体片段发生了易位\nC: 妻子无法形成染色体数目和结构正常的卵细胞\nD: 染色体异常可能不会导致个体表现出疾病性状\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一对表现型正常的夫妻进行了生育遗传咨询和染色体检查, 结果发现妻子的染色体出现了异常。如图是该个体染色体异常的形成过程示意图 (其余染色体正常)。下列分析错误的是 ( )\n\n[图1]\n\nA: 妻子的体细胞染色体数目是 45 条\nB: 异常染色体的形成原因是染色体片段发生了易位\nC: 妻子无法形成染色体数目和结构正常的卵细胞\nD: 染色体异常可能不会导致个体表现出疾病性状\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-22.jpg?height=472&width=1428&top_left_y=158&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_269",
        "problem": "Boron (B) is an essential micronutrient for plants. Its uptake, transport, and function in plants appear to be dependent on the formation of B complexes with sugars, such as sorbitol in phloem sap (Fig. 1) and a specific dimerised form of pectin in growing cell walls. Yokota and Konishi (1999) studied the effect of various exogenously supplied sugars including sucrose, glucose, and fructose on promotion of pollen tube growth by formation of sugar-borate complexes. Pollens were cultivated at different concentrations of B for 20 hours. Effect of sugars on the $\\mathrm{pH}$ of the media is shown in Table 1, and the length of pollen tubes incubated with the various sugars is shown in Fig. 2.\n\n[figure1]\n\nFigure 1. Boron-Sorbitol complex\n\nTable 1. Changes of the $\\mathrm{pH}$ values of the pollen culture media containing different sugars after adding boric acid at different concentrations (first row).\n\n| $\\mathbf{H}_{3} \\mathbf{B O}_{3}(\\mathbf{m M})$ | $\\mathbf{0}$ | $\\mathbf{5}$ | $\\mathbf{1 0}$ | $\\mathbf{2 0}$ |\n| :--- | :---: | :---: | :---: | :---: |\n| Sucrose | 5.2 | 5.2 | 5.0 | 4.9 |\n| Glucose | 5.2 | 4.7 | 4.5 | 4.3 |\n| Fructose | 5.0 | 3.7 | 3.5 | 3.4 |\n\n[figure2]\nA: Growth inhibition of pollen tubes by high concentrations of B is more pronounced in medium containing sucrose as compared to fructose.\nB: At low B concentrations $(5 \\mathrm{mM})$, growth inhibition of pollen tubes was more pronounced in medium containing glucose as compared to the other sugars.\nC: Inhibitory effects of fructose on pollen tube growth increased with increasing concentrations of $\\mathrm{B}$.\nD: Based on effects of sugar on $\\mathrm{pH}$ of the media, relative levels of sugar-B complex formation is as follows: Sucrose $<$ Glucose $<$ Fructose.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBoron (B) is an essential micronutrient for plants. Its uptake, transport, and function in plants appear to be dependent on the formation of B complexes with sugars, such as sorbitol in phloem sap (Fig. 1) and a specific dimerised form of pectin in growing cell walls. Yokota and Konishi (1999) studied the effect of various exogenously supplied sugars including sucrose, glucose, and fructose on promotion of pollen tube growth by formation of sugar-borate complexes. Pollens were cultivated at different concentrations of B for 20 hours. Effect of sugars on the $\\mathrm{pH}$ of the media is shown in Table 1, and the length of pollen tubes incubated with the various sugars is shown in Fig. 2.\n\n[figure1]\n\nFigure 1. Boron-Sorbitol complex\n\nTable 1. Changes of the $\\mathrm{pH}$ values of the pollen culture media containing different sugars after adding boric acid at different concentrations (first row).\n\n| $\\mathbf{H}_{3} \\mathbf{B O}_{3}(\\mathbf{m M})$ | $\\mathbf{0}$ | $\\mathbf{5}$ | $\\mathbf{1 0}$ | $\\mathbf{2 0}$ |\n| :--- | :---: | :---: | :---: | :---: |\n| Sucrose | 5.2 | 5.2 | 5.0 | 4.9 |\n| Glucose | 5.2 | 4.7 | 4.5 | 4.3 |\n| Fructose | 5.0 | 3.7 | 3.5 | 3.4 |\n\n[figure2]\n\nA: Growth inhibition of pollen tubes by high concentrations of B is more pronounced in medium containing sucrose as compared to fructose.\nB: At low B concentrations $(5 \\mathrm{mM})$, growth inhibition of pollen tubes was more pronounced in medium containing glucose as compared to the other sugars.\nC: Inhibitory effects of fructose on pollen tube growth increased with increasing concentrations of $\\mathrm{B}$.\nD: Based on effects of sugar on $\\mathrm{pH}$ of the media, relative levels of sugar-B complex formation is as follows: Sucrose $<$ Glucose $<$ Fructose.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-32.jpg?height=285&width=506&top_left_y=757&top_left_x=775",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-32.jpg?height=531&width=728&top_left_y=1528&top_left_x=664"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_670",
        "problem": "果蝇的灰身和黑身由位于常染色体上的 1 对等位基因控制; 黑身果蝇的体色深度受\n\n另 1 对等位基因的影响。现将纯合黑身雌蝇和灰身雄蝇杂交, $F_{1}$ 全为灰身, $F_{1}$ 雌雄果蝇随机交配, $F_{2}$ 中灰身：黑身：深黑身=12: 3: 1, 其中黑身果蝇中雌雄数量比为 $2: 1$ 。下列说法错误的是（ ）\nA: $F_{2}$ 灰身果蝇中, 雌雄果蝇数量比为 1: 1\nB: 若进行亲本正反交实验, $\\mathrm{F}_{1}$ 表型相同\nC: $F_{2}$ 雌蝇有灰色、黑色和深黑色三种体色\nD: $F_{2}$ 灰身雌雄果蝇自由交配后代中灰身占 $8 / 9$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的灰身和黑身由位于常染色体上的 1 对等位基因控制; 黑身果蝇的体色深度受\n\n另 1 对等位基因的影响。现将纯合黑身雌蝇和灰身雄蝇杂交, $F_{1}$ 全为灰身, $F_{1}$ 雌雄果蝇随机交配, $F_{2}$ 中灰身：黑身：深黑身=12: 3: 1, 其中黑身果蝇中雌雄数量比为 $2: 1$ 。下列说法错误的是（ ）\n\nA: $F_{2}$ 灰身果蝇中, 雌雄果蝇数量比为 1: 1\nB: 若进行亲本正反交实验, $\\mathrm{F}_{1}$ 表型相同\nC: $F_{2}$ 雌蝇有灰色、黑色和深黑色三种体色\nD: $F_{2}$ 灰身雌雄果蝇自由交配后代中灰身占 $8 / 9$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_615",
        "problem": "某自花传粉植物体内有三种物质（甲、乙、丙），其代谢过程如图所示，三种酶均由染色体上的显性基因控制合成。为培育生产乙物质的优良品种, 科学家利用野生型植株和两种突变植株 $\\left(T_{1} 、 T_{2}\\right)$ 进行自交，结果如表所示（多或少指三种物质含量的多或少)。下列分析正确的是（）\n\n[图1]\nA: 野生型、 $\\mathrm{T}_{1} 、 \\mathrm{~T}_{2}$ 基因型分别为 AABBDD、AaBbDd、AaBbDD\nB: $\\mathrm{T}_{1} 、 \\mathrm{~T}_{2}$ 自交 $\\mathrm{F}_{1}$ 代中（甲少、乙少、丙多）个体的基因型各有 2 种\nC: $\\mathrm{T}_{1} 、 \\mathrm{~T}_{2}$ 自交 $\\mathrm{F}_{1}$ 代中（甲少、乙多、丙少）个体的基因型完全相同\nD: 理论上 $T_{2}$ 自交 $F_{1}$ 代中能稳定遗传的目标植株与 $T_{1}$ 自交 $F_{1}$ 代中一样多\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某自花传粉植物体内有三种物质（甲、乙、丙），其代谢过程如图所示，三种酶均由染色体上的显性基因控制合成。为培育生产乙物质的优良品种, 科学家利用野生型植株和两种突变植株 $\\left(T_{1} 、 T_{2}\\right)$ 进行自交，结果如表所示（多或少指三种物质含量的多或少)。下列分析正确的是（）\n\n[图1]\n\nA: 野生型、 $\\mathrm{T}_{1} 、 \\mathrm{~T}_{2}$ 基因型分别为 AABBDD、AaBbDd、AaBbDD\nB: $\\mathrm{T}_{1} 、 \\mathrm{~T}_{2}$ 自交 $\\mathrm{F}_{1}$ 代中（甲少、乙少、丙多）个体的基因型各有 2 种\nC: $\\mathrm{T}_{1} 、 \\mathrm{~T}_{2}$ 自交 $\\mathrm{F}_{1}$ 代中（甲少、乙多、丙少）个体的基因型完全相同\nD: 理论上 $T_{2}$ 自交 $F_{1}$ 代中能稳定遗传的目标植株与 $T_{1}$ 自交 $F_{1}$ 代中一样多\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-55.jpg?height=956&width=1428&top_left_y=1138&top_left_x=317"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_99",
        "problem": "The flowering plants show great diversity in their flower structure reflecting evolutionary changes in course of time. One method to illustrate the flower morphology is the usage of floral diagrams implementing various shapes and symbols to show the structure as exact as possible.\n[figure1]\nA: Diagram A can represents a basal herbaceous dicot.\nB: Diagram B may represent a species from the family Orchidaceae with 2 sterile stamens.\nC: The most primitive flower among the illustrated diagrams in this task is $\\mathrm{C}$.\nD: Diagram $\\mathrm{D}$ belong to a same family as $\\mathrm{C}$.\nE: The perianth in diagram $B$ is composed of 3 whorls.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe flowering plants show great diversity in their flower structure reflecting evolutionary changes in course of time. One method to illustrate the flower morphology is the usage of floral diagrams implementing various shapes and symbols to show the structure as exact as possible.\n[figure1]\n\nA: Diagram A can represents a basal herbaceous dicot.\nB: Diagram B may represent a species from the family Orchidaceae with 2 sterile stamens.\nC: The most primitive flower among the illustrated diagrams in this task is $\\mathrm{C}$.\nD: Diagram $\\mathrm{D}$ belong to a same family as $\\mathrm{C}$.\nE: The perianth in diagram $B$ is composed of 3 whorls.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-46.jpg?height=796&width=856&top_left_y=546&top_left_x=607"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_357",
        "problem": "大麻为雌雄异株 XY 型性别决定植物, 其雌株存在性反转现象, 等位基因 E、e 与性反转有关, 当 e 基因纯合时, 雌株发生性反转表型变为雄株, 而雄株不存在性反转现象。遗传过程不考虑其它变异, 下列说法错误的是（ ）\nA: 若 E、e 位于常染色体, 两株未发生性反转的杂合植株杂交子代的雌雄比例为 3: 5\nB: 性反转形成的雄株与杂合正常雌株杂交, 子代表型一定都是雌株\nC: 若 E、e 位于常染色体, 未发生性反转的杂合雌株与隐性雄株杂交子代雌雄比例可能为 1: 3\nD: 若未发生性反转的两纯合植株杂交, 其子代表型雌雄比例为 1: 1, 则不能判断 E、e 基因是否位于 X 染色体上\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n大麻为雌雄异株 XY 型性别决定植物, 其雌株存在性反转现象, 等位基因 E、e 与性反转有关, 当 e 基因纯合时, 雌株发生性反转表型变为雄株, 而雄株不存在性反转现象。遗传过程不考虑其它变异, 下列说法错误的是（ ）\n\nA: 若 E、e 位于常染色体, 两株未发生性反转的杂合植株杂交子代的雌雄比例为 3: 5\nB: 性反转形成的雄株与杂合正常雌株杂交, 子代表型一定都是雌株\nC: 若 E、e 位于常染色体, 未发生性反转的杂合雌株与隐性雄株杂交子代雌雄比例可能为 1: 3\nD: 若未发生性反转的两纯合植株杂交, 其子代表型雌雄比例为 1: 1, 则不能判断 E、e 基因是否位于 X 染色体上\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_577",
        "problem": "果蝇的黑身、灰身由一对等位基因（B、b）控制，另一对同源染色体上的等位基因\n\n（R、r）会影响黑身果蝇的体色深度。现有黑身雌果蝇与灰身雄果蝇一对亲本杂交, $F_{1}$全为灰身, $F_{1}$ 随机交配, $F_{2}$ 雌蝇为灰身:黑身 $=3: 1$, 雄蝇为灰身:黑身:深黑身 $=6: 1: 1$ 。下列叙述错误的是（）\nA: 果蝇的灰身基因是 B\nB: 基因 $\\mathrm{R}$ 使黑身果蝇体色加深\nC: 亲本灰身雄果蝇的基因型为 $B B X^{r} Y$\nD: $F_{2}$ 灰身雌果蝇中杂合子占 $5 / 6$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的黑身、灰身由一对等位基因（B、b）控制，另一对同源染色体上的等位基因\n\n（R、r）会影响黑身果蝇的体色深度。现有黑身雌果蝇与灰身雄果蝇一对亲本杂交, $F_{1}$全为灰身, $F_{1}$ 随机交配, $F_{2}$ 雌蝇为灰身:黑身 $=3: 1$, 雄蝇为灰身:黑身:深黑身 $=6: 1: 1$ 。下列叙述错误的是（）\n\nA: 果蝇的灰身基因是 B\nB: 基因 $\\mathrm{R}$ 使黑身果蝇体色加深\nC: 亲本灰身雄果蝇的基因型为 $B B X^{r} Y$\nD: $F_{2}$ 灰身雌果蝇中杂合子占 $5 / 6$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_963",
        "problem": "Which of the following terms describes the closing of leaves or flowers at night?\nA: Epinasty\nB: Nyctinasty\nC: Seismonasty\nD: Thermonasty\nE: Thigmonasty\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following terms describes the closing of leaves or flowers at night?\n\nA: Epinasty\nB: Nyctinasty\nC: Seismonasty\nD: Thermonasty\nE: Thigmonasty\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_984",
        "problem": "A male child with Klinefelter's syndrome is found to be color-blind (a recessive X-linked trait). Both the mother and father have normal vision. How can this be explained?\nA: Nondisjunction during the mother's meiosis I\nB: Nondisjunction during the mother's meiosis II\nC: Nondisjunction during the father's meiosis I\nD: Nondisjunction during the father's meiosis II\nE: Failure of $\\mathrm{X}$ inactivation in the child\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA male child with Klinefelter's syndrome is found to be color-blind (a recessive X-linked trait). Both the mother and father have normal vision. How can this be explained?\n\nA: Nondisjunction during the mother's meiosis I\nB: Nondisjunction during the mother's meiosis II\nC: Nondisjunction during the father's meiosis I\nD: Nondisjunction during the father's meiosis II\nE: Failure of $\\mathrm{X}$ inactivation in the child\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_246",
        "problem": "A wildtype female Drosophila was mated with a wildtype male that had been $X$ ray irradiated. One of $F_{1}$ females was mated with a male that had recessive phenotype (caused by recessive allele a). Progenies of the second mating were unusual in two aspects. Firstly, there were twice as many females as males. Secondly, while all males were wild type, $1 / 2$ females were wild type, and the other $1 / 2$ exhibited the recessive phenotype a.\nA: X-rays converted a dominant allele (A) on chromosome $\\mathrm{X}$ coding for the wild type to a recessive allele (a).\nB: X-rays produced a chromosomal translocation.\nC: A loop could be seen on chromosome during prophase of meiosis 1 .\nD: If a female from the second mating exhibiting recessive phenotype (a) was crossed to a wild type male then her progenies compose of females and males at the ratio of 2 females to 1 male.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA wildtype female Drosophila was mated with a wildtype male that had been $X$ ray irradiated. One of $F_{1}$ females was mated with a male that had recessive phenotype (caused by recessive allele a). Progenies of the second mating were unusual in two aspects. Firstly, there were twice as many females as males. Secondly, while all males were wild type, $1 / 2$ females were wild type, and the other $1 / 2$ exhibited the recessive phenotype a.\n\nA: X-rays converted a dominant allele (A) on chromosome $\\mathrm{X}$ coding for the wild type to a recessive allele (a).\nB: X-rays produced a chromosomal translocation.\nC: A loop could be seen on chromosome during prophase of meiosis 1 .\nD: If a female from the second mating exhibiting recessive phenotype (a) was crossed to a wild type male then her progenies compose of females and males at the ratio of 2 females to 1 male.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1150",
        "problem": "Which one of the following pieces of information provides the most direct evidence that the puffs, seen on the giant (polytene) chromosomes of fly larvae, are regions of active gene transcription?\nA: Substances such as chloramphenicol, which inhibit protein synthesis in the cytoplasm, do not reduce the formation of chromosome puffs\nB: During metamorphosis, puffs appear in new regions of the chromosomes. Injecting ecdysone (insect moulting hormone) induces similar changes\nC: In cells supplied with radioactive uridine (a uracil-containing nucleotide), the giant chromosomes become most densely labelled in the region of the puffs\nD: Substances such as actinomycin D, which inhibit RNA synthesis, do not prevent puffing of giant chromosomes\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following pieces of information provides the most direct evidence that the puffs, seen on the giant (polytene) chromosomes of fly larvae, are regions of active gene transcription?\n\nA: Substances such as chloramphenicol, which inhibit protein synthesis in the cytoplasm, do not reduce the formation of chromosome puffs\nB: During metamorphosis, puffs appear in new regions of the chromosomes. Injecting ecdysone (insect moulting hormone) induces similar changes\nC: In cells supplied with radioactive uridine (a uracil-containing nucleotide), the giant chromosomes become most densely labelled in the region of the puffs\nD: Substances such as actinomycin D, which inhibit RNA synthesis, do not prevent puffing of giant chromosomes\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_838",
        "problem": "某两性花二倍体植物的花色由 3 对等位基因控制, 基因 A 控制紫色, a 无控制色素合成的功能。基因 $\\mathrm{B}$ 控制红色, $\\mathrm{b}$ 控制蓝色。基因 $\\mathrm{I}$ 不影响上述 2 对基因的功能，但 $\\mathrm{i}$基因纯合的个体为白色花。所有基因型的植株都能正常生长和繁殖, 基因型为 A_B_I 和 A_bbI_的个体分别表现紫红色花和靛蓝色花。现有该植物的 3 个不同纯种品系甲、乙、丙, 花色分别为靛蓝色、白色和红色。不考虑突变, 根据表中杂交结果, 下列推断正确的是 ( )\n\n| 杂交组合 |  | $F_{1}$ 表型 | $F_{2}$ 表型及比例 |\n| :---: | :---: | :---: | :---: |\n| I | 甲×乙 | 紫红色 1 | 紫红色:靛蓝色:白色 $=9: 3: 4$ |\n| II | 乙×丙 | 紫红色(2) | 紫红色:红色:白色 $=9: 3: 4$ |\n| III | 甲×丙 | 紫红色(3 | $?$ |\nA: 上述表格中出现的白色花植株的基因型共有 9 种, 其中杂合子基因型有 5 种\nB: 让杂交组合 $I 中 F_{2}$ 的紫红色花植株自交一代, 白色花植株在子代中的比例为 $1 / 9$\nC: 杂交组合III中, $F_{2}$ 的表型及比例为 9 紫红色: 3 靛蓝色: 3 红色: 1 蓝色\nD: 让紫红色(1)和紫红色(2)杂交, 则子代表型及比例为紫红色:白色 $=3: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某两性花二倍体植物的花色由 3 对等位基因控制, 基因 A 控制紫色, a 无控制色素合成的功能。基因 $\\mathrm{B}$ 控制红色, $\\mathrm{b}$ 控制蓝色。基因 $\\mathrm{I}$ 不影响上述 2 对基因的功能，但 $\\mathrm{i}$基因纯合的个体为白色花。所有基因型的植株都能正常生长和繁殖, 基因型为 A_B_I 和 A_bbI_的个体分别表现紫红色花和靛蓝色花。现有该植物的 3 个不同纯种品系甲、乙、丙, 花色分别为靛蓝色、白色和红色。不考虑突变, 根据表中杂交结果, 下列推断正确的是 ( )\n\n| 杂交组合 |  | $F_{1}$ 表型 | $F_{2}$ 表型及比例 |\n| :---: | :---: | :---: | :---: |\n| I | 甲×乙 | 紫红色 1 | 紫红色:靛蓝色:白色 $=9: 3: 4$ |\n| II | 乙×丙 | 紫红色(2) | 紫红色:红色:白色 $=9: 3: 4$ |\n| III | 甲×丙 | 紫红色(3 | $?$ |\n\nA: 上述表格中出现的白色花植株的基因型共有 9 种, 其中杂合子基因型有 5 种\nB: 让杂交组合 $I 中 F_{2}$ 的紫红色花植株自交一代, 白色花植株在子代中的比例为 $1 / 9$\nC: 杂交组合III中, $F_{2}$ 的表型及比例为 9 紫红色: 3 靛蓝色: 3 红色: 1 蓝色\nD: 让紫红色(1)和紫红色(2)杂交, 则子代表型及比例为紫红色:白色 $=3: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1509",
        "problem": "Peaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nWhich results are expected if fuzz is environmentally controlled?\nA: $\\quad \\mathrm{A}$\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: D and A\nE: $\\quad$ E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPeaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nWhich results are expected if fuzz is environmentally controlled?\n\nA: $\\quad \\mathrm{A}$\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: D and A\nE: $\\quad$ E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-45.jpg?height=782&width=1231&top_left_y=474&top_left_x=241"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1072",
        "problem": "Discontinuous growth is a characteristic of arthropods, which have exoskeleton. It also leads to phenotypic variation. Several factors affect the growth of larvae and the size at maturity can vary. In Daphnia, sizes at birth and maturity of $6^{\\text {th }}$ instar and $7^{\\text {th }}$ instar larvae (6 larvae each) are plotted.\n\n[figure1]\n\nSituation 1: The observed offspring size of Daphnia population was in the range of $0.9-$ $1.3 \\mathrm{~mm}$.\n\nSituation 2: The observed offspring size of Daphnia population was in the range of 0.7 $1.1 \\mathrm{~mm}$.\n\nBased on the information provided in the graph, which of the following is correct?\nA: Size at maturity increases proportionately as the size at birth increases.\nB: $7^{\\text {th }}$ instar larvae give individuals with larger body size at maturity than $6^{\\text {th }}$ instar larvae.\nC: In situation 1, if a new predator preferably eats Daphnia larger than approx. $3.3 \\mathrm{~mm}$ size over several generations, stabilizing selection will be underway on the daphnia size at birth.\nD: In situation 2, if a new predator preferably eats Daphnia larger than approx. $3.3 \\mathrm{~mm}$ size over several generations, it will lead to disruptive selection on the size of daphnia at birth.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nDiscontinuous growth is a characteristic of arthropods, which have exoskeleton. It also leads to phenotypic variation. Several factors affect the growth of larvae and the size at maturity can vary. In Daphnia, sizes at birth and maturity of $6^{\\text {th }}$ instar and $7^{\\text {th }}$ instar larvae (6 larvae each) are plotted.\n\n[figure1]\n\nSituation 1: The observed offspring size of Daphnia population was in the range of $0.9-$ $1.3 \\mathrm{~mm}$.\n\nSituation 2: The observed offspring size of Daphnia population was in the range of 0.7 $1.1 \\mathrm{~mm}$.\n\nBased on the information provided in the graph, which of the following is correct?\n\nA: Size at maturity increases proportionately as the size at birth increases.\nB: $7^{\\text {th }}$ instar larvae give individuals with larger body size at maturity than $6^{\\text {th }}$ instar larvae.\nC: In situation 1, if a new predator preferably eats Daphnia larger than approx. $3.3 \\mathrm{~mm}$ size over several generations, stabilizing selection will be underway on the daphnia size at birth.\nD: In situation 2, if a new predator preferably eats Daphnia larger than approx. $3.3 \\mathrm{~mm}$ size over several generations, it will lead to disruptive selection on the size of daphnia at birth.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-30.jpg?height=851&width=832&top_left_y=1336&top_left_x=495"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_93",
        "problem": "The territorial behavior of the red-whiskered bulbul (Pycnonotus jocosus) was studied during pre-nesting and nesting periods. Songs of decoys were played from ten different locations around a farmhouse, where the test birds lived, and the time taken to display territorial behavior by the resident male measured at the different locations (Table Q.31). The resident bird defended the territory with aggressive calls, threat display and tried to attack the decoy.\n\n\n\nTable Q. 31\n\n|  |  |  |  | (code | Stat <br> distan | ons <br> e, dire | tion) |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | $\\mathrm{C}$ <br> $20 \\mathrm{~m}$ <br> $\\mathrm{~N}$ | I <br> $50 \\mathrm{~m}$ <br> $\\mathrm{~N}$ | $\\mathrm{K}$ <br> 110 <br> $\\mathrm{~m}$ <br> $\\mathrm{~N}$ | $\\mathrm{G}$ <br> $30 \\mathrm{~m}$ <br> $\\mathrm{~S}$ | $\\mathrm{H}$ <br> $50 \\mathrm{~m}$ <br> $\\mathrm{~S}$ | $\\mathrm{B}$ <br> $19 \\mathrm{~m}$ <br> $\\mathrm{E}$ | $F$ <br> $46 \\mathrm{~m}$ <br> $E$ | $\\mathrm{J}$ <br> $72 \\mathrm{~m}$ <br> $\\mathrm{E}$ | D <br> $29 \\mathrm{~m}$ <br> $\\mathrm{~W}$ | $E$ <br> $38 \\mathrm{~m}$ <br> $W$ |\n| Pre-nesting <br> Territory <br> defense <br> Time spent | + <br> 1 | + <br> 3 | - <br> $>10$ | + <br> 5 | - <br> $>10$ | + <br> 1 | + <br> 3 | - <br> $>10$ | + <br> 4 | - <br> $>10$ |\n| Nesting <br> Territory <br> defense <br> Time spent | + <br> 0.2 | + <br> 1 | - <br> $>10$ | + <br> 0.7 | - <br> $>10$ | + <br> 0.3 | + <br> 2.5 | - <br> $>10$ | + <br> 1 | - <br> $>10$ |\n\nMean times spent $=$ time taken to defend territory (min). (+) signifies territory defense whereas ( - ) stands for \"no territory defense\". $N=$ north, $S=$ south, $E=$ east and $W=$ west. (Sotthibandhu 2003)\nA: The territory size during the pre-nesting period was smaller than that during the nesting period.\nB: The quadrangle made by GDIF stations marks the territory size of the bird.\nC: The male responded more rapidly to the decoy within its territory during the nesting period.\nD: The intensity of territorial behavior displayed is a dependent on seasonal fluctuation of hormones.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe territorial behavior of the red-whiskered bulbul (Pycnonotus jocosus) was studied during pre-nesting and nesting periods. Songs of decoys were played from ten different locations around a farmhouse, where the test birds lived, and the time taken to display territorial behavior by the resident male measured at the different locations (Table Q.31). The resident bird defended the territory with aggressive calls, threat display and tried to attack the decoy.\n\n\n\nTable Q. 31\n\n|  |  |  |  | (code | Stat <br> distan | ons <br> e, dire | tion) |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | $\\mathrm{C}$ <br> $20 \\mathrm{~m}$ <br> $\\mathrm{~N}$ | I <br> $50 \\mathrm{~m}$ <br> $\\mathrm{~N}$ | $\\mathrm{K}$ <br> 110 <br> $\\mathrm{~m}$ <br> $\\mathrm{~N}$ | $\\mathrm{G}$ <br> $30 \\mathrm{~m}$ <br> $\\mathrm{~S}$ | $\\mathrm{H}$ <br> $50 \\mathrm{~m}$ <br> $\\mathrm{~S}$ | $\\mathrm{B}$ <br> $19 \\mathrm{~m}$ <br> $\\mathrm{E}$ | $F$ <br> $46 \\mathrm{~m}$ <br> $E$ | $\\mathrm{J}$ <br> $72 \\mathrm{~m}$ <br> $\\mathrm{E}$ | D <br> $29 \\mathrm{~m}$ <br> $\\mathrm{~W}$ | $E$ <br> $38 \\mathrm{~m}$ <br> $W$ |\n| Pre-nesting <br> Territory <br> defense <br> Time spent | + <br> 1 | + <br> 3 | - <br> $>10$ | + <br> 5 | - <br> $>10$ | + <br> 1 | + <br> 3 | - <br> $>10$ | + <br> 4 | - <br> $>10$ |\n| Nesting <br> Territory <br> defense <br> Time spent | + <br> 0.2 | + <br> 1 | - <br> $>10$ | + <br> 0.7 | - <br> $>10$ | + <br> 0.3 | + <br> 2.5 | - <br> $>10$ | + <br> 1 | - <br> $>10$ |\n\nMean times spent $=$ time taken to defend territory (min). (+) signifies territory defense whereas ( - ) stands for \"no territory defense\". $N=$ north, $S=$ south, $E=$ east and $W=$ west. (Sotthibandhu 2003)\n\nA: The territory size during the pre-nesting period was smaller than that during the nesting period.\nB: The quadrangle made by GDIF stations marks the territory size of the bird.\nC: The male responded more rapidly to the decoy within its territory during the nesting period.\nD: The intensity of territorial behavior displayed is a dependent on seasonal fluctuation of hormones.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-070.jpg?height=751&width=937&top_left_y=1638&top_left_x=865"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1096",
        "problem": "Elk (a type of deer) and bison (large wild cattle) are herbivores that forage in the same area. The figure below depicts changes in the populations of these two species before and after the introduction of wolves (predator species) in their habitat.\n\n[figure1]\n\nWhich of the following is correct?\nA: The decrease in elk population is a result of predation by wolves as well as an increase in bison population which (being a large size cattle) consumes a larger fraction of vegetation.\nB: In the initial years of introduction of wolves, the high predation on elk by wolf reduced the predation pressure on bison calves increasing the survival rate of young.\nC: There is a likelihood of competitive dietary overlap between the elk and bison population.\nD: The fluctuations in the elk and bison populations indicate that wolves have exclusively fed on elks.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nElk (a type of deer) and bison (large wild cattle) are herbivores that forage in the same area. The figure below depicts changes in the populations of these two species before and after the introduction of wolves (predator species) in their habitat.\n\n[figure1]\n\nWhich of the following is correct?\n\nA: The decrease in elk population is a result of predation by wolves as well as an increase in bison population which (being a large size cattle) consumes a larger fraction of vegetation.\nB: In the initial years of introduction of wolves, the high predation on elk by wolf reduced the predation pressure on bison calves increasing the survival rate of young.\nC: There is a likelihood of competitive dietary overlap between the elk and bison population.\nD: The fluctuations in the elk and bison populations indicate that wolves have exclusively fed on elks.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-31.jpg?height=694&width=851&top_left_y=1350&top_left_x=580"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_346",
        "problem": "控制棉花纤维长度的三对等位基因 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b} 、 \\mathrm{C} / \\mathrm{c}$ 对长度的作用相等, 分别位于三对同源染色体上。已知基因型为 aabbcc 的棉花纤维长度为 6 厘米, 每个显性基因增加纤维长度 2 厘米。棉花植株甲 (AABbcc) 与乙 ( $\\mathrm{aaBbCc})$杂交, 则 $\\mathrm{F}_{1}$ 的棉花纤维长度范围是\nA: $6 \\sim 14$ 厘米\nB: $6 \\sim 16$ 厘米\nC: $8 \\sim 14$ 厘米\nD: $8 \\sim 16$ 厘米\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n控制棉花纤维长度的三对等位基因 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b} 、 \\mathrm{C} / \\mathrm{c}$ 对长度的作用相等, 分别位于三对同源染色体上。已知基因型为 aabbcc 的棉花纤维长度为 6 厘米, 每个显性基因增加纤维长度 2 厘米。棉花植株甲 (AABbcc) 与乙 ( $\\mathrm{aaBbCc})$杂交, 则 $\\mathrm{F}_{1}$ 的棉花纤维长度范围是\n\nA: $6 \\sim 14$ 厘米\nB: $6 \\sim 16$ 厘米\nC: $8 \\sim 14$ 厘米\nD: $8 \\sim 16$ 厘米\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_539",
        "problem": "某种牛, 基因型为 $\\mathrm{AA}$ 的体色是红褐色, aa 是红色, 基因型为 $\\mathrm{Aa}$ 的雄性红褐色,雌性红色。现有多只红褐色雄牛和多只红色雌牛进行随机交配, 子代雄性中红褐色: 红色=13：3, 雌性中红褐色：红色=3：13。下列叙述错误的是（ ）\nA: 亲本红褐色雄牛有两种基因型, $\\mathrm{AA}: \\mathrm{Aa}=1: 1$\nB: 亲本红色雌牛有两种基因型, Aa: $a a=1 ： 1$\nC: 子代中基因型 $\\mathrm{AA}: \\mathrm{Aa}=3$ : 10\nD: 让子代中红色雄牛与红色雌牛随机交配, $\\mathrm{F}_{2}$ 雌牛中红褐色：红色是 5:8\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种牛, 基因型为 $\\mathrm{AA}$ 的体色是红褐色, aa 是红色, 基因型为 $\\mathrm{Aa}$ 的雄性红褐色,雌性红色。现有多只红褐色雄牛和多只红色雌牛进行随机交配, 子代雄性中红褐色: 红色=13：3, 雌性中红褐色：红色=3：13。下列叙述错误的是（ ）\n\nA: 亲本红褐色雄牛有两种基因型, $\\mathrm{AA}: \\mathrm{Aa}=1: 1$\nB: 亲本红色雌牛有两种基因型, Aa: $a a=1 ： 1$\nC: 子代中基因型 $\\mathrm{AA}: \\mathrm{Aa}=3$ : 10\nD: 让子代中红色雄牛与红色雌牛随机交配, $\\mathrm{F}_{2}$ 雌牛中红褐色：红色是 5:8\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_371",
        "problem": "为平衡两性中 $\\mathrm{X}$ 染色体基因表达水平, 雌性哺乳动物进化出一套独特的遗传系统,\n即在体细胞中随机失活其中一条 $\\mathrm{X}$ 染色体, 失活的 X 染色体称为巴氏小体。巴氏小体的形成受 Xist 基因调控, 由于卵原细胞中 Xist 基因是沉默的，因此已形成的巴氏小体会重新变成原始的状态, 进行正常的减数分裂。红绿色盲是由对应颜色的视雉细胞中的特定光敏色素的功能障碍或缺失造成的。下列相关叙述正确的是（）\nA: 女性一生排出的卵细胞中有一半含巴氏小体\nB: 巴氏小体的形成可能是由表观遗传控制的\nC: 红绿色盲基因携带者的视雉细胞中的特定光敏色素功能正常\nD: 红绿色盲基因携带者的女性后代不会出现巴氏小体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n为平衡两性中 $\\mathrm{X}$ 染色体基因表达水平, 雌性哺乳动物进化出一套独特的遗传系统,\n即在体细胞中随机失活其中一条 $\\mathrm{X}$ 染色体, 失活的 X 染色体称为巴氏小体。巴氏小体的形成受 Xist 基因调控, 由于卵原细胞中 Xist 基因是沉默的，因此已形成的巴氏小体会重新变成原始的状态, 进行正常的减数分裂。红绿色盲是由对应颜色的视雉细胞中的特定光敏色素的功能障碍或缺失造成的。下列相关叙述正确的是（）\n\nA: 女性一生排出的卵细胞中有一半含巴氏小体\nB: 巴氏小体的形成可能是由表观遗传控制的\nC: 红绿色盲基因携带者的视雉细胞中的特定光敏色素功能正常\nD: 红绿色盲基因携带者的女性后代不会出现巴氏小体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1012",
        "problem": "Many metabolic pathways involve multi-step reactions. Consider the following pathway, where E represents different enzymes, and A, B, C, D and F represent substrates and products of the pathway.\n\n$$\n\\begin{aligned}\n& \\begin{array}{llll}\n\\mathrm{E}_{1} & \\mathrm{E}_{2} & \\mathrm{E}_{3} & \\mathrm{E}_{4}\n\\end{array} \\\\\n& \\mathrm{~A} \\rightarrow \\mathrm{B} \\rightarrow \\mathrm{C} \\rightarrow \\mathrm{D} \\rightarrow \\mathrm{F}\n\\end{aligned}\n$$\n\nIn the example above, assume that $D$ is an allosteric inhibitor of the enzyme $\\mathrm{E}_{2}$. D would\nA: Compete with $\\mathrm{B}$ for binding to the $\\mathrm{E}_{2}$ active site\nB: Compete with $\\mathrm{F}$ for binding with $\\mathrm{E}_{2}$\nC: Bind directly to the substrate $\\mathrm{B}$ and prevent it from entering the $\\mathrm{E}_{2}$ active site\nD: Bind $\\mathrm{E}_{2}$ at a site different from the active site but change the shape of the active site so $\\mathrm{B}$ can no longer bind\nE: Bind $\\mathrm{E}_{2}$ at the active site, change its shape and prevent $\\mathrm{B}$ from binding the active site\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMany metabolic pathways involve multi-step reactions. Consider the following pathway, where E represents different enzymes, and A, B, C, D and F represent substrates and products of the pathway.\n\n$$\n\\begin{aligned}\n& \\begin{array}{llll}\n\\mathrm{E}_{1} & \\mathrm{E}_{2} & \\mathrm{E}_{3} & \\mathrm{E}_{4}\n\\end{array} \\\\\n& \\mathrm{~A} \\rightarrow \\mathrm{B} \\rightarrow \\mathrm{C} \\rightarrow \\mathrm{D} \\rightarrow \\mathrm{F}\n\\end{aligned}\n$$\n\nIn the example above, assume that $D$ is an allosteric inhibitor of the enzyme $\\mathrm{E}_{2}$. D would\n\nA: Compete with $\\mathrm{B}$ for binding to the $\\mathrm{E}_{2}$ active site\nB: Compete with $\\mathrm{F}$ for binding with $\\mathrm{E}_{2}$\nC: Bind directly to the substrate $\\mathrm{B}$ and prevent it from entering the $\\mathrm{E}_{2}$ active site\nD: Bind $\\mathrm{E}_{2}$ at a site different from the active site but change the shape of the active site so $\\mathrm{B}$ can no longer bind\nE: Bind $\\mathrm{E}_{2}$ at the active site, change its shape and prevent $\\mathrm{B}$ from binding the active site\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_622",
        "problem": "果蝇细胞分裂过程中染色体数量变化的部分模型如图所示。下列说法错误的是\n\n[图1]\nA: $\\mathrm{AB}$ 段不可能含有同源染色体\nB: $\\mathrm{BC}$ 段的变化为 $8 \\rightarrow 16$ 或 $4 \\rightarrow 8$\nC: CD 段时, 核 DNA 和染色体数相等\nD: 该模型也可表示染色体组的数量变化\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇细胞分裂过程中染色体数量变化的部分模型如图所示。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{AB}$ 段不可能含有同源染色体\nB: $\\mathrm{BC}$ 段的变化为 $8 \\rightarrow 16$ 或 $4 \\rightarrow 8$\nC: CD 段时, 核 DNA 和染色体数相等\nD: 该模型也可表示染色体组的数量变化\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-16.jpg?height=288&width=534&top_left_y=1718&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_457",
        "problem": "甲基转移酶介导的 $\\mathrm{N}_{6}$-甲基腺苷（ $\\mathrm{m}^{6} \\mathrm{~A}$ ）修饰是小鼠肝脏发育所必需的, 具体机制如图 1 所示, 其中 $\\mathrm{Mettl}_{3}$ 基团是甲基转移酶复合物的重要组成元件, Hnf4a 是核心转录因子, $\\mathrm{Igf}_{2} \\mathrm{bpl}$ 是在该过程中发挥重要作用的一种有机物分子, $\\mathrm{ADOC}_{3}$ 是与肝脏发育和成熟密切相关的基因。科学家分析了 $\\mathrm{mRNA}$ 不同区域被 $\\mathrm{m}^{6} \\mathrm{~A}$ 修饰的频率, 结果如图 2 所示。下列说法错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 甲基转移酶复合物催化 mRNA 上的腺苷发生 $m^{6} \\mathrm{~A}$ 修饰, 影响 $\\mathrm{ADOC}_{3}$ 基因的转录过程\nB: $\\operatorname{Igf}_{2} \\mathrm{bpl}$ 发挥的作用可能是通过与 $\\mathrm{m}^{6} \\mathrm{~A}$ 特异性结合, 维持 $\\mathrm{Hnf}_{4} \\mathrm{a} \\mathrm{mRNA}$ 的稳定\nC: 若 $\\mathrm{m}^{6} \\mathrm{~A}$ 修饰的碱基序列是 GGAC\", 理论上 $\\mathrm{mRNA}$ 上发生 $\\mathrm{m}^{6} \\mathrm{~A}$ 修饰的概率是 $1 / 256$\nD: $\\mathrm{m}^{6} \\mathrm{~A}$ 修饰的峰值发生在 mRNA 上的终止密码子附近\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n甲基转移酶介导的 $\\mathrm{N}_{6}$-甲基腺苷（ $\\mathrm{m}^{6} \\mathrm{~A}$ ）修饰是小鼠肝脏发育所必需的, 具体机制如图 1 所示, 其中 $\\mathrm{Mettl}_{3}$ 基团是甲基转移酶复合物的重要组成元件, Hnf4a 是核心转录因子, $\\mathrm{Igf}_{2} \\mathrm{bpl}$ 是在该过程中发挥重要作用的一种有机物分子, $\\mathrm{ADOC}_{3}$ 是与肝脏发育和成熟密切相关的基因。科学家分析了 $\\mathrm{mRNA}$ 不同区域被 $\\mathrm{m}^{6} \\mathrm{~A}$ 修饰的频率, 结果如图 2 所示。下列说法错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 甲基转移酶复合物催化 mRNA 上的腺苷发生 $m^{6} \\mathrm{~A}$ 修饰, 影响 $\\mathrm{ADOC}_{3}$ 基因的转录过程\nB: $\\operatorname{Igf}_{2} \\mathrm{bpl}$ 发挥的作用可能是通过与 $\\mathrm{m}^{6} \\mathrm{~A}$ 特异性结合, 维持 $\\mathrm{Hnf}_{4} \\mathrm{a} \\mathrm{mRNA}$ 的稳定\nC: 若 $\\mathrm{m}^{6} \\mathrm{~A}$ 修饰的碱基序列是 GGAC\", 理论上 $\\mathrm{mRNA}$ 上发生 $\\mathrm{m}^{6} \\mathrm{~A}$ 修饰的概率是 $1 / 256$\nD: $\\mathrm{m}^{6} \\mathrm{~A}$ 修饰的峰值发生在 mRNA 上的终止密码子附近\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-26.jpg?height=660&width=597&top_left_y=1780&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-26.jpg?height=502&width=645&top_left_y=1939&top_left_x=997"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_431",
        "problem": "紫花亘宿 $(2 n=32)$ 是应用较为广泛的豆科牧草, 但易造成家畜鼓胀病。百脉根 $(2 n=12)$ 富含单宁, 单宁可与植物蛋白质结合, 不会引起家畜采食后鼓胀。科研人员利用野生型清水紫花苜宿和里奥百脉根为材料培育抗鼓胀病苜宿新品种。研究流程如图 (注: IOA 可抑制植物细胞呼吸第一阶段, $\\mathrm{R}-6 \\mathrm{G}$ 可阻止线粒体的呼吸作用), 下列说法错误的是（ ）\n\n[图1]\nA: 将两种细胞分别置于较高渗透压环境下，有利于去除细胞壁获得两种植物的原生质体\nB: (2)过程需要利用选择培养基筛选出杂种愈伤组织\nC: (3)过程需要在有光照、植物激素的作用才能再分化形成完整植株\nD: 只考虑两个细胞融合的情况下, (2)过程的细胞中染色体数目最多有 88 条, (3)过程产生的再生植株是可育的多倍体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n紫花亘宿 $(2 n=32)$ 是应用较为广泛的豆科牧草, 但易造成家畜鼓胀病。百脉根 $(2 n=12)$ 富含单宁, 单宁可与植物蛋白质结合, 不会引起家畜采食后鼓胀。科研人员利用野生型清水紫花苜宿和里奥百脉根为材料培育抗鼓胀病苜宿新品种。研究流程如图 (注: IOA 可抑制植物细胞呼吸第一阶段, $\\mathrm{R}-6 \\mathrm{G}$ 可阻止线粒体的呼吸作用), 下列说法错误的是（ ）\n\n[图1]\n\nA: 将两种细胞分别置于较高渗透压环境下，有利于去除细胞壁获得两种植物的原生质体\nB: (2)过程需要利用选择培养基筛选出杂种愈伤组织\nC: (3)过程需要在有光照、植物激素的作用才能再分化形成完整植株\nD: 只考虑两个细胞融合的情况下, (2)过程的细胞中染色体数目最多有 88 条, (3)过程产生的再生植株是可育的多倍体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-12.jpg?height=362&width=1140&top_left_y=2189&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_233",
        "problem": "BDNF, a brain protein, is crucial for neuron activities. BDNF is activated during the activation of neurons of the limbic system. An important function of BDNF in mammals is shown in Figure Q.28. The signal via the catecholamine receptor ( $\\beta-A R)$ is specific for the browning process and thermogenesis.\n\n[figure1]\n\nNote: $\\uparrow:$ increase\n\nFigure Q.28.\nA: Regularly learning and memorizing activities help to increase the number of brown fat cells.\nB: Inhibition of BDNF expression reduces the size of white adipose tissue.\nC: Psychological anxiety increases the manifestation of the $\\beta-\\mathrm{AR}$.\nD: $\\beta$-AR gene-deleted mice will display decreases in the plasma levels of free fatty acids.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBDNF, a brain protein, is crucial for neuron activities. BDNF is activated during the activation of neurons of the limbic system. An important function of BDNF in mammals is shown in Figure Q.28. The signal via the catecholamine receptor ( $\\beta-A R)$ is specific for the browning process and thermogenesis.\n\n[figure1]\n\nNote: $\\uparrow:$ increase\n\nFigure Q.28.\n\nA: Regularly learning and memorizing activities help to increase the number of brown fat cells.\nB: Inhibition of BDNF expression reduces the size of white adipose tissue.\nC: Psychological anxiety increases the manifestation of the $\\beta-\\mathrm{AR}$.\nD: $\\beta$-AR gene-deleted mice will display decreases in the plasma levels of free fatty acids.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-063.jpg?height=1141&width=1598&top_left_y=707&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_657",
        "problem": "图 1 为加热杀死的 $S$ 型细菌与 $\\mathrm{R}$ 型活细菌混合注射到小体内后, 两种细菌的含量变化。图 2 是用 DNA 测序仪测出的 R 型细菌的一个 DNA 分子片段上被标记一条脱氧核苷酸链的碱基排列顺序 (TGCGTATTGG)。图 3 为 $\\mathrm{S}$ 型细菌的一个 DNA 分子片段上被标记一条脱氧核苷酸链的碱基排列顺序。下列有关说法错误的是（ ）\n\n[图1]\n\n图1\n\n## 注射后时间\n\n[图2]\n\n测序结果图图2\n\n[图3]\n\n碱基序列\nA: 图 1 中的实线代表 $\\mathrm{R}$ 型细菌, 虚线代表 $\\mathrm{S}$ 型细菌\nB: CD 段细菌含量增加的原因是 $\\mathrm{S}$ 型细菌增加，破坏了小鼠的免疫系统\nC: R 型细菌的此 DNA 双链片段含有 1 个腺嘌呤\nD: $\\mathrm{S}$ 型细菌的此 DNA 分子片段碱基排列顺序为 CCAGTGCGCC\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 1 为加热杀死的 $S$ 型细菌与 $\\mathrm{R}$ 型活细菌混合注射到小体内后, 两种细菌的含量变化。图 2 是用 DNA 测序仪测出的 R 型细菌的一个 DNA 分子片段上被标记一条脱氧核苷酸链的碱基排列顺序 (TGCGTATTGG)。图 3 为 $\\mathrm{S}$ 型细菌的一个 DNA 分子片段上被标记一条脱氧核苷酸链的碱基排列顺序。下列有关说法错误的是（ ）\n\n[图1]\n\n图1\n\n## 注射后时间\n\n[图2]\n\n测序结果图图2\n\n[图3]\n\n碱基序列\n\nA: 图 1 中的实线代表 $\\mathrm{R}$ 型细菌, 虚线代表 $\\mathrm{S}$ 型细菌\nB: CD 段细菌含量增加的原因是 $\\mathrm{S}$ 型细菌增加，破坏了小鼠的免疫系统\nC: R 型细菌的此 DNA 双链片段含有 1 个腺嘌呤\nD: $\\mathrm{S}$ 型细菌的此 DNA 分子片段碱基排列顺序为 CCAGTGCGCC\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-91.jpg?height=419&width=533&top_left_y=1027&top_left_x=356",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-91.jpg?height=394&width=411&top_left_y=1551&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-91.jpg?height=394&width=146&top_left_y=1551&top_left_x=772"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_601",
        "problem": "调查中发现某个家系关于甲、乙两种遗传病的系谱图如下图所示，已知控制甲、乙遗传病的基因分别为 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$, 两对基因独立遗传，且甲病在人群中的发病率为 1/10000。不考虑基因位于 X、Y 染色体同源区段的情况，据图分析，下列说法错误的是 ( )\n\nII\n\nIII\n\n[图1]\nA: 甲病的遗传方式为常染色体隐性遗传\nB: 若 $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生育一个孩子, 其患甲病的概率为 0\nC: 若 $I_{1}$ 携带乙病致病基因, 则 $I_{1}$ 和 $I_{2}$ 生育正常孩子的概率为 9/16\nD: 若 $\\mathrm{I}_{1}$ 不携带乙病致病基因, 则 $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生一个男孩患乙病的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n调查中发现某个家系关于甲、乙两种遗传病的系谱图如下图所示，已知控制甲、乙遗传病的基因分别为 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$, 两对基因独立遗传，且甲病在人群中的发病率为 1/10000。不考虑基因位于 X、Y 染色体同源区段的情况，据图分析，下列说法错误的是 ( )\n\nII\n\nIII\n\n[图1]\n\nA: 甲病的遗传方式为常染色体隐性遗传\nB: 若 $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生育一个孩子, 其患甲病的概率为 0\nC: 若 $I_{1}$ 携带乙病致病基因, 则 $I_{1}$ 和 $I_{2}$ 生育正常孩子的概率为 9/16\nD: 若 $\\mathrm{I}_{1}$ 不携带乙病致病基因, 则 $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生一个男孩患乙病的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-03.jpg?height=434&width=912&top_left_y=2279&top_left_x=346"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_449",
        "problem": "玉米种子的颜色与糊粉层细胞含有的色素种类有关。已知玉米糊粉层颜色受三对等\n\n位基因（D/d、 $\\mathrm{E} / \\mathrm{e} 、 \\mathrm{~F} / \\mathrm{f}$ ) 控制，其中 $\\mathrm{E}$ 决定红色素的合成， $\\mathrm{F}$ 决定紫色素的合成， $\\mathrm{D}$ 基因可抑制前体物质（a）的合成，从而抑制色素的合成，基因与色素的合成关系如图所示。纯合白色玉米 (甲) 与纯合紫色玉米 (乙) 杂交, 得 $F_{1}, F_{1}$ 自交, $F_{2}$ 中白色:红色: 紫色=52: 3: 9。下列叙述错误的是（ ）\n\n[图1]\nA: 白色玉米甲的基因型为 DDeeff\nB: $\\mathrm{F}_{2}$ 中白色玉米共有 21 种基因型\nC: $\\mathrm{F}_{2}$ 中红色玉米相互传粉会产生一定数量的紫色玉米\nD: $\\mathrm{F}_{2}$ 紫色玉米中纯合子所占比例为 $1 / 9$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n玉米种子的颜色与糊粉层细胞含有的色素种类有关。已知玉米糊粉层颜色受三对等\n\n位基因（D/d、 $\\mathrm{E} / \\mathrm{e} 、 \\mathrm{~F} / \\mathrm{f}$ ) 控制，其中 $\\mathrm{E}$ 决定红色素的合成， $\\mathrm{F}$ 决定紫色素的合成， $\\mathrm{D}$ 基因可抑制前体物质（a）的合成，从而抑制色素的合成，基因与色素的合成关系如图所示。纯合白色玉米 (甲) 与纯合紫色玉米 (乙) 杂交, 得 $F_{1}, F_{1}$ 自交, $F_{2}$ 中白色:红色: 紫色=52: 3: 9。下列叙述错误的是（ ）\n\n[图1]\n\nA: 白色玉米甲的基因型为 DDeeff\nB: $\\mathrm{F}_{2}$ 中白色玉米共有 21 种基因型\nC: $\\mathrm{F}_{2}$ 中红色玉米相互传粉会产生一定数量的紫色玉米\nD: $\\mathrm{F}_{2}$ 紫色玉米中纯合子所占比例为 $1 / 9$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-61.jpg?height=220&width=831&top_left_y=1289&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_632",
        "problem": "女娄菜是 XY 型雌雄异株的二倍体植物, 其叶片形状有宽叶和窄叶两种类型（假设控制该相对性状的基因为 $\\mathrm{A} / \\mathrm{a}$ ), 某学习小组欲探究该相对性状的遗传规律, 设计并完成了三组杂交实验，并将实验结果记录在下表中。\n\n| 组别 | 父本 | 母本 | $\\mathrm{F}_{1}$ 表现型及比例 |\n| :---: | :---: | :---: | :---: |\n| 一 | 窄叶 | 宽叶 | 全为宽叶 |\n| 二 | 窄叶 | 宽叶 | 宽叶雄株:窄叶雄株 $=1: 1$ |\n| 三 | 宽叶 | 宽叶 | 宽叶雌株:宽叶雄株:窄叶雄株 $=2: 1: 1$ |\n\n下列有关叙述不正确的是（ ）\nA: 与踠豆相比, 女娄菜在杂交过程中不需要去雄\nB: 可通过实验组别一、二、三判断出宽叶为显性性状\nC: 实验二无雌性植株的原因可能是含 $\\mathrm{X}^{\\mathrm{a}}$ 的雄配子致死\nD: 由实验二、三可推断女娄菜控制宽叶和窄叶的基因仅位于 X 染色体上\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n女娄菜是 XY 型雌雄异株的二倍体植物, 其叶片形状有宽叶和窄叶两种类型（假设控制该相对性状的基因为 $\\mathrm{A} / \\mathrm{a}$ ), 某学习小组欲探究该相对性状的遗传规律, 设计并完成了三组杂交实验，并将实验结果记录在下表中。\n\n| 组别 | 父本 | 母本 | $\\mathrm{F}_{1}$ 表现型及比例 |\n| :---: | :---: | :---: | :---: |\n| 一 | 窄叶 | 宽叶 | 全为宽叶 |\n| 二 | 窄叶 | 宽叶 | 宽叶雄株:窄叶雄株 $=1: 1$ |\n| 三 | 宽叶 | 宽叶 | 宽叶雌株:宽叶雄株:窄叶雄株 $=2: 1: 1$ |\n\n下列有关叙述不正确的是（ ）\n\nA: 与踠豆相比, 女娄菜在杂交过程中不需要去雄\nB: 可通过实验组别一、二、三判断出宽叶为显性性状\nC: 实验二无雌性植株的原因可能是含 $\\mathrm{X}^{\\mathrm{a}}$ 的雄配子致死\nD: 由实验二、三可推断女娄菜控制宽叶和窄叶的基因仅位于 X 染色体上\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_683",
        "problem": "肺炎链球菌转化实验中, $\\mathrm{R}$ 型细菌会释放多种酶: 一些酶将 $\\mathrm{S}$ 型细菌的 DNA 降解为片段; 另一些酶使 $\\mathrm{S}$ 型细菌 DNA 片段的双链打开, 一条链被降解, 另一条链进入细胞与 R 型细菌 DNA 上的同源区段配对, 切除并替换相应的单链片段, 形成一个杂种 DNA 区段。下列有关叙述不合理的是（ ）\nA: 推测该过程中需要限制酶、解旋酶、DNA 酶等共同参与\nB: 加热杀死的 S 型细菌仍可保持转化活性的原因是 DNA 未发生变性\nC: $\\mathrm{R}$ 型细菌转化为 $\\mathrm{S}$ 型细菌时摄入的片段包含英膜基因, 该变异属于基因重组\nD: 含杂种 DNA 区段的细菌增殖后会出现 $R$ 型和 S 型两种细菌\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n肺炎链球菌转化实验中, $\\mathrm{R}$ 型细菌会释放多种酶: 一些酶将 $\\mathrm{S}$ 型细菌的 DNA 降解为片段; 另一些酶使 $\\mathrm{S}$ 型细菌 DNA 片段的双链打开, 一条链被降解, 另一条链进入细胞与 R 型细菌 DNA 上的同源区段配对, 切除并替换相应的单链片段, 形成一个杂种 DNA 区段。下列有关叙述不合理的是（ ）\n\nA: 推测该过程中需要限制酶、解旋酶、DNA 酶等共同参与\nB: 加热杀死的 S 型细菌仍可保持转化活性的原因是 DNA 未发生变性\nC: $\\mathrm{R}$ 型细菌转化为 $\\mathrm{S}$ 型细菌时摄入的片段包含英膜基因, 该变异属于基因重组\nD: 含杂种 DNA 区段的细菌增殖后会出现 $R$ 型和 S 型两种细菌\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1177",
        "problem": "In a breeding experiment, two white-flowered sweet peas were crossed and all the offspring had purple flowers. This result could be explained if the parent plants\nA: both carried the genes for the synthesis of purple pigment.\nB: had different genotypes, one carrying the genes for purple pigment and the other carrying the genes for the enzyme activating them.\nC: were both lacking one of the genes controlling the enzymes on the pathway synthesising purple pigment.\nD: had the same genotype and were both heterozygous for the genes responsible for synthesising purple pigment.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn a breeding experiment, two white-flowered sweet peas were crossed and all the offspring had purple flowers. This result could be explained if the parent plants\n\nA: both carried the genes for the synthesis of purple pigment.\nB: had different genotypes, one carrying the genes for purple pigment and the other carrying the genes for the enzyme activating them.\nC: were both lacking one of the genes controlling the enzymes on the pathway synthesising purple pigment.\nD: had the same genotype and were both heterozygous for the genes responsible for synthesising purple pigment.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_97",
        "problem": "There are two hypotheses that try to explain the origin of the Indo-European language family.\n\nThe Steppe hypothesis traces back the origin of Indo-European language to the Pontic steppes region north of the Caspian Sea. Archaeological records provide some clues regarding to expansion from this area about 6000 years ago, however the implied models of such expansion remain untested.\n\nAlternatively, it has been claimed that the languages spread from Anatolia with the development of agricultural activities about 8000 to 9500 years ago (the Anatolian hypothesis). The agricultural expansion reached the edge of Western Europe by 5000 years ago and had run its course about 4000 years ago.\n\nBouckaert and his colleagues (2018) assessed the two hypotheses with Bayesian phylogeographic approaches using vocabulary information from the ancient and contemporary Indo-European languages.\n\n[figure1]\n\nMaximum clade credibility tree indicating the divergence of the major Indo-European subfamilies. The tree demonstrates the timing of the appearance of the major branches and their subsequent variation. The area of each triangle represents the relative number of languages in each subfamily. Tocharian and Armenian are originated from steppe region.\nA: The result does not favour an Anatolian origin over a steppe origin.\nB: It is unlikely that the spread of agriculture serves as the sole driver of language expansion on the continent.\nC: Rate of language diversification in Balto-Slavic subfamily was higher than Germanic subfamily.\nD: Drawing such trees is based on the assumption that languages does not affect each other when they come in contact.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThere are two hypotheses that try to explain the origin of the Indo-European language family.\n\nThe Steppe hypothesis traces back the origin of Indo-European language to the Pontic steppes region north of the Caspian Sea. Archaeological records provide some clues regarding to expansion from this area about 6000 years ago, however the implied models of such expansion remain untested.\n\nAlternatively, it has been claimed that the languages spread from Anatolia with the development of agricultural activities about 8000 to 9500 years ago (the Anatolian hypothesis). The agricultural expansion reached the edge of Western Europe by 5000 years ago and had run its course about 4000 years ago.\n\nBouckaert and his colleagues (2018) assessed the two hypotheses with Bayesian phylogeographic approaches using vocabulary information from the ancient and contemporary Indo-European languages.\n\n[figure1]\n\nMaximum clade credibility tree indicating the divergence of the major Indo-European subfamilies. The tree demonstrates the timing of the appearance of the major branches and their subsequent variation. The area of each triangle represents the relative number of languages in each subfamily. Tocharian and Armenian are originated from steppe region.\n\nA: The result does not favour an Anatolian origin over a steppe origin.\nB: It is unlikely that the spread of agriculture serves as the sole driver of language expansion on the continent.\nC: Rate of language diversification in Balto-Slavic subfamily was higher than Germanic subfamily.\nD: Drawing such trees is based on the assumption that languages does not affect each other when they come in contact.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-55.jpg?height=611&width=1148&top_left_y=937&top_left_x=454"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_949",
        "problem": "脊髓小脑性共济失调（SCA）属于单基因遗传病。大部分 SCA 患者在中青年时发病, 患者常表现出行走不稳、吐字不清、吞咽困难、容易呛咳等症状。该病发病原因为三核苷酸（CAG）重复扩增突变导致多聚谷氨酰胺（其密码子为 $\\mathrm{CAG}$ ）毒性聚集及蛋白质错误重叠。下图为某一 SCA 患者的家系图（相关基因不位于 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段），下列相关叙述正确的是（）\n\n[图1]\nA: SCA 的致病基因不可能位于性染色体, 该病最可能是隐性遗传病\nB: 该实例说明基因可以通过控制蛋白质的结构控制生物体的性状\nC: SCA 患者的致病基因中增加了含多个 CAG 序列的核苷酸单链片段\nD: 若咨询者的一个初级卵母细胞分裂产生的极体中含 SCA 致病基因, 则该细胞产生的卵细胞不含致病基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n脊髓小脑性共济失调（SCA）属于单基因遗传病。大部分 SCA 患者在中青年时发病, 患者常表现出行走不稳、吐字不清、吞咽困难、容易呛咳等症状。该病发病原因为三核苷酸（CAG）重复扩增突变导致多聚谷氨酰胺（其密码子为 $\\mathrm{CAG}$ ）毒性聚集及蛋白质错误重叠。下图为某一 SCA 患者的家系图（相关基因不位于 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段），下列相关叙述正确的是（）\n\n[图1]\n\nA: SCA 的致病基因不可能位于性染色体, 该病最可能是隐性遗传病\nB: 该实例说明基因可以通过控制蛋白质的结构控制生物体的性状\nC: SCA 患者的致病基因中增加了含多个 CAG 序列的核苷酸单链片段\nD: 若咨询者的一个初级卵母细胞分裂产生的极体中含 SCA 致病基因, 则该细胞产生的卵细胞不含致病基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-77.jpg?height=354&width=1085&top_left_y=571&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1518",
        "problem": "The 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nHere are some observations:\n\n1. The average speed with which allele frequency changes\n2. Some alleles takeover very large populations even when they have impacts too small to measure in laboratories\n3. The probability an allele will takeover a population\n4. The speed an allele's frequency changes in any one population at any one time\n5. The frequency at which an allele is initially introduced alters the probability it will takeover a population\n6. More fit alleles sometimes go extinct before less fit alleles\n\n\nWhich are caused by both natural selection and genetic drift\nA: 1 and 2\nB: 3 and 4\nC: 5 and 6\nD: 1 only\nE: 2, 3 and 4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nHere are some observations:\n\n1. The average speed with which allele frequency changes\n2. Some alleles takeover very large populations even when they have impacts too small to measure in laboratories\n3. The probability an allele will takeover a population\n4. The speed an allele's frequency changes in any one population at any one time\n5. The frequency at which an allele is initially introduced alters the probability it will takeover a population\n6. More fit alleles sometimes go extinct before less fit alleles\n\n\nWhich are caused by both natural selection and genetic drift\n\nA: 1 and 2\nB: 3 and 4\nC: 5 and 6\nD: 1 only\nE: 2, 3 and 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=970&width=704&top_left_y=778&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=966&width=706&top_left_y=778&top_left_x=1118"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1465",
        "problem": "Review the following map and answer the 5 subsequent questions.\n\n[figure1]\n\nEcologists have been studying the small archipelago shown in the map above. They have been comparing the ecology of the mainland with those on the nearby islands and have come up with a few discoveries. The area has not been known to have been inhabited by humans, so it remains very natural. The mainland contains a low mountain range approximately $50 \\mathrm{~km}$ inland from the coast (highest peak approximately $400 \\mathrm{~m}$ above sea level) and the prevailing winds are westerlies.\n\nAfter investigating all of the vegetation types on each region the ecologists identified an area that would fit the description of a rain shadow. Where would the rain shadow be most likely to have been identified?\nA: East of the mountain range on the mainland\nB: West of the mountain range on the mainland\nC: Island 3\nD: Island 6\nE: Island 1\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nReview the following map and answer the 5 subsequent questions.\n\n[figure1]\n\nEcologists have been studying the small archipelago shown in the map above. They have been comparing the ecology of the mainland with those on the nearby islands and have come up with a few discoveries. The area has not been known to have been inhabited by humans, so it remains very natural. The mainland contains a low mountain range approximately $50 \\mathrm{~km}$ inland from the coast (highest peak approximately $400 \\mathrm{~m}$ above sea level) and the prevailing winds are westerlies.\n\nAfter investigating all of the vegetation types on each region the ecologists identified an area that would fit the description of a rain shadow. Where would the rain shadow be most likely to have been identified?\n\nA: East of the mountain range on the mainland\nB: West of the mountain range on the mainland\nC: Island 3\nD: Island 6\nE: Island 1\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-39.jpg?height=925&width=1479&top_left_y=451&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_43",
        "problem": "Cell walls provide plant cells with a substantial degree of volume homeostasis relative to the large changes in water potential that they experience as the everyday consequence of the transpirational water loss. Water potential $\\left(\\Psi_{\\mathrm{w}}\\right)$ of a plant cell is composed of solute potential $\\left(\\Psi_{\\mathrm{s}}\\right.$ ) and turgor pressure potential ( $\\Psi_{\\mathrm{p}}$ ) (Fig.Q11). Relative cell volume is correlated with cell water potential and its components as described in the graph below.\n\n[figure1]\n\nFig.Q11\nA: Alterations of cell water potential are generally accompanied by a large change in turgor pressure and in cell volume.\nB: Disappearance of turgor pressure indicates the ending point of cell plasmolysis with reduction of approximately by $15 \\%$ cell volume.\nC: As the cell volume decreases by $15 \\%$, most of the change in cell water potential is caused by the drop in cell solute potential together with little change in turgor pressure.\nD: During cell rehydration, cell wall expansion stops when cell wall generates pressure equivalent to turgor pressure and the water potential of the cell reach zero.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nCell walls provide plant cells with a substantial degree of volume homeostasis relative to the large changes in water potential that they experience as the everyday consequence of the transpirational water loss. Water potential $\\left(\\Psi_{\\mathrm{w}}\\right)$ of a plant cell is composed of solute potential $\\left(\\Psi_{\\mathrm{s}}\\right.$ ) and turgor pressure potential ( $\\Psi_{\\mathrm{p}}$ ) (Fig.Q11). Relative cell volume is correlated with cell water potential and its components as described in the graph below.\n\n[figure1]\n\nFig.Q11\n\nA: Alterations of cell water potential are generally accompanied by a large change in turgor pressure and in cell volume.\nB: Disappearance of turgor pressure indicates the ending point of cell plasmolysis with reduction of approximately by $15 \\%$ cell volume.\nC: As the cell volume decreases by $15 \\%$, most of the change in cell water potential is caused by the drop in cell solute potential together with little change in turgor pressure.\nD: During cell rehydration, cell wall expansion stops when cell wall generates pressure equivalent to turgor pressure and the water potential of the cell reach zero.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-026.jpg?height=913&width=1183&top_left_y=747&top_left_x=481"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1381",
        "problem": "The following graph illustrates the change in membrane potential $(\\mathrm{mV})$ during an action potential.\n\n[figure1]\n\nChanges in the membrane potential that do not reach the threshold voltage are known as graded potentials. Which of the following statements is false?\nA: The size of a graded potential is proportional to the stimulus, whereas the size of an action potential is the same regardless of the stimulus size.\nB: Graded potentials may be depolarising or hyperpolarising, whereas all action potentials are depolarising.\nC: Graded potentials propagate bidirectionally in a neuron, whereas action potentials are unidirectional.\nD: Both graded potentials and action potentials result in the opening of voltagegated ion channels.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph illustrates the change in membrane potential $(\\mathrm{mV})$ during an action potential.\n\n[figure1]\n\nChanges in the membrane potential that do not reach the threshold voltage are known as graded potentials. Which of the following statements is false?\n\nA: The size of a graded potential is proportional to the stimulus, whereas the size of an action potential is the same regardless of the stimulus size.\nB: Graded potentials may be depolarising or hyperpolarising, whereas all action potentials are depolarising.\nC: Graded potentials propagate bidirectionally in a neuron, whereas action potentials are unidirectional.\nD: Both graded potentials and action potentials result in the opening of voltagegated ion channels.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-14.jpg?height=651&width=694&top_left_y=443&top_left_x=270"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_513",
        "problem": "相互易位是一种非同源染色体互相交换部分等长的染色体片段的现象。如图所示,\n\n一个相互易位的杂合体在联会时会出现特征性的十字形图像, 随着分裂的进行, 十字形的配对同源染色体逐渐打开, 成为圆形或 8 字形。在相互易位杂合体的花粉母细胞中,大约有 50\\%的配对同源染色体呈圆形, $50 \\%$ 呈 8 字形, 说明配对同源染色体的 4 个着丝粒是随机移向两极的。已知某种植物形成的配子中染色体片段有重复或缺失可引起配子不育，下列相关叙述错误的是（ ）\n\n(a)\n\n(a)\n\n[图1]\n\n(b)\n[图2]\nA: 该种植物相互易位的杂合体通过减数分裂形成的花粉粒只有一半是可育的\nB: 该种植物相互易位的杂合体与正常个体杂交, 有 $1 / 4$ 的后代是相互易位杂合体\nC: 该变异可能改变了染色体上的基因排列顺序，但对该个体基因的种类没有影响\nD: 图 a 可出现在减数分裂I前期, 图 b 和图 c 都出现在减数分裂I的后期\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n相互易位是一种非同源染色体互相交换部分等长的染色体片段的现象。如图所示,\n\n一个相互易位的杂合体在联会时会出现特征性的十字形图像, 随着分裂的进行, 十字形的配对同源染色体逐渐打开, 成为圆形或 8 字形。在相互易位杂合体的花粉母细胞中,大约有 50\\%的配对同源染色体呈圆形, $50 \\%$ 呈 8 字形, 说明配对同源染色体的 4 个着丝粒是随机移向两极的。已知某种植物形成的配子中染色体片段有重复或缺失可引起配子不育，下列相关叙述错误的是（ ）\n\n(a)\n\n(a)\n\n[图1]\n\n(b)\n[图2]\n\nA: 该种植物相互易位的杂合体通过减数分裂形成的花粉粒只有一半是可育的\nB: 该种植物相互易位的杂合体与正常个体杂交, 有 $1 / 4$ 的后代是相互易位杂合体\nC: 该变异可能改变了染色体上的基因排列顺序，但对该个体基因的种类没有影响\nD: 图 a 可出现在减数分裂I前期, 图 b 和图 c 都出现在减数分裂I的后期\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-010.jpg?height=192&width=349&top_left_y=1863&top_left_x=636",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-010.jpg?height=410&width=984&top_left_y=2038&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_380",
        "problem": "某种小麦的株高由等位基因 $\\mathrm{G} 、 \\mathrm{~g}$ 决定，纯种高秆（GG）与纯种矮秆（gg）杂交的子代均为中秆。利用基因工程技术，导入另一种可在小麦细胞中正常表达的高秆基因\n\n- B 基因， B、G 基因在小麦的株高上有相同的效果，且具有累积效应。现培育出甲、乙、丙、丁四种转基因小麦（下图示），B 基因均成功插入染色体并正常表达（已知序列被破坏的基因无法正常表达)，下列相关叙述正确的是（注：不考虑染色体互换）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\n\n[图4]\n\n丁\nA: 此种小麦 $\\mathrm{G}$ 基因对 $\\mathrm{g}$ 基因为不完全显性，与孟德尔一对相对性状的踠豆杂交实验中高茎（D）、矮茎（d）基因的组合效果相同\nB: 图示四种小麦甲、乙、丙、丁株高一致\nC: 丁自交, 子代只有一种表型\nD: 甲自交的子代中，高度最高的植株占 $9 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种小麦的株高由等位基因 $\\mathrm{G} 、 \\mathrm{~g}$ 决定，纯种高秆（GG）与纯种矮秆（gg）杂交的子代均为中秆。利用基因工程技术，导入另一种可在小麦细胞中正常表达的高秆基因\n\n- B 基因， B、G 基因在小麦的株高上有相同的效果，且具有累积效应。现培育出甲、乙、丙、丁四种转基因小麦（下图示），B 基因均成功插入染色体并正常表达（已知序列被破坏的基因无法正常表达)，下列相关叙述正确的是（注：不考虑染色体互换）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\n\n[图4]\n\n丁\n\nA: 此种小麦 $\\mathrm{G}$ 基因对 $\\mathrm{g}$ 基因为不完全显性，与孟德尔一对相对性状的踠豆杂交实验中高茎（D）、矮茎（d）基因的组合效果相同\nB: 图示四种小麦甲、乙、丙、丁株高一致\nC: 丁自交, 子代只有一种表型\nD: 甲自交的子代中，高度最高的植株占 $9 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-84.jpg?height=291&width=303&top_left_y=163&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-84.jpg?height=285&width=300&top_left_y=163&top_left_x=638",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-84.jpg?height=277&width=280&top_left_y=170&top_left_x=931",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-84.jpg?height=272&width=266&top_left_y=178&top_left_x=1226"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1448",
        "problem": "The following graph illustrates the antimicrobial effect of antibiotics in vitro.\n\n[figure1]\n\nWhich of the following statements is supported?\nA: In people who are immunocompromised, bacteriostatic drugs are less effective than bactericidal drugs.\nB: When given to treat bacterial infection, bactericidal drugs lead to a quicker recovery than bacteriostatic drugs.\nC: Bactericidal drugs tend to cause more adverse effects than bacteriostatic drugs.\nD: If a patient is only given bacteriostatic drugs for a bacterial infection, they will never be cured of the infection.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph illustrates the antimicrobial effect of antibiotics in vitro.\n\n[figure1]\n\nWhich of the following statements is supported?\n\nA: In people who are immunocompromised, bacteriostatic drugs are less effective than bactericidal drugs.\nB: When given to treat bacterial infection, bactericidal drugs lead to a quicker recovery than bacteriostatic drugs.\nC: Bactericidal drugs tend to cause more adverse effects than bacteriostatic drugs.\nD: If a patient is only given bacteriostatic drugs for a bacterial infection, they will never be cured of the infection.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-17.jpg?height=574&width=1037&top_left_y=364&top_left_x=244"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1134",
        "problem": "A female with normal genotype ( $X X, 44)$ showed the presence of testis determining factor gene (TDF gene) on the $X$ chromosome. This is most likely a result of :\nA: X chromosome inactivation\nB: Dosage compensation effect\nC: Mutation\nD: Meiotic recombination\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA female with normal genotype ( $X X, 44)$ showed the presence of testis determining factor gene (TDF gene) on the $X$ chromosome. This is most likely a result of :\n\nA: X chromosome inactivation\nB: Dosage compensation effect\nC: Mutation\nD: Meiotic recombination\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1555",
        "problem": "The 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nHere are some observations:\n\n1. The average speed with which allele frequency changes\n2. Some alleles takeover very large populations even when they have impacts too small to measure in laboratories\n3. The probability an allele will takeover a population\n4. The speed an allele's frequency changes in any one population at any one time\n5. The frequency at which an allele is initially introduced alters the probability it will takeover a population\n6. More fit alleles sometimes go extinct before less fit alleles\n\n\nWhich are caused by drift alone\nA: 1 and 2\nB: 3 and 4\nC: 5 and 6\nD: 1 only\nE: 2, 3 and 4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nHere are some observations:\n\n1. The average speed with which allele frequency changes\n2. Some alleles takeover very large populations even when they have impacts too small to measure in laboratories\n3. The probability an allele will takeover a population\n4. The speed an allele's frequency changes in any one population at any one time\n5. The frequency at which an allele is initially introduced alters the probability it will takeover a population\n6. More fit alleles sometimes go extinct before less fit alleles\n\n\nWhich are caused by drift alone\n\nA: 1 and 2\nB: 3 and 4\nC: 5 and 6\nD: 1 only\nE: 2, 3 and 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=970&width=704&top_left_y=778&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=966&width=706&top_left_y=778&top_left_x=1118"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_782",
        "problem": "半乳糖血症是由半乳糖代谢途径中酶缺陷造成的遗传代谢病, 该病在人群中的发病率约为 $1 / 40000$ 。图 2 是图 1 中 $I_{1} 、 I_{2} 、 I_{3} 、 I_{4}$ 的基因电泳检测结果（不考虑 $X 、 Y$ 染色体的同源区段)。下列说法正确的是（ ）\n\n[图1]\n\n男性患者正常女性正常男性未知性别及性状\n\n图1\n\n[图2]\nA: 半乳糖血症的遗传方式是伴 X 染色体隐性遗传\nB: 图 1 中的 $\\mathrm{III}_{1}$ 为半乳糖血症男孩的概率是 $1 / 4$\nC: 若要判断胎儿III $I_{1}$ 是否患半乳糖血症, 需进行基因检测或孕妇血细胞检查\nD: 若 $\\mathrm{II}_{4}$ 与一位红绿色盲女性婚配, 则所生男孩兼患两病的概率为 $1 / 804$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n半乳糖血症是由半乳糖代谢途径中酶缺陷造成的遗传代谢病, 该病在人群中的发病率约为 $1 / 40000$ 。图 2 是图 1 中 $I_{1} 、 I_{2} 、 I_{3} 、 I_{4}$ 的基因电泳检测结果（不考虑 $X 、 Y$ 染色体的同源区段)。下列说法正确的是（ ）\n\n[图1]\n\n男性患者正常女性正常男性未知性别及性状\n\n图1\n\n[图2]\n\nA: 半乳糖血症的遗传方式是伴 X 染色体隐性遗传\nB: 图 1 中的 $\\mathrm{III}_{1}$ 为半乳糖血症男孩的概率是 $1 / 4$\nC: 若要判断胎儿III $I_{1}$ 是否患半乳糖血症, 需进行基因检测或孕妇血细胞检查\nD: 若 $\\mathrm{II}_{4}$ 与一位红绿色盲女性婚配, 则所生男孩兼患两病的概率为 $1 / 804$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-38.jpg?height=346&width=460&top_left_y=181&top_left_x=364",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-38.jpg?height=323&width=414&top_left_y=198&top_left_x=1204",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-38.jpg?height=48&width=1197&top_left_y=1415&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_943",
        "problem": "下图是某单基因遗传病的系谱图, 表是对该家系中 1 4 号个体进行相关基因检测 (先用某种限制酶切割样品 DNA, 再进行电泳) 得到的电泳结果 (电泳时不同大小的 DNA 片段移动速率不同)。已知编号 $\\mathrm{a}$ 对应的样品来自图中 4 号个体, 下列有关叙述错误的是 ( )\n\n[图1]\n[图2]\nA: 该病一定不是伴性遗传病\nB: 由图、表可知致病基因可以被限制酶切割为两种片段\nC: 8 号个体基因检测的电泳结果可能是编号 $\\mathrm{b}$ 或 $\\mathrm{c}$ 的条带类型\nD: 9 号与该病基因携带者结婚生一个正常男孩的概率为 5/6\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图是某单基因遗传病的系谱图, 表是对该家系中 1 4 号个体进行相关基因检测 (先用某种限制酶切割样品 DNA, 再进行电泳) 得到的电泳结果 (电泳时不同大小的 DNA 片段移动速率不同)。已知编号 $\\mathrm{a}$ 对应的样品来自图中 4 号个体, 下列有关叙述错误的是 ( )\n\n[图1]\n[图2]\n\nA: 该病一定不是伴性遗传病\nB: 由图、表可知致病基因可以被限制酶切割为两种片段\nC: 8 号个体基因检测的电泳结果可能是编号 $\\mathrm{b}$ 或 $\\mathrm{c}$ 的条带类型\nD: 9 号与该病基因携带者结婚生一个正常男孩的概率为 5/6\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-089.jpg?height=711&width=1234&top_left_y=2032&top_left_x=317",
            "https://i.postimg.cc/2SS4yMyd/image.png",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-090.jpg?height=54&width=934&top_left_y=2126&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1197",
        "problem": "The graph shows the birth rate and death rate for a population over a 100-year period.\n\n[figure1]\n\nAssuming equal emigration and immigration, from 1900 to 2000 , the population has\nA: Increased\nB: Decreased\nC: Stayed the same\nD: Increased until 1930, then decreased\nE: Fluctuated over the years\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows the birth rate and death rate for a population over a 100-year period.\n\n[figure1]\n\nAssuming equal emigration and immigration, from 1900 to 2000 , the population has\n\nA: Increased\nB: Decreased\nC: Stayed the same\nD: Increased until 1930, then decreased\nE: Fluctuated over the years\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-17.jpg?height=734&width=1140&top_left_y=338&top_left_x=158"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_464",
        "problem": "在搭建 DNA 分子模型的实验中, 若有 4 种碱基塑料片共 20 个, 其中 4 个 $\\mathrm{C}, 6$ 个 $\\mathrm{G}, 3$ 个 $\\mathrm{A}, 7$ 个 $\\mathrm{T}$, 脱氧核糖和磷酸之间的连接物 14 个, 脱氧核糖塑料片 40 个, 磷酸塑料片 100 个, 代表氢键的连接物若干, 脱氧核糖和碱基之间的连接物若干，则(\nA: 能搭建出 20 个脱氧核苷酸\nB: 所搭建的 DNA 分子片段最长为 7 碱基对\nC: 能搭建出 $4^{10}$ 种不同的 DNA 分子模型\nD: 能搭建出一个 4 碱基对的 DNA 分子片段\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在搭建 DNA 分子模型的实验中, 若有 4 种碱基塑料片共 20 个, 其中 4 个 $\\mathrm{C}, 6$ 个 $\\mathrm{G}, 3$ 个 $\\mathrm{A}, 7$ 个 $\\mathrm{T}$, 脱氧核糖和磷酸之间的连接物 14 个, 脱氧核糖塑料片 40 个, 磷酸塑料片 100 个, 代表氢键的连接物若干, 脱氧核糖和碱基之间的连接物若干，则(\n\nA: 能搭建出 20 个脱氧核苷酸\nB: 所搭建的 DNA 分子片段最长为 7 碱基对\nC: 能搭建出 $4^{10}$ 种不同的 DNA 分子模型\nD: 能搭建出一个 4 碱基对的 DNA 分子片段\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1253",
        "problem": "SAVING THE MAUI'S DOLPHINS - A STORY OF BIOLOGY, POLICY \\& CONSERVATION\n\n[figure1]\n\nWWF-New Zealand\n\nhttp://wwf.panda.org/wwf news/?206249/NZ-govt-fails-Mauisdolphins-on-global-stage\n\n[figure2]\n\nhttp://uww.doc.govt.nz/conservation/native-animals/marinemammals/dolphins/hectors-dolphin/docs-work/hectors-and-mauisdolphin-incident-databasel\n\nMaui's dolphin (Cephalorhynchus hectori maui) is one of the world's rarest dolphins and is found only on the west coast of the North Island of New Zealand (Resource Pack, Figure 1). It is a sub-species of Hector's dolphin (Cephalorhynchus hectori). The Maui's dolphin is protected by the West Coast North Island Marine Mammal Sanctuary (WCNIMMS) which restricts seabed mining activities, acoustic seismic survey work and commercial and recreational set netting. The boundaries of this sanctuary extend alongshore from Maunganui Bluff in Northland to Oakura Beach, Taranaki, in the south and from mean high water springs to the 12 nautical mile (nm) territorial sea limit. The total area of the sanctuary is approximately 1,200,086 hectares covering 2,164 km of coastline. (Resource Pack, Figure 1).\n\nMaui's dolphin is 'critically endangered' (IUCN Red List), with the population dropping from around 1000 individuals in 1970 to 111 in 2004 according to research by Assoc. Prof. Dr Liz Slooten and others from Otago University. Recent research by the Department of Conservation (DOC) suggests there are now even fewer Maui's dolphins remaining. Maui's dolphins are relatively short-lived (approximately 25 years), and are slow breeders. Females do not have their first calf until they are about seven or eight years old, and have a new calf only every two to four years. This means the species may be threatened by even occasional deaths caused by human activity. Fishing, particularly set netting, is the greatest known human threat to Maui's dolphins and thought to be responsible for about 75 per cent of reported deaths with a known cause. Other human threats include marine tourism, vessel traffic, mining, construction, coastal development pollution, sedimentation, oil spills, plastic bags, marine farming and climate change.\n\nIn recent weeks the need to develop effective management strategies for this species to prevent its extinction has hit the news. Liz Slooten presented the latest research to the International Whaling Commission in May this year, showing that the current protection measures are not sufficient to avoid the extinction of Maui's dolphin. WWF-New Zealand, Greenpeace, and Forest \\& Bird and international conservation groups including NABU and WDC in 2012 and 2013 are all campaigning to protect the Maui's dolphin and hundreds of angry protesters marched to the office of the Energy and Resources Minister Simon Bridges' office in Tauranga to protest the government allowing oil exploration within the West Coast North Island Marine Mammal Sanctuary in their annual tender process 'Block Offer 2014' (Resource Pack, Figure 3). This government decision came just two weeks after the International Whaling Committee (IWC) criticised New Zealand for not taking the necessary steps to save the Maui's dolphin. The IWC noted that a $350 \\mathrm{~km}^{2}$ set net restriction had been added to the WCNIMMS but commented that these measures fell significantly short of those required to reverse the Maui's dolphin decline as recommended by the IWC in 2012 and 2013. The IWC reiterated its extreme concern about the continued decline of such a small population \"as the humaninduced death of even one dolphin would increase the extinction risk for this subspecies\". In 2013 it strongly recommended that the NZ government should:\n\n\"take immediate management actions that will eliminate bycatch of Maui's dolphins. This includes full closures of any fisheries within the range of Maui's dolphins that are known to pose a risk of bycatch of small cetaceans (i.e. set net and trawl fisheries)\". Ensuring full protection of Maui's dolphins in all areas throughout their habitat, together with an ample buffer zone, would minimise the risk of bycatch and maximise the chances of population increase\". and \"commit to specific population increase targets and timelines\".\n\nBecause this recommendation was ignored by the NZ government, in 2014, the IWC recommended that the protected area should be extended south to Whanganui, offshore to 20 nautical miles and should include the harbours.\n\nThe resource pack contains maps of Maui's and Hector's dolphin sightings from 1970 - July 2013, Protection measures for Maui's dolphins on the West Coast North Island and the 'Block Offer 2014' for the Offshore Release Area: Taranaki Basin 14TAR-R1. It also has a table of all Maui's dolphin sightings from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti. Much of this area is included in the Block Offer 2014' Taranaki Basin 14TAR-R1 area. This table is modified from the Maui's dolphin sightings database:\n\nhttp://www.doc.govt.nz/conservation/native-animals/marine-mammals/dolphins/mauis-dolphin/docs-work/mauisdolphin-sightings/\n\nConservation Minister Nick Smith was questioned in parliament about the decision to open part of the sanctuary for exploration. \"The block offer is nowhere near where the Maui's live,\" Dr Smith said. \"There hasn't been a single observation of a Maui's dolphin, and the oil and gas industry hasn't been involved in a single Maui's dolphin incident in Taranaki over the past 40 years despite 23 wells being drilled\" (Source: http://www.3news.co.nz/Greenpeace-get23000-signatures-in-petition-to-sack-Bridges/tabid/423/articleID/349373/Default.aspx).Determining the distribution and population size of Maui's dolphins from sightings has a number of problems. These include:\nA: The sampling effort is unequal so the lack of presence of a species could relate to a real absence or simply to a lack of sampling effort.\nB: Low or zero sampling effort could easily miss the presence of a low-density species such as Maui's dolphin.\nC: Misidentification of species, or misreporting of locations, can confound such studies.\nD: Environmental factors such as rough seas can decrease sampling effort and result in dolphins being 'missed'.\nE: All of the problems above (A-D) are inherent in using sighting data to determine the distribution and population size of Maui's dolphins.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nSAVING THE MAUI'S DOLPHINS - A STORY OF BIOLOGY, POLICY \\& CONSERVATION\n\n[figure1]\n\nWWF-New Zealand\n\nhttp://wwf.panda.org/wwf news/?206249/NZ-govt-fails-Mauisdolphins-on-global-stage\n\n[figure2]\n\nhttp://uww.doc.govt.nz/conservation/native-animals/marinemammals/dolphins/hectors-dolphin/docs-work/hectors-and-mauisdolphin-incident-databasel\n\nMaui's dolphin (Cephalorhynchus hectori maui) is one of the world's rarest dolphins and is found only on the west coast of the North Island of New Zealand (Resource Pack, Figure 1). It is a sub-species of Hector's dolphin (Cephalorhynchus hectori). The Maui's dolphin is protected by the West Coast North Island Marine Mammal Sanctuary (WCNIMMS) which restricts seabed mining activities, acoustic seismic survey work and commercial and recreational set netting. The boundaries of this sanctuary extend alongshore from Maunganui Bluff in Northland to Oakura Beach, Taranaki, in the south and from mean high water springs to the 12 nautical mile (nm) territorial sea limit. The total area of the sanctuary is approximately 1,200,086 hectares covering 2,164 km of coastline. (Resource Pack, Figure 1).\n\nMaui's dolphin is 'critically endangered' (IUCN Red List), with the population dropping from around 1000 individuals in 1970 to 111 in 2004 according to research by Assoc. Prof. Dr Liz Slooten and others from Otago University. Recent research by the Department of Conservation (DOC) suggests there are now even fewer Maui's dolphins remaining. Maui's dolphins are relatively short-lived (approximately 25 years), and are slow breeders. Females do not have their first calf until they are about seven or eight years old, and have a new calf only every two to four years. This means the species may be threatened by even occasional deaths caused by human activity. Fishing, particularly set netting, is the greatest known human threat to Maui's dolphins and thought to be responsible for about 75 per cent of reported deaths with a known cause. Other human threats include marine tourism, vessel traffic, mining, construction, coastal development pollution, sedimentation, oil spills, plastic bags, marine farming and climate change.\n\nIn recent weeks the need to develop effective management strategies for this species to prevent its extinction has hit the news. Liz Slooten presented the latest research to the International Whaling Commission in May this year, showing that the current protection measures are not sufficient to avoid the extinction of Maui's dolphin. WWF-New Zealand, Greenpeace, and Forest \\& Bird and international conservation groups including NABU and WDC in 2012 and 2013 are all campaigning to protect the Maui's dolphin and hundreds of angry protesters marched to the office of the Energy and Resources Minister Simon Bridges' office in Tauranga to protest the government allowing oil exploration within the West Coast North Island Marine Mammal Sanctuary in their annual tender process 'Block Offer 2014' (Resource Pack, Figure 3). This government decision came just two weeks after the International Whaling Committee (IWC) criticised New Zealand for not taking the necessary steps to save the Maui's dolphin. The IWC noted that a $350 \\mathrm{~km}^{2}$ set net restriction had been added to the WCNIMMS but commented that these measures fell significantly short of those required to reverse the Maui's dolphin decline as recommended by the IWC in 2012 and 2013. The IWC reiterated its extreme concern about the continued decline of such a small population \"as the humaninduced death of even one dolphin would increase the extinction risk for this subspecies\". In 2013 it strongly recommended that the NZ government should:\n\n\"take immediate management actions that will eliminate bycatch of Maui's dolphins. This includes full closures of any fisheries within the range of Maui's dolphins that are known to pose a risk of bycatch of small cetaceans (i.e. set net and trawl fisheries)\". Ensuring full protection of Maui's dolphins in all areas throughout their habitat, together with an ample buffer zone, would minimise the risk of bycatch and maximise the chances of population increase\". and \"commit to specific population increase targets and timelines\".\n\nBecause this recommendation was ignored by the NZ government, in 2014, the IWC recommended that the protected area should be extended south to Whanganui, offshore to 20 nautical miles and should include the harbours.\n\nThe resource pack contains maps of Maui's and Hector's dolphin sightings from 1970 - July 2013, Protection measures for Maui's dolphins on the West Coast North Island and the 'Block Offer 2014' for the Offshore Release Area: Taranaki Basin 14TAR-R1. It also has a table of all Maui's dolphin sightings from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti. Much of this area is included in the Block Offer 2014' Taranaki Basin 14TAR-R1 area. This table is modified from the Maui's dolphin sightings database:\n\nhttp://www.doc.govt.nz/conservation/native-animals/marine-mammals/dolphins/mauis-dolphin/docs-work/mauisdolphin-sightings/\n\nConservation Minister Nick Smith was questioned in parliament about the decision to open part of the sanctuary for exploration. \"The block offer is nowhere near where the Maui's live,\" Dr Smith said. \"There hasn't been a single observation of a Maui's dolphin, and the oil and gas industry hasn't been involved in a single Maui's dolphin incident in Taranaki over the past 40 years despite 23 wells being drilled\" (Source: http://www.3news.co.nz/Greenpeace-get23000-signatures-in-petition-to-sack-Bridges/tabid/423/articleID/349373/Default.aspx).\n\nproblem:\nDetermining the distribution and population size of Maui's dolphins from sightings has a number of problems. These include:\n\nA: The sampling effort is unequal so the lack of presence of a species could relate to a real absence or simply to a lack of sampling effort.\nB: Low or zero sampling effort could easily miss the presence of a low-density species such as Maui's dolphin.\nC: Misidentification of species, or misreporting of locations, can confound such studies.\nD: Environmental factors such as rough seas can decrease sampling effort and result in dolphins being 'missed'.\nE: All of the problems above (A-D) are inherent in using sighting data to determine the distribution and population size of Maui's dolphins.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-30.jpg?height=503&width=682&top_left_y=277&top_left_x=113",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-30.jpg?height=497&width=848&top_left_y=277&top_left_x=1049"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_162",
        "problem": "Plasmodium is a parasitic protozoan that is prevalent in tropical countries. Once the parasites invade host red blood cells, they multiply within $24 \\mathrm{~h} . \\mathrm{Fe}^{2+}$ in red blood cell can react with free $\\mathrm{O}_{2}$ as well as $\\mathrm{H}_{2} \\mathrm{O}_{2}$, causing formation of free radicals that can damage parasite cells.\n\nUnlike host cells, the parasites lack several enzymes of antioxidant defence systems. Scientists analysed the proteins from the parasite cytosol using two dimensional gel electrophoresis (see image below).\n\nThen, using mass spectrometry and peptide mass fingerprinting, they identified 6 proteins (spots 1 to 6) corresponding to human peroxiredoxin, while one protein (spot 7) corresponding to Plasmodium peroxiredoxin.\n\n[figure1]\nA: Based on the presented data, Plasmodium peroxiredoxin is a multimeric protein.\nB: All human peroxiredoxin proteins have a positive net charge at physiological $\\mathrm{pH}$.\nC: Gel filtration chromatography is suitable for separation and purification of the six human peroxiredoxin proteins.\nD: Immunoaffinity chromatography can be used for separation of Plasmodium peroxiredoxin from other Plasmodium cytosolic proteins.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPlasmodium is a parasitic protozoan that is prevalent in tropical countries. Once the parasites invade host red blood cells, they multiply within $24 \\mathrm{~h} . \\mathrm{Fe}^{2+}$ in red blood cell can react with free $\\mathrm{O}_{2}$ as well as $\\mathrm{H}_{2} \\mathrm{O}_{2}$, causing formation of free radicals that can damage parasite cells.\n\nUnlike host cells, the parasites lack several enzymes of antioxidant defence systems. Scientists analysed the proteins from the parasite cytosol using two dimensional gel electrophoresis (see image below).\n\nThen, using mass spectrometry and peptide mass fingerprinting, they identified 6 proteins (spots 1 to 6) corresponding to human peroxiredoxin, while one protein (spot 7) corresponding to Plasmodium peroxiredoxin.\n\n[figure1]\n\nA: Based on the presented data, Plasmodium peroxiredoxin is a multimeric protein.\nB: All human peroxiredoxin proteins have a positive net charge at physiological $\\mathrm{pH}$.\nC: Gel filtration chromatography is suitable for separation and purification of the six human peroxiredoxin proteins.\nD: Immunoaffinity chromatography can be used for separation of Plasmodium peroxiredoxin from other Plasmodium cytosolic proteins.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-22.jpg?height=460&width=1250&top_left_y=844&top_left_x=403"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_494",
        "problem": "$\\beta$-地中海贫血症是一种由 $\\beta$-珠蛋白基因突变引起的单基因遗传病, 部分患者该基因的编码链中与 $\\beta$-珠蛋白第 71-72 位氨基酸有关的编码序列间插入了一个 $\\mathrm{A}$ （如图）, UAG/UGA 为终止密码子。下列叙述正确的是（ ）\n\n正常序列 GCC TTT AGT GAT\n\n| 70 | 71 | 72 | 73 |\n| :--- | :--- | :--- | :--- |\n\n异常序列 1 GCC TAT TAG TGA\n\n异常序列 2 GCC TTA TAG TGA\n\n异常序列 3 GCC TTT AAG TGA\n\n异常序列 4 GCC TTT AGA TGA\nA: 突变后参与该基因表达的 tRNA 种类可能减少\nB: 编码链中插入 $A$ 会导致 mRNA 对应位点插入 $U$\nC: mRNA 在细胞核内加工时对应的异常序列会被剪切 D．该异常序列不会影响基因表达产物的空间结构\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\beta$-地中海贫血症是一种由 $\\beta$-珠蛋白基因突变引起的单基因遗传病, 部分患者该基因的编码链中与 $\\beta$-珠蛋白第 71-72 位氨基酸有关的编码序列间插入了一个 $\\mathrm{A}$ （如图）, UAG/UGA 为终止密码子。下列叙述正确的是（ ）\n\n正常序列 GCC TTT AGT GAT\n\n| 70 | 71 | 72 | 73 |\n| :--- | :--- | :--- | :--- |\n\n异常序列 1 GCC TAT TAG TGA\n\n异常序列 2 GCC TTA TAG TGA\n\n异常序列 3 GCC TTT AAG TGA\n\n异常序列 4 GCC TTT AGA TGA\n\nA: 突变后参与该基因表达的 tRNA 种类可能减少\nB: 编码链中插入 $A$ 会导致 mRNA 对应位点插入 $U$\nC: mRNA 在细胞核内加工时对应的异常序列会被剪切 D．该异常序列不会影响基因表达产物的空间结构\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_503",
        "problem": "致死基因的存在可影响后代性状分离比。现有基因型为 $\\mathrm{AaBb}$ 的个体，两对等位基因独立遗传, 但具有某种基因型的配子或个体致死, 不考虑环境因素对表现型的影响,若该个体自交，下列说法不正确的是\nA: 后代分离比为 $6: 3: 2: 1$, 则推测原因可能是某对基因显性纯合致死\nB: 后代分离比为 5:3:3:1, 则推测原因可能是基因型为 $\\mathrm{AB}$ 的雄配子或雌配子致死\nC: 后代分离比为 7:3:1:1, 则推测原因可能是基因型为 $\\mathrm{Ab}$ 的雄配子或雌配子致死\nD: 后代分离比为 $9: 3: 3$, 则推测原因可能是基因型为 $a B$ 的雄配子或雌配子致死\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n致死基因的存在可影响后代性状分离比。现有基因型为 $\\mathrm{AaBb}$ 的个体，两对等位基因独立遗传, 但具有某种基因型的配子或个体致死, 不考虑环境因素对表现型的影响,若该个体自交，下列说法不正确的是\n\nA: 后代分离比为 $6: 3: 2: 1$, 则推测原因可能是某对基因显性纯合致死\nB: 后代分离比为 5:3:3:1, 则推测原因可能是基因型为 $\\mathrm{AB}$ 的雄配子或雌配子致死\nC: 后代分离比为 7:3:1:1, 则推测原因可能是基因型为 $\\mathrm{Ab}$ 的雄配子或雌配子致死\nD: 后代分离比为 $9: 3: 3$, 则推测原因可能是基因型为 $a B$ 的雄配子或雌配子致死\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_329",
        "problem": "糖酵解是将葡萄糖降解为丙酮酸并伴随着 ATP 生成的一系列化学反应, 其中甘油醛-3-磷酸脱氢酶 (GAPDH) 是糖酻解途径中的一个关键酶。研究发现, 即使在氧浓度充足条件下, 癌细胞的能量供应仍主要依赖糖酵解途径, 并产生大量乳酸。下列相关叙述正确的是（ ）\nA: 细胞发生癌变的根本原因是基因的选择性表达\nB: 癌细胞中产生 $\\mathrm{CO}_{2}$ 的主要场所是细胞质基质\nC: 与正常细胞相比，癌细胞对葡萄糖的摄取量较大\nD: 可用 GAPDH 活性抑制剂治疗癌症, 且副作用较小\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n糖酵解是将葡萄糖降解为丙酮酸并伴随着 ATP 生成的一系列化学反应, 其中甘油醛-3-磷酸脱氢酶 (GAPDH) 是糖酻解途径中的一个关键酶。研究发现, 即使在氧浓度充足条件下, 癌细胞的能量供应仍主要依赖糖酵解途径, 并产生大量乳酸。下列相关叙述正确的是（ ）\n\nA: 细胞发生癌变的根本原因是基因的选择性表达\nB: 癌细胞中产生 $\\mathrm{CO}_{2}$ 的主要场所是细胞质基质\nC: 与正常细胞相比，癌细胞对葡萄糖的摄取量较大\nD: 可用 GAPDH 活性抑制剂治疗癌症, 且副作用较小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1240",
        "problem": "Evolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).How many species of elephant birds are known to have existed?\nA: Scientists are uncertain of the number\nB: 2\nC: 3\nD: 4\nE: 5\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nEvolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).\n\nproblem:\nHow many species of elephant birds are known to have existed?\n\nA: Scientists are uncertain of the number\nB: 2\nC: 3\nD: 4\nE: 5\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-28.jpg?height=1868&width=1627&top_left_y=748&top_left_x=206"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1440",
        "problem": "Pathogens are agents which infect humans and cause disease. Molecular mimicry occurs when foreign pathogens contain molecules that look like human molecules. When infected by such a pathogen, the body's immune system attacks the pathogenic molecule.\n\nHowever, after the infection is cleared, the immune system may then mistakenly attack our own self-molecules.\n\nMolecular mimicry could occur because:\nA: pathogen proteins resemble host proteins\nB: pathogens activate specific host cytokines\nC: pathogens activate specific host cell metabolism\nD: pathogens mimic specific host genes\nE: pathogens change their molecules to look like human molecules during an infection\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPathogens are agents which infect humans and cause disease. Molecular mimicry occurs when foreign pathogens contain molecules that look like human molecules. When infected by such a pathogen, the body's immune system attacks the pathogenic molecule.\n\nHowever, after the infection is cleared, the immune system may then mistakenly attack our own self-molecules.\n\nMolecular mimicry could occur because:\n\nA: pathogen proteins resemble host proteins\nB: pathogens activate specific host cytokines\nC: pathogens activate specific host cell metabolism\nD: pathogens mimic specific host genes\nE: pathogens change their molecules to look like human molecules during an infection\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_650",
        "problem": "某植物属于雌雄同花、自花传粉的作物, 其矮秆 $(R)$ 与高秆 $(r)$ 、雄性不育 $(M)$与雄性可育 $(\\mathrm{m}$ ) 是两对相对性状。科研工作者进行了实验, 构建了用于杂交育种的矮败（矮秆雄性不育）品系。部分杂交组合如下:\n\n杂交一:矮秆雄性可育 $\\mathrm{X}$ 高秆雄性不育 $\\rightarrow \\mathrm{F}_{1}$ 矮秆雄性不育: 矮秆雄性可育 $=1: 1$;\n\n杂交二: 选取 $\\mathrm{F}_{1}$ 中的矮秆雄性不育 $\\mathrm{X}$ 高秆雄性可育 $\\rightarrow \\mathrm{F}_{2}$ 矮秆雄性可育: 高秆雄性不育 $=1: 1$, 科研人员经过多次实验, 在 $F_{2}$ 中偶然发现一株矮秆雄性不育的植株甲。下列叙述错误的是 ( )\nA: 从杂交一可知, $\\mathrm{F}_{1}$ 中矮秆雄性不育的基因型是 $\\mathrm{RrMm}$\nB: 若 $F_{2}$ 的全部植株（不包括甲）自然繁殖, 子代中高秆雄性可育植株占 $1 / 4$\nC: 若甲为矮秆雄性不育三体植株, 则其不会产生基因型为 Rrmm 的配子\nD: 若甲为矮秆雄性不育二倍体植株, 则可能是父本发生了基因突变\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某植物属于雌雄同花、自花传粉的作物, 其矮秆 $(R)$ 与高秆 $(r)$ 、雄性不育 $(M)$与雄性可育 $(\\mathrm{m}$ ) 是两对相对性状。科研工作者进行了实验, 构建了用于杂交育种的矮败（矮秆雄性不育）品系。部分杂交组合如下:\n\n杂交一:矮秆雄性可育 $\\mathrm{X}$ 高秆雄性不育 $\\rightarrow \\mathrm{F}_{1}$ 矮秆雄性不育: 矮秆雄性可育 $=1: 1$;\n\n杂交二: 选取 $\\mathrm{F}_{1}$ 中的矮秆雄性不育 $\\mathrm{X}$ 高秆雄性可育 $\\rightarrow \\mathrm{F}_{2}$ 矮秆雄性可育: 高秆雄性不育 $=1: 1$, 科研人员经过多次实验, 在 $F_{2}$ 中偶然发现一株矮秆雄性不育的植株甲。下列叙述错误的是 ( )\n\nA: 从杂交一可知, $\\mathrm{F}_{1}$ 中矮秆雄性不育的基因型是 $\\mathrm{RrMm}$\nB: 若 $F_{2}$ 的全部植株（不包括甲）自然繁殖, 子代中高秆雄性可育植株占 $1 / 4$\nC: 若甲为矮秆雄性不育三体植株, 则其不会产生基因型为 Rrmm 的配子\nD: 若甲为矮秆雄性不育二倍体植株, 则可能是父本发生了基因突变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_203",
        "problem": "Phytochrome is one of the plant photoreceptors involved in photoperiodism. It exists in two spectophotometrically different forms: red-light absorbing $\\mathrm{P}_{\\mathrm{r}}$ and far-red light absorbing $\\mathrm{P}_{\\mathrm{fr}}$. An investigation explored how plant flowering was affected by different light flashes [white (W), red (R), or far-red (FR) light] applied during the dark period or darkness in the light-period of plant growth. The figure below shows the experimental results.\n\n[figure1]\n\nBased on this experiment, find the most accurate explanation or expectation for the light control of flowering in this plant,\nA: This plant flowers whenever the total night length exceeds a $12 \\mathrm{hr}$ threshold (out of the $24 \\mathrm{hr}$ night/light period) with or without light interruption.\nB: This plant is likely to be a short-day plant that requires a certain length of uninterrupted light period for flowering.\nC: The plant in experiment 3 will flower if it is irradiated with a flash of far-red light, instead of white..\nD: The plant in experiment 4 will flower.\nE: The plant in experiment 5 will not flower.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPhytochrome is one of the plant photoreceptors involved in photoperiodism. It exists in two spectophotometrically different forms: red-light absorbing $\\mathrm{P}_{\\mathrm{r}}$ and far-red light absorbing $\\mathrm{P}_{\\mathrm{fr}}$. An investigation explored how plant flowering was affected by different light flashes [white (W), red (R), or far-red (FR) light] applied during the dark period or darkness in the light-period of plant growth. The figure below shows the experimental results.\n\n[figure1]\n\nBased on this experiment, find the most accurate explanation or expectation for the light control of flowering in this plant,\n\nA: This plant flowers whenever the total night length exceeds a $12 \\mathrm{hr}$ threshold (out of the $24 \\mathrm{hr}$ night/light period) with or without light interruption.\nB: This plant is likely to be a short-day plant that requires a certain length of uninterrupted light period for flowering.\nC: The plant in experiment 3 will flower if it is irradiated with a flash of far-red light, instead of white..\nD: The plant in experiment 4 will flower.\nE: The plant in experiment 5 will not flower.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-15.jpg?height=508&width=1482&top_left_y=817&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_392",
        "problem": "某果蝇精原细胞的一对常染色体上有五对等位基因, 其中一条染色体 b、c、d 基因所在的片段发生了染色体结构变异倒位。该精原细胞减数分裂联会时, 倒位片段会扭曲与同源片段形成倒位环，且联会后两条非姐妹染色单体在基因 $B 、 C （ b 、 c)$ 之间或基因 C、D（c、d）之间发生了一次交换，交换后形成了无着丝粒片段和双着丝粒桥，无着丝粒片段会因水解而丢失, 双着丝粒桥会发生随机断裂, 假设断裂处发生在相邻两基因之间。最终该精原细胞完成减数分裂形成四个配子, 整个过程如图所示。不考虑基因突变, 下列叙述正确的是（）\n\n[图1]\n\n复制 联会\n\n[图2]\nA: 若一缺失配子基因型为 $\\mathrm{ad}$ ，则另一缺失配子基因型为 $\\mathrm{ABc}$\nB: 若一缺失配子基因型为 $\\mathrm{ABC}$ ，则交换位点发生在基因 $\\mathrm{B} 、 \\mathrm{C}(\\mathrm{b} 、 \\mathrm{c}$ ）之间\nC: 若一缺失配子基因型为 $a$ ，则双着丝粒桥断裂位点一定发生在 $a 、 d$ 之间\nD: 若该精原细胞无穷多, 且均按图示过程完成减数分裂, 则可形成 14 种配子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某果蝇精原细胞的一对常染色体上有五对等位基因, 其中一条染色体 b、c、d 基因所在的片段发生了染色体结构变异倒位。该精原细胞减数分裂联会时, 倒位片段会扭曲与同源片段形成倒位环，且联会后两条非姐妹染色单体在基因 $B 、 C （ b 、 c)$ 之间或基因 C、D（c、d）之间发生了一次交换，交换后形成了无着丝粒片段和双着丝粒桥，无着丝粒片段会因水解而丢失, 双着丝粒桥会发生随机断裂, 假设断裂处发生在相邻两基因之间。最终该精原细胞完成减数分裂形成四个配子, 整个过程如图所示。不考虑基因突变, 下列叙述正确的是（）\n\n[图1]\n\n复制 联会\n\n[图2]\n\nA: 若一缺失配子基因型为 $\\mathrm{ad}$ ，则另一缺失配子基因型为 $\\mathrm{ABc}$\nB: 若一缺失配子基因型为 $\\mathrm{ABC}$ ，则交换位点发生在基因 $\\mathrm{B} 、 \\mathrm{C}(\\mathrm{b} 、 \\mathrm{c}$ ）之间\nC: 若一缺失配子基因型为 $a$ ，则双着丝粒桥断裂位点一定发生在 $a 、 d$ 之间\nD: 若该精原细胞无穷多, 且均按图示过程完成减数分裂, 则可形成 14 种配子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-078.jpg?height=146&width=599&top_left_y=178&top_left_x=363",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-078.jpg?height=423&width=1403&top_left_y=314&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1486",
        "problem": "Peaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nWhich results are expected if fuzz is produced by many alleles in combination?\nA: $\\quad \\mathrm{A}$\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: $\\quad \\mathrm{D}$\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPeaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nWhich results are expected if fuzz is produced by many alleles in combination?\n\nA: $\\quad \\mathrm{A}$\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: $\\quad \\mathrm{D}$\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-45.jpg?height=782&width=1231&top_left_y=474&top_left_x=241"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1231",
        "problem": "A count was made of the number of cells showing different stages of mitosis in a zone of an onion root tip. The following results were obtained.\n\n| Stage | Percentage <br> of total <br> number of <br> dividing <br> cells |\n| :--- | :--- |\n| prophase | 85.0 |\n| metaphase | 7.7 |\n| anaphase | 2.9 |\n| telophase | 4.4 |\n\nFrom this it can be deduced that:\nA: Prophase takes much longer than the other stages.\nB: The division process was just starting.\nC: Telophase is the shortest phase in mitosis.\nD: The sample used for the count was too small.\nE: The area investigated was very close to the root tip.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA count was made of the number of cells showing different stages of mitosis in a zone of an onion root tip. The following results were obtained.\n\n| Stage | Percentage <br> of total <br> number of <br> dividing <br> cells |\n| :--- | :--- |\n| prophase | 85.0 |\n| metaphase | 7.7 |\n| anaphase | 2.9 |\n| telophase | 4.4 |\n\nFrom this it can be deduced that:\n\nA: Prophase takes much longer than the other stages.\nB: The division process was just starting.\nC: Telophase is the shortest phase in mitosis.\nD: The sample used for the count was too small.\nE: The area investigated was very close to the root tip.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1030",
        "problem": "Roots absorb cations from the soil by a process called:\nA: Cation absorption\nB: Cation exchange\nC: Cation transfer\nD: Cation cotransport\nE: Cation reduction\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRoots absorb cations from the soil by a process called:\n\nA: Cation absorption\nB: Cation exchange\nC: Cation transfer\nD: Cation cotransport\nE: Cation reduction\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_265",
        "problem": "The inheritance pattern of two rare traits, represented as vertical and horizontal lines, is shown in the pedigree below.\n\n[figure1]\nA: Inheritance of one of the traits is Y-linked.\nB: The data presented is consistent with autosomal recessive inheritance for both traits.\nC: The phenotype of III-2 may be due to non-disjunction in II-2.\nD: The phenotype of III-2 may be due to cross over in II-2.\nE: The probability of the phenotype of III-2 being due to a mutation that occurred in an egg cell of II-2 before fertilization is twice the probability of the phenotype of III-4 being due to a mutation that occurred in an egg cell of II-2 before fertilization.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe inheritance pattern of two rare traits, represented as vertical and horizontal lines, is shown in the pedigree below.\n\n[figure1]\n\nA: Inheritance of one of the traits is Y-linked.\nB: The data presented is consistent with autosomal recessive inheritance for both traits.\nC: The phenotype of III-2 may be due to non-disjunction in II-2.\nD: The phenotype of III-2 may be due to cross over in II-2.\nE: The probability of the phenotype of III-2 being due to a mutation that occurred in an egg cell of II-2 before fertilization is twice the probability of the phenotype of III-4 being due to a mutation that occurred in an egg cell of II-2 before fertilization.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-24.jpg?height=622&width=1079&top_left_y=503&top_left_x=494"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_362",
        "problem": "苋菜的野生型品系对除草剂敏感, 突变品系对除草剂有抗性。某科研小组研究了苋菜叶绿体基因 pbsA 野生型品系和突变品系的部分 DNA 碱基序列和氨基酸位置, 结果如表所示。下列叙述正确的是（ ）\n\n| 野生型品系 | CGA 丙氨 <br> 酸 | AGT 丝氨 <br> 酸 | AAG 苯丙氨 <br> 酸 | TT 天冬氨 <br> 酸 |\n| :---: | :---: | :---: | :---: | :---: |\n| 突变品系 | CGT 丙氨 <br> 酸 | GGT 脯氨 <br> 酸 | AAG 苯丙氨 <br> 酸 | TTA 天冬氨 <br> 酸 |\n\n\n| 氨基酸位置 | 227 | 228 | 229 | 230 |\n| :--- | :--- | :--- | :--- | :--- |\nA: 基因 pbsA 发生了碱基增添导致植株出现抗性\nB: 基因突变导致 228 号位的脯氨酸被丝氨酸取代\nC: 基因突变后第 228 号位氨基酸的密码子转变为 CCA\nD: 基因突变后，其子代出现抗性的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n苋菜的野生型品系对除草剂敏感, 突变品系对除草剂有抗性。某科研小组研究了苋菜叶绿体基因 pbsA 野生型品系和突变品系的部分 DNA 碱基序列和氨基酸位置, 结果如表所示。下列叙述正确的是（ ）\n\n| 野生型品系 | CGA 丙氨 <br> 酸 | AGT 丝氨 <br> 酸 | AAG 苯丙氨 <br> 酸 | TT 天冬氨 <br> 酸 |\n| :---: | :---: | :---: | :---: | :---: |\n| 突变品系 | CGT 丙氨 <br> 酸 | GGT 脯氨 <br> 酸 | AAG 苯丙氨 <br> 酸 | TTA 天冬氨 <br> 酸 |\n\n\n| 氨基酸位置 | 227 | 228 | 229 | 230 |\n| :--- | :--- | :--- | :--- | :--- |\n\nA: 基因 pbsA 发生了碱基增添导致植株出现抗性\nB: 基因突变导致 228 号位的脯氨酸被丝氨酸取代\nC: 基因突变后第 228 号位氨基酸的密码子转变为 CCA\nD: 基因突变后，其子代出现抗性的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1475",
        "problem": "The following graph depicts the results of a study where the number of tadpoles within a particular habitat (finite resources) were experimentally varied.\n\nNote that the numbers next to the lines are the numbers of individuals in each treatment.\n\n[figure1]\n\nThis figure provides support for:\nA: density-dependent population regulation\nB: density-independent population regulation\nC: neither of these density population effects\nD: both of these density population effects\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph depicts the results of a study where the number of tadpoles within a particular habitat (finite resources) were experimentally varied.\n\nNote that the numbers next to the lines are the numbers of individuals in each treatment.\n\n[figure1]\n\nThis figure provides support for:\n\nA: density-dependent population regulation\nB: density-independent population regulation\nC: neither of these density population effects\nD: both of these density population effects\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-07.jpg?height=602&width=1205&top_left_y=1561&top_left_x=266"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1347",
        "problem": "## BIRDS OF A FEATHER - EVOLUTIONARY RELATIONSHIPS AMONGST THE RATITES\n\n[figure1]\n\nOur national bird, the kiwi, is a ratite, a group of flightless birds that includes the emu and cassowary in Australia and New Guinea, the ostrich in Africa, and the rhea in South America. There are also two recently extinct groups that include the largest birds ever known: our own moa and the elephant birds from Madagascar who reached heights of up to $3 \\mathrm{~m}$. Ratites and tinamous (found in South America and weak fliers) belong to an ancestral group of birds called \"palaeognaths\" and are the sister group (closest relatives) to all other living birds, the \"neognaths\".\n\nThe evolutionary relationships within the ratites have been the subject of considerable research as these birds are believed to have originated through vicariant speciation driven by the continental breakup of the supercontinent Gondwana. Vicariant speciation is the process by which new species are formed from the separation of the original population into two or more populations by a geographic barrier. Researchers at the Australian Centre for Ancient DNA, and the Allan Wilson Centre for Molecular Ecology in New Zealand have recently published a study in Science which examines ancient DNA and clarifies ratite evolution.\n\nThe maps at right show the position of continents during the Late Cretaceous and Tertiary. Continental landmasses are coloured according to the order in which they broke away from the remaining Gondwanan landmass: Africa and Madagascar (dark gray) split 100 to 130 Million years ago (Ma), followed by New Zealand (red; 60 to $80 \\mathrm{Ma}$ ), then finally Australia, Antarctica, and South America (green; 30 to $50 \\mathrm{Ma})$.\n\nSource: Mitchell et al. Science 344, 898 (2014).\n\n[figure2]Scientists discover a \"new\" fossil palaeognath from Antarctica, dated at 60 Million years old. Under a vicariant speciation model you would expect this fossil to be most closely related to ratites from:\nA: Africa and Madagascar only.\nB: Australia and South America only.\nC: New Zealand only.\nD: Both $A$ and $B$.\nE: Both $\\mathrm{A}$ and $\\mathrm{C}$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## BIRDS OF A FEATHER - EVOLUTIONARY RELATIONSHIPS AMONGST THE RATITES\n\n[figure1]\n\nOur national bird, the kiwi, is a ratite, a group of flightless birds that includes the emu and cassowary in Australia and New Guinea, the ostrich in Africa, and the rhea in South America. There are also two recently extinct groups that include the largest birds ever known: our own moa and the elephant birds from Madagascar who reached heights of up to $3 \\mathrm{~m}$. Ratites and tinamous (found in South America and weak fliers) belong to an ancestral group of birds called \"palaeognaths\" and are the sister group (closest relatives) to all other living birds, the \"neognaths\".\n\nThe evolutionary relationships within the ratites have been the subject of considerable research as these birds are believed to have originated through vicariant speciation driven by the continental breakup of the supercontinent Gondwana. Vicariant speciation is the process by which new species are formed from the separation of the original population into two or more populations by a geographic barrier. Researchers at the Australian Centre for Ancient DNA, and the Allan Wilson Centre for Molecular Ecology in New Zealand have recently published a study in Science which examines ancient DNA and clarifies ratite evolution.\n\nThe maps at right show the position of continents during the Late Cretaceous and Tertiary. Continental landmasses are coloured according to the order in which they broke away from the remaining Gondwanan landmass: Africa and Madagascar (dark gray) split 100 to 130 Million years ago (Ma), followed by New Zealand (red; 60 to $80 \\mathrm{Ma}$ ), then finally Australia, Antarctica, and South America (green; 30 to $50 \\mathrm{Ma})$.\n\nSource: Mitchell et al. Science 344, 898 (2014).\n\n[figure2]\n\nproblem:\nScientists discover a \"new\" fossil palaeognath from Antarctica, dated at 60 Million years old. Under a vicariant speciation model you would expect this fossil to be most closely related to ratites from:\n\nA: Africa and Madagascar only.\nB: Australia and South America only.\nC: New Zealand only.\nD: Both $A$ and $B$.\nE: Both $\\mathrm{A}$ and $\\mathrm{C}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-27.jpg?height=759&width=559&top_left_y=246&top_left_x=126",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-27.jpg?height=1477&width=802&top_left_y=1072&top_left_x=1118"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_887",
        "problem": "科研小组将两个红色苂光蛋白基因随机整合到番茄体细胞的染色体上，出现如下三种情况, 若不考虑基因突变、交叉互换和染色体畸变, 分析下列相关叙述错误的是\n\n[图1]\n\n(1)\n\n[图2]\n\n(2)\n\n[图3]\n\n注: ・表示红色荧光蛋白基因\nA: 植株(3)产生的精细胞中最多有 2 个红色荧光蛋白基因\nB: 通过与非转基因植株杂交可以判断红色荧光蛋白基因的位置\nC: 有丝分裂后期含 4 个红色苂光蛋白基因的细胞只可来自植株(2)(3)\nD: 减数第二次分裂后期含 4 个红色苂光蛋白基因的细胞只可来自植株(1)(3)\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n科研小组将两个红色苂光蛋白基因随机整合到番茄体细胞的染色体上，出现如下三种情况, 若不考虑基因突变、交叉互换和染色体畸变, 分析下列相关叙述错误的是\n\n[图1]\n\n(1)\n\n[图2]\n\n(2)\n\n[图3]\n\n注: ・表示红色荧光蛋白基因\n\nA: 植株(3)产生的精细胞中最多有 2 个红色荧光蛋白基因\nB: 通过与非转基因植株杂交可以判断红色荧光蛋白基因的位置\nC: 有丝分裂后期含 4 个红色苂光蛋白基因的细胞只可来自植株(2)(3)\nD: 减数第二次分裂后期含 4 个红色苂光蛋白基因的细胞只可来自植株(1)(3)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-89.jpg?height=237&width=254&top_left_y=181&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-89.jpg?height=231&width=254&top_left_y=187&top_left_x=615",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-89.jpg?height=228&width=255&top_left_y=186&top_left_x=889"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_971",
        "problem": "Which of the following is an example of habituation?\nA: Chickadees learning new songs when they shift from living in small groups to living in large winter flocks.\nB: Lion cubs stalking and attacking litter mates\nC: Hydra initially contracting when gently touched, and then soon stopping to respond\nD: Golden plovers migrating yearly from Arctic breeding grounds to southeastern South America\nE: A mother eagle flying beneath her young to prevent them from falling\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is an example of habituation?\n\nA: Chickadees learning new songs when they shift from living in small groups to living in large winter flocks.\nB: Lion cubs stalking and attacking litter mates\nC: Hydra initially contracting when gently touched, and then soon stopping to respond\nD: Golden plovers migrating yearly from Arctic breeding grounds to southeastern South America\nE: A mother eagle flying beneath her young to prevent them from falling\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_682",
        "problem": "某种二倍体植物不含性染色体，但花的发育受等位基因 $G 、 g$ 的调控，当基因 $G$ 存在时发育为雄花序（雄株，仅有基因 $\\mathrm{g}$ 时发育为雌花序（雌株）。该植物的宽叶（M）对窄叶（m）为显性，上述两对基因独立遗传。下列相关说法错误的是（）\nA: 宽叶雄株的基因型有 2 种, 窄叶雌株的基因型有 1 种\nB: 若子代中既有宽叶又有窄叶，则父本的基因型有 4 种\nC: 窄叶雄株与杂合的宽叶雌株杂交, 子代中窄叶雄株占 $1 / 4$\nD: 两宽叶雌、雄植株杂交，子代中宽叶雌株与宽叶雄株所占比例相同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种二倍体植物不含性染色体，但花的发育受等位基因 $G 、 g$ 的调控，当基因 $G$ 存在时发育为雄花序（雄株，仅有基因 $\\mathrm{g}$ 时发育为雌花序（雌株）。该植物的宽叶（M）对窄叶（m）为显性，上述两对基因独立遗传。下列相关说法错误的是（）\n\nA: 宽叶雄株的基因型有 2 种, 窄叶雌株的基因型有 1 种\nB: 若子代中既有宽叶又有窄叶，则父本的基因型有 4 种\nC: 窄叶雄株与杂合的宽叶雌株杂交, 子代中窄叶雄株占 $1 / 4$\nD: 两宽叶雌、雄植株杂交，子代中宽叶雌株与宽叶雄株所占比例相同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_797",
        "problem": "如图为一种单基因遗传病的家系图（控制该病的致病基因不在 $\\mathrm{X}$ 染色体与 $\\mathrm{Y}$ 染色体的同源区段上), 在不考虑变异的情况下, 下列叙述正确的是\n\n[图1]\nA: 若 $\\mathrm{II}_{1}$ 体细胞中不含该病的致病基因, 则该病在人群中男性的发病率高于女性\nB: 若 $\\mathrm{I}_{2}$ 体细胞中含有一个该病的致病基因, 则 $\\mathrm{II}_{2}$ 与正常女性婚配后不可能生出患该病的男孩\nC: 若 $\\mathrm{I}_{1}$ 体细胞中不含该病的致病基因, $\\mathrm{II}_{1}$ 体细胞中含该病的致病基因, 则 $\\mathrm{II}_{2}$ 的外祖父不是该病的患者\nD: 若 $\\mathrm{I}_{1}$ 和 $\\mathrm{II}_{1}$ 体细胞中含该病的致病基因, 则该病为常染色体隐性遗传病\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图为一种单基因遗传病的家系图（控制该病的致病基因不在 $\\mathrm{X}$ 染色体与 $\\mathrm{Y}$ 染色体的同源区段上), 在不考虑变异的情况下, 下列叙述正确的是\n\n[图1]\n\nA: 若 $\\mathrm{II}_{1}$ 体细胞中不含该病的致病基因, 则该病在人群中男性的发病率高于女性\nB: 若 $\\mathrm{I}_{2}$ 体细胞中含有一个该病的致病基因, 则 $\\mathrm{II}_{2}$ 与正常女性婚配后不可能生出患该病的男孩\nC: 若 $\\mathrm{I}_{1}$ 体细胞中不含该病的致病基因, $\\mathrm{II}_{1}$ 体细胞中含该病的致病基因, 则 $\\mathrm{II}_{2}$ 的外祖父不是该病的患者\nD: 若 $\\mathrm{I}_{1}$ 和 $\\mathrm{II}_{1}$ 体细胞中含该病的致病基因, 则该病为常染色体隐性遗传病\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-036.jpg?height=191&width=445&top_left_y=567&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_753",
        "problem": "最新研究表明, 在黑色素瘤斑马鱼模型中找到了控制黑色素瘤形成的“必需基因”一ATAD2 基因。若敲除 ATAD2 基因, 即使存在诸多致癌突变, 细胞依旧无法癌变,而后添加 ATAD2 基因, 细胞则重新获得了癌变的能力。下列有关细胞癌变的叙述, 错误的是 ( )\nA: 细胞中的某些特定基因, 与多基因突变协同作用才能使正常细胞癌变\nB: 黑色素瘤的形成说明 ATAD2 基因可能直接或间接控制斑马鱼的性状\nC: 基因突变是癌细胞形成的关键驱动因素, 但并非发生突变的细胞就会癌变\nD: 一些理化因素导致 ATAD2 基因突变不断积累, 最终形成了癌细胞和肿瘤\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n最新研究表明, 在黑色素瘤斑马鱼模型中找到了控制黑色素瘤形成的“必需基因”一ATAD2 基因。若敲除 ATAD2 基因, 即使存在诸多致癌突变, 细胞依旧无法癌变,而后添加 ATAD2 基因, 细胞则重新获得了癌变的能力。下列有关细胞癌变的叙述, 错误的是 ( )\n\nA: 细胞中的某些特定基因, 与多基因突变协同作用才能使正常细胞癌变\nB: 黑色素瘤的形成说明 ATAD2 基因可能直接或间接控制斑马鱼的性状\nC: 基因突变是癌细胞形成的关键驱动因素, 但并非发生突变的细胞就会癌变\nD: 一些理化因素导致 ATAD2 基因突变不断积累, 最终形成了癌细胞和肿瘤\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_664",
        "problem": "人类颅面骨发育不全症是一种单基因显性遗传病, 携带致病基因的个体有一定概率不表现症状。下图是某家族该病的系谱图, 其中II-1 无致病基因。下列相关叙述错误的是 ( )\n\nII\n\nIII\n\n[图1]\n\n $\\square \\bigcirc$ 正常男女\n\n患病男女\nA: 该病致病基因位于常染色体上\nB: 个体 $\\mathrm{II}_{2}$ 和 $\\mathrm{II}_{3}$ 基因型可能相同\nC: 个体 $\\mathrm{II}_{2}$ 和 $\\mathrm{III} 4$ 均为杂合子\nD: 个体 III $_{2}$ 和正常男性婚配后代患病概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n人类颅面骨发育不全症是一种单基因显性遗传病, 携带致病基因的个体有一定概率不表现症状。下图是某家族该病的系谱图, 其中II-1 无致病基因。下列相关叙述错误的是 ( )\n\nII\n\nIII\n\n[图1]\n\n $\\square \\bigcirc$ 正常男女\n\n患病男女\n\nA: 该病致病基因位于常染色体上\nB: 个体 $\\mathrm{II}_{2}$ 和 $\\mathrm{II}_{3}$ 基因型可能相同\nC: 个体 $\\mathrm{II}_{2}$ 和 $\\mathrm{III} 4$ 均为杂合子\nD: 个体 III $_{2}$ 和正常男性婚配后代患病概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-090.jpg?height=539&width=688&top_left_y=2209&top_left_x=473"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_567",
        "problem": "已知果蝇的长翅和截翅由一对等位基因控制。多只长翅果蝇进行单对交配(每个瓶中有 1 只雌果蝇和 1 只雄果蝇), 子代果蝇中长翅 : 截翅 $=3: 1$ 。据此无法判断的是 ( )\nA: 长翅是显性性状还是隐性性状\nB: 亲代雌蝇是杂合子还是纯合子\nC: 该等位基因位于常染色体还是 X染色体上\nD: 该等位基因在雌蝇体细胞中是否成对存在\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知果蝇的长翅和截翅由一对等位基因控制。多只长翅果蝇进行单对交配(每个瓶中有 1 只雌果蝇和 1 只雄果蝇), 子代果蝇中长翅 : 截翅 $=3: 1$ 。据此无法判断的是 ( )\n\nA: 长翅是显性性状还是隐性性状\nB: 亲代雌蝇是杂合子还是纯合子\nC: 该等位基因位于常染色体还是 X染色体上\nD: 该等位基因在雌蝇体细胞中是否成对存在\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1004",
        "problem": "One of your classmates has symptoms of protein deficiency. You suspect that she might have a mineral deficiency due to her diet of nothing but canned spaghetti. Which of the following minerals should you test for first?\nA: Iron\nB: Copper\nC: Manganese\nD: Molybdenum\nE: Sulfur\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOne of your classmates has symptoms of protein deficiency. You suspect that she might have a mineral deficiency due to her diet of nothing but canned spaghetti. Which of the following minerals should you test for first?\n\nA: Iron\nB: Copper\nC: Manganese\nD: Molybdenum\nE: Sulfur\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_295",
        "problem": "It is important that you learn how to \"read\" a phylogenetic tree correctly. \n[figure1]\nA: Clade (a) and the clade containing (b), (c), (d) and (e) originated at the same time.\nB: The degree of similarity among extant organisms of taxa (b) and (c) is larger than that between (c) and (d).\nC: Taxon (b) is more closely related to taxon (a) than to taxon (e).\nD: The lineage leading to taxon (a) was the first to diverge from the other lineages.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIt is important that you learn how to \"read\" a phylogenetic tree correctly. \n[figure1]\n\nA: Clade (a) and the clade containing (b), (c), (d) and (e) originated at the same time.\nB: The degree of similarity among extant organisms of taxa (b) and (c) is larger than that between (c) and (d).\nC: Taxon (b) is more closely related to taxon (a) than to taxon (e).\nD: The lineage leading to taxon (a) was the first to diverge from the other lineages.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-104.jpg?height=411&width=899&top_left_y=640&top_left_x=721"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_820",
        "problem": "重叠基因是指两个或两个以上的基因共有一段 DNA 序列。感染大肠杆菌的 $\\varphi X 174$噬菌体的遗传物质为环状单链 DNA, 含有 6870 个碱基,其中 $A 、 G 、 T 、 C$ 四种碱基比例依次为 $20 \\% 、 10 \\% 、 30 \\% 、 40 \\%$ 。图甲表示该噬菌体复制的部分过程，(1) (4)表示相应生理过程。图乙表示该噬菌体部分 DNA 的碱基排列顺序(图中数字表示对应氨基酸的编号)。下列选项中说法错误的是（ ）\n\n[图1]\nA: 图甲中发生碱基互补配对的过程有(1)(2)(3)(4)\nB: 完成过程 (1)需要消耗 2061 个含 A\nC: 据图乙推断, D 基因和 $\\mathrm{E}$ 基因的重叠部分指导合成的氨基酸序列相同\nD: 重叠基因在遗传学上的意义是可增大遗传信息\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n重叠基因是指两个或两个以上的基因共有一段 DNA 序列。感染大肠杆菌的 $\\varphi X 174$噬菌体的遗传物质为环状单链 DNA, 含有 6870 个碱基,其中 $A 、 G 、 T 、 C$ 四种碱基比例依次为 $20 \\% 、 10 \\% 、 30 \\% 、 40 \\%$ 。图甲表示该噬菌体复制的部分过程，(1) (4)表示相应生理过程。图乙表示该噬菌体部分 DNA 的碱基排列顺序(图中数字表示对应氨基酸的编号)。下列选项中说法错误的是（ ）\n\n[图1]\n\nA: 图甲中发生碱基互补配对的过程有(1)(2)(3)(4)\nB: 完成过程 (1)需要消耗 2061 个含 A\nC: 据图乙推断, D 基因和 $\\mathrm{E}$ 基因的重叠部分指导合成的氨基酸序列相同\nD: 重叠基因在遗传学上的意义是可增大遗传信息\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-47.jpg?height=784&width=1467&top_left_y=1136&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1079",
        "problem": "A $5 \\mathrm{~kb}$ circular plasmid has three EcoRI sites at positions 500,3200 and 4100 as shown below. Numbering starts from the nucleotide denoted by 1.\n\n[figure1]\n\nThis plasmid was partially digested with EcoRI. Partial digestion means that only some of the sites are cut by the enzyme. Assume that all plasmids are cut at least at one site.\n\nThe total number of bands visible in the gel after the restriction digestion will be:",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nA $5 \\mathrm{~kb}$ circular plasmid has three EcoRI sites at positions 500,3200 and 4100 as shown below. Numbering starts from the nucleotide denoted by 1.\n\n[figure1]\n\nThis plasmid was partially digested with EcoRI. Partial digestion means that only some of the sites are cut by the enzyme. Assume that all plasmids are cut at least at one site.\n\nThe total number of bands visible in the gel after the restriction digestion will be:\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-16.jpg?height=433&width=563&top_left_y=2074&top_left_x=735"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_363",
        "problem": "基因 $\\mathrm{A} / \\mathrm{a}$ 控制果蝇的长翅与残翅, 且长翅对残翅为显性, 基因 $\\mathrm{D} / \\mathrm{d}$ 控制果蝇的刚毛与截毛, 刚毛对截毛为显性, 两对等位基因独立遗传。研究人员以某只雄果蝇为材料,用以上 2 对等位基因为引物, 以单个精子的 DNA 为模板进行 PCR 扩增后, 对其产物进行电泳分析, 结果如图所示。研究人员让该只雄果蝇与残翅截毛雌果蝇杂交, $F_{1}$ 中雌、雄果蝇均表现为长翅: 残翅 $=1$ ： 1 , 但雌果蝇全表现为截毛, 雄果蝇全表现为刚毛。不考虑突变和染色体互换，下列分析正确的是（）\n\n## 精子的基因型\n\n[图1]\nA: 该只雄果蝇的表型是长翅刚毛，其细胞中不含有基因 d\nB: 两对等位基因的遗传遵循自由组合定律，且基因 $\\mathrm{D} / \\mathrm{d}$ 位于 $\\mathrm{X}$ 染色体上\nC: 只考虑基因 $\\mathrm{D} / \\mathrm{d}$, 本实验中的雄果蝇的基因型均相同, 雌果蝇的基因型均相同\nD: $F_{1}$ 雌、雄果蝇随机交配, $F_{2}$ 中雄果蝇均表现为刚毛, 雌果蝇表现为刚毛或截毛\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n基因 $\\mathrm{A} / \\mathrm{a}$ 控制果蝇的长翅与残翅, 且长翅对残翅为显性, 基因 $\\mathrm{D} / \\mathrm{d}$ 控制果蝇的刚毛与截毛, 刚毛对截毛为显性, 两对等位基因独立遗传。研究人员以某只雄果蝇为材料,用以上 2 对等位基因为引物, 以单个精子的 DNA 为模板进行 PCR 扩增后, 对其产物进行电泳分析, 结果如图所示。研究人员让该只雄果蝇与残翅截毛雌果蝇杂交, $F_{1}$ 中雌、雄果蝇均表现为长翅: 残翅 $=1$ ： 1 , 但雌果蝇全表现为截毛, 雄果蝇全表现为刚毛。不考虑突变和染色体互换，下列分析正确的是（）\n\n## 精子的基因型\n\n[图1]\n\nA: 该只雄果蝇的表型是长翅刚毛，其细胞中不含有基因 d\nB: 两对等位基因的遗传遵循自由组合定律，且基因 $\\mathrm{D} / \\mathrm{d}$ 位于 $\\mathrm{X}$ 染色体上\nC: 只考虑基因 $\\mathrm{D} / \\mathrm{d}$, 本实验中的雄果蝇的基因型均相同, 雌果蝇的基因型均相同\nD: $F_{1}$ 雌、雄果蝇随机交配, $F_{2}$ 中雄果蝇均表现为刚毛, 雌果蝇表现为刚毛或截毛\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-71.jpg?height=328&width=871&top_left_y=1909&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_414",
        "problem": "某作物的穗分枝正常和穗分枝减少由两对等位基因 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 共同控制，两对基因独立遗传。基因型为 $\\mathrm{AaBb}$ 的个体自交，后代中穗分枝正常：穗分枝减少 $=15$ ：1。现利用甲（AABB）、乙 ( aaBb)、丙（aabb）三种植株进行杂交实验。下列相关推测错误的是 ( )\nA: 穗分枝正常的基因型有 8 种， aabb 植株的表型为穗分枝减少\nB: 甲与乙或甲与丙杂交，两组后代植株的表型均为穗分枝正常\nC: 乙与丙杂交, $F_{1}$ 自交所得 $F_{2}$ 中穗分枝正常: 穗分枝减少 $=3: 5$\nD: 甲与乙杂交, $F_{1}$ 自交所得 $F_{2}$ 中穗分枝正常植株中纯合子占比为 $12 / 31$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某作物的穗分枝正常和穗分枝减少由两对等位基因 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 共同控制，两对基因独立遗传。基因型为 $\\mathrm{AaBb}$ 的个体自交，后代中穗分枝正常：穗分枝减少 $=15$ ：1。现利用甲（AABB）、乙 ( aaBb)、丙（aabb）三种植株进行杂交实验。下列相关推测错误的是 ( )\n\nA: 穗分枝正常的基因型有 8 种， aabb 植株的表型为穗分枝减少\nB: 甲与乙或甲与丙杂交，两组后代植株的表型均为穗分枝正常\nC: 乙与丙杂交, $F_{1}$ 自交所得 $F_{2}$ 中穗分枝正常: 穗分枝减少 $=3: 5$\nD: 甲与乙杂交, $F_{1}$ 自交所得 $F_{2}$ 中穗分枝正常植株中纯合子占比为 $12 / 31$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1288",
        "problem": "Which of the following factors is likely to increase species diversity in a forest community?\nA: Moderate levels of disturbance\nB: Stable conditions with low disturbance\nC: Human protection of the community to prevent disturbance\nD: Intensive human disturbance\nE: Frequent natural disturbance such as fires and floods\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following factors is likely to increase species diversity in a forest community?\n\nA: Moderate levels of disturbance\nB: Stable conditions with low disturbance\nC: Human protection of the community to prevent disturbance\nD: Intensive human disturbance\nE: Frequent natural disturbance such as fires and floods\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_281",
        "problem": "In an experimental study on magpies Pica pica, conducted in Sweden, Goran Högstedt manipulated the original clutch size (shown on the right of the graph) of experimental birds to generate a number of different clutch sizes for each category of bird, as shown in the X-axis. The number of young fledged successfully by the birds under these different conditions is indicated by the $\\mathrm{Y}$-axis. Food abundance and quality of territory are thought to be associated with clutch size. Predation is lower in large clutches.\n\n[figure1]\nA: The birds, in general, did better with experimentally manipulated, larger broods.\nB: The reproductive rate of birds is closest to that which maximises individual breeding success.\nC: Birds in high quality territory tend to have larger clutches.\nD: Experimentally-manipulated clutches experience higher starvation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn an experimental study on magpies Pica pica, conducted in Sweden, Goran Högstedt manipulated the original clutch size (shown on the right of the graph) of experimental birds to generate a number of different clutch sizes for each category of bird, as shown in the X-axis. The number of young fledged successfully by the birds under these different conditions is indicated by the $\\mathrm{Y}$-axis. Food abundance and quality of territory are thought to be associated with clutch size. Predation is lower in large clutches.\n\n[figure1]\n\nA: The birds, in general, did better with experimentally manipulated, larger broods.\nB: The reproductive rate of birds is closest to that which maximises individual breeding success.\nC: Birds in high quality territory tend to have larger clutches.\nD: Experimentally-manipulated clutches experience higher starvation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-072.jpg?height=879&width=1190&top_left_y=875&top_left_x=401"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_759",
        "problem": "有丝分裂的核心事件是染色体分离, 需满足姐妹染色单体分离的两个条件之一是连接姐妹染色单体的黏连蛋白降解。当“黏连”一对姐妹染色单体的黏连蛋白被分离酶切割后, 姐妹染色单体分开, 成为两条染色体 (如图所示)。研究发现, PATRONUS 蛋白是分离酶抑制剂。下列相关叙述错误的是（）\n\n[图1]\nA: PATRONUS 蛋白突变体中, 姐妹染色单体间的黏连蛋白可能提前降解\nB: 秋水仙素和 PATRONUS 蛋白的作用机理相似, 都能使分裂的细胞停留在特定 时期\nC: 分离酶发挥作用的时期为有丝分裂后期和减数分裂 II 后期\nD: 姐妹染色单体在黏连蛋白“黏连”阶段, 细胞中染色体数: 染色单体数: 核 DNA 数 $=1: 2: 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n有丝分裂的核心事件是染色体分离, 需满足姐妹染色单体分离的两个条件之一是连接姐妹染色单体的黏连蛋白降解。当“黏连”一对姐妹染色单体的黏连蛋白被分离酶切割后, 姐妹染色单体分开, 成为两条染色体 (如图所示)。研究发现, PATRONUS 蛋白是分离酶抑制剂。下列相关叙述错误的是（）\n\n[图1]\n\nA: PATRONUS 蛋白突变体中, 姐妹染色单体间的黏连蛋白可能提前降解\nB: 秋水仙素和 PATRONUS 蛋白的作用机理相似, 都能使分裂的细胞停留在特定 时期\nC: 分离酶发挥作用的时期为有丝分裂后期和减数分裂 II 后期\nD: 姐妹染色单体在黏连蛋白“黏连”阶段, 细胞中染色体数: 染色单体数: 核 DNA 数 $=1: 2: 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-085.jpg?height=285&width=649&top_left_y=2316&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_316",
        "problem": "豌豆子叶的颜色黄色 (A) 对绿色 (a) 为显性, 种子的形状圆粒 (B) 对皱粒 (b)为显性, 两对基因 位于两对染色体上。现有一批黄色圆粒种子进行测交，子代表现型及比例见下表。则这批种子的基因型及 比例为\n\n| 亲本 | 交配方式 | 子 代 |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  |  | 黄色圆粒 | 黄色皱粒 | 绿色圆粒 | 绿色皱粒 |\n| 黄色圆粒 | 测交 | 4978 | 5106 | 1031 | 989 |\n|  |  |  |  |  |  |\nA: 全为 $\\mathrm{AaBb}$\nB: $\\mathrm{AaBb}: \\mathrm{AABb}=1: 2$\nC: $\\mathrm{AaBb}: \\mathrm{AABb}=1: 3$\nD: $\\mathrm{AaBb}: \\mathrm{AABb}=1: 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n豌豆子叶的颜色黄色 (A) 对绿色 (a) 为显性, 种子的形状圆粒 (B) 对皱粒 (b)为显性, 两对基因 位于两对染色体上。现有一批黄色圆粒种子进行测交，子代表现型及比例见下表。则这批种子的基因型及 比例为\n\n| 亲本 | 交配方式 | 子 代 |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  |  | 黄色圆粒 | 黄色皱粒 | 绿色圆粒 | 绿色皱粒 |\n| 黄色圆粒 | 测交 | 4978 | 5106 | 1031 | 989 |\n|  |  |  |  |  |  |\n\nA: 全为 $\\mathrm{AaBb}$\nB: $\\mathrm{AaBb}: \\mathrm{AABb}=1: 2$\nC: $\\mathrm{AaBb}: \\mathrm{AABb}=1: 3$\nD: $\\mathrm{AaBb}: \\mathrm{AABb}=1: 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_946",
        "problem": "已知家兔中 $\\mathrm{C}$ 基因控制青色, $\\mathrm{c}$ 基因控制白化性状, $\\mathrm{L} / 1$ 基因决定毛的长短, 现选择青色短毛与白化长毛兔杂交, $F_{1}$ 均为短毛兔, 青色与白化兔的比例为 $1: 1$, 不同类型的兔中雌雄比例相等, 选择 $F_{1}$ 中不同类型的兔进行交配, 子代数量足够多且两对基因独立遗传。下列关于子代的分析正确的是（）\nA: 亲代中控制毛长短的基因均纯合且短毛为显性性状, $F_{1}$ 的个体中一半为纯合子\nB: $F_{1}$ 中所有个体自由交配, $F_{2}$ 中能稳定遗传的个体有四种表型且所占比例为 $5 / 16$\nC: 选择 $F_{1}$ 青色短毛兔个体自由交配, $F_{2}$ 白化兔共有 3 种基因型, 其中 $1 / 4$ 可稳定遗传\nD: 选择白化长毛兔与 $\\mathrm{F}_{2}$ 中任何类型的兔杂交, 均可判断其产生的配子的数量以及比例\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知家兔中 $\\mathrm{C}$ 基因控制青色, $\\mathrm{c}$ 基因控制白化性状, $\\mathrm{L} / 1$ 基因决定毛的长短, 现选择青色短毛与白化长毛兔杂交, $F_{1}$ 均为短毛兔, 青色与白化兔的比例为 $1: 1$, 不同类型的兔中雌雄比例相等, 选择 $F_{1}$ 中不同类型的兔进行交配, 子代数量足够多且两对基因独立遗传。下列关于子代的分析正确的是（）\n\nA: 亲代中控制毛长短的基因均纯合且短毛为显性性状, $F_{1}$ 的个体中一半为纯合子\nB: $F_{1}$ 中所有个体自由交配, $F_{2}$ 中能稳定遗传的个体有四种表型且所占比例为 $5 / 16$\nC: 选择 $F_{1}$ 青色短毛兔个体自由交配, $F_{2}$ 白化兔共有 3 种基因型, 其中 $1 / 4$ 可稳定遗传\nD: 选择白化长毛兔与 $\\mathrm{F}_{2}$ 中任何类型的兔杂交, 均可判断其产生的配子的数量以及比例\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_496",
        "problem": "某 ZW 型二倍体昆虫的裂翅 (A) 对正常翅 (a) 为性, 科研人员将一个 sDNA 导入实验一中 $F_{1}$ 裂翅雄性个体的细胞, 获得转基因克隆昆虫, 再将上述转基因昆虫与 $F_{1}$正常翅雌性杂交。通过 sDNA 具体的插入位置分析后代表型，下列说法错误的是（）\n\n## 实验二\n\n[图1]\n\n注：SDNA不控制具体性状，但会抑制A基因的表达，使个体表现为正常翅, a基因的表达不受影响。若受精卵无完整的A或a基因, 胚胎致死。\nA: 若 sDNA 导入的染色体上不含 A 或 $\\mathrm{a}$ 基因, 则子代裂翅和正常翅的比例为 $1: 3$\nB: 若 sDNA 的导入破坏了 $\\mathrm{A}$ 基因, 则子代雌雄比例为 $1: 2$\nC: 若 sDNA 的导入破坏了 $\\mathrm{a}$ 基因, 则子代裂翅和正常翅的比例为 $2: 1$\nD: 若 $\\mathrm{sDNA}$ 导入 $\\mathrm{A}$ 基因所在的染色体，但不破坏 $\\mathrm{A}$ 基因, 则子代裂翅和正常翅的比例为 $1: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某 ZW 型二倍体昆虫的裂翅 (A) 对正常翅 (a) 为性, 科研人员将一个 sDNA 导入实验一中 $F_{1}$ 裂翅雄性个体的细胞, 获得转基因克隆昆虫, 再将上述转基因昆虫与 $F_{1}$正常翅雌性杂交。通过 sDNA 具体的插入位置分析后代表型，下列说法错误的是（）\n\n## 实验二\n\n[图1]\n\n注：SDNA不控制具体性状，但会抑制A基因的表达，使个体表现为正常翅, a基因的表达不受影响。若受精卵无完整的A或a基因, 胚胎致死。\n\nA: 若 sDNA 导入的染色体上不含 A 或 $\\mathrm{a}$ 基因, 则子代裂翅和正常翅的比例为 $1: 3$\nB: 若 sDNA 的导入破坏了 $\\mathrm{A}$ 基因, 则子代雌雄比例为 $1: 2$\nC: 若 sDNA 的导入破坏了 $\\mathrm{a}$ 基因, 则子代裂翅和正常翅的比例为 $2: 1$\nD: 若 $\\mathrm{sDNA}$ 导入 $\\mathrm{A}$ 基因所在的染色体，但不破坏 $\\mathrm{A}$ 基因, 则子代裂翅和正常翅的比例为 $1: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-35.jpg?height=191&width=531&top_left_y=681&top_left_x=480"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1054",
        "problem": "During blood circulation, the amount of blood reaching every organ will depend on several factors such as total weight of the organ, the type of metabolic function it performs etc. When body is at rest, the percentage of total blood flowing to different organs will vary. Which of the following is correct?\nA: At rest, the blood supply to skin in cool weather will be greater than in hot weather.\nB: Flow of blood $(\\mathrm{ml} / \\mathrm{min})$ per unit gram of tissue will be greater to kidneys than to the brain, at rest.\nC: At rest, flow of blood ( $\\mathrm{ml} / \\mathrm{min})$ to liver tissue will be much greater than heart.\nD: When a person is at rest, a small percent ( $<5 \\%$ ) blood will be found in bronchi.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nDuring blood circulation, the amount of blood reaching every organ will depend on several factors such as total weight of the organ, the type of metabolic function it performs etc. When body is at rest, the percentage of total blood flowing to different organs will vary. Which of the following is correct?\n\nA: At rest, the blood supply to skin in cool weather will be greater than in hot weather.\nB: Flow of blood $(\\mathrm{ml} / \\mathrm{min})$ per unit gram of tissue will be greater to kidneys than to the brain, at rest.\nC: At rest, flow of blood ( $\\mathrm{ml} / \\mathrm{min})$ to liver tissue will be much greater than heart.\nD: When a person is at rest, a small percent ( $<5 \\%$ ) blood will be found in bronchi.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_497",
        "problem": "下图所示的五个（1 5) 家庭所患遗传病类型各不相同, 且致病基因均位于性染色体上。下列有关说法正确的是（）\n\n[图1]\nA: 1 号家庭所患遗传病，在人群中女性发病率大于男性\nB: 2 号家庭遗传病的致病基因应位于 X、Y 染色体同源区段, 此区段基因所控制的遗传与性别无关\nC: 从优生角度考虑, 3 号家庭应优先生男孩, 4 号家庭相反\nD: 控制 5 号家庭遗传病的基因用 $\\mathrm{B} 、 \\mathrm{~b}$ 表示，则该夫妇的基因型是 $\\mathrm{X}^{\\mathrm{b}} Y^{\\mathrm{B}} 、 \\mathrm{X}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图所示的五个（1 5) 家庭所患遗传病类型各不相同, 且致病基因均位于性染色体上。下列有关说法正确的是（）\n\n[图1]\n\nA: 1 号家庭所患遗传病，在人群中女性发病率大于男性\nB: 2 号家庭遗传病的致病基因应位于 X、Y 染色体同源区段, 此区段基因所控制的遗传与性别无关\nC: 从优生角度考虑, 3 号家庭应优先生男孩, 4 号家庭相反\nD: 控制 5 号家庭遗传病的基因用 $\\mathrm{B} 、 \\mathrm{~b}$ 表示，则该夫妇的基因型是 $\\mathrm{X}^{\\mathrm{b}} Y^{\\mathrm{B}} 、 \\mathrm{X}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-045.jpg?height=243&width=1082&top_left_y=1152&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_319",
        "problem": "某家系甲病和乙病的系谱图如图所示。已知两病独立遗传, 各由一对等位基因控制,且基因不位于 Y 染色体。下列叙述错误的是（ ）\n\n[图1]\nA: 甲病的遗传方式为常染色体隐性遗传\nB: 乙病的遗传方式可能为常染色体显性遗传\nC: 乙病的遗传方式可能为伴 X 染色体显性遗传\nD: 同时考虑两种病, $\\mathrm{III}_{3}$ 个体是纯合子的概率为 $1 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某家系甲病和乙病的系谱图如图所示。已知两病独立遗传, 各由一对等位基因控制,且基因不位于 Y 染色体。下列叙述错误的是（ ）\n\n[图1]\n\nA: 甲病的遗传方式为常染色体隐性遗传\nB: 乙病的遗传方式可能为常染色体显性遗传\nC: 乙病的遗传方式可能为伴 X 染色体显性遗传\nD: 同时考虑两种病, $\\mathrm{III}_{3}$ 个体是纯合子的概率为 $1 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-30.jpg?height=391&width=1100&top_left_y=827&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1306",
        "problem": "An investigator developed a scoring system that enabled her to predict an individual's body mass index (BMI) based on information about what they ate and how much. Information is collected from a small sample of subjects in order to compute their \"diet score,\" and the weight and height of each subject is measured in order to compute their BMI. The graph on the left shows the relationship between the new \"diet score\" and BMI, and it suggests that the \"diet score\" is not a very good predictor of BMI, (i.e. there is little if any association between the two). She then identified the age and gender of the subjects and these data are presented in the graph on the right.\n[figure1]\n\nhttp://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/BS704-EP713 MultivariableMethods/BS704-EP713 MultivariableMethods print.html\n\nConsidering these data what is the best conclusion about the relationship between \"diet score\" and BMI?\nA: There is no clear association between the two variables.\nB: Age is confounding the association between diet score and BMI.\nC: Gender is confounding the association between diet score and BMI.\nD: Both age and gender have an effect on BMI.\nE: Diet, age, and gender each have an independent effect on BMI.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn investigator developed a scoring system that enabled her to predict an individual's body mass index (BMI) based on information about what they ate and how much. Information is collected from a small sample of subjects in order to compute their \"diet score,\" and the weight and height of each subject is measured in order to compute their BMI. The graph on the left shows the relationship between the new \"diet score\" and BMI, and it suggests that the \"diet score\" is not a very good predictor of BMI, (i.e. there is little if any association between the two). She then identified the age and gender of the subjects and these data are presented in the graph on the right.\n[figure1]\n\nhttp://sphweb.bumc.bu.edu/otlt/MPH-Modules/BS/BS704-EP713 MultivariableMethods/BS704-EP713 MultivariableMethods print.html\n\nConsidering these data what is the best conclusion about the relationship between \"diet score\" and BMI?\n\nA: There is no clear association between the two variables.\nB: Age is confounding the association between diet score and BMI.\nC: Gender is confounding the association between diet score and BMI.\nD: Both age and gender have an effect on BMI.\nE: Diet, age, and gender each have an independent effect on BMI.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-11.jpg?height=572&width=1674&top_left_y=546&top_left_x=178"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_646",
        "problem": "家猫体色由 $\\mathrm{X}$ 染色体上一对等位基因 $\\mathrm{B} 、 \\mathrm{~b}$ 控制，只含基因 $\\mathrm{B}$ 的个体为黑猫，只含基因 $\\mathrm{b}$ 的个体为黄猫，其他个体为玳瑁猫，下列说法正确的是（）\nA: 玳瑁猫互交的后代中有 $25 \\%$ 雄性黄猫\nB: 玳瑁猫与黄猫杂交后代中玳瑁猫占 $50 \\%$\nC: 为持续高效地繁育玳瑁猫, 应逐代淘汰其他体色的猫\nD: 只有用黑猫和黄猫杂交，才能获得最大比例的玳琩猫\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n家猫体色由 $\\mathrm{X}$ 染色体上一对等位基因 $\\mathrm{B} 、 \\mathrm{~b}$ 控制，只含基因 $\\mathrm{B}$ 的个体为黑猫，只含基因 $\\mathrm{b}$ 的个体为黄猫，其他个体为玳瑁猫，下列说法正确的是（）\n\nA: 玳瑁猫互交的后代中有 $25 \\%$ 雄性黄猫\nB: 玳瑁猫与黄猫杂交后代中玳瑁猫占 $50 \\%$\nC: 为持续高效地繁育玳瑁猫, 应逐代淘汰其他体色的猫\nD: 只有用黑猫和黄猫杂交，才能获得最大比例的玳琩猫\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_857",
        "problem": "某昆虫体色的黄色对黑色为显性, 翅形的长翅对残翅为显性, 两种性状受两对独立遗传的等位基因控制。现有两纯合亲本杂交得 $F_{1}, F_{1}$ 雌雄个体间相互交配得 $F_{2}, F_{2}$ 的表现型及比例为黄色长翅:黄色残翅:黑色长翅:黑色残翅 $=2: 3: 3: 1$, 下列相关分析, 正确的是 ( )\nA: 该昆虫种群中存在控制黄色和长翅的基因纯合致死现象\nB: $F_{1}$ 产生的具有受精能力的雌、雄配子的种类不同\nC: $\\mathrm{F}_{2}$ 个体存在 5 种基因型, 其中纯合子所占比例为 $1 / 3$\nD: $F_{2}$ 黄色长翅个体与黑色残翅个体杂交后代有 3 种表现型, 比例为 $1: 1: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某昆虫体色的黄色对黑色为显性, 翅形的长翅对残翅为显性, 两种性状受两对独立遗传的等位基因控制。现有两纯合亲本杂交得 $F_{1}, F_{1}$ 雌雄个体间相互交配得 $F_{2}, F_{2}$ 的表现型及比例为黄色长翅:黄色残翅:黑色长翅:黑色残翅 $=2: 3: 3: 1$, 下列相关分析, 正确的是 ( )\n\nA: 该昆虫种群中存在控制黄色和长翅的基因纯合致死现象\nB: $F_{1}$ 产生的具有受精能力的雌、雄配子的种类不同\nC: $\\mathrm{F}_{2}$ 个体存在 5 种基因型, 其中纯合子所占比例为 $1 / 3$\nD: $F_{2}$ 黄色长翅个体与黑色残翅个体杂交后代有 3 种表现型, 比例为 $1: 1: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_492",
        "problem": "某昆虫（性别决定为 XY 型）的眼色由两对等位基因控制。现有一只白眼雄昆虫与纯合红眼雌昆虫杂交, 后代雄雄个体均为红眼, 让子一代雄雄个体相互交配, 子二代红眼: 朱砂眼: 白眼=12:3:1, 且子二代中的白眼和朱砂眼个体均为雄性。下列相关分析错误的是 ( )\nA: 该昆虫眼色的遗传与 X 染色体上的等位基因有关\nB: 子二代红眼雄昆虫中纯合子所占的比例为 $1 / 4$\nC: 若子二代出现一只白眼雌昆虫, 则是亲本产生配子时发生基因突变所致\nD: 子二代朱砂眼与子一代红眼尾虫回交，后代的性状分离比为 $6: 5: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某昆虫（性别决定为 XY 型）的眼色由两对等位基因控制。现有一只白眼雄昆虫与纯合红眼雌昆虫杂交, 后代雄雄个体均为红眼, 让子一代雄雄个体相互交配, 子二代红眼: 朱砂眼: 白眼=12:3:1, 且子二代中的白眼和朱砂眼个体均为雄性。下列相关分析错误的是 ( )\n\nA: 该昆虫眼色的遗传与 X 染色体上的等位基因有关\nB: 子二代红眼雄昆虫中纯合子所占的比例为 $1 / 4$\nC: 若子二代出现一只白眼雌昆虫, 则是亲本产生配子时发生基因突变所致\nD: 子二代朱砂眼与子一代红眼尾虫回交，后代的性状分离比为 $6: 5: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_535",
        "problem": "在某严格自花传粉的二倍体植物中, 发现甲、乙两类矮生突变体 (如图所示), 矮化植株无 A 基因, 矮化程度与 $\\mathrm{a}$ 基因的数量呈正相关。丙为花粉不育突变体, 含 $\\mathrm{b}$ 基因的花粉败育。甲、乙、丙均为纯合体。下列叙述错误的是（）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\nA: 甲类变异属于基因突变，乙类变异是在甲类变异的基础上发生了染色体结构变异\nB: 乙减数分裂产生 2 种花粉, 在分裂中期, 一个次级精母细胞最多带有 4 个 $\\mathrm{a}$ 基因\nC: 乙的自交后代中, $\\mathrm{F}_{1}$ 有 3 种矮化类型, 植株矮化程度由低到高, 数量比为 1 : 2: 1\nD: 若 $\\mathrm{a}$ 与 $\\mathrm{b}$ 位于同一对染色体上, 丙 $(\\rho) \\times$ 甲 $\\left(\\mathrm{O}^{\\mathrm{O}}\\right)$ 得 $\\mathrm{F}_{1}, \\mathrm{~F}_{1}$ 自交后代中只有矮生类型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在某严格自花传粉的二倍体植物中, 发现甲、乙两类矮生突变体 (如图所示), 矮化植株无 A 基因, 矮化程度与 $\\mathrm{a}$ 基因的数量呈正相关。丙为花粉不育突变体, 含 $\\mathrm{b}$ 基因的花粉败育。甲、乙、丙均为纯合体。下列叙述错误的是（）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\n\nA: 甲类变异属于基因突变，乙类变异是在甲类变异的基础上发生了染色体结构变异\nB: 乙减数分裂产生 2 种花粉, 在分裂中期, 一个次级精母细胞最多带有 4 个 $\\mathrm{a}$ 基因\nC: 乙的自交后代中, $\\mathrm{F}_{1}$ 有 3 种矮化类型, 植株矮化程度由低到高, 数量比为 1 : 2: 1\nD: 若 $\\mathrm{a}$ 与 $\\mathrm{b}$ 位于同一对染色体上, 丙 $(\\rho) \\times$ 甲 $\\left(\\mathrm{O}^{\\mathrm{O}}\\right)$ 得 $\\mathrm{F}_{1}, \\mathrm{~F}_{1}$ 自交后代中只有矮生类型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-87.jpg?height=251&width=283&top_left_y=203&top_left_x=361",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-87.jpg?height=254&width=302&top_left_y=204&top_left_x=700",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-87.jpg?height=251&width=297&top_left_y=203&top_left_x=1048"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1484",
        "problem": "Peaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nWhich results is expected if fuzz is produced by a single dominant allele?\nA: $\\mathrm{A}$\nB: $\\quad$ B\nC: $\\quad \\mathrm{C}$\nD: B and D\nE: $\\quad$ E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPeaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nWhich results is expected if fuzz is produced by a single dominant allele?\n\nA: $\\mathrm{A}$\nB: $\\quad$ B\nC: $\\quad \\mathrm{C}$\nD: B and D\nE: $\\quad$ E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-45.jpg?height=782&width=1231&top_left_y=474&top_left_x=241"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_572",
        "problem": "CIB 品系果蝇具有一条正常的 $\\mathrm{X}$ 染色体 $\\left(\\mathrm{X}^{+}\\right)$和一条含 CIB 区段的 $\\mathrm{X}$ 染色体 $\\left(\\mathrm{X}^{\\mathrm{CIB}}\\right)$,其中 $\\mathrm{C}$ 表示染色体上的倒位区; $\\mathrm{I}$ 为雄性果蝇胚胎致死的隐性基因; $\\mathrm{B}$ 为显性棒眼基因。为探究某射线诱变处理的正常眼雄果蝇是否发生 X 染色体新的隐性突变及该突变是否致死，设计了如下图所示的检测技术路线。不考虑 X 染色体非姐妹染色单体互换。下列分析正确的是（ ）\n\n[图1]\nA: 可以在自然界中找到基因型为 $\\mathrm{X}^{\\mathrm{CIB}} \\mathrm{X}^{\\mathrm{CIB}}$ 的棒眼雌果蝇\nB: CIB 品系果蝇 C 区发生的变异不可以在光学下显微镜观察到\nC: $F_{1}$ 群体 I 中棒眼果蝇占比为 $1 / 2$ 时说明亲本雄果蝇发生了隐性突变\nD: $F_{2}$ 中不出现雄果蝇可说明亲本雄果蝇 $\\mathrm{X}$ 染色体上出现隐性致死突变\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nCIB 品系果蝇具有一条正常的 $\\mathrm{X}$ 染色体 $\\left(\\mathrm{X}^{+}\\right)$和一条含 CIB 区段的 $\\mathrm{X}$ 染色体 $\\left(\\mathrm{X}^{\\mathrm{CIB}}\\right)$,其中 $\\mathrm{C}$ 表示染色体上的倒位区; $\\mathrm{I}$ 为雄性果蝇胚胎致死的隐性基因; $\\mathrm{B}$ 为显性棒眼基因。为探究某射线诱变处理的正常眼雄果蝇是否发生 X 染色体新的隐性突变及该突变是否致死，设计了如下图所示的检测技术路线。不考虑 X 染色体非姐妹染色单体互换。下列分析正确的是（ ）\n\n[图1]\n\nA: 可以在自然界中找到基因型为 $\\mathrm{X}^{\\mathrm{CIB}} \\mathrm{X}^{\\mathrm{CIB}}$ 的棒眼雌果蝇\nB: CIB 品系果蝇 C 区发生的变异不可以在光学下显微镜观察到\nC: $F_{1}$ 群体 I 中棒眼果蝇占比为 $1 / 2$ 时说明亲本雄果蝇发生了隐性突变\nD: $F_{2}$ 中不出现雄果蝇可说明亲本雄果蝇 $\\mathrm{X}$ 染色体上出现隐性致死突变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-19.jpg?height=560&width=1085&top_left_y=1393&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_424",
        "problem": "已知控制果蝇眼色和翅型的基因均位于 X 染色体上, 研究人员在纯合红眼果蝇品系中发现了一只缺刻翅变异的雌果蝇。研究人员进行如下杂交实验, 下列关于该性状出现原因的分析最可能是（ ）\n\n亲代 红眼缺刻翅 $9 \\times$ 白眼正常翅 $\\hat{3}$\n\n## 子代 白眼缺刻翅 9 红眼正常翅 9 红眼正常翅 $\\hat{\\delta}$\n\n1 : 1 : 1\nA: 控制眼色的基因与控制翅型的基因之间不会发生基因重组\nB: 只要控制翅型的基因发生了显性突变即可解释缺刻翅的出现\nC: 一条 X 染色体部分缺失的雌果蝇发育成缺刻翅, 且含该 $\\mathrm{X}$ 染色体的雄果蝇不能存活\nD: 若为 $\\mathrm{X}$ 染色体部分缺失导致, 则 $\\mathrm{X}$ 染色体缺失部位不包含眼色基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知控制果蝇眼色和翅型的基因均位于 X 染色体上, 研究人员在纯合红眼果蝇品系中发现了一只缺刻翅变异的雌果蝇。研究人员进行如下杂交实验, 下列关于该性状出现原因的分析最可能是（ ）\n\n亲代 红眼缺刻翅 $9 \\times$ 白眼正常翅 $\\hat{3}$\n\n## 子代 白眼缺刻翅 9 红眼正常翅 9 红眼正常翅 $\\hat{\\delta}$\n\n1 : 1 : 1\n\nA: 控制眼色的基因与控制翅型的基因之间不会发生基因重组\nB: 只要控制翅型的基因发生了显性突变即可解释缺刻翅的出现\nC: 一条 X 染色体部分缺失的雌果蝇发育成缺刻翅, 且含该 $\\mathrm{X}$ 染色体的雄果蝇不能存活\nD: 若为 $\\mathrm{X}$ 染色体部分缺失导致, 则 $\\mathrm{X}$ 染色体缺失部位不包含眼色基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_195",
        "problem": "Rubisco is an enzyme crucial for carbon fixation in plants. In addition to the predominant carboxylation reaction, this enzyme catalyzes an oxidation reaction as well. For an aquatic plant, the frequency of the oxidation reaction depends on the relative concentrations of the reagents $\\mathrm{CO}_{2}$ and $\\mathrm{O}_{2}$ in the aquatic solution, which in turn are coupled to temperature. The figures show the absolute (a) and relative (b) concentrations of $\\mathrm{CO}_{2}$ and $\\mathrm{O}_{2}$ dissolved in water that is at equilibrium with the atmosphere.\n[figure1]\n\nChoose the following statement that is correct.\nA: The frequency of the oxidation reaction decreases with increasing temperature.\nB: In water at equilibrium with the atmosphere, the relative concentration change with temperature of $\\mathrm{CO}_{2}$ is larger than of $\\mathrm{O}_{2}$.\nC: Rubisco has a higher affinity for $\\mathrm{O}_{2}$ than for $\\mathrm{CO}_{2}$.\nD: At a temperature of $90^{\\circ} \\mathrm{C}$, Rubisco catalyzes only one of the above two reactions in vascular plants.\nE: This sensitivity to temperature matters for submerged aquatic plants only.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRubisco is an enzyme crucial for carbon fixation in plants. In addition to the predominant carboxylation reaction, this enzyme catalyzes an oxidation reaction as well. For an aquatic plant, the frequency of the oxidation reaction depends on the relative concentrations of the reagents $\\mathrm{CO}_{2}$ and $\\mathrm{O}_{2}$ in the aquatic solution, which in turn are coupled to temperature. The figures show the absolute (a) and relative (b) concentrations of $\\mathrm{CO}_{2}$ and $\\mathrm{O}_{2}$ dissolved in water that is at equilibrium with the atmosphere.\n[figure1]\n\nChoose the following statement that is correct.\n\nA: The frequency of the oxidation reaction decreases with increasing temperature.\nB: In water at equilibrium with the atmosphere, the relative concentration change with temperature of $\\mathrm{CO}_{2}$ is larger than of $\\mathrm{O}_{2}$.\nC: Rubisco has a higher affinity for $\\mathrm{O}_{2}$ than for $\\mathrm{CO}_{2}$.\nD: At a temperature of $90^{\\circ} \\mathrm{C}$, Rubisco catalyzes only one of the above two reactions in vascular plants.\nE: This sensitivity to temperature matters for submerged aquatic plants only.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-18.jpg?height=424&width=1464&top_left_y=854&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_872",
        "problem": "用 ${ }^{15} \\mathrm{~N}$ 标记某基因型为 $\\mathrm{AaX} X^{\\mathrm{b}} \\mathrm{Y}$ 的动物精原细胞甲（2n=10）的所有 DNA 双链, 将甲放在不含 ${ }^{15} \\mathrm{~N}$ 的培养基中进行培养, 发现了一个含 XY 的异常精子, 若无其他突变和染色体互换发生，下列叙述正确的是（）\nA: 与异常精子同时产生的另外三个精子的基因型可能是 $A^{b}$ Y $、 A 、 a$\nB: 产生该异常精子的次级精母细胞中被标记的核 DNA 分子最多有 10 个\nC: 若甲连续进行两次正常的有丝分裂, 则一个子细胞中含 ${ }^{15} \\mathrm{~N}$ 的染色体数为 10 条\nD: 若初级精母细胞中含 ${ }^{15} \\mathrm{~N}$ 的 DNA 分子占一半，则甲至少进行了一次有丝分裂\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n用 ${ }^{15} \\mathrm{~N}$ 标记某基因型为 $\\mathrm{AaX} X^{\\mathrm{b}} \\mathrm{Y}$ 的动物精原细胞甲（2n=10）的所有 DNA 双链, 将甲放在不含 ${ }^{15} \\mathrm{~N}$ 的培养基中进行培养, 发现了一个含 XY 的异常精子, 若无其他突变和染色体互换发生，下列叙述正确的是（）\n\nA: 与异常精子同时产生的另外三个精子的基因型可能是 $A^{b}$ Y $、 A 、 a$\nB: 产生该异常精子的次级精母细胞中被标记的核 DNA 分子最多有 10 个\nC: 若甲连续进行两次正常的有丝分裂, 则一个子细胞中含 ${ }^{15} \\mathrm{~N}$ 的染色体数为 10 条\nD: 若初级精母细胞中含 ${ }^{15} \\mathrm{~N}$ 的 DNA 分子占一半，则甲至少进行了一次有丝分裂\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_967",
        "problem": "Fixation is a lab procedure used to prevent biological tissues from decay. In instances where perfect morphology of the whole animal is desired, cardiac fixation is the preferred method. Where should the researcher inject the fixative to ensure the best results?\nA: Right atrium\nB: Abdominal cavity\nC: Right ventricle\nD: Left atrium\nE: Left ventricle\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFixation is a lab procedure used to prevent biological tissues from decay. In instances where perfect morphology of the whole animal is desired, cardiac fixation is the preferred method. Where should the researcher inject the fixative to ensure the best results?\n\nA: Right atrium\nB: Abdominal cavity\nC: Right ventricle\nD: Left atrium\nE: Left ventricle\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_699",
        "problem": "某实验室发现一种 $\\mathrm{X}$ 基因缺失的突变体酵母菌, 该突变体中会有部分个体减数分裂后产生染色体数目异常的单倍体孢子, 这样的孢子活力降低。下图为酵母菌减数分裂过程示意图 (只标注了其中一对染色体), MI 前期同源染色体间通过蛋白质为主要成分的链状结构——联会复合体（SC）连接。下列相关叙述错误的是（ ）\n\n[图1]\nA: 突变体在减数第一次分裂时期 SC 降解缓慢, 导致染色体行为异常\nB: 减数第二次分裂过程中, 突变体染色体的非姐妹染色单体进入同一个细胞\nC: 推测野生型酵母菌中 X 基因的功能是促进 SC 降解, 促进同源染色体联会后分 离\nD: 若突变体酵母菌产生的单倍体孢子中约 $14.4 \\%$ 染色体数目异常, 则说明有 $7.2 \\%$的亲代酵母细胞减数分裂异常\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某实验室发现一种 $\\mathrm{X}$ 基因缺失的突变体酵母菌, 该突变体中会有部分个体减数分裂后产生染色体数目异常的单倍体孢子, 这样的孢子活力降低。下图为酵母菌减数分裂过程示意图 (只标注了其中一对染色体), MI 前期同源染色体间通过蛋白质为主要成分的链状结构——联会复合体（SC）连接。下列相关叙述错误的是（ ）\n\n[图1]\n\nA: 突变体在减数第一次分裂时期 SC 降解缓慢, 导致染色体行为异常\nB: 减数第二次分裂过程中, 突变体染色体的非姐妹染色单体进入同一个细胞\nC: 推测野生型酵母菌中 X 基因的功能是促进 SC 降解, 促进同源染色体联会后分 离\nD: 若突变体酵母菌产生的单倍体孢子中约 $14.4 \\%$ 染色体数目异常, 则说明有 $7.2 \\%$的亲代酵母细胞减数分裂异常\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-55.jpg?height=623&width=968&top_left_y=1639&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_766",
        "problem": "果蝇的有眼和无眼、红眼和白眼分别由等位基因 $\\mathrm{A} / \\mathrm{a}$ 和 $\\mathrm{B} / \\mathrm{b}$ 控制。现有一只纯合红眼雄蝇和一只纯合无眼雌蝇杂交, $F_{1}$ 雌蝇全为红眼、雄蝇全为白眼。让 $F_{1}$ 雌蝇、雄蝇随机交配得到 $\\mathrm{F}_{2}, \\mathrm{~F}_{2}$ 雌蝇、雄蝇均表现为红眼: 白眼: 无眼 $=2: 3: 1$. 下列叙述正确的是 ( )\nA: 亲本基因型为 $B B X^{A} Y$ 和 $b b X^{a} X^{a}$\nB: $F_{2}$ 的表型及比例异常, 是因为 $F_{1}$ 中雌配子 $a X^{B}$ 致死\nC: $F_{2}$ 中红眼雌蝇与白眼雄蝇随机交配, $F_{3}$ 无眼果蝇的基因型共 4 种\nD: $\\mathrm{F}_{1}$ 雄蝇的一个次级精母细胞中含有 0 或 1 个白眼基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的有眼和无眼、红眼和白眼分别由等位基因 $\\mathrm{A} / \\mathrm{a}$ 和 $\\mathrm{B} / \\mathrm{b}$ 控制。现有一只纯合红眼雄蝇和一只纯合无眼雌蝇杂交, $F_{1}$ 雌蝇全为红眼、雄蝇全为白眼。让 $F_{1}$ 雌蝇、雄蝇随机交配得到 $\\mathrm{F}_{2}, \\mathrm{~F}_{2}$ 雌蝇、雄蝇均表现为红眼: 白眼: 无眼 $=2: 3: 1$. 下列叙述正确的是 ( )\n\nA: 亲本基因型为 $B B X^{A} Y$ 和 $b b X^{a} X^{a}$\nB: $F_{2}$ 的表型及比例异常, 是因为 $F_{1}$ 中雌配子 $a X^{B}$ 致死\nC: $F_{2}$ 中红眼雌蝇与白眼雄蝇随机交配, $F_{3}$ 无眼果蝇的基因型共 4 种\nD: $\\mathrm{F}_{1}$ 雄蝇的一个次级精母细胞中含有 0 或 1 个白眼基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_114",
        "problem": "Scientists grew cucumber plants in different nutrient conditions to generate obtain either super ovary or normal ovary types. They labelled flowers when they emerged and observed the development of flowers. Based on the color and shape, corolla development was divided into four stages: green bud (G), green-yellow bud (GY), yellow bud (Y), and flowering $(F)$. They also measured plant growth regulator concentrations in different flower developmental stages.\n[figure1]\n\nFig.Q68-1. Morphological characterization of the normal ovary and super ovary types.\n\n(A)\n[figure2]\n\n(B)\n\n[figure3]\n\n(E)\n\n[figure4]\n\n(C)\n\n[figure5]\n\n(F)\n\n[figure6]\n\nFig.Q68-2. Concentration of cytokinins (IPA, ZR, DHZR), gibberellins (GA3, GA4) and auxin (IAA) in different flower developmental stages. In super ovary type, two substages were included. ** indicates statistically significant differences within one developmental stage.\nA: The corolla progression between stages was much delayed in the super ovary\nB: The ovary at anthesis was on average much longer in the super ovary than the normal ovary, while at the same time courses after labelling, fruit length was not different between two types.\nC: Gibberellins were increased in the super ovary during the early stages of corolla development, which corresponds to the enlarged corolla size\nD: Cytokinins appear to be the primary regulator for flower opening in cucumber, whereas auxin is probably involved in the size control of corolla and fruit.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nScientists grew cucumber plants in different nutrient conditions to generate obtain either super ovary or normal ovary types. They labelled flowers when they emerged and observed the development of flowers. Based on the color and shape, corolla development was divided into four stages: green bud (G), green-yellow bud (GY), yellow bud (Y), and flowering $(F)$. They also measured plant growth regulator concentrations in different flower developmental stages.\n[figure1]\n\nFig.Q68-1. Morphological characterization of the normal ovary and super ovary types.\n\n(A)\n[figure2]\n\n(B)\n\n[figure3]\n\n(E)\n\n[figure4]\n\n(C)\n\n[figure5]\n\n(F)\n\n[figure6]\n\nFig.Q68-2. Concentration of cytokinins (IPA, ZR, DHZR), gibberellins (GA3, GA4) and auxin (IAA) in different flower developmental stages. In super ovary type, two substages were included. ** indicates statistically significant differences within one developmental stage.\n\nA: The corolla progression between stages was much delayed in the super ovary\nB: The ovary at anthesis was on average much longer in the super ovary than the normal ovary, while at the same time courses after labelling, fruit length was not different between two types.\nC: Gibberellins were increased in the super ovary during the early stages of corolla development, which corresponds to the enlarged corolla size\nD: Cytokinins appear to be the primary regulator for flower opening in cucumber, whereas auxin is probably involved in the size control of corolla and fruit.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-045.jpg?height=1016&width=1494&top_left_y=824&top_left_x=286",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-046.jpg?height=730&width=590&top_left_y=274&top_left_x=216",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-046.jpg?height=329&width=537&top_left_y=293&top_left_x=798",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-046.jpg?height=344&width=560&top_left_y=660&top_left_x=795",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-046.jpg?height=327&width=523&top_left_y=291&top_left_x=1317",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-046.jpg?height=338&width=503&top_left_y=666&top_left_x=1327"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1289",
        "problem": "SAVING THE MAUI'S DOLPHINS - A STORY OF BIOLOGY, POLICY \\& CONSERVATION\n\n[figure1]\n\nWWF-New Zealand\n\nhttp://wwf.panda.org/wwf news/?206249/NZ-govt-fails-Mauisdolphins-on-global-stage\n\n[figure2]\n\nhttp://uww.doc.govt.nz/conservation/native-animals/marinemammals/dolphins/hectors-dolphin/docs-work/hectors-and-mauisdolphin-incident-databasel\n\nMaui's dolphin (Cephalorhynchus hectori maui) is one of the world's rarest dolphins and is found only on the west coast of the North Island of New Zealand (Resource Pack, Figure 1). It is a sub-species of Hector's dolphin (Cephalorhynchus hectori). The Maui's dolphin is protected by the West Coast North Island Marine Mammal Sanctuary (WCNIMMS) which restricts seabed mining activities, acoustic seismic survey work and commercial and recreational set netting. The boundaries of this sanctuary extend alongshore from Maunganui Bluff in Northland to Oakura Beach, Taranaki, in the south and from mean high water springs to the 12 nautical mile (nm) territorial sea limit. The total area of the sanctuary is approximately 1,200,086 hectares covering 2,164 km of coastline. (Resource Pack, Figure 1).\n\nMaui's dolphin is 'critically endangered' (IUCN Red List), with the population dropping from around 1000 individuals in 1970 to 111 in 2004 according to research by Assoc. Prof. Dr Liz Slooten and others from Otago University. Recent research by the Department of Conservation (DOC) suggests there are now even fewer Maui's dolphins remaining. Maui's dolphins are relatively short-lived (approximately 25 years), and are slow breeders. Females do not have their first calf until they are about seven or eight years old, and have a new calf only every two to four years. This means the species may be threatened by even occasional deaths caused by human activity. Fishing, particularly set netting, is the greatest known human threat to Maui's dolphins and thought to be responsible for about 75 per cent of reported deaths with a known cause. Other human threats include marine tourism, vessel traffic, mining, construction, coastal development pollution, sedimentation, oil spills, plastic bags, marine farming and climate change.\n\nIn recent weeks the need to develop effective management strategies for this species to prevent its extinction has hit the news. Liz Slooten presented the latest research to the International Whaling Commission in May this year, showing that the current protection measures are not sufficient to avoid the extinction of Maui's dolphin. WWF-New Zealand, Greenpeace, and Forest \\& Bird and international conservation groups including NABU and WDC in 2012 and 2013 are all campaigning to protect the Maui's dolphin and hundreds of angry protesters marched to the office of the Energy and Resources Minister Simon Bridges' office in Tauranga to protest the government allowing oil exploration within the West Coast North Island Marine Mammal Sanctuary in their annual tender process 'Block Offer 2014' (Resource Pack, Figure 3). This government decision came just two weeks after the International Whaling Committee (IWC) criticised New Zealand for not taking the necessary steps to save the Maui's dolphin. The IWC noted that a $350 \\mathrm{~km}^{2}$ set net restriction had been added to the WCNIMMS but commented that these measures fell significantly short of those required to reverse the Maui's dolphin decline as recommended by the IWC in 2012 and 2013. The IWC reiterated its extreme concern about the continued decline of such a small population \"as the humaninduced death of even one dolphin would increase the extinction risk for this subspecies\". In 2013 it strongly recommended that the NZ government should:\n\n\"take immediate management actions that will eliminate bycatch of Maui's dolphins. This includes full closures of any fisheries within the range of Maui's dolphins that are known to pose a risk of bycatch of small cetaceans (i.e. set net and trawl fisheries)\". Ensuring full protection of Maui's dolphins in all areas throughout their habitat, together with an ample buffer zone, would minimise the risk of bycatch and maximise the chances of population increase\". and \"commit to specific population increase targets and timelines\".\n\nBecause this recommendation was ignored by the NZ government, in 2014, the IWC recommended that the protected area should be extended south to Whanganui, offshore to 20 nautical miles and should include the harbours.\n\nThe resource pack contains maps of Maui's and Hector's dolphin sightings from 1970 - July 2013, Protection measures for Maui's dolphins on the West Coast North Island and the 'Block Offer 2014' for the Offshore Release Area: Taranaki Basin 14TAR-R1. It also has a table of all Maui's dolphin sightings from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti. Much of this area is included in the Block Offer 2014' Taranaki Basin 14TAR-R1 area. This table is modified from the Maui's dolphin sightings database:\n\nhttp://www.doc.govt.nz/conservation/native-animals/marine-mammals/dolphins/mauis-dolphin/docs-work/mauisdolphin-sightings/\n\nConservation Minister Nick Smith was questioned in parliament about the decision to open part of the sanctuary for exploration. \"The block offer is nowhere near where the Maui's live,\" Dr Smith said. \"There hasn't been a single observation of a Maui's dolphin, and the oil and gas industry hasn't been involved in a single Maui's dolphin incident in Taranaki over the past 40 years despite 23 wells being drilled\" (Source: http://www.3news.co.nz/Greenpeace-get23000-signatures-in-petition-to-sack-Bridges/tabid/423/articleID/349373/Default.aspx).Considering the data in Table 1 in the Resource Pack, how many Maui's dolphins have been sighted from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti\nA: 78\nB: 89\nC: 92\nD: 95\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nSAVING THE MAUI'S DOLPHINS - A STORY OF BIOLOGY, POLICY \\& CONSERVATION\n\n[figure1]\n\nWWF-New Zealand\n\nhttp://wwf.panda.org/wwf news/?206249/NZ-govt-fails-Mauisdolphins-on-global-stage\n\n[figure2]\n\nhttp://uww.doc.govt.nz/conservation/native-animals/marinemammals/dolphins/hectors-dolphin/docs-work/hectors-and-mauisdolphin-incident-databasel\n\nMaui's dolphin (Cephalorhynchus hectori maui) is one of the world's rarest dolphins and is found only on the west coast of the North Island of New Zealand (Resource Pack, Figure 1). It is a sub-species of Hector's dolphin (Cephalorhynchus hectori). The Maui's dolphin is protected by the West Coast North Island Marine Mammal Sanctuary (WCNIMMS) which restricts seabed mining activities, acoustic seismic survey work and commercial and recreational set netting. The boundaries of this sanctuary extend alongshore from Maunganui Bluff in Northland to Oakura Beach, Taranaki, in the south and from mean high water springs to the 12 nautical mile (nm) territorial sea limit. The total area of the sanctuary is approximately 1,200,086 hectares covering 2,164 km of coastline. (Resource Pack, Figure 1).\n\nMaui's dolphin is 'critically endangered' (IUCN Red List), with the population dropping from around 1000 individuals in 1970 to 111 in 2004 according to research by Assoc. Prof. Dr Liz Slooten and others from Otago University. Recent research by the Department of Conservation (DOC) suggests there are now even fewer Maui's dolphins remaining. Maui's dolphins are relatively short-lived (approximately 25 years), and are slow breeders. Females do not have their first calf until they are about seven or eight years old, and have a new calf only every two to four years. This means the species may be threatened by even occasional deaths caused by human activity. Fishing, particularly set netting, is the greatest known human threat to Maui's dolphins and thought to be responsible for about 75 per cent of reported deaths with a known cause. Other human threats include marine tourism, vessel traffic, mining, construction, coastal development pollution, sedimentation, oil spills, plastic bags, marine farming and climate change.\n\nIn recent weeks the need to develop effective management strategies for this species to prevent its extinction has hit the news. Liz Slooten presented the latest research to the International Whaling Commission in May this year, showing that the current protection measures are not sufficient to avoid the extinction of Maui's dolphin. WWF-New Zealand, Greenpeace, and Forest \\& Bird and international conservation groups including NABU and WDC in 2012 and 2013 are all campaigning to protect the Maui's dolphin and hundreds of angry protesters marched to the office of the Energy and Resources Minister Simon Bridges' office in Tauranga to protest the government allowing oil exploration within the West Coast North Island Marine Mammal Sanctuary in their annual tender process 'Block Offer 2014' (Resource Pack, Figure 3). This government decision came just two weeks after the International Whaling Committee (IWC) criticised New Zealand for not taking the necessary steps to save the Maui's dolphin. The IWC noted that a $350 \\mathrm{~km}^{2}$ set net restriction had been added to the WCNIMMS but commented that these measures fell significantly short of those required to reverse the Maui's dolphin decline as recommended by the IWC in 2012 and 2013. The IWC reiterated its extreme concern about the continued decline of such a small population \"as the humaninduced death of even one dolphin would increase the extinction risk for this subspecies\". In 2013 it strongly recommended that the NZ government should:\n\n\"take immediate management actions that will eliminate bycatch of Maui's dolphins. This includes full closures of any fisheries within the range of Maui's dolphins that are known to pose a risk of bycatch of small cetaceans (i.e. set net and trawl fisheries)\". Ensuring full protection of Maui's dolphins in all areas throughout their habitat, together with an ample buffer zone, would minimise the risk of bycatch and maximise the chances of population increase\". and \"commit to specific population increase targets and timelines\".\n\nBecause this recommendation was ignored by the NZ government, in 2014, the IWC recommended that the protected area should be extended south to Whanganui, offshore to 20 nautical miles and should include the harbours.\n\nThe resource pack contains maps of Maui's and Hector's dolphin sightings from 1970 - July 2013, Protection measures for Maui's dolphins on the West Coast North Island and the 'Block Offer 2014' for the Offshore Release Area: Taranaki Basin 14TAR-R1. It also has a table of all Maui's dolphin sightings from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti. Much of this area is included in the Block Offer 2014' Taranaki Basin 14TAR-R1 area. This table is modified from the Maui's dolphin sightings database:\n\nhttp://www.doc.govt.nz/conservation/native-animals/marine-mammals/dolphins/mauis-dolphin/docs-work/mauisdolphin-sightings/\n\nConservation Minister Nick Smith was questioned in parliament about the decision to open part of the sanctuary for exploration. \"The block offer is nowhere near where the Maui's live,\" Dr Smith said. \"There hasn't been a single observation of a Maui's dolphin, and the oil and gas industry hasn't been involved in a single Maui's dolphin incident in Taranaki over the past 40 years despite 23 wells being drilled\" (Source: http://www.3news.co.nz/Greenpeace-get23000-signatures-in-petition-to-sack-Bridges/tabid/423/articleID/349373/Default.aspx).\n\nproblem:\nConsidering the data in Table 1 in the Resource Pack, how many Maui's dolphins have been sighted from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti\n\nA: 78\nB: 89\nC: 92\nD: 95\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-30.jpg?height=503&width=682&top_left_y=277&top_left_x=113",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-30.jpg?height=497&width=848&top_left_y=277&top_left_x=1049"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1168",
        "problem": "The diagram shows the pedigree of a human family in which the individuals marked with shaded symbols are deaf.\n\nDeafness in this family is inherited as\n\n[figure1]\nA: an autosomal dominant characteristic\nB: an autosomal recessive characteristic\nC: an X-linked dominant characteristic\nD: an X-linked recessive characteristic\nE: a Y-linked characteristic\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram shows the pedigree of a human family in which the individuals marked with shaded symbols are deaf.\n\nDeafness in this family is inherited as\n\n[figure1]\n\nA: an autosomal dominant characteristic\nB: an autosomal recessive characteristic\nC: an X-linked dominant characteristic\nD: an X-linked recessive characteristic\nE: a Y-linked characteristic\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-13.jpg?height=355&width=916&top_left_y=1910&top_left_x=602"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1023",
        "problem": "Which of the following statements about plant stomata is TRUE?\nA: When the guard cells are filled with water and turgid, the stoma is closed.\nB: Cellulose microfibrils are oriented along the longer axis of guard cells.\nC: After a certain point, stomata closure is irreversible.\nD: Stomata open with the active uptake of $\\mathrm{K}^{+}$ions by the guard cells.\nE: Only $\\mathrm{CO}_{2}$ depletion contributes to stomatal opening.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following statements about plant stomata is TRUE?\n\nA: When the guard cells are filled with water and turgid, the stoma is closed.\nB: Cellulose microfibrils are oriented along the longer axis of guard cells.\nC: After a certain point, stomata closure is irreversible.\nD: Stomata open with the active uptake of $\\mathrm{K}^{+}$ions by the guard cells.\nE: Only $\\mathrm{CO}_{2}$ depletion contributes to stomatal opening.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_752",
        "problem": "已知小麦的耐盐对不耐盐为显性, 多粒对少粒为显性, 分别由等位基因 $\\mathrm{A} / \\mathrm{a}, \\mathrm{B} / \\mathrm{b}$控制。已知含有某种基因的花粉 $1 / 3$ 致死，现有一株表现为耐盐多粒的小麦，以其为父本进行测交，测交后代 $F_{1}$ 的 4 种表现型为耐盐多粒:耐盐少粒:不耐盐多粒:不耐盐少粒 $=3: 2: 3: 2$ 。下列叙述错误的是 ( )\nA: 这两对等位基因的遗传遵循自由组合定律\nB: 取 $F_{1}$ 的耐盐多程小麦和耐盐少粒小麦各一株杂交, 后代不耐盐多粒占 $1 / 8$ 或 $3 / 20$\nC: 若以该植株为母本进行测交, 后代上述 4 种表现型比例为 $1: 1: 1: 1$\nD: 若该植株进行自交, 后代上述 4 种表现型比例为 $12: 4: 3: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知小麦的耐盐对不耐盐为显性, 多粒对少粒为显性, 分别由等位基因 $\\mathrm{A} / \\mathrm{a}, \\mathrm{B} / \\mathrm{b}$控制。已知含有某种基因的花粉 $1 / 3$ 致死，现有一株表现为耐盐多粒的小麦，以其为父本进行测交，测交后代 $F_{1}$ 的 4 种表现型为耐盐多粒:耐盐少粒:不耐盐多粒:不耐盐少粒 $=3: 2: 3: 2$ 。下列叙述错误的是 ( )\n\nA: 这两对等位基因的遗传遵循自由组合定律\nB: 取 $F_{1}$ 的耐盐多程小麦和耐盐少粒小麦各一株杂交, 后代不耐盐多粒占 $1 / 8$ 或 $3 / 20$\nC: 若以该植株为母本进行测交, 后代上述 4 种表现型比例为 $1: 1: 1: 1$\nD: 若该植株进行自交, 后代上述 4 种表现型比例为 $12: 4: 3: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_794",
        "problem": "图 1 为两个家族关于甲、乙两种遗传病的系谱图。甲病（由 $\\mathrm{A} / \\mathrm{a}$ 基因控制）是某单基因突变引起的遗传病, 男性和女性群体中致病基因的基因频率均为 $1 / 100$, 图 2 为图 1 中部分成员关于甲病基因 DNA 酶切并电泳后的条带情况; 乙病由 $\\mathrm{B} / \\mathrm{b}$ 基因控制，甲病和乙病致病基因均不在 $\\mathrm{X} 、 \\mathrm{Y}$ 同源区段上。下列叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 甲病是伴 X 染色体隐性遗传病、乙病是常染色体显性遗传病\nB: II- 2 和II- 3 生育患病孩子的概率是 $1 / 4$\nC: 若III-1 是个女性, 其基因型与其母亲相同的概率是 3/8\nD: 若III- 1 是个健康男性, 与一个正常女性结婚生育患甲病孩子的概率为 $1 / 202$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 1 为两个家族关于甲、乙两种遗传病的系谱图。甲病（由 $\\mathrm{A} / \\mathrm{a}$ 基因控制）是某单基因突变引起的遗传病, 男性和女性群体中致病基因的基因频率均为 $1 / 100$, 图 2 为图 1 中部分成员关于甲病基因 DNA 酶切并电泳后的条带情况; 乙病由 $\\mathrm{B} / \\mathrm{b}$ 基因控制，甲病和乙病致病基因均不在 $\\mathrm{X} 、 \\mathrm{Y}$ 同源区段上。下列叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 甲病是伴 X 染色体隐性遗传病、乙病是常染色体显性遗传病\nB: II- 2 和II- 3 生育患病孩子的概率是 $1 / 4$\nC: 若III-1 是个女性, 其基因型与其母亲相同的概率是 3/8\nD: 若III- 1 是个健康男性, 与一个正常女性结婚生育患甲病孩子的概率为 $1 / 202$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-92.jpg?height=357&width=525&top_left_y=1238&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-92.jpg?height=269&width=574&top_left_y=1279&top_left_x=1209"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_819",
        "problem": "神经性耳聋、腓骨肌萎缩症是两种单基因遗传病，分别由 B. b 和 D. d 基因控制。下图 1 为某家族相关的遗传系谱图, 其中已死亡个体无法知道其性状, 经检测 $\\mathrm{IV}_{21}$ 不携带致病基因。为了确定腓骨肌萎缩症基因在染色体上的分布，科研人员对 $\\mathrm{III}_{9} \\sim \\mathrm{III}_{13}$ 个体含相关基因的 DNA 片段扩增后用某种限制酶处理, 并进行电泳分析, 结果如图 2。下列说法正确的是（ ）\n[图1]\n\n图 1\n\n[图2]\n\n图 2\nA: 神经性耳聋的遗传方式为常染色体隐性, 腓骨肌萎缩症的遗传方式为伴 X 染色体显性\nB: $\\mathrm{IV}_{19}$ 的致病基因只能来自于 $\\mathrm{II}_{5}$ 或者 $\\mathrm{II}_{6}$\nC: $\\mathrm{III}_{13}$ 的基因型为 $B b X^{D} X^{d}$ 或 $B B X^{D} X^{d}$\nD: $V_{22}$ 与 $V_{23}$ 婚配, 他们后代中出现不患病女孩的概率是 $7 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n神经性耳聋、腓骨肌萎缩症是两种单基因遗传病，分别由 B. b 和 D. d 基因控制。下图 1 为某家族相关的遗传系谱图, 其中已死亡个体无法知道其性状, 经检测 $\\mathrm{IV}_{21}$ 不携带致病基因。为了确定腓骨肌萎缩症基因在染色体上的分布，科研人员对 $\\mathrm{III}_{9} \\sim \\mathrm{III}_{13}$ 个体含相关基因的 DNA 片段扩增后用某种限制酶处理, 并进行电泳分析, 结果如图 2。下列说法正确的是（ ）\n[图1]\n\n图 1\n\n[图2]\n\n图 2\n\nA: 神经性耳聋的遗传方式为常染色体隐性, 腓骨肌萎缩症的遗传方式为伴 X 染色体显性\nB: $\\mathrm{IV}_{19}$ 的致病基因只能来自于 $\\mathrm{II}_{5}$ 或者 $\\mathrm{II}_{6}$\nC: $\\mathrm{III}_{13}$ 的基因型为 $B b X^{D} X^{d}$ 或 $B B X^{D} X^{d}$\nD: $V_{22}$ 与 $V_{23}$ 婚配, 他们后代中出现不患病女孩的概率是 $7 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-070.jpg?height=431&width=831&top_left_y=1372&top_left_x=361",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-070.jpg?height=183&width=500&top_left_y=1622&top_left_x=1232"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_788",
        "problem": "如图为某卵原细胞进行减数分裂的过程, 卵原细胞的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$, 且在减数分裂过程中仅发生过一次异常（无基因突变），甲 庚表示细胞，(1) (4)表示过程。下列相关叙述正确的是（ ）\n\n[图1]\nA: 过程(1)会发生染色体复制和基因重组, 过程(2)染色体数目减半\nB: 若丁细胞基因型为 $a X^{b} X^{b}$ ，原因可能是减数分裂 $I$ 异常所致\nC: 若己细胞基因型为 $\\mathrm{A}$, 则丁和戊细胞的基因型均可能为 $\\mathrm{aX}{ }^{\\mathrm{B}}$\nD: 若将卵原细胞的 DNA 用 ${ }^{32} \\mathrm{P}$ 标记后并正常培养, 则庚细胞的 DNA 有一半含 ${ }^{32} \\mathrm{P}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图为某卵原细胞进行减数分裂的过程, 卵原细胞的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$, 且在减数分裂过程中仅发生过一次异常（无基因突变），甲 庚表示细胞，(1) (4)表示过程。下列相关叙述正确的是（ ）\n\n[图1]\n\nA: 过程(1)会发生染色体复制和基因重组, 过程(2)染色体数目减半\nB: 若丁细胞基因型为 $a X^{b} X^{b}$ ，原因可能是减数分裂 $I$ 异常所致\nC: 若己细胞基因型为 $\\mathrm{A}$, 则丁和戊细胞的基因型均可能为 $\\mathrm{aX}{ }^{\\mathrm{B}}$\nD: 若将卵原细胞的 DNA 用 ${ }^{32} \\mathrm{P}$ 标记后并正常培养, 则庚细胞的 DNA 有一半含 ${ }^{32} \\mathrm{P}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-80.jpg?height=503&width=619&top_left_y=662&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1520",
        "problem": "The graph shows pressure of pulses of blood flowing through different vessels.\n\n## Blood pressure\n\n[figure1]\n\nWhy is the average pressure of blood reduced as it flows away from the heart?\nA: Fluid leaks out of vessels so there is less volume further from the heart.\nB: The combined cross-sectional area of vessels is greater further from the heart so blood flows with less resistance.\nC: Muscles of arteries are constricted, pressing on the blood.\nD: The heart sucks on the blood in veins, reducing the pressure in them.\nE: Blood flow is very slow through capillaries in organs.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows pressure of pulses of blood flowing through different vessels.\n\n## Blood pressure\n\n[figure1]\n\nWhy is the average pressure of blood reduced as it flows away from the heart?\n\nA: Fluid leaks out of vessels so there is less volume further from the heart.\nB: The combined cross-sectional area of vessels is greater further from the heart so blood flows with less resistance.\nC: Muscles of arteries are constricted, pressing on the blood.\nD: The heart sucks on the blood in veins, reducing the pressure in them.\nE: Blood flow is very slow through capillaries in organs.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-13.jpg?height=948&width=1213&top_left_y=491&top_left_x=590"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_506",
        "problem": "棉花是雌雄同花的经济作物, 棉纤维的长线和短线(由 $\\mathrm{A} 、 \\mathrm{a}$ 基因控制)、白色和红色(由 $\\mathrm{B} 、 \\mathrm{~b}$ 基因控制)为两对相对性状。为了获得长线线红棉新品种, 育种专家对长线深红棉 $\\mathrm{M}$ 的萌发种子进行电离辐射处理, 得到下图所示的两种变异植株 M1 和 M2。下列相关叙述错误的是（ ）\n\n[图1]\n长线浅红棉植株 M1 长线深红棉植株M 长线浅红棉植株 M2\nA: 电离辐射处理 M 的种子得到植株 M1 和 M2 的变异类型是基因突变\nB: M2 中控制棉纤维长度和颜色的两对基因在遗传时遵循自由组合定律\nC: 将变异植物 M1 进行自花传粉, 子代中出现长线浅红棉的概率为 $3 / 4$\nD: 将变异植物 M2 与 $\\mathrm{M}$ 进行杂交，子代中出现长线浅红棉的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n棉花是雌雄同花的经济作物, 棉纤维的长线和短线(由 $\\mathrm{A} 、 \\mathrm{a}$ 基因控制)、白色和红色(由 $\\mathrm{B} 、 \\mathrm{~b}$ 基因控制)为两对相对性状。为了获得长线线红棉新品种, 育种专家对长线深红棉 $\\mathrm{M}$ 的萌发种子进行电离辐射处理, 得到下图所示的两种变异植株 M1 和 M2。下列相关叙述错误的是（ ）\n\n[图1]\n长线浅红棉植株 M1 长线深红棉植株M 长线浅红棉植株 M2\n\nA: 电离辐射处理 M 的种子得到植株 M1 和 M2 的变异类型是基因突变\nB: M2 中控制棉纤维长度和颜色的两对基因在遗传时遵循自由组合定律\nC: 将变异植物 M1 进行自花传粉, 子代中出现长线浅红棉的概率为 $3 / 4$\nD: 将变异植物 M2 与 $\\mathrm{M}$ 进行杂交，子代中出现长线浅红棉的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-100.jpg?height=314&width=1179&top_left_y=154&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_14",
        "problem": "Part of the sequence of vector A, which is for protein expression using Escherichia coli as a host, is shown. It was planned to express a plant-derived gene $\\mathrm{X}$ using vector $\\mathrm{A}$. Vector $\\mathrm{A}$ is a plasmid vector that expresses a protein fused to the N-terminus His-tag, which enables efficient purification of the expressed protein. As shown in Figure 1, translation of the protein occurs from the start codon immediately before the His tag with six consecutive His residues. The DNA sequences of the $5^{\\prime}$ and 3' regions of gene $\\mathrm{X}$ are shown in Figure 2. We planned to clone gene X using restriction enzyme sites, EcoRI, SmaI, or Sall in vector A. When the gene $\\mathrm{X}$ is amplified by PCR, a fragment with a restriction enzyme site at the end can be amplified using the primer with a restriction enzyme site. Since the restriction enzyme site is not recognized if it is located at the end of the DNA fragment, three \"Cs\" were also attached in addition to the restriction enzyme site.\n\nFor example, in order to add the EcoRI site to 5'-XXXXXXXXXX----, the primer is designed as below.\n\n## 5'-CCCGAATTCXXXXXXXXXX----,\n\n[figure1]\n\nFigure 1. DNA sequence of the cloning region of vector $\\mathrm{A}$ (double strands).\n\n[figure2]\n\nStop codon\n\nFigure 2. DNA sequence of the gene $X$ showing 5 ' region and 3' region: 1566 base pair\nA: Forward primer ( $\\square$ : the start codon) 5'-CCC GAA TTC ATG AAG TTA TTG AGC AAT A-3' EcoRI site\nB: Forward primer ( $\\square$ : the start codon) 5'- CCC CCC GGG ATG AAG TTA TTG AGC AAT A-3' SmaI site\nC: Reverse primer ( $\\square$ : the stop codon) 5'-CCC GTC GAC TCA AGA CGA AGC ATA TGA T-3' SalI site\nD: Reverse primer ( $\\square$ : the stop codon) 5'-CCC AAG CTT GTA GGT AGT ATA CGA AGC AGA ACT -3' Hind III site\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPart of the sequence of vector A, which is for protein expression using Escherichia coli as a host, is shown. It was planned to express a plant-derived gene $\\mathrm{X}$ using vector $\\mathrm{A}$. Vector $\\mathrm{A}$ is a plasmid vector that expresses a protein fused to the N-terminus His-tag, which enables efficient purification of the expressed protein. As shown in Figure 1, translation of the protein occurs from the start codon immediately before the His tag with six consecutive His residues. The DNA sequences of the $5^{\\prime}$ and 3' regions of gene $\\mathrm{X}$ are shown in Figure 2. We planned to clone gene X using restriction enzyme sites, EcoRI, SmaI, or Sall in vector A. When the gene $\\mathrm{X}$ is amplified by PCR, a fragment with a restriction enzyme site at the end can be amplified using the primer with a restriction enzyme site. Since the restriction enzyme site is not recognized if it is located at the end of the DNA fragment, three \"Cs\" were also attached in addition to the restriction enzyme site.\n\nFor example, in order to add the EcoRI site to 5'-XXXXXXXXXX----, the primer is designed as below.\n\n## 5'-CCCGAATTCXXXXXXXXXX----,\n\n[figure1]\n\nFigure 1. DNA sequence of the cloning region of vector $\\mathrm{A}$ (double strands).\n\n[figure2]\n\nStop codon\n\nFigure 2. DNA sequence of the gene $X$ showing 5 ' region and 3' region: 1566 base pair\n\nA: Forward primer ( $\\square$ : the start codon) 5'-CCC GAA TTC ATG AAG TTA TTG AGC AAT A-3' EcoRI site\nB: Forward primer ( $\\square$ : the start codon) 5'- CCC CCC GGG ATG AAG TTA TTG AGC AAT A-3' SmaI site\nC: Reverse primer ( $\\square$ : the stop codon) 5'-CCC GTC GAC TCA AGA CGA AGC ATA TGA T-3' SalI site\nD: Reverse primer ( $\\square$ : the stop codon) 5'-CCC AAG CTT GTA GGT AGT ATA CGA AGC AGA ACT -3' Hind III site\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-21.jpg?height=385&width=1556&top_left_y=1321&top_left_x=244",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-21.jpg?height=335&width=1339&top_left_y=1892&top_left_x=262"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_275",
        "problem": "An experiment was set up to observe cell cycle length of a strain of yeast. Activated yeast cells were subcultured into a new medium with an initial concentration of $10^{6}$ cells $/ \\mathrm{mL}$. After $40 \\mathrm{~h}$, the number of cells increased to $4 \\times 10^{6}$ cells $/ \\mathrm{mL}$. A portion of the culture was taken for a separate experiment. In this experiment, cells were incubated for $15 \\mathrm{~min}$ into a media containing radioactive thymidine before washing and re-grown on a new media containing non-radioactive thymidine. Cell samples were then taken periodically to measure the percentage of mitotic cells containing radioactive thymidine. Fig.Q. 10 shows the result obtained from the experiment. At each sampling, about $1 \\%$ of the total cells sampled were undergoing mitosis.\n\n[figure1]\n\nFig.Q.10. Experiment result of yeast cell culture.\nA: G2 phase of the cell cycle takes approximatively 10 hours.\nB: Most of the yeast cells in the culture are at G1.\nC: M phase of the cell cycle takes longer than 1 hour.\nD: Most of the radioactive thymidine is assimilated in the $\\mathrm{S}$ phase of the cell cycle.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAn experiment was set up to observe cell cycle length of a strain of yeast. Activated yeast cells were subcultured into a new medium with an initial concentration of $10^{6}$ cells $/ \\mathrm{mL}$. After $40 \\mathrm{~h}$, the number of cells increased to $4 \\times 10^{6}$ cells $/ \\mathrm{mL}$. A portion of the culture was taken for a separate experiment. In this experiment, cells were incubated for $15 \\mathrm{~min}$ into a media containing radioactive thymidine before washing and re-grown on a new media containing non-radioactive thymidine. Cell samples were then taken periodically to measure the percentage of mitotic cells containing radioactive thymidine. Fig.Q. 10 shows the result obtained from the experiment. At each sampling, about $1 \\%$ of the total cells sampled were undergoing mitosis.\n\n[figure1]\n\nFig.Q.10. Experiment result of yeast cell culture.\n\nA: G2 phase of the cell cycle takes approximatively 10 hours.\nB: Most of the yeast cells in the culture are at G1.\nC: M phase of the cell cycle takes longer than 1 hour.\nD: Most of the radioactive thymidine is assimilated in the $\\mathrm{S}$ phase of the cell cycle.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-024.jpg?height=778&width=1123&top_left_y=1001&top_left_x=541"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_514",
        "problem": "家蝇 $Y$ 染色体由于某种影响断成两段, 含 $s$ 基因的小片段移接到常染色体获得 $X Y^{\\prime}$个体，不含 $\\mathrm{s}$ 基因的大片段丢失。含 $\\mathrm{s}$ 基因的家蝇发育为雄性，只含一条 $\\mathrm{X}$ 染色体的雌蝇胚胎致死, 其他均可存活且繁殖力相同。 $\\mathrm{M} 、 \\mathrm{~m}$ 是控制家蝇体色的基因, 灰色基因 $\\mathrm{M}$对黑色基因 $m$ 为完全显性。如图所示的两亲本杂交获得 $F_{1}$, 从 $F_{1}$ 开始逐代随机交配获得 $F_{n}$ 。不考虑交换和其他突变, 关于 $F_{1}$ 至 $F_{n}$, 下列说法错误的是 ( )\n\n[图1]\n\n雌性亲本(XX)\n\n[图2]\n\n雄性亲本 $\\left(\\mathrm{XY}^{\\prime}\\right)$\nA: 题中变异类型包括染色体结构和数目变异\nB: 子代所有个体均可由体色判断性别, 故 M、m 基因的遗传属于伴性遗传\nC: 雄性亲本可以产生四种类型的配子，比例为 $1: 1: 1: 1$\nD: $\\mathrm{F}_{1}$ 至 $\\mathrm{F}$ 各代中雌性个体均为 XXmm，雄性个体有两种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n家蝇 $Y$ 染色体由于某种影响断成两段, 含 $s$ 基因的小片段移接到常染色体获得 $X Y^{\\prime}$个体，不含 $\\mathrm{s}$ 基因的大片段丢失。含 $\\mathrm{s}$ 基因的家蝇发育为雄性，只含一条 $\\mathrm{X}$ 染色体的雌蝇胚胎致死, 其他均可存活且繁殖力相同。 $\\mathrm{M} 、 \\mathrm{~m}$ 是控制家蝇体色的基因, 灰色基因 $\\mathrm{M}$对黑色基因 $m$ 为完全显性。如图所示的两亲本杂交获得 $F_{1}$, 从 $F_{1}$ 开始逐代随机交配获得 $F_{n}$ 。不考虑交换和其他突变, 关于 $F_{1}$ 至 $F_{n}$, 下列说法错误的是 ( )\n\n[图1]\n\n雌性亲本(XX)\n\n[图2]\n\n雄性亲本 $\\left(\\mathrm{XY}^{\\prime}\\right)$\n\nA: 题中变异类型包括染色体结构和数目变异\nB: 子代所有个体均可由体色判断性别, 故 M、m 基因的遗传属于伴性遗传\nC: 雄性亲本可以产生四种类型的配子，比例为 $1: 1: 1: 1$\nD: $\\mathrm{F}_{1}$ 至 $\\mathrm{F}$ 各代中雌性个体均为 XXmm，雄性个体有两种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-74.jpg?height=252&width=274&top_left_y=1296&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-74.jpg?height=248&width=262&top_left_y=1298&top_left_x=680"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1361",
        "problem": "Every day, some 200 litres of blood are filtered by your kidneys. Most of that fluid is reabsorbed back into the body; the rest of it becomes urine. The basic unit of structure in the kidney is the nephron. Blood enters the nephron through the afferent arteriole which branches into a network of glomerular capillaries. A series of hydrostatic and osmotic pressures then draw fluid in or drive fluid out of these capillaries. The filtrate is collected in the capsular space which drains into the proximal convoluted tubule (PCT) of the nephron. This is the basis of renal filtration.\n\nGlomerular hydrostatic pressure (GHP) is the pressure exerted by blood on the walls of the glomerular capillaries. Capsular hydrostatic pressure (CHP) is the pressure exerted by the filtrate in the capsular space. Blood colloid osmotic pressure (BCOP) is the pressure exerted by the concentration of proteins in the blood. No proteins are expected to found in the capsular space as the glomerular capillary walls do not allow them to pass. The net filtration pressure (NFP) is the sum of all these pressures drawing fluids in or out of the glomerular capillaries.\n\nThe arrows in the diagram represent fluid moving in and out of the capillaries under the influence of these pressures.\n\n[figure1]\n\nWhich option expresses the correct relationship between these four pressures?\nA: $\\quad \\mathrm{NFP}=\\mathrm{CHP}+\\mathrm{BCOP}-\\mathrm{GHP}$\nB: $\\quad \\mathrm{NFP}=\\mathrm{CHP}-\\mathrm{BCOP}-\\mathrm{GHP}$\nC: $\\mathrm{NFP}=\\mathrm{GHP}+\\mathrm{CHP}+\\mathrm{BCOP}$\nD: $\\quad \\mathrm{NFP}=\\mathrm{GHP}-\\mathrm{CHP}-\\mathrm{BCOP}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEvery day, some 200 litres of blood are filtered by your kidneys. Most of that fluid is reabsorbed back into the body; the rest of it becomes urine. The basic unit of structure in the kidney is the nephron. Blood enters the nephron through the afferent arteriole which branches into a network of glomerular capillaries. A series of hydrostatic and osmotic pressures then draw fluid in or drive fluid out of these capillaries. The filtrate is collected in the capsular space which drains into the proximal convoluted tubule (PCT) of the nephron. This is the basis of renal filtration.\n\nGlomerular hydrostatic pressure (GHP) is the pressure exerted by blood on the walls of the glomerular capillaries. Capsular hydrostatic pressure (CHP) is the pressure exerted by the filtrate in the capsular space. Blood colloid osmotic pressure (BCOP) is the pressure exerted by the concentration of proteins in the blood. No proteins are expected to found in the capsular space as the glomerular capillary walls do not allow them to pass. The net filtration pressure (NFP) is the sum of all these pressures drawing fluids in or out of the glomerular capillaries.\n\nThe arrows in the diagram represent fluid moving in and out of the capillaries under the influence of these pressures.\n\n[figure1]\n\nWhich option expresses the correct relationship between these four pressures?\n\nA: $\\quad \\mathrm{NFP}=\\mathrm{CHP}+\\mathrm{BCOP}-\\mathrm{GHP}$\nB: $\\quad \\mathrm{NFP}=\\mathrm{CHP}-\\mathrm{BCOP}-\\mathrm{GHP}$\nC: $\\mathrm{NFP}=\\mathrm{GHP}+\\mathrm{CHP}+\\mathrm{BCOP}$\nD: $\\quad \\mathrm{NFP}=\\mathrm{GHP}-\\mathrm{CHP}-\\mathrm{BCOP}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-25.jpg?height=633&width=851&top_left_y=1314&top_left_x=497"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_407",
        "problem": "图 1 是某细胞 (染色体数量为 $2 \\mathrm{n}$ )的细胞周期示意图。周期蛋白影响细胞周期的进行, 其中周期蛋白 cyclinB 与蛋白激酶 CDK1 结合形成复合物 MPF 后, 激活的 CDK1 促进细胞由 $\\mathrm{G}$ 期进入 $\\mathrm{M}$ 期, 周期蛋白 cyclinE 与蛋白激酶 CDK2 结合后, 激活的 CDK2 促进细胞由 G1 期进入 S 期。MPF 的活性和周期蛋白的浓度变化如图 2, 下列说法错误\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 图中 $\\mathrm{G} 1$ 期细胞的核 DNA 数量为 $2 \\mathrm{n}, \\mathrm{M}$ 期细胞的染色体条数为 $2 \\mathrm{n}$ 或 $4 \\mathrm{n}$ 条\nB: 激活的 CDK1 可能具有促进染色体和纺锤体形成的作用\nC: 激活的 CDK2 可能参与调控 RNA 聚合酶和解旋酶合成\nD: 加入 DNA 合成抑制剂, $6 \\mathrm{~h}$ 后不是所有细胞都停留在 G1/S 交界处\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 1 是某细胞 (染色体数量为 $2 \\mathrm{n}$ )的细胞周期示意图。周期蛋白影响细胞周期的进行, 其中周期蛋白 cyclinB 与蛋白激酶 CDK1 结合形成复合物 MPF 后, 激活的 CDK1 促进细胞由 $\\mathrm{G}$ 期进入 $\\mathrm{M}$ 期, 周期蛋白 cyclinE 与蛋白激酶 CDK2 结合后, 激活的 CDK2 促进细胞由 G1 期进入 S 期。MPF 的活性和周期蛋白的浓度变化如图 2, 下列说法错误\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 图中 $\\mathrm{G} 1$ 期细胞的核 DNA 数量为 $2 \\mathrm{n}, \\mathrm{M}$ 期细胞的染色体条数为 $2 \\mathrm{n}$ 或 $4 \\mathrm{n}$ 条\nB: 激活的 CDK1 可能具有促进染色体和纺锤体形成的作用\nC: 激活的 CDK2 可能参与调控 RNA 聚合酶和解旋酶合成\nD: 加入 DNA 合成抑制剂, $6 \\mathrm{~h}$ 后不是所有细胞都停留在 G1/S 交界处\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-90.jpg?height=431&width=443&top_left_y=296&top_left_x=381",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-90.jpg?height=457&width=805&top_left_y=294&top_left_x=957"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1045",
        "problem": "Birds and mammals share all the following characteristics EXCEPT:\nA: Amniotic eggs\nB: A notochord\nC: Bi-directional breathing\nD: Endothermy\nE: A four-chambered heart\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBirds and mammals share all the following characteristics EXCEPT:\n\nA: Amniotic eggs\nB: A notochord\nC: Bi-directional breathing\nD: Endothermy\nE: A four-chambered heart\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_553",
        "problem": "图 1 为甲、乙两种单基因遗传病的系谱图，两种致病基因位于非同源染色体上，甲病在人群中的患病率为 $5 / 81$ 。用某种限制酶对图 1 中部分个体的甲病相关基因切割后电泳，甲病相关基因的酶切位点及电泳结果如图 2 所示。下列叙述正确的是（ ）\n\n[图1]\n\n图1图2\nA: 乙病为常染色体隐性遗传病, 7 号个体的乙病基因来自 3 号和 4 号\nB: 8 号个体只携带一种致病基因的概率为 $1 / 2$\nC: 5 号个体与正常女性婚配生育患甲病孩子的概率是 $8 / 81$\nD: 若 9 号个体为正常女性, 其甲病相关基因切割后电泳可得 4 种长度的片段\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n图 1 为甲、乙两种单基因遗传病的系谱图，两种致病基因位于非同源染色体上，甲病在人群中的患病率为 $5 / 81$ 。用某种限制酶对图 1 中部分个体的甲病相关基因切割后电泳，甲病相关基因的酶切位点及电泳结果如图 2 所示。下列叙述正确的是（ ）\n\n[图1]\n\n图1图2\n\nA: 乙病为常染色体隐性遗传病, 7 号个体的乙病基因来自 3 号和 4 号\nB: 8 号个体只携带一种致病基因的概率为 $1 / 2$\nC: 5 号个体与正常女性婚配生育患甲病孩子的概率是 $8 / 81$\nD: 若 9 号个体为正常女性, 其甲病相关基因切割后电泳可得 4 种长度的片段\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-081.jpg?height=297&width=1454&top_left_y=1142&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_240",
        "problem": "An experiment was conducted to examine the relative effect of pollinators during the night and in the daytime on the reproductive success of golden rod flowers. Pollinators cannot visit the bagged flowers. The figure shows the number of viable seeds produced (mean $\\pm$ standard deviation) by flowers that were not bagged (1), those bagged during the night (2), those bagged in the daytime (3), those bagged during both day and night (4), and those that underwent enforced pollination by an experimenter (5).\n\n[figure1]\nA: Nighttime pollinators contribute to about $60 \\%$ of the total seed production.\nB: The flowers may be capable of self-pollination.\nC: The contribution of daytime pollinators has a greater variability than nighttime pollinators.\nD: There are no limitations to pollination under natural conditions.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn experiment was conducted to examine the relative effect of pollinators during the night and in the daytime on the reproductive success of golden rod flowers. Pollinators cannot visit the bagged flowers. The figure shows the number of viable seeds produced (mean $\\pm$ standard deviation) by flowers that were not bagged (1), those bagged during the night (2), those bagged in the daytime (3), those bagged during both day and night (4), and those that underwent enforced pollination by an experimenter (5).\n\n[figure1]\n\nA: Nighttime pollinators contribute to about $60 \\%$ of the total seed production.\nB: The flowers may be capable of self-pollination.\nC: The contribution of daytime pollinators has a greater variability than nighttime pollinators.\nD: There are no limitations to pollination under natural conditions.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-62.jpg?height=842&width=1285&top_left_y=941&top_left_x=405"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_602",
        "problem": "图 1 是某细胞的细胞周期示意图, 细胞周期可分为分裂间期 (包括 $\\mathrm{G}_{1}$ 期、 $\\mathrm{S}$ 期和 $\\mathrm{G}_{2}$期）和分裂期（M 期），S 期进行 DNA 复制。周期蛋白影响细胞周期的进行，其中周期蛋白 cyclinB 与蛋白激酶 CDK1 结合形成复合物 MPF 后, 激活的 CDK1 促进细胞由 $\\mathrm{G}_{2}$期进入 M 期 周期蛋白 cyclinE 与蛋白激酶 CDK2 结合后, 激活的 CDK2 促进细胞由 $\\mathrm{G}_{1}$期进人 S 期。MPF 的活性和周期蛋白的浓度变化如图 2, 下列说法错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\nA: 若将 $\\mathrm{G}_{2}$ 期和 $M$ 期细胞融合，则融合后细胞进入 $M$ 期的时间会提前\nB: 激活的 CDK1 可能具有促进染色质螺旋形成染色体的作用\nC: 激活的 CDK2 可能参与调控 DNA 聚合酶和解旋酶合成\nD: 加入 DNA 合成抑制剂, $6 \\mathrm{~h}$ 后所有细胞停留在 $\\mathrm{G}_{1} / \\mathrm{S}$ 交界处\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 1 是某细胞的细胞周期示意图, 细胞周期可分为分裂间期 (包括 $\\mathrm{G}_{1}$ 期、 $\\mathrm{S}$ 期和 $\\mathrm{G}_{2}$期）和分裂期（M 期），S 期进行 DNA 复制。周期蛋白影响细胞周期的进行，其中周期蛋白 cyclinB 与蛋白激酶 CDK1 结合形成复合物 MPF 后, 激活的 CDK1 促进细胞由 $\\mathrm{G}_{2}$期进入 M 期 周期蛋白 cyclinE 与蛋白激酶 CDK2 结合后, 激活的 CDK2 促进细胞由 $\\mathrm{G}_{1}$期进人 S 期。MPF 的活性和周期蛋白的浓度变化如图 2, 下列说法错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\nA: 若将 $\\mathrm{G}_{2}$ 期和 $M$ 期细胞融合，则融合后细胞进入 $M$ 期的时间会提前\nB: 激活的 CDK1 可能具有促进染色质螺旋形成染色体的作用\nC: 激活的 CDK2 可能参与调控 DNA 聚合酶和解旋酶合成\nD: 加入 DNA 合成抑制剂, $6 \\mathrm{~h}$ 后所有细胞停留在 $\\mathrm{G}_{1} / \\mathrm{S}$ 交界处\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-76.jpg?height=440&width=580&top_left_y=1773&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-77.jpg?height=497&width=1060&top_left_y=174&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_895",
        "problem": "杂交水稻的无融合生殖指的是不发生雌、雄配子的细胞核融合而产生种子的一种无性繁殖方式。无融合生殖过程主要由两个基因控制一含基因 A 的植株形成雌配子时,减数第一次分裂异常, 导致雌配子染色体数目加倍; 含基因 P 的植株产生的雌配子不经过受精作用, 直接发育成个体。雄配子的发育不受基因 A、P 的影响。下列与之相关的说法中错误的是\nA: 利用无融合生殖技术可以获得母本单倍体植株\nB: 利用无融合生殖技术可以保持作物的杂种优势\nC: 基因型为 Aapp 的水稻自交，子代基因型为 Aaappp\nD: 基因型为 $\\mathrm{AaPp}$ 的水稻自交, 子代基因型与亲代相同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n杂交水稻的无融合生殖指的是不发生雌、雄配子的细胞核融合而产生种子的一种无性繁殖方式。无融合生殖过程主要由两个基因控制一含基因 A 的植株形成雌配子时,减数第一次分裂异常, 导致雌配子染色体数目加倍; 含基因 P 的植株产生的雌配子不经过受精作用, 直接发育成个体。雄配子的发育不受基因 A、P 的影响。下列与之相关的说法中错误的是\n\nA: 利用无融合生殖技术可以获得母本单倍体植株\nB: 利用无融合生殖技术可以保持作物的杂种优势\nC: 基因型为 Aapp 的水稻自交，子代基因型为 Aaappp\nD: 基因型为 $\\mathrm{AaPp}$ 的水稻自交, 子代基因型与亲代相同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1034",
        "problem": "A variety of different techniques are used to separate, analyze, and purify different chemical compounds. Select ALL of the following techniques that can be used to estimate the molecular weight of a protein?\nA: Isoelectric focusing.\nB: Electrophoresis.\nC: Size exclusion gel chromatography.\nD: Ion-exchange chromatography.\nE: Affinity chromatography.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA variety of different techniques are used to separate, analyze, and purify different chemical compounds. Select ALL of the following techniques that can be used to estimate the molecular weight of a protein?\n\nA: Isoelectric focusing.\nB: Electrophoresis.\nC: Size exclusion gel chromatography.\nD: Ion-exchange chromatography.\nE: Affinity chromatography.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_436",
        "problem": "研究发现, 线粒体内的部分代谢产物如丙酮酸经氧化脱羧形成的乙酰辅酶 $\\mathrm{A}$ 可调控细胞核内某些基因的表达进而调控细胞的功能。如图为 $\\mathrm{T}$ 淋巴细胞的线粒体部分调控关系的示意图, 其中 NFAT 蛋白为 $\\mathrm{T}$ 细胞核活化因子，在与免疫反应有关的重要基因转录中起关键作用。下列说法正确的是（ ）\n\n[图1]\nA: 乙酰辅酶 $\\mathrm{A}$ 经过三羧酸循环彻底分解成 $\\mathrm{CO}_{2}$ 和 $\\mathrm{NADPH}$\nB: 有氧呼吸前两个阶段产生的 $\\mathrm{NADH}$ 在线粒体基质中与 $\\mathrm{O}_{2}$ 结合产生水并释放大量能量\nC: 图中相关基因的组蛋白乙酰化, 激活其转录, 此过程中该基因的碱基序列没有发生变化\nD: 线粒体内产生的自由基可进入细胞核调控细胞因子相关基因的转录, 提高机体免疫力\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究发现, 线粒体内的部分代谢产物如丙酮酸经氧化脱羧形成的乙酰辅酶 $\\mathrm{A}$ 可调控细胞核内某些基因的表达进而调控细胞的功能。如图为 $\\mathrm{T}$ 淋巴细胞的线粒体部分调控关系的示意图, 其中 NFAT 蛋白为 $\\mathrm{T}$ 细胞核活化因子，在与免疫反应有关的重要基因转录中起关键作用。下列说法正确的是（ ）\n\n[图1]\n\nA: 乙酰辅酶 $\\mathrm{A}$ 经过三羧酸循环彻底分解成 $\\mathrm{CO}_{2}$ 和 $\\mathrm{NADPH}$\nB: 有氧呼吸前两个阶段产生的 $\\mathrm{NADH}$ 在线粒体基质中与 $\\mathrm{O}_{2}$ 结合产生水并释放大量能量\nC: 图中相关基因的组蛋白乙酰化, 激活其转录, 此过程中该基因的碱基序列没有发生变化\nD: 线粒体内产生的自由基可进入细胞核调控细胞因子相关基因的转录, 提高机体免疫力\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-61.jpg?height=837&width=1427&top_left_y=572&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1534",
        "problem": "Blue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nBased on fossils, whales seem to have gotten larger overtime. Satellite images suggest that the largest individual animals ever to live are currently alive.\n\nAre satellites better at estimating the size of whales or the total population?\nA: Change in population\nB: Total population.\nC: Both\nD: Neither\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBlue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nBased on fossils, whales seem to have gotten larger overtime. Satellite images suggest that the largest individual animals ever to live are currently alive.\n\nAre satellites better at estimating the size of whales or the total population?\n\nA: Change in population\nB: Total population.\nC: Both\nD: Neither\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-36.jpg?height=948&width=1673&top_left_y=657&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_802",
        "problem": "下图是大肠杆菌控制乳糖代谢的相关 DNA 结构。调节基因编码的阻遏物蛋白能与操作基因结合, 从而阻碍 RNA 聚合酶与启动子的结合, 而该阻遏物蛋白会与乳糖分子结合而失去活性。CAP 结合区能与被 cAMP 激活后的 CAP 蛋白结合, 从而促进 RNA 聚合酶与启动子的结合，但葡萄糖能显著抑制 cAMP 的合成。启动子、操纵基因、结构基因 Z、结构基因 Y、结构基因 A 依次紧密相连。结构基因 Z、结构基因 Y、结构基因 A 共用启动子，分别编码与乳糖分解有关的酶。下列叙述错误的是（ ）\n\n[图1]\nA: 若想大量获得结构基因 $\\mathrm{Y}$ 的表达产物可将大肠杆菌培养在乳糖为唯一碳源的培养基中\nB: 若 CAP 结合区、启动子、操作基因中的碱基发生甲基化，可能会增强或降低大肠杆菌对乳糖的利用能力\nC: 结构基因 $\\mathrm{Z}$ 、结构基因 $\\mathrm{Y}$ 、结构基因 $\\mathrm{A}$ 表达时分别转录形成三条 mRNA, 有各自的起始密码子和终止密码子\nD: 在葡萄糖、乳糖均大量存在时, 人为增加细胞内 cAMP 含量也能促使结构基因 Z 正常表达\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图是大肠杆菌控制乳糖代谢的相关 DNA 结构。调节基因编码的阻遏物蛋白能与操作基因结合, 从而阻碍 RNA 聚合酶与启动子的结合, 而该阻遏物蛋白会与乳糖分子结合而失去活性。CAP 结合区能与被 cAMP 激活后的 CAP 蛋白结合, 从而促进 RNA 聚合酶与启动子的结合，但葡萄糖能显著抑制 cAMP 的合成。启动子、操纵基因、结构基因 Z、结构基因 Y、结构基因 A 依次紧密相连。结构基因 Z、结构基因 Y、结构基因 A 共用启动子，分别编码与乳糖分解有关的酶。下列叙述错误的是（ ）\n\n[图1]\n\nA: 若想大量获得结构基因 $\\mathrm{Y}$ 的表达产物可将大肠杆菌培养在乳糖为唯一碳源的培养基中\nB: 若 CAP 结合区、启动子、操作基因中的碱基发生甲基化，可能会增强或降低大肠杆菌对乳糖的利用能力\nC: 结构基因 $\\mathrm{Z}$ 、结构基因 $\\mathrm{Y}$ 、结构基因 $\\mathrm{A}$ 表达时分别转录形成三条 mRNA, 有各自的起始密码子和终止密码子\nD: 在葡萄糖、乳糖均大量存在时, 人为增加细胞内 cAMP 含量也能促使结构基因 Z 正常表达\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-01.jpg?height=168&width=1357&top_left_y=1001&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1232",
        "problem": "SAVING THE MAUI'S DOLPHINS - A STORY OF BIOLOGY, POLICY \\& CONSERVATION\n\n[figure1]\n\nWWF-New Zealand\n\nhttp://wwf.panda.org/wwf news/?206249/NZ-govt-fails-Mauisdolphins-on-global-stage\n\n[figure2]\n\nhttp://uww.doc.govt.nz/conservation/native-animals/marinemammals/dolphins/hectors-dolphin/docs-work/hectors-and-mauisdolphin-incident-databasel\n\nMaui's dolphin (Cephalorhynchus hectori maui) is one of the world's rarest dolphins and is found only on the west coast of the North Island of New Zealand (Resource Pack, Figure 1). It is a sub-species of Hector's dolphin (Cephalorhynchus hectori). The Maui's dolphin is protected by the West Coast North Island Marine Mammal Sanctuary (WCNIMMS) which restricts seabed mining activities, acoustic seismic survey work and commercial and recreational set netting. The boundaries of this sanctuary extend alongshore from Maunganui Bluff in Northland to Oakura Beach, Taranaki, in the south and from mean high water springs to the 12 nautical mile (nm) territorial sea limit. The total area of the sanctuary is approximately 1,200,086 hectares covering 2,164 km of coastline. (Resource Pack, Figure 1).\n\nMaui's dolphin is 'critically endangered' (IUCN Red List), with the population dropping from around 1000 individuals in 1970 to 111 in 2004 according to research by Assoc. Prof. Dr Liz Slooten and others from Otago University. Recent research by the Department of Conservation (DOC) suggests there are now even fewer Maui's dolphins remaining. Maui's dolphins are relatively short-lived (approximately 25 years), and are slow breeders. Females do not have their first calf until they are about seven or eight years old, and have a new calf only every two to four years. This means the species may be threatened by even occasional deaths caused by human activity. Fishing, particularly set netting, is the greatest known human threat to Maui's dolphins and thought to be responsible for about 75 per cent of reported deaths with a known cause. Other human threats include marine tourism, vessel traffic, mining, construction, coastal development pollution, sedimentation, oil spills, plastic bags, marine farming and climate change.\n\nIn recent weeks the need to develop effective management strategies for this species to prevent its extinction has hit the news. Liz Slooten presented the latest research to the International Whaling Commission in May this year, showing that the current protection measures are not sufficient to avoid the extinction of Maui's dolphin. WWF-New Zealand, Greenpeace, and Forest \\& Bird and international conservation groups including NABU and WDC in 2012 and 2013 are all campaigning to protect the Maui's dolphin and hundreds of angry protesters marched to the office of the Energy and Resources Minister Simon Bridges' office in Tauranga to protest the government allowing oil exploration within the West Coast North Island Marine Mammal Sanctuary in their annual tender process 'Block Offer 2014' (Resource Pack, Figure 3). This government decision came just two weeks after the International Whaling Committee (IWC) criticised New Zealand for not taking the necessary steps to save the Maui's dolphin. The IWC noted that a $350 \\mathrm{~km}^{2}$ set net restriction had been added to the WCNIMMS but commented that these measures fell significantly short of those required to reverse the Maui's dolphin decline as recommended by the IWC in 2012 and 2013. The IWC reiterated its extreme concern about the continued decline of such a small population \"as the humaninduced death of even one dolphin would increase the extinction risk for this subspecies\". In 2013 it strongly recommended that the NZ government should:\n\n\"take immediate management actions that will eliminate bycatch of Maui's dolphins. This includes full closures of any fisheries within the range of Maui's dolphins that are known to pose a risk of bycatch of small cetaceans (i.e. set net and trawl fisheries)\". Ensuring full protection of Maui's dolphins in all areas throughout their habitat, together with an ample buffer zone, would minimise the risk of bycatch and maximise the chances of population increase\". and \"commit to specific population increase targets and timelines\".\n\nBecause this recommendation was ignored by the NZ government, in 2014, the IWC recommended that the protected area should be extended south to Whanganui, offshore to 20 nautical miles and should include the harbours.\n\nThe resource pack contains maps of Maui's and Hector's dolphin sightings from 1970 - July 2013, Protection measures for Maui's dolphins on the West Coast North Island and the 'Block Offer 2014' for the Offshore Release Area: Taranaki Basin 14TAR-R1. It also has a table of all Maui's dolphin sightings from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti. Much of this area is included in the Block Offer 2014' Taranaki Basin 14TAR-R1 area. This table is modified from the Maui's dolphin sightings database:\n\nhttp://www.doc.govt.nz/conservation/native-animals/marine-mammals/dolphins/mauis-dolphin/docs-work/mauisdolphin-sightings/\n\nConservation Minister Nick Smith was questioned in parliament about the decision to open part of the sanctuary for exploration. \"The block offer is nowhere near where the Maui's live,\" Dr Smith said. \"There hasn't been a single observation of a Maui's dolphin, and the oil and gas industry hasn't been involved in a single Maui's dolphin incident in Taranaki over the past 40 years despite 23 wells being drilled\" (Source: http://www.3news.co.nz/Greenpeace-get23000-signatures-in-petition-to-sack-Bridges/tabid/423/articleID/349373/Default.aspx).Considering the recorded sightings of Maui's dolphins since 1970 (Resource Pack, Figure 1). What valid conclusion could Dr Smith have drawn?\nA: The block offer is nowhere near where the Maui's live.\nB: There has not been a single observation of a Maui's dolphin in the block offer area.\nC: The block offer covers some of the most important areas where Maui's dolphins live.\nD: The block offer overlaps with the southern end of the Maui's dolphin range.\nE: The block offer covers most of the area where the Maui's dolphins live.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nSAVING THE MAUI'S DOLPHINS - A STORY OF BIOLOGY, POLICY \\& CONSERVATION\n\n[figure1]\n\nWWF-New Zealand\n\nhttp://wwf.panda.org/wwf news/?206249/NZ-govt-fails-Mauisdolphins-on-global-stage\n\n[figure2]\n\nhttp://uww.doc.govt.nz/conservation/native-animals/marinemammals/dolphins/hectors-dolphin/docs-work/hectors-and-mauisdolphin-incident-databasel\n\nMaui's dolphin (Cephalorhynchus hectori maui) is one of the world's rarest dolphins and is found only on the west coast of the North Island of New Zealand (Resource Pack, Figure 1). It is a sub-species of Hector's dolphin (Cephalorhynchus hectori). The Maui's dolphin is protected by the West Coast North Island Marine Mammal Sanctuary (WCNIMMS) which restricts seabed mining activities, acoustic seismic survey work and commercial and recreational set netting. The boundaries of this sanctuary extend alongshore from Maunganui Bluff in Northland to Oakura Beach, Taranaki, in the south and from mean high water springs to the 12 nautical mile (nm) territorial sea limit. The total area of the sanctuary is approximately 1,200,086 hectares covering 2,164 km of coastline. (Resource Pack, Figure 1).\n\nMaui's dolphin is 'critically endangered' (IUCN Red List), with the population dropping from around 1000 individuals in 1970 to 111 in 2004 according to research by Assoc. Prof. Dr Liz Slooten and others from Otago University. Recent research by the Department of Conservation (DOC) suggests there are now even fewer Maui's dolphins remaining. Maui's dolphins are relatively short-lived (approximately 25 years), and are slow breeders. Females do not have their first calf until they are about seven or eight years old, and have a new calf only every two to four years. This means the species may be threatened by even occasional deaths caused by human activity. Fishing, particularly set netting, is the greatest known human threat to Maui's dolphins and thought to be responsible for about 75 per cent of reported deaths with a known cause. Other human threats include marine tourism, vessel traffic, mining, construction, coastal development pollution, sedimentation, oil spills, plastic bags, marine farming and climate change.\n\nIn recent weeks the need to develop effective management strategies for this species to prevent its extinction has hit the news. Liz Slooten presented the latest research to the International Whaling Commission in May this year, showing that the current protection measures are not sufficient to avoid the extinction of Maui's dolphin. WWF-New Zealand, Greenpeace, and Forest \\& Bird and international conservation groups including NABU and WDC in 2012 and 2013 are all campaigning to protect the Maui's dolphin and hundreds of angry protesters marched to the office of the Energy and Resources Minister Simon Bridges' office in Tauranga to protest the government allowing oil exploration within the West Coast North Island Marine Mammal Sanctuary in their annual tender process 'Block Offer 2014' (Resource Pack, Figure 3). This government decision came just two weeks after the International Whaling Committee (IWC) criticised New Zealand for not taking the necessary steps to save the Maui's dolphin. The IWC noted that a $350 \\mathrm{~km}^{2}$ set net restriction had been added to the WCNIMMS but commented that these measures fell significantly short of those required to reverse the Maui's dolphin decline as recommended by the IWC in 2012 and 2013. The IWC reiterated its extreme concern about the continued decline of such a small population \"as the humaninduced death of even one dolphin would increase the extinction risk for this subspecies\". In 2013 it strongly recommended that the NZ government should:\n\n\"take immediate management actions that will eliminate bycatch of Maui's dolphins. This includes full closures of any fisheries within the range of Maui's dolphins that are known to pose a risk of bycatch of small cetaceans (i.e. set net and trawl fisheries)\". Ensuring full protection of Maui's dolphins in all areas throughout their habitat, together with an ample buffer zone, would minimise the risk of bycatch and maximise the chances of population increase\". and \"commit to specific population increase targets and timelines\".\n\nBecause this recommendation was ignored by the NZ government, in 2014, the IWC recommended that the protected area should be extended south to Whanganui, offshore to 20 nautical miles and should include the harbours.\n\nThe resource pack contains maps of Maui's and Hector's dolphin sightings from 1970 - July 2013, Protection measures for Maui's dolphins on the West Coast North Island and the 'Block Offer 2014' for the Offshore Release Area: Taranaki Basin 14TAR-R1. It also has a table of all Maui's dolphin sightings from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti. Much of this area is included in the Block Offer 2014' Taranaki Basin 14TAR-R1 area. This table is modified from the Maui's dolphin sightings database:\n\nhttp://www.doc.govt.nz/conservation/native-animals/marine-mammals/dolphins/mauis-dolphin/docs-work/mauisdolphin-sightings/\n\nConservation Minister Nick Smith was questioned in parliament about the decision to open part of the sanctuary for exploration. \"The block offer is nowhere near where the Maui's live,\" Dr Smith said. \"There hasn't been a single observation of a Maui's dolphin, and the oil and gas industry hasn't been involved in a single Maui's dolphin incident in Taranaki over the past 40 years despite 23 wells being drilled\" (Source: http://www.3news.co.nz/Greenpeace-get23000-signatures-in-petition-to-sack-Bridges/tabid/423/articleID/349373/Default.aspx).\n\nproblem:\nConsidering the recorded sightings of Maui's dolphins since 1970 (Resource Pack, Figure 1). What valid conclusion could Dr Smith have drawn?\n\nA: The block offer is nowhere near where the Maui's live.\nB: There has not been a single observation of a Maui's dolphin in the block offer area.\nC: The block offer covers some of the most important areas where Maui's dolphins live.\nD: The block offer overlaps with the southern end of the Maui's dolphin range.\nE: The block offer covers most of the area where the Maui's dolphins live.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-30.jpg?height=503&width=682&top_left_y=277&top_left_x=113",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-30.jpg?height=497&width=848&top_left_y=277&top_left_x=1049"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1382",
        "problem": "Betsy and Bob are both high-risk carriers of a hereditary illness, $X$. After receiving genetic counselling, they were told that the probability of any child they have inheriting $X$ is $12 \\%$. The illness affects boys and girls equally, both in terms of severity and in the probability of inheritance. Betsy and Bob plan to have two children.\n\nCalculate the probability that they will have two healthy (non-affected) sons.\nA: $11.4 \\%$\nB: $12 \\%$\nC: $19.4 \\%$\nD: $44 \\%$\nE: $77.4 \\%$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBetsy and Bob are both high-risk carriers of a hereditary illness, $X$. After receiving genetic counselling, they were told that the probability of any child they have inheriting $X$ is $12 \\%$. The illness affects boys and girls equally, both in terms of severity and in the probability of inheritance. Betsy and Bob plan to have two children.\n\nCalculate the probability that they will have two healthy (non-affected) sons.\n\nA: $11.4 \\%$\nB: $12 \\%$\nC: $19.4 \\%$\nD: $44 \\%$\nE: $77.4 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1171",
        "problem": "Geometric isomers are molecules which have the same molecular formula with the same connectivity between atoms but which have a different orientation across a double bond. The diagram below shows the different types of geometric isomers.\n\n## Geometric isomers\n\n[figure1]\n\nCis\n\n[figure2]\n\n(Z)\n\n[figure3]\n\nTrans\n\n[figure4]\n\n(E)\n\nWhen both sides of the double bond contains the same 2 groups, then cis and trans is used. Cis = same side, Trans $=$ opposite sides.\n\nWhen different groups are attached to either side, $Z$ and $E$ is used. $Z$ is when the higher priority groups (ranked according to the Cahn-Ingold-Prelog rules) are orientated on the same side across the double bond. Zusammen is the German word for together. $\\mathrm{E}$ is when the higher priority groups are orientated on different sides across the double bond. Entgegen is the German word for opposed.\n\nThe isomerization of a molecule may have significant effects on its biological function. Tamoxifen is an antiestrogenic drug used to treat and prevent a form of breast cancer without the negative side effects of traditional chemotherapy. It exists as (Z)-Tamoxifen and its structure is shown below (1). It also exists less commonly as another isomer (2) that has estrogenic activity and therefore promotes the growth of breast cancer.(2)\n\nEstrogenic activity\n\nIsomer 2 is best described as:\nA: Cis-Tamoxifen\nB: Trans-Tamoxifen\nC: E-Tamoxifen\nD: Cis(Z)-Tamoxifen\nE: None of the above\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGeometric isomers are molecules which have the same molecular formula with the same connectivity between atoms but which have a different orientation across a double bond. The diagram below shows the different types of geometric isomers.\n\n## Geometric isomers\n\n[figure1]\n\nCis\n\n[figure2]\n\n(Z)\n\n[figure3]\n\nTrans\n\n[figure4]\n\n(E)\n\nWhen both sides of the double bond contains the same 2 groups, then cis and trans is used. Cis = same side, Trans $=$ opposite sides.\n\nWhen different groups are attached to either side, $Z$ and $E$ is used. $Z$ is when the higher priority groups (ranked according to the Cahn-Ingold-Prelog rules) are orientated on the same side across the double bond. Zusammen is the German word for together. $\\mathrm{E}$ is when the higher priority groups are orientated on different sides across the double bond. Entgegen is the German word for opposed.\n\nThe isomerization of a molecule may have significant effects on its biological function. Tamoxifen is an antiestrogenic drug used to treat and prevent a form of breast cancer without the negative side effects of traditional chemotherapy. It exists as (Z)-Tamoxifen and its structure is shown below (1). It also exists less commonly as another isomer (2) that has estrogenic activity and therefore promotes the growth of breast cancer.(2)\n\nEstrogenic activity\n\nIsomer 2 is best described as:\n\nA: Cis-Tamoxifen\nB: Trans-Tamoxifen\nC: E-Tamoxifen\nD: Cis(Z)-Tamoxifen\nE: None of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-20.jpg?height=214&width=411&top_left_y=453&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-20.jpg?height=220&width=400&top_left_y=775&top_left_x=128",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-20.jpg?height=214&width=403&top_left_y=453&top_left_x=541",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-20.jpg?height=222&width=412&top_left_y=777&top_left_x=545"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_621",
        "problem": "SREBP 前体由 $\\mathrm{S}$ 蛋白协助从内质网转运到高尔基体, 经酶切后产生具有转录调节活性的结构域，随后转运到细胞核激活胆固醇合成相关基因的表达。白华醋醇能特异性结合 S 蛋白并抑制其活化。下列相关叙述错误的是（）\nA: 胆固醇不溶于水, 在人体内参与血液中脂质的运输\nB: SREBP 前体常以囊泡形式从内质网转运到高尔基体加工\nC: S 蛋白可以调节胆固醇合成酶基因在细胞核内的转录\nD: 白桦醋醇能抑制胆固醇合成并降低血液中胆固醇含量\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nSREBP 前体由 $\\mathrm{S}$ 蛋白协助从内质网转运到高尔基体, 经酶切后产生具有转录调节活性的结构域，随后转运到细胞核激活胆固醇合成相关基因的表达。白华醋醇能特异性结合 S 蛋白并抑制其活化。下列相关叙述错误的是（）\n\nA: 胆固醇不溶于水, 在人体内参与血液中脂质的运输\nB: SREBP 前体常以囊泡形式从内质网转运到高尔基体加工\nC: S 蛋白可以调节胆固醇合成酶基因在细胞核内的转录\nD: 白桦醋醇能抑制胆固醇合成并降低血液中胆固醇含量\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1174",
        "problem": "Which one of the following conditions is essential to maintain a large, balanced fresh-water community?\nA: The number of autotrophs must be greater than the number of heterotrophs\nB: The productivity of the plants must be greater than the productivity of the animals\nC: The energy lost by the plants must not exceed the energy lost by the animals\nD: The number of carnivores must not exceed the number of herbivores\nE: The total biomass of the autotrophs must exceed the total biomass of the heterotrophs\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following conditions is essential to maintain a large, balanced fresh-water community?\n\nA: The number of autotrophs must be greater than the number of heterotrophs\nB: The productivity of the plants must be greater than the productivity of the animals\nC: The energy lost by the plants must not exceed the energy lost by the animals\nD: The number of carnivores must not exceed the number of herbivores\nE: The total biomass of the autotrophs must exceed the total biomass of the heterotrophs\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1145",
        "problem": "GNS scientist, James Crampton, and his research team have been using NZ's extensive, though imperfect, fossil record of molluscs (clams and snails) together with molecular data of living species to study evolution and the patterns of marine biodiversity change over the past 50 million years in the New Zealand region.\n\n[figure1]\n\nThe graph shows the diversity of New Zealand molluscs over the past 50 million years in relation to rock outcrop area.This graph demonstrates that\nA: Diversity has increased over the last 50 million years.\nB: Diversity has remained more-or-less constant throughout the last 50 million years.\nC: Diversity has fluctuated significantly over the last 50 million years.\nD: Diversity has fluctuated widely but with an underlying increase over the last 50 million years.\nE: Diversity remained relatively constant for much of the last 50 million years but declined over the past few million.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nGNS scientist, James Crampton, and his research team have been using NZ's extensive, though imperfect, fossil record of molluscs (clams and snails) together with molecular data of living species to study evolution and the patterns of marine biodiversity change over the past 50 million years in the New Zealand region.\n\n[figure1]\n\nThe graph shows the diversity of New Zealand molluscs over the past 50 million years in relation to rock outcrop area.\n\nproblem:\nThis graph demonstrates that\n\nA: Diversity has increased over the last 50 million years.\nB: Diversity has remained more-or-less constant throughout the last 50 million years.\nC: Diversity has fluctuated significantly over the last 50 million years.\nD: Diversity has fluctuated widely but with an underlying increase over the last 50 million years.\nE: Diversity remained relatively constant for much of the last 50 million years but declined over the past few million.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-22.jpg?height=543&width=965&top_left_y=368&top_left_x=134"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1363",
        "problem": "The following graph depicts the results of a study where the number of tadpoles within a particular habitat (finite resources) were experimentally varied.\n\nNote that the numbers next to the lines are the numbers of individuals in each treatment.\n\n[figure1]\n\nThis figure indicates that:\nA: access to resources decreases as population size decreases\nB: access to resources decreases as population size increases\nC: access to resources is not affected by population size\nD: carrying capacity does not affect population size\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph depicts the results of a study where the number of tadpoles within a particular habitat (finite resources) were experimentally varied.\n\nNote that the numbers next to the lines are the numbers of individuals in each treatment.\n\n[figure1]\n\nThis figure indicates that:\n\nA: access to resources decreases as population size decreases\nB: access to resources decreases as population size increases\nC: access to resources is not affected by population size\nD: carrying capacity does not affect population size\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-07.jpg?height=602&width=1205&top_left_y=1561&top_left_x=266"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_889",
        "problem": "现有一只基因型为 $b b X^{\\mathrm{R}} X^{\\mathrm{R}}$ 的雌果蝇（ $\\left.2 \\mathrm{n}=8\\right)$, 其性原细胞中的 I、II号染色体发生图示变异。变异细胞在减数分裂时, 所有染色体同源区段均须联会且相互分离, 才能形成可育配子。下列叙述错误的是（ ）\n\n[图1]\nA: 该变异细胞到 MI 后期时有 7 条染色体，到 MII后期时有 6 条或者 8 条染色体\nB: 图示变异不改变基因中碱基排列顺序\nC: 该果蝇可产生基因组成为 $\\mathrm{bX} \\mathrm{X}^{\\mathrm{R}}$ 的可育配子\nD: 该果蝇与 $B B X^{r} Y$ 的雄蝇杂交, $F_{1}$ 雄蝇产生 $b Y$ 精子的概率为 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n现有一只基因型为 $b b X^{\\mathrm{R}} X^{\\mathrm{R}}$ 的雌果蝇（ $\\left.2 \\mathrm{n}=8\\right)$, 其性原细胞中的 I、II号染色体发生图示变异。变异细胞在减数分裂时, 所有染色体同源区段均须联会且相互分离, 才能形成可育配子。下列叙述错误的是（ ）\n\n[图1]\n\nA: 该变异细胞到 MI 后期时有 7 条染色体，到 MII后期时有 6 条或者 8 条染色体\nB: 图示变异不改变基因中碱基排列顺序\nC: 该果蝇可产生基因组成为 $\\mathrm{bX} \\mathrm{X}^{\\mathrm{R}}$ 的可育配子\nD: 该果蝇与 $B B X^{r} Y$ 的雄蝇杂交, $F_{1}$ 雄蝇产生 $b Y$ 精子的概率为 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-60.jpg?height=374&width=791&top_left_y=578&top_left_x=375"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_102",
        "problem": "Taxonomy has traditionally been based on morphology. DNA Barcoding is a new approach that aims to allow accurate and relatively simple species identification based on the nucleotide sequence of a 650 -bp fragment of the mitochondrial COI gene. The Kimura-2 parameter (K2P) distance is a widely accepted index that reflects the divergence between two DNA sequences.\n\nSelected results relating to 16 species from five barcoding studies from different regions of the world are presented below. Three types of divergence values were calculated, global, intraregional, and interregional distances. Global divergence which is commonly used was the average of all pairwise comparisons of sequences belonging to the same species regardless of location of origin.\n\nIntraregional divergences were calculated by averaging the distances of all sequences belonging to the same species from the same location. Finally, interregional distances were calculated by averaging all distance values obtained from comparing each of the sequences from one location with all sequences of the same species in a second location.\n\nResults of three calculations (i, ii, and iii) are given below:\n\ni. The average standard deviation (SD) of 17 intraregional comparisons pertaining to 10 species was $0.11 \\%$ (minimum: $0 \\%$; maximum: $0.3 \\%$ ). The SD of divergences of an 18 th intraregional comparison pertaining to one of these species was $1.26 \\%$.\n\nii. The intraregional divergence of Argyrops spinifer specimens from India was $0.20 \\%$, and the interregional divergence of specimens from India and South Africa was $0.13 \\%$.\n\niii. Global divergence for Platycephalus indicus was $9.46 \\%$. Interregional divergences for this species were as follows:\n\nIndia/China $15.78 \\%$ China/Australia $12.05 \\%$\n\nIndia/Australia $10.61 \\%$ China/ S Africa $16.05 \\%$\n\nIndia/S Africa $4.05 \\%$ Australia/S Africa $10.95 \\%$\nA: The SD value of the 18th comparison in calculation (i) above is consistent with the suggestion that the fish compared in fact do not all belong to the same species.\nB: The K2P divergence values reported in calculation (ii) is consistent with the proposal that A. spinifer populations of India and South Africa arose from a common source population, but that there is greater variation in the ecological niches of India as compared to South Africa.\nC: The divergence values in calculation (iii) show that, as compared to interregional divergence values, global divergence values are more informative of extent of divergence that exists for P. indicus in the world.\nD: With reference to calculation (iii), the difference between global divergence value $(9.46 \\%)$ and average of interregional divergence values (11.58) can be\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nTaxonomy has traditionally been based on morphology. DNA Barcoding is a new approach that aims to allow accurate and relatively simple species identification based on the nucleotide sequence of a 650 -bp fragment of the mitochondrial COI gene. The Kimura-2 parameter (K2P) distance is a widely accepted index that reflects the divergence between two DNA sequences.\n\nSelected results relating to 16 species from five barcoding studies from different regions of the world are presented below. Three types of divergence values were calculated, global, intraregional, and interregional distances. Global divergence which is commonly used was the average of all pairwise comparisons of sequences belonging to the same species regardless of location of origin.\n\nIntraregional divergences were calculated by averaging the distances of all sequences belonging to the same species from the same location. Finally, interregional distances were calculated by averaging all distance values obtained from comparing each of the sequences from one location with all sequences of the same species in a second location.\n\nResults of three calculations (i, ii, and iii) are given below:\n\ni. The average standard deviation (SD) of 17 intraregional comparisons pertaining to 10 species was $0.11 \\%$ (minimum: $0 \\%$; maximum: $0.3 \\%$ ). The SD of divergences of an 18 th intraregional comparison pertaining to one of these species was $1.26 \\%$.\n\nii. The intraregional divergence of Argyrops spinifer specimens from India was $0.20 \\%$, and the interregional divergence of specimens from India and South Africa was $0.13 \\%$.\n\niii. Global divergence for Platycephalus indicus was $9.46 \\%$. Interregional divergences for this species were as follows:\n\nIndia/China $15.78 \\%$ China/Australia $12.05 \\%$\n\nIndia/Australia $10.61 \\%$ China/ S Africa $16.05 \\%$\n\nIndia/S Africa $4.05 \\%$ Australia/S Africa $10.95 \\%$\n\nA: The SD value of the 18th comparison in calculation (i) above is consistent with the suggestion that the fish compared in fact do not all belong to the same species.\nB: The K2P divergence values reported in calculation (ii) is consistent with the proposal that A. spinifer populations of India and South Africa arose from a common source population, but that there is greater variation in the ecological niches of India as compared to South Africa.\nC: The divergence values in calculation (iii) show that, as compared to interregional divergence values, global divergence values are more informative of extent of divergence that exists for P. indicus in the world.\nD: With reference to calculation (iii), the difference between global divergence value $(9.46 \\%)$ and average of interregional divergence values (11.58) can be\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_555",
        "problem": "将核 DNA 全部用 ${ }^{32} \\mathrm{P}$ 标记的果蝇精原细胞置于含 ${ }^{31} \\mathrm{P}$ 的培养液中培养。经过连续两次细胞分裂后产生 4 个子细胞，检测子细胞及细胞内染色体的标记情况。下列分析不正确的是 ( )\nA: 若进行有丝分裂, 则第二次分裂中期含 ${ }^{32} \\mathrm{P}$ 的染色单体有 8 条\nB: 若进行减数分裂, 则第二次分裂中期含 ${ }^{32} \\mathrm{P}$ 的染色单体有 8 条\nC: 含 ${ }^{32} \\mathrm{P}$ 染色体的子细胞比例为 1 , 则一定进行减数分裂\nD: 含 ${ }^{32} \\mathrm{P}$ 染色体的子细胞比例为 $1 / 2$, 则一定进行有丝分裂\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将核 DNA 全部用 ${ }^{32} \\mathrm{P}$ 标记的果蝇精原细胞置于含 ${ }^{31} \\mathrm{P}$ 的培养液中培养。经过连续两次细胞分裂后产生 4 个子细胞，检测子细胞及细胞内染色体的标记情况。下列分析不正确的是 ( )\n\nA: 若进行有丝分裂, 则第二次分裂中期含 ${ }^{32} \\mathrm{P}$ 的染色单体有 8 条\nB: 若进行减数分裂, 则第二次分裂中期含 ${ }^{32} \\mathrm{P}$ 的染色单体有 8 条\nC: 含 ${ }^{32} \\mathrm{P}$ 染色体的子细胞比例为 1 , 则一定进行减数分裂\nD: 含 ${ }^{32} \\mathrm{P}$ 染色体的子细胞比例为 $1 / 2$, 则一定进行有丝分裂\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_53",
        "problem": "Basilica of St Mark in Venice is an architectural marvel built in the 11th century CE. When examining the architecture, it is hard to avoid the spandrels in the corners of the ceiling, like the one in the figure (the triangular shape delineated by the dotted lines).\n\nGould \\& Lewontin (1979) argue that these spandrels are the result of simple geometric constraints one encounters when trying to hold a dome with column and arches. If we were to use the tools of evolutionary biology we would at first consider the spandrels as \"adaptations\" for the purpose of having the decorations, but in fact they are non-adaptive features, and we cannot count them as adaptations.\n\n[figure1]\nchoose all the false statements\nA: If the feather characteristic of birds first evolved in dinosaurs for the purpose of thermal regulation, that feather is similar to spandrels.\nB: When investigating the evolution of small regulatory RNAs (where the phenotype is closely tied to the genotype), biologists should not presume\nC: Biological \"spandrels\" are more common in species with historically high population size.\nD: The evolution of big brain in humans would be a biological spandrel if its evolutionary trajectory is constrained by developmental process in primates.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBasilica of St Mark in Venice is an architectural marvel built in the 11th century CE. When examining the architecture, it is hard to avoid the spandrels in the corners of the ceiling, like the one in the figure (the triangular shape delineated by the dotted lines).\n\nGould \\& Lewontin (1979) argue that these spandrels are the result of simple geometric constraints one encounters when trying to hold a dome with column and arches. If we were to use the tools of evolutionary biology we would at first consider the spandrels as \"adaptations\" for the purpose of having the decorations, but in fact they are non-adaptive features, and we cannot count them as adaptations.\n\n[figure1]\nchoose all the false statements\n\nA: If the feather characteristic of birds first evolved in dinosaurs for the purpose of thermal regulation, that feather is similar to spandrels.\nB: When investigating the evolution of small regulatory RNAs (where the phenotype is closely tied to the genotype), biologists should not presume\nC: Biological \"spandrels\" are more common in species with historically high population size.\nD: The evolution of big brain in humans would be a biological spandrel if its evolutionary trajectory is constrained by developmental process in primates.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-52.jpg?height=365&width=512&top_left_y=780&top_left_x=772"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1225",
        "problem": "The table shows the recombination (crossover) values found in an investigation of three linked alleles.\n\n| Between alleles | \\% recombination |\n| :---: | :---: |\n| $P r$ and $b$ | 6.1 |\n| $P r$ and $v g$ | 10.2 |\n| $B$ and $v g$ | 16.3 |\n\nWhich one of the following is the correct linkage map?\n\n[figure1]\nA: A\nB: B\nC: C\nD: D\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe table shows the recombination (crossover) values found in an investigation of three linked alleles.\n\n| Between alleles | \\% recombination |\n| :---: | :---: |\n| $P r$ and $b$ | 6.1 |\n| $P r$ and $v g$ | 10.2 |\n| $B$ and $v g$ | 16.3 |\n\nWhich one of the following is the correct linkage map?\n\n[figure1]\n\nA: A\nB: B\nC: C\nD: D\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-18.jpg?height=529&width=485&top_left_y=1161&top_left_x=820"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_456",
        "problem": "人类某遗传病病因是体内缺乏物质 $\\mathrm{C}$, 物质 $\\mathrm{C}$ 合成途径如图所示, 已知基因 $\\mathrm{A} 、 \\mathrm{~B}$均位于 $\\mathrm{X}$ 染色体上, $\\mathrm{I}_{2}$ 基因型为 $\\mathrm{X}^{\\mathrm{Ab}} \\mathrm{Y}$, 不考虑变异。根据系谱图, 下列分析错误的是\n\n[图1]\nA: 基因 $A 、 B$ 同时存在时, 个体才不患病\nB: $\\mathrm{I}_{2}$ 与 $\\mathrm{II}_{2}$ 基因型不可能相同\nC: $\\mathrm{II}_{3}$ 的基因型一定为 $\\mathrm{X}^{\\mathrm{Ab}} \\mathrm{X}^{\\mathrm{aB}}$\nD: $\\mathrm{III}_{2}$ 与 $\\mathrm{III}_{3}$ 再生一个患病男孩的概率为 $1 / 8$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人类某遗传病病因是体内缺乏物质 $\\mathrm{C}$, 物质 $\\mathrm{C}$ 合成途径如图所示, 已知基因 $\\mathrm{A} 、 \\mathrm{~B}$均位于 $\\mathrm{X}$ 染色体上, $\\mathrm{I}_{2}$ 基因型为 $\\mathrm{X}^{\\mathrm{Ab}} \\mathrm{Y}$, 不考虑变异。根据系谱图, 下列分析错误的是\n\n[图1]\n\nA: 基因 $A 、 B$ 同时存在时, 个体才不患病\nB: $\\mathrm{I}_{2}$ 与 $\\mathrm{II}_{2}$ 基因型不可能相同\nC: $\\mathrm{II}_{3}$ 的基因型一定为 $\\mathrm{X}^{\\mathrm{Ab}} \\mathrm{X}^{\\mathrm{aB}}$\nD: $\\mathrm{III}_{2}$ 与 $\\mathrm{III}_{3}$ 再生一个患病男孩的概率为 $1 / 8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-16.jpg?height=431&width=1376&top_left_y=1212&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_85",
        "problem": "The figure below shows the nucleotide sequence of the mouse $\\beta$-globin gene. The DNA nucleotide sequence represents the coding strand, and the 3-letter abbreviations below represent the amino acid sequence. The 79th cAp marked with the black arrow is the 5' capping site, and the 1467th pA is the site where the poly-A tail is attached.\n\n[figure1]\n\nWhich of the following statements about this gene structure is correct?\nA: This gene has 3 introns and 4 exons.\nB: The size of the mature mRNA, not including the poly-A tail, is about $1389 \\mathrm{nt}$.\nC: Transcription starts at nucleotide 132.\nD: The region between the nucleotide sequence 1336 and 1467 is the $3^{\\prime}$ untranslated region of the mRNA.\nE: The promoter of this gene resides in the region up to nucleotide 131 .\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe figure below shows the nucleotide sequence of the mouse $\\beta$-globin gene. The DNA nucleotide sequence represents the coding strand, and the 3-letter abbreviations below represent the amino acid sequence. The 79th cAp marked with the black arrow is the 5' capping site, and the 1467th pA is the site where the poly-A tail is attached.\n\n[figure1]\n\nWhich of the following statements about this gene structure is correct?\n\nA: This gene has 3 introns and 4 exons.\nB: The size of the mature mRNA, not including the poly-A tail, is about $1389 \\mathrm{nt}$.\nC: Transcription starts at nucleotide 132.\nD: The region between the nucleotide sequence 1336 and 1467 is the $3^{\\prime}$ untranslated region of the mRNA.\nE: The promoter of this gene resides in the region up to nucleotide 131 .\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-10.jpg?height=1205&width=1679&top_left_y=665&top_left_x=246"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1273",
        "problem": "The diagram below represents the nitrogen cycle.\n\n[figure1]\n\nWhich one of $\\mathbf{A}$ to $\\mathbf{E}$ correctly identifies the events happening at points $\\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$ in the cycle?\nA: $\\mathrm{X}$: Non-biological nitrogen fixation, $\\mathrm{Y}$: Decay by microorganisms, $\\mathrm{Z}$: Nitrification\nB: $\\mathrm{X}$: Biological nitrogen fixation, $\\mathrm{Y}$: Nitrification, $\\mathrm{Z}$: Denitrification\nC: $\\mathrm{X}$: Non-biological nitrogen fixation, $\\mathrm{Y}$: Decay by microorganisms, $\\mathrm{Z}$: Denitrification\nD: $\\mathrm{X}$: Non-biological nitrogen fixation, $\\mathrm{Y}$: Nitrification, $\\mathrm{Z}$: Denitrification\nE: $\\mathrm{X}$: Biological nitrogen fixation, $\\mathrm{Y}$: Decay by microorganisms, $\\mathrm{Z}$: Nitrification\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram below represents the nitrogen cycle.\n\n[figure1]\n\nWhich one of $\\mathbf{A}$ to $\\mathbf{E}$ correctly identifies the events happening at points $\\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$ in the cycle?\n\nA: $\\mathrm{X}$: Non-biological nitrogen fixation, $\\mathrm{Y}$: Decay by microorganisms, $\\mathrm{Z}$: Nitrification\nB: $\\mathrm{X}$: Biological nitrogen fixation, $\\mathrm{Y}$: Nitrification, $\\mathrm{Z}$: Denitrification\nC: $\\mathrm{X}$: Non-biological nitrogen fixation, $\\mathrm{Y}$: Decay by microorganisms, $\\mathrm{Z}$: Denitrification\nD: $\\mathrm{X}$: Non-biological nitrogen fixation, $\\mathrm{Y}$: Nitrification, $\\mathrm{Z}$: Denitrification\nE: $\\mathrm{X}$: Biological nitrogen fixation, $\\mathrm{Y}$: Decay by microorganisms, $\\mathrm{Z}$: Nitrification\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-04.jpg?height=621&width=1523&top_left_y=286&top_left_x=298"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_235",
        "problem": "A researcher aimed to induce undifferentiated cells by expressing multiple genes in human fibroblasts. They focused on 24 genes that were identified as highly expressed in embryonic stem (ES) cells. It was found that when all 24 genes were simultaneously introduced in fibroblasts, the colony formation characteristics of undifferentiated cells occurs. Next the researchers tried to find the minimum set of genes that induce undifferentiated cells. The graph shows the colony formation when 23 genes except one were introduced into fibroblast cells.\n\n[figure1]\n\n(\"All\" on the $\\mathrm{X}$ axis shows that the 24 genes are all introduced.)\nA: These results show that colonies can be formed through the introduction of genes 14,15 , and 20 into the fibroblast together.\nB: These results show that gene 14, gene15, and gene 20 are required for colony formation.\nC: These results show that the colony number is the highest when gene 9 is introduced into the fibroblast.\nD: This experiment alone is not sufficient to find the minimum gene set needed to induce colonies.\nE: These results show that genes 14,15 , and 20 are expressed in fibroblast cells.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA researcher aimed to induce undifferentiated cells by expressing multiple genes in human fibroblasts. They focused on 24 genes that were identified as highly expressed in embryonic stem (ES) cells. It was found that when all 24 genes were simultaneously introduced in fibroblasts, the colony formation characteristics of undifferentiated cells occurs. Next the researchers tried to find the minimum set of genes that induce undifferentiated cells. The graph shows the colony formation when 23 genes except one were introduced into fibroblast cells.\n\n[figure1]\n\n(\"All\" on the $\\mathrm{X}$ axis shows that the 24 genes are all introduced.)\n\nA: These results show that colonies can be formed through the introduction of genes 14,15 , and 20 into the fibroblast together.\nB: These results show that gene 14, gene15, and gene 20 are required for colony formation.\nC: These results show that the colony number is the highest when gene 9 is introduced into the fibroblast.\nD: This experiment alone is not sufficient to find the minimum gene set needed to induce colonies.\nE: These results show that genes 14,15 , and 20 are expressed in fibroblast cells.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-37.jpg?height=582&width=1487&top_left_y=1017&top_left_x=296"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1158",
        "problem": "Which one of the following can only be explained by mass flow of solutes through the phloem?\nA: Sap is exuded when a sieve tube element is pierced\nB: The phloem conducts heat longitudinally only in the direction in which solutes are being transported\nC: Movement through the phloem ceases when living cells in the stem are killed by heat \nD: Metabolic inhibitors stop solute flow through the phloem\nE: A sudden reduction in sap pressure causes precipitation of p-protein\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following can only be explained by mass flow of solutes through the phloem?\n\nA: Sap is exuded when a sieve tube element is pierced\nB: The phloem conducts heat longitudinally only in the direction in which solutes are being transported\nC: Movement through the phloem ceases when living cells in the stem are killed by heat \nD: Metabolic inhibitors stop solute flow through the phloem\nE: A sudden reduction in sap pressure causes precipitation of p-protein\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_877",
        "problem": "果蝇有 3 对常染色体 (分别编号为II、III、IV) 和 1 对性染色体。科研人员培育出果蝇甲品系, 其 4 种突变性状（多翅脉、卷曲翅、短刚毛、针圆平衡棒）分别由一种显性突变基因控制，位置如下图，突变基因纯合时胚胎致死（不考虑交叉互换）。下列叙述正确的是（）\n\n[图1]\n\nII 号染色体\n\n[图2]\n\nIII号染色体\nA: 果蝇甲品系的雌、雄个体间相互交配，子代果蝇的成活率为 $25 \\%$\nB: 雄性果蝇的 1 个染色体组含有 3 条常染色体以及 $\\mathrm{X}$ 和 $\\mathrm{Y}$ 两条性染色体\nC: 果蝇甲品系的 4 种显性突变基因都是由于基因中碱基对的替换引起的\nD: 摩尔根用果蝇证明了基因位于染色体上, 并提出了遗传的染色体学说\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇有 3 对常染色体 (分别编号为II、III、IV) 和 1 对性染色体。科研人员培育出果蝇甲品系, 其 4 种突变性状（多翅脉、卷曲翅、短刚毛、针圆平衡棒）分别由一种显性突变基因控制，位置如下图，突变基因纯合时胚胎致死（不考虑交叉互换）。下列叙述正确的是（）\n\n[图1]\n\nII 号染色体\n\n[图2]\n\nIII号染色体\n\nA: 果蝇甲品系的雌、雄个体间相互交配，子代果蝇的成活率为 $25 \\%$\nB: 雄性果蝇的 1 个染色体组含有 3 条常染色体以及 $\\mathrm{X}$ 和 $\\mathrm{Y}$ 两条性染色体\nC: 果蝇甲品系的 4 种显性突变基因都是由于基因中碱基对的替换引起的\nD: 摩尔根用果蝇证明了基因位于染色体上, 并提出了遗传的染色体学说\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-21.jpg?height=185&width=391&top_left_y=2369&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-21.jpg?height=163&width=439&top_left_y=2374&top_left_x=840"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1119",
        "problem": "After initial labeling of plants in an old field community, accumulation of radioactive phosphorus in different trophic levels was monitored over the next 25 days. The results are shown in the graph. P,Q, R \\& S respectively most likely represent:\n\n[figure1]\nA: Detrivore, herbivore, primary carnivore, secondary carnivore.\nB: Detrivore, omnivore, herbivore, carnivore.\nC: Primary carnivore, detrivore, herbivore, omnivore.\nD: Herbivore, omnivore, primary carnivore, secondary carnivore.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAfter initial labeling of plants in an old field community, accumulation of radioactive phosphorus in different trophic levels was monitored over the next 25 days. The results are shown in the graph. P,Q, R \\& S respectively most likely represent:\n\n[figure1]\n\nA: Detrivore, herbivore, primary carnivore, secondary carnivore.\nB: Detrivore, omnivore, herbivore, carnivore.\nC: Primary carnivore, detrivore, herbivore, omnivore.\nD: Herbivore, omnivore, primary carnivore, secondary carnivore.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-10.jpg?height=626&width=979&top_left_y=733&top_left_x=432"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1239",
        "problem": "The light and dark forms of the peppered moth Biston betularia illustrate discontinuous variation. This means that\nA: there are no intermediate forms\nB: they are confined to certain areas\nC: selection by predation has eliminated further variation\nD: they are unlikely to vary in nature\nE: the frequency of the rarer form is lower than that which could be accounted for by recurrent mutation\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe light and dark forms of the peppered moth Biston betularia illustrate discontinuous variation. This means that\n\nA: there are no intermediate forms\nB: they are confined to certain areas\nC: selection by predation has eliminated further variation\nD: they are unlikely to vary in nature\nE: the frequency of the rarer form is lower than that which could be accounted for by recurrent mutation\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1082",
        "problem": "Metamerism is a salient feature found in Annelids, dividing the body into a series of similar segments, each containing repeated arrangements of many of the internal organ systems. One of the following systems which is not metameric is the:\nA: integumentary system.\nB: nervous system.\nC: digestive system.\nD: excretory system.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMetamerism is a salient feature found in Annelids, dividing the body into a series of similar segments, each containing repeated arrangements of many of the internal organ systems. One of the following systems which is not metameric is the:\n\nA: integumentary system.\nB: nervous system.\nC: digestive system.\nD: excretory system.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_415",
        "problem": "下图为某红绿色盲家族系谱图，相关基因用 $\\mathrm{X}^{\\mathrm{B}} 、 \\mathrm{X}^{\\mathrm{b}}$ 表示。人的 $\\mathrm{MN}$ 血型基因位于常染色体上, 基因型有 3 种: $L^{M} L^{M}$ ( $M$ 型)、 $L^{N} L^{N}$ ( $N$ 型)、 $L$ （型）已知I-1、I-3 为 $M$型, I-2、 I-4 为 $\\mathrm{N}$ 型。下列叙述正确的是 ( )\n\n[图1]\nA: II-3 的基因型可能为 $\\mathrm{L}^{\\mathrm{M}} \\mathrm{L}^{\\mathrm{N}} X^{\\mathrm{B}} X^{\\mathrm{B}}$\nB: II-4 的血型可能为 M 型或 MN 型\nC: II-2 是红绿色盲基因携带者的概率为 $1 / 2$\nD: III- 1 携带的的 $\\mathrm{X}^{\\mathrm{b}}$ 概率为 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下图为某红绿色盲家族系谱图，相关基因用 $\\mathrm{X}^{\\mathrm{B}} 、 \\mathrm{X}^{\\mathrm{b}}$ 表示。人的 $\\mathrm{MN}$ 血型基因位于常染色体上, 基因型有 3 种: $L^{M} L^{M}$ ( $M$ 型)、 $L^{N} L^{N}$ ( $N$ 型)、 $L$ （型）已知I-1、I-3 为 $M$型, I-2、 I-4 为 $\\mathrm{N}$ 型。下列叙述正确的是 ( )\n\n[图1]\n\nA: II-3 的基因型可能为 $\\mathrm{L}^{\\mathrm{M}} \\mathrm{L}^{\\mathrm{N}} X^{\\mathrm{B}} X^{\\mathrm{B}}$\nB: II-4 的血型可能为 M 型或 MN 型\nC: II-2 是红绿色盲基因携带者的概率为 $1 / 2$\nD: III- 1 携带的的 $\\mathrm{X}^{\\mathrm{b}}$ 概率为 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-096.jpg?height=340&width=917&top_left_y=1555&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1050",
        "problem": "Some genes evolve rapidly while some others are highly conserved. The nucleotide substitution that occurs in a gene is of two types: Synonymous and Nonsynonymous. They are defined as follows:\n\nSynonymous substitution: Change in a single nucleotide that does not alter the encoded amino acid.\n\nNon-synonymous substitution: Change in a single nucleotide that alters the encoded amino acid.\n\nWhich of the following is correct?\nA: Rate of non-synonymous mutation will always be higher than that of synonymous one if the gene is highly conserved.\nB: Rate of non-synonymous and synonymous mutations could be comparable for a rapidly evolving gene.\nC: Rate of non-synonymous and synonymous mutations will depend on the number of amino acids present in the protein.\nD: Greater rate of non-synonymous mutation will indicate that the gene is more tolerant to nucleotide changes.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nSome genes evolve rapidly while some others are highly conserved. The nucleotide substitution that occurs in a gene is of two types: Synonymous and Nonsynonymous. They are defined as follows:\n\nSynonymous substitution: Change in a single nucleotide that does not alter the encoded amino acid.\n\nNon-synonymous substitution: Change in a single nucleotide that alters the encoded amino acid.\n\nWhich of the following is correct?\n\nA: Rate of non-synonymous mutation will always be higher than that of synonymous one if the gene is highly conserved.\nB: Rate of non-synonymous and synonymous mutations could be comparable for a rapidly evolving gene.\nC: Rate of non-synonymous and synonymous mutations will depend on the number of amino acids present in the protein.\nD: Greater rate of non-synonymous mutation will indicate that the gene is more tolerant to nucleotide changes.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_52",
        "problem": "Some fruit flies (Drosophila melanogaster) have a mutation that makes them shake. These fruit flies are called 'shakers'.\n\nAn experimental cross is shown below:\n\n[figure1]\n\nWhat kind of inheritance best explains the inheritance pattern for the shaker gene?\nA: Somatic dominant.\nB: Somatic recessive.\nC: X-linked dominant.\nD: X-linked recessive.\nE: Y-linked dominant.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSome fruit flies (Drosophila melanogaster) have a mutation that makes them shake. These fruit flies are called 'shakers'.\n\nAn experimental cross is shown below:\n\n[figure1]\n\nWhat kind of inheritance best explains the inheritance pattern for the shaker gene?\n\nA: Somatic dominant.\nB: Somatic recessive.\nC: X-linked dominant.\nD: X-linked recessive.\nE: Y-linked dominant.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-49.jpg?height=671&width=1256&top_left_y=595&top_left_x=500"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1048",
        "problem": "How many root meristems do you estimate are on this plant?\n\n[figure1]\nA: 1\nB: 2\nC: 12\nD: 24\nE: 48\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many root meristems do you estimate are on this plant?\n\n[figure1]\n\nA: 1\nB: 2\nC: 12\nD: 24\nE: 48\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_c11579df664240d6158ag-05.jpg?height=558&width=391&top_left_y=369&top_left_x=951"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1264",
        "problem": "## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAAYou will note when looking at the dissected bolus' that coloured plastic is more common than transparent and that much of the plastic is red or orange, rather than blue. What is the LEAST LIKELY inference that can be drawn from these observations?\nA: Albatross select specific colours, mistaking them for prey items such as crustaceans.\nB: Albatross are non-selective feeders and consume plastic debris in proportion to its abundance.\nC: Transparent items are harder to see and are therefore seldom consumed by albatross.\nD: Coloured plastic items are more easily seen and therefore consumed by albatross.\nE: Blue plastic items are harder to see and are therefore seldom consumed by albatross.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAA\n\nproblem:\nYou will note when looking at the dissected bolus' that coloured plastic is more common than transparent and that much of the plastic is red or orange, rather than blue. What is the LEAST LIKELY inference that can be drawn from these observations?\n\nA: Albatross select specific colours, mistaking them for prey items such as crustaceans.\nB: Albatross are non-selective feeders and consume plastic debris in proportion to its abundance.\nC: Transparent items are harder to see and are therefore seldom consumed by albatross.\nD: Coloured plastic items are more easily seen and therefore consumed by albatross.\nE: Blue plastic items are harder to see and are therefore seldom consumed by albatross.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=800&width=617&top_left_y=1299&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=817&width=1097&top_left_y=1299&top_left_x=845"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1499",
        "problem": "Dynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\nHow many different alignments have the highest score?\nA: 1\nB: 2\nC: 3\nD: 4\nE: 5\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\nHow many different alignments have the highest score?\n\nA: 1\nB: 2\nC: 3\nD: 4\nE: 5\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=925&width=928&top_left_y=354&top_left_x=238",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=922&width=928&top_left_y=1458&top_left_x=238"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_450",
        "problem": "研究表明, 在人体细胞的细胞核内, 甲状腺激素受体与 DNA 上的某些片段结合,抑制 $\\mathrm{A}$ 蛋白基因（指导合成 $\\mathrm{A}$ 蛋白）的表达。当甲状腺激素与甲状腺激素受体结合后会解除该受体对 $\\mathrm{A}$ 蛋白基因表达的抑制。当垂体释放促甲状腺激素（TSH）的量增加之后，下列说法不正确的是（ ）\nA: 人体内 $\\mathrm{TSH}$ 含量与 $\\mathrm{A}$ 蛋白的含量成反比\nB: 甲状腺激素受体对 A 蛋白基因表达的抑制减弱\nC: 促甲状腺激素可作为信号分子来调控基因的表达\nD: 甲状腺激素对下丘脑和垂体分泌活动的抑制作用增强\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究表明, 在人体细胞的细胞核内, 甲状腺激素受体与 DNA 上的某些片段结合,抑制 $\\mathrm{A}$ 蛋白基因（指导合成 $\\mathrm{A}$ 蛋白）的表达。当甲状腺激素与甲状腺激素受体结合后会解除该受体对 $\\mathrm{A}$ 蛋白基因表达的抑制。当垂体释放促甲状腺激素（TSH）的量增加之后，下列说法不正确的是（ ）\n\nA: 人体内 $\\mathrm{TSH}$ 含量与 $\\mathrm{A}$ 蛋白的含量成反比\nB: 甲状腺激素受体对 A 蛋白基因表达的抑制减弱\nC: 促甲状腺激素可作为信号分子来调控基因的表达\nD: 甲状腺激素对下丘脑和垂体分泌活动的抑制作用增强\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_840",
        "problem": "雄性家蚕生命力强, 饲料利用效率高, 茧丝质量好。现有家蚕品种甲和乙, 它们的试卷第 36 页，共 102 页\n性染色体、10 号染色体及卵色基因组成如图所示。已知蚕卵（受精卵）同时具有 A、B 基因时为黑色, 否则为白色。育种工作者期望利用甲、乙品种获得新品种，以实现通过选择卵色只养雄蚕（不考虑其他变异）, 下列说法正确的是（）\n\n[图1]\n\n甲 $(Z W ，$ P $)$\n\n[图2]\n\n乙 $(\\mathrm{ZZ}, \\hat{O})$\nA: 甲品种家蚕 $\\mathrm{W}$ 染色体上有来自 10 号染色体的片段, 该变异属于基因重组\nB: $\\mathrm{F}_{1}$ 代雌蚕的基因型为 $A a B Z W{ }^{A B}$ ，可以产生 8 种配子\nC: 让甲、乙品种家蚕杂交, 在 $F_{2}$ 代的雄蚕中, 卵色为白色的概率为 $1 / 16$\nD: 在 $F_{2}$ 代中，选择白色蚕卵孵化即可实现只养雄蚕\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n雄性家蚕生命力强, 饲料利用效率高, 茧丝质量好。现有家蚕品种甲和乙, 它们的试卷第 36 页，共 102 页\n性染色体、10 号染色体及卵色基因组成如图所示。已知蚕卵（受精卵）同时具有 A、B 基因时为黑色, 否则为白色。育种工作者期望利用甲、乙品种获得新品种，以实现通过选择卵色只养雄蚕（不考虑其他变异）, 下列说法正确的是（）\n\n[图1]\n\n甲 $(Z W ，$ P $)$\n\n[图2]\n\n乙 $(\\mathrm{ZZ}, \\hat{O})$\n\nA: 甲品种家蚕 $\\mathrm{W}$ 染色体上有来自 10 号染色体的片段, 该变异属于基因重组\nB: $\\mathrm{F}_{1}$ 代雌蚕的基因型为 $A a B Z W{ }^{A B}$ ，可以产生 8 种配子\nC: 让甲、乙品种家蚕杂交, 在 $F_{2}$ 代的雄蚕中, 卵色为白色的概率为 $1 / 16$\nD: 在 $F_{2}$ 代中，选择白色蚕卵孵化即可实现只养雄蚕\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-037.jpg?height=520&width=531&top_left_y=408&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-037.jpg?height=517&width=531&top_left_y=410&top_left_x=1042"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1483",
        "problem": "The adult cells of baobab trees have 168 chromosomes, compared to 46 in humans. These contain four copies of the genome, compared to two copies in humans.\n\n[figure1]\n\nHow many chromosomes need to be sequenced to get one complete copy of the baobab genome?\nA: 168\nB: 84\nC: 63\nD: 42\nE: 21\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe adult cells of baobab trees have 168 chromosomes, compared to 46 in humans. These contain four copies of the genome, compared to two copies in humans.\n\n[figure1]\n\nHow many chromosomes need to be sequenced to get one complete copy of the baobab genome?\n\nA: 168\nB: 84\nC: 63\nD: 42\nE: 21\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-49.jpg?height=1228&width=925&top_left_y=474&top_left_x=240"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_181",
        "problem": "0\nA: A completely recessive allele is lethal in homozygous condition and if its dominant allele mutates to recessive allele at a rate of $10^{-6}$, then frequency of the lethal allele when the population reaches mutation-selection equilibrium is 0.001 .\nB: If frequency of a completely recessive lethal allele is 0.2 and it remains unchanged from generation to generation due to the superior fitness of heterozygotes, then the intensity of selection against the dominant homozygotes should be 0.025 .\nC: Selection for recessive alleles is less effective than selection against recessive alleles.\nD: In a large, randomly mating population, the frequency of an autosomal recessive lethal allele is 0.2 . Frequency of this allele in the next generation will be 0.07 if the lethality takes place before reproduction.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n0\n\nA: A completely recessive allele is lethal in homozygous condition and if its dominant allele mutates to recessive allele at a rate of $10^{-6}$, then frequency of the lethal allele when the population reaches mutation-selection equilibrium is 0.001 .\nB: If frequency of a completely recessive lethal allele is 0.2 and it remains unchanged from generation to generation due to the superior fitness of heterozygotes, then the intensity of selection against the dominant homozygotes should be 0.025 .\nC: Selection for recessive alleles is less effective than selection against recessive alleles.\nD: In a large, randomly mating population, the frequency of an autosomal recessive lethal allele is 0.2 . Frequency of this allele in the next generation will be 0.07 if the lethality takes place before reproduction.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_519",
        "problem": "为研究某种植物 3 种营养成分 (A、B 和 C) 含量的遗传机制, 先采用 CRISPR/Cas9\n基因编辑技术，对野生型进行基因敲除突变实验，经分子鉴定获得 3 个突变植株（M1、\n\nM2、M3)，其自交一代结果见下表，表中高或低指营养成分含量高或低。\n\n| 植株（表现型） | 自交一代植株数目（表现型） |\n| :---: | :---: |\n| 野生型（A 低 B 低 C <br> 高） | 150（A 低 B 低 C 高） |\n| Ml（A 低 B 低 C 高） | $60 （ \\mathrm{~A}$ 高 B 低 C 低）、181（A 低 B 低 C 高）、79（A <br> 低 B 低 C 低） |\n| M2（A 低 B 低 C 高） | 122 (A 高 $\\mathrm{B}$ 低 C 低)、91（A 低 B 高 C 低）、272（A <br> 低 B 低 C 高） |\n| M3（A 低 B 低 C 高） | $59 （ \\mathrm{~A}$ 低 B 高 C 低）、179（A 低 B 低 C 高）、80（A <br> 低 B 低 C 低） |\n\n下列叙述正确的是 ( )\nA: 从 $\\mathrm{Ml}$ 自交一代中取纯合的（A 高 B 低 C 低）植株, 与 M2 基因型相同的植株杂交, 理论上其杂交一代中只出现（A 高 B 低 C 低）和（A 低 B 低 C 高）两种表现型, 且比例一定是 1:1\nB: 从 M2 自交一代中取纯合的（A 低 B 高 C 低）植株, 与 M3 基因型相同的植株杂交，理论上其杂交一代中，纯合基因型个体数：杂合基因型个体数一定是 1：1 C. M3 在产生花粉的减数分裂过程中, 某对同源染色体有一小段没有配对, 说明其中一个同源染色体上一定是由于基因敲除缺失了一个片段\nC: 可从突变植株自交一代中取 A 高植株与 B 高植株杂交, 从后代中选取 A 和 B 两种成分均高的植株, 再与 $\\mathrm{C}$ 高植株杂交, 从杂交后代中能选到 $\\mathrm{A} 、 \\mathrm{~B}$ 和 $\\mathrm{C}$ 三种成分均高的植株\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n为研究某种植物 3 种营养成分 (A、B 和 C) 含量的遗传机制, 先采用 CRISPR/Cas9\n基因编辑技术，对野生型进行基因敲除突变实验，经分子鉴定获得 3 个突变植株（M1、\n\nM2、M3)，其自交一代结果见下表，表中高或低指营养成分含量高或低。\n\n| 植株（表现型） | 自交一代植株数目（表现型） |\n| :---: | :---: |\n| 野生型（A 低 B 低 C <br> 高） | 150（A 低 B 低 C 高） |\n| Ml（A 低 B 低 C 高） | $60 （ \\mathrm{~A}$ 高 B 低 C 低）、181（A 低 B 低 C 高）、79（A <br> 低 B 低 C 低） |\n| M2（A 低 B 低 C 高） | 122 (A 高 $\\mathrm{B}$ 低 C 低)、91（A 低 B 高 C 低）、272（A <br> 低 B 低 C 高） |\n| M3（A 低 B 低 C 高） | $59 （ \\mathrm{~A}$ 低 B 高 C 低）、179（A 低 B 低 C 高）、80（A <br> 低 B 低 C 低） |\n\n下列叙述正确的是 ( )\n\nA: 从 $\\mathrm{Ml}$ 自交一代中取纯合的（A 高 B 低 C 低）植株, 与 M2 基因型相同的植株杂交, 理论上其杂交一代中只出现（A 高 B 低 C 低）和（A 低 B 低 C 高）两种表现型, 且比例一定是 1:1\nB: 从 M2 自交一代中取纯合的（A 低 B 高 C 低）植株, 与 M3 基因型相同的植株杂交，理论上其杂交一代中，纯合基因型个体数：杂合基因型个体数一定是 1：1 C. M3 在产生花粉的减数分裂过程中, 某对同源染色体有一小段没有配对, 说明其中一个同源染色体上一定是由于基因敲除缺失了一个片段\nC: 可从突变植株自交一代中取 A 高植株与 B 高植株杂交, 从后代中选取 A 和 B 两种成分均高的植株, 再与 $\\mathrm{C}$ 高植株杂交, 从杂交后代中能选到 $\\mathrm{A} 、 \\mathrm{~B}$ 和 $\\mathrm{C}$ 三种成分均高的植株\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-028.jpg?height=59&width=1276&top_left_y=2272&top_left_x=344",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-029.jpg?height=66&width=1382&top_left_y=1069&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_845",
        "problem": "图 1 为人类某种单基因遗传病的系谱图（相关基因用 $\\mathrm{E} 、 \\mathrm{e}$ 表示), 此病是由于正常基因中间缺失一段较大的 DNA 片段所致。图 2 是对图 1 中的部分个体的相关基因进行扩增和电泳的结果（离点样侧距离远的 DNA 片段分子量较小，不考虑致死、突变和 X、 Y 染色体同源区段遗传)。下列叙述错误的是（）\n\nI\n\nII\n[图1]\n\nII\n\n[图2]\n\n图1点样侧\n\n[图3]\n\n图2\nA: 根据图 1 推测, 此病可能为伴 X 染色体显性遗传病\nB: 根据图 2 推测, 片段乙为致病基因, a 个体的基因型为 ee\nC: 若 $\\mathrm{I}_{3}$ 的电泳结果为 $\\mathrm{c}$, 则 $\\mathrm{II}_{4}$ 电泳的结果为 $\\mathrm{b}$\nD: 若 $\\mathrm{II}_{3}$ 的电泳结果为 $\\mathrm{b}$, 则 $\\mathrm{III}_{2}$ 与正常男性结婚生育患病男孩的概率为 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n图 1 为人类某种单基因遗传病的系谱图（相关基因用 $\\mathrm{E} 、 \\mathrm{e}$ 表示), 此病是由于正常基因中间缺失一段较大的 DNA 片段所致。图 2 是对图 1 中的部分个体的相关基因进行扩增和电泳的结果（离点样侧距离远的 DNA 片段分子量较小，不考虑致死、突变和 X、 Y 染色体同源区段遗传)。下列叙述错误的是（）\n\nI\n\nII\n[图1]\n\nII\n\n[图2]\n\n图1点样侧\n\n[图3]\n\n图2\n\nA: 根据图 1 推测, 此病可能为伴 X 染色体显性遗传病\nB: 根据图 2 推测, 片段乙为致病基因, a 个体的基因型为 ee\nC: 若 $\\mathrm{I}_{3}$ 的电泳结果为 $\\mathrm{c}$, 则 $\\mathrm{II}_{4}$ 电泳的结果为 $\\mathrm{b}$\nD: 若 $\\mathrm{II}_{3}$ 的电泳结果为 $\\mathrm{b}$, 则 $\\mathrm{III}_{2}$ 与正常男性结婚生育患病男孩的概率为 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-085.jpg?height=268&width=190&top_left_y=914&top_left_x=401",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-085.jpg?height=350&width=593&top_left_y=996&top_left_x=503",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-085.jpg?height=342&width=529&top_left_y=994&top_left_x=1089"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_624",
        "problem": "桔树抗病性状由基因 $Q$ 控制, 易感病性状由基因 q 控制, 细胞中另一对等位基因 $N / n$对抗 病基因的表达有影响, $\\mathrm{Nn}$ 使抗病性减弱。现用两纯合亲本杂交, $\\mathrm{F}_{1}$ 全为弱抗病, $F_{1}$ 自交得 $F_{2}, F_{2}$ 中易感病：弱抗病：抗病 $=7: 6: 3$ 。下列叙述正确的是 $(\\quad)$\nA: 上述两对等位基因的遗传不遵循自由组合定律\nB: $F_{2}$ 易感病植株的基因型有 4 种\nC: $F_{2}$ 中的弱抗病植株有纯合子和杂合子\nD: $F_{2}$ 中抗病植株自交, 后代抗病植株占 $5 / 6$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n桔树抗病性状由基因 $Q$ 控制, 易感病性状由基因 q 控制, 细胞中另一对等位基因 $N / n$对抗 病基因的表达有影响, $\\mathrm{Nn}$ 使抗病性减弱。现用两纯合亲本杂交, $\\mathrm{F}_{1}$ 全为弱抗病, $F_{1}$ 自交得 $F_{2}, F_{2}$ 中易感病：弱抗病：抗病 $=7: 6: 3$ 。下列叙述正确的是 $(\\quad)$\n\nA: 上述两对等位基因的遗传不遵循自由组合定律\nB: $F_{2}$ 易感病植株的基因型有 4 种\nC: $F_{2}$ 中的弱抗病植株有纯合子和杂合子\nD: $F_{2}$ 中抗病植株自交, 后代抗病植株占 $5 / 6$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_24",
        "problem": "Inflammatory caspases such as caspase 1 are activated in response to microbial infection and stress signals. When activated, they cleave human gasdermin D (GSDMD) after Asp275, to generate an Nterminal cleavage product (GSDMD-NT) and a C-terminal fragment (GSDMD-CT). GSDMD-NT kills bacteria and induces a form of programmed cell death called pyroptosis in human cells. The details of the mechanism by which GSDMD-NT causes cell death are unknown. Mutation of two evolutionarily conserved positively charged residues to alanine produces a mutant form of the protein known as GSDMD-NT 2A, and mutation of four conserved positively charged residues to alanine produces GSDMD-NT 4A. Fig. 1 shows result of non-reducing gel electrophoresis of equal amounts of N-terminal cleavage product of the wild type and mutated forms of GSDMD.\n\nIn an experiment, E. coli and Staphylococcus aureus bacteria were exposed to nanomolar concentrations of recombinant forms of GSDMD, GSDMD-NT, GSDMD-CT, GSDMD-NT 4A and granulysin (a known cytotoxic lymphocyte pore-forming protein) and the antibacterial effect of these molecules was assessed by measuring reduction of colony formation (Colony Forming Unit, CFU) (Fig. 2). Other experiments showed the same relative effects of the wild type and mutant forms on pyroptosis in human cells.\n\n[figure1]\n\nFigure 1\n\n[figure2]\n\n| $(\\mathrm{nM})$ | - | 50 | 100 | 200 | 400 | - | - | - | - | - | - | - | - |\n| ---: | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| GSDM-NT 4A (nM) | - | - | - | - | - | 400 | - | - | - | - | - | - | - |\n| GSDMD-CT (nM) | - | - | - | - | - | - | 400 | - | - | - | - | - | - |\n| GSDMD (nM) | - | - | - | - | - | - | - | 400 | - | - | - | - | - |\n| Granulysin (nM) | - | - | - | - | - | - | - | - | 50100 | 2004001000 |  |  |  |\n\nFigure 2\nA: The data shows that GSDMD-NT oligomerization is involved in pyroptosis.\nB: GSDMD-NT is a more effective anti-bacterial agent than granulysin.\nC: By mutation of evolutionarily conserved basic residues to Ala, the number of disulfide bonds between monomers of GSDMD does not change.\nD: GSDMD-NT 2A mutant is likely less effective than GSDMD-NT for induction of pyroptosis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nInflammatory caspases such as caspase 1 are activated in response to microbial infection and stress signals. When activated, they cleave human gasdermin D (GSDMD) after Asp275, to generate an Nterminal cleavage product (GSDMD-NT) and a C-terminal fragment (GSDMD-CT). GSDMD-NT kills bacteria and induces a form of programmed cell death called pyroptosis in human cells. The details of the mechanism by which GSDMD-NT causes cell death are unknown. Mutation of two evolutionarily conserved positively charged residues to alanine produces a mutant form of the protein known as GSDMD-NT 2A, and mutation of four conserved positively charged residues to alanine produces GSDMD-NT 4A. Fig. 1 shows result of non-reducing gel electrophoresis of equal amounts of N-terminal cleavage product of the wild type and mutated forms of GSDMD.\n\nIn an experiment, E. coli and Staphylococcus aureus bacteria were exposed to nanomolar concentrations of recombinant forms of GSDMD, GSDMD-NT, GSDMD-CT, GSDMD-NT 4A and granulysin (a known cytotoxic lymphocyte pore-forming protein) and the antibacterial effect of these molecules was assessed by measuring reduction of colony formation (Colony Forming Unit, CFU) (Fig. 2). Other experiments showed the same relative effects of the wild type and mutant forms on pyroptosis in human cells.\n\n[figure1]\n\nFigure 1\n\n[figure2]\n\n| $(\\mathrm{nM})$ | - | 50 | 100 | 200 | 400 | - | - | - | - | - | - | - | - |\n| ---: | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| GSDM-NT 4A (nM) | - | - | - | - | - | 400 | - | - | - | - | - | - | - |\n| GSDMD-CT (nM) | - | - | - | - | - | - | 400 | - | - | - | - | - | - |\n| GSDMD (nM) | - | - | - | - | - | - | - | 400 | - | - | - | - | - |\n| Granulysin (nM) | - | - | - | - | - | - | - | - | 50100 | 2004001000 |  |  |  |\n\nFigure 2\n\nA: The data shows that GSDMD-NT oligomerization is involved in pyroptosis.\nB: GSDMD-NT is a more effective anti-bacterial agent than granulysin.\nC: By mutation of evolutionarily conserved basic residues to Ala, the number of disulfide bonds between monomers of GSDMD does not change.\nD: GSDMD-NT 2A mutant is likely less effective than GSDMD-NT for induction of pyroptosis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-07.jpg?height=568&width=417&top_left_y=1092&top_left_x=822",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-07.jpg?height=425&width=1191&top_left_y=1758&top_left_x=495"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_231",
        "problem": "A researcher recorded neurotransmitter responses from a neurosecretory neuron in the hypothalamus. Gammaaminobutyric acid (GABA) is well-known as the neurotransmitter at most inhibitory synapses in the brain. The researcher found that the application of GABA to this neuron induced more action potentials (Figure 1). Then, the researcher measured GABA-induced chloride current responses of the neuron under various experimentally controlled membrane potentials (from -50 to $0 \\mathrm{mV}$ at $10 \\mathrm{mV}$ steps; Figure 2). They also plotted maximum current amplitudes (current differences before and after the GABA application) against membrane potentials (Figure 3). A downward deflection of a current trace is referred to as an inward current and reflects the movement of Clions out of the cell (Figure 4). Table 1 shows the intra- and extracellular concentrations and the equilibrium potentials of sodium, potassium, and chloride ions calculated by Nernst's equation.\n\n[figure1]\n\nFigure 1\n\n[figure2]\n\nFigure 2\n\n[figure3]\n\nFigure 3\n\n\n\n[figure4]\n\nFigure 4\n\n| Ion | Concentration (mM) |  | Equilibrium <br> Potential <br> $(\\mathrm{mV})$ |\n| :---: | :---: | :---: | :---: |\n|  | Inner | Outer |  |\n| $\\mathrm{Na}^{+}$ | 15 | 150 | 58 |\n| $\\mathrm{K}^{+}$ | 140 | 7 | -75 |\n| $\\mathrm{Cl}^{-}$ | 40 | 120 | -28 |\n\nTable 1\nA: When the membrane potential was $-10 \\mathrm{mV}$, the application of GABA induced the depolarization of the recorded neuron.\nB: The equilibrium potential of chloride ions was more positive (less negative) than the resting membrane potential of the recorded neuron.\nC: Under the presence of tetrodotoxin (pufferfish toxin that blocks the generation of action potentials), the higher concentration of GABA depolarized the neuron more positively than $0 \\mathrm{mV}$.\nD: The researcher recorded other neurons. The neurons hyperpolarized their membrane potentials by GABA. If the resting membrane potential of both neurons are the same, intracellular chloride ion concentration of the hyperpolarized neurons is lower than those of the neurons observed in Figure 1-4.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA researcher recorded neurotransmitter responses from a neurosecretory neuron in the hypothalamus. Gammaaminobutyric acid (GABA) is well-known as the neurotransmitter at most inhibitory synapses in the brain. The researcher found that the application of GABA to this neuron induced more action potentials (Figure 1). Then, the researcher measured GABA-induced chloride current responses of the neuron under various experimentally controlled membrane potentials (from -50 to $0 \\mathrm{mV}$ at $10 \\mathrm{mV}$ steps; Figure 2). They also plotted maximum current amplitudes (current differences before and after the GABA application) against membrane potentials (Figure 3). A downward deflection of a current trace is referred to as an inward current and reflects the movement of Clions out of the cell (Figure 4). Table 1 shows the intra- and extracellular concentrations and the equilibrium potentials of sodium, potassium, and chloride ions calculated by Nernst's equation.\n\n[figure1]\n\nFigure 1\n\n[figure2]\n\nFigure 2\n\n[figure3]\n\nFigure 3\n\n\n\n[figure4]\n\nFigure 4\n\n| Ion | Concentration (mM) |  | Equilibrium <br> Potential <br> $(\\mathrm{mV})$ |\n| :---: | :---: | :---: | :---: |\n|  | Inner | Outer |  |\n| $\\mathrm{Na}^{+}$ | 15 | 150 | 58 |\n| $\\mathrm{K}^{+}$ | 140 | 7 | -75 |\n| $\\mathrm{Cl}^{-}$ | 40 | 120 | -28 |\n\nTable 1\n\nA: When the membrane potential was $-10 \\mathrm{mV}$, the application of GABA induced the depolarization of the recorded neuron.\nB: The equilibrium potential of chloride ions was more positive (less negative) than the resting membrane potential of the recorded neuron.\nC: Under the presence of tetrodotoxin (pufferfish toxin that blocks the generation of action potentials), the higher concentration of GABA depolarized the neuron more positively than $0 \\mathrm{mV}$.\nD: The researcher recorded other neurons. The neurons hyperpolarized their membrane potentials by GABA. If the resting membrane potential of both neurons are the same, intracellular chloride ion concentration of the hyperpolarized neurons is lower than those of the neurons observed in Figure 1-4.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-32.jpg?height=451&width=1122&top_left_y=1199&top_left_x=387",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-32.jpg?height=411&width=536&top_left_y=1873&top_left_x=366",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-32.jpg?height=497&width=694&top_left_y=1870&top_left_x=952",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-33.jpg?height=371&width=756&top_left_y=271&top_left_x=313"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1482",
        "problem": "The NDMAR protein is a membrane channel found on neurones. It binds glutamate then opens to allow ions into the cell. Several drugs (NO, ethanol, ketamine) inhibit NDMAR and can cause powerful psychological dissociation. The effect of several inhibitors is shown.\n\n$$\n\\text { Ion entry }\n$$\n\n$$\n\\text { into neurone }\n$$\n[figure1]\n\nWhich inhibitor binds elsewhere but still prevents glutamate binding?\nA: A\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: All of them\nE: None of them\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe NDMAR protein is a membrane channel found on neurones. It binds glutamate then opens to allow ions into the cell. Several drugs (NO, ethanol, ketamine) inhibit NDMAR and can cause powerful psychological dissociation. The effect of several inhibitors is shown.\n\n$$\n\\text { Ion entry }\n$$\n\n$$\n\\text { into neurone }\n$$\n[figure1]\n\nWhich inhibitor binds elsewhere but still prevents glutamate binding?\n\nA: A\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: All of them\nE: None of them\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-10.jpg?height=960&width=1476&top_left_y=591&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_886",
        "problem": "28. 如图是某家族的遗传系谱图。甲、乙两种病均为单基因遗传病, 其中一对基因位于性染色体上， $\\mathrm{II}_{1}$ 不携带致病基因。理论上，下列相关分析错误的是（）\n\n[图1]\nA: $\\mathrm{I}_{2} 、 \\mathrm{III}_{1}$ 和 $^{2} \\mathrm{II}_{2}$ 的基因型相同\nB: $\\mathrm{II}_{2}$ 的次级精母细胞中最多含 4 个致病基因 \nC: [图2]\nD: $I_{1}$ 产生不含致病基因的精子的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n28. 如图是某家族的遗传系谱图。甲、乙两种病均为单基因遗传病, 其中一对基因位于性染色体上， $\\mathrm{II}_{1}$ 不携带致病基因。理论上，下列相关分析错误的是（）\n\n[图1]\n\nA: $\\mathrm{I}_{2} 、 \\mathrm{III}_{1}$ 和 $^{2} \\mathrm{II}_{2}$ 的基因型相同\nB: $\\mathrm{II}_{2}$ 的次级精母细胞中最多含 4 个致病基因 \nC: [图2]\nD: $I_{1}$ 产生不含致病基因的精子的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-029.jpg?height=300&width=740&top_left_y=1655&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-029.jpg?height=54&width=713&top_left_y=2143&top_left_x=403"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_318",
        "problem": "由于缺乏的凝血因子不同, 血友病存在甲和乙两种类型。控制甲型血友病的基因为 $a$, 位于 $X$ 染色体上, 控制乙型血友病的基因为 $b$, 位于常染色体上。图 1 表示某家系血友病的遗传图谱, 图 2 表示该家系部分成员与血友病有关的基因的电泳结果（A、B、 $a, b$ 基因均只电泳出一个条带)。下列叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 条带(1)表示 B 基因, 条带(2)表示 A 基因\nB: 若对 5 号进行电泳, 可能含有条带(1)(2)(4)\nC: 5 号与基因型和 3 号相同的女性婚配, 他们生育患病女儿的概率是 $1 / 6$\nD: 某些染色体结构的变异可能导致机体患血友病\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n由于缺乏的凝血因子不同, 血友病存在甲和乙两种类型。控制甲型血友病的基因为 $a$, 位于 $X$ 染色体上, 控制乙型血友病的基因为 $b$, 位于常染色体上。图 1 表示某家系血友病的遗传图谱, 图 2 表示该家系部分成员与血友病有关的基因的电泳结果（A、B、 $a, b$ 基因均只电泳出一个条带)。下列叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 条带(1)表示 B 基因, 条带(2)表示 A 基因\nB: 若对 5 号进行电泳, 可能含有条带(1)(2)(4)\nC: 5 号与基因型和 3 号相同的女性婚配, 他们生育患病女儿的概率是 $1 / 6$\nD: 某些染色体结构的变异可能导致机体患血友病\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-04.jpg?height=346&width=402&top_left_y=2283&top_left_x=333",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-04.jpg?height=360&width=1013&top_left_y=2264&top_left_x=727"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1359",
        "problem": "Cellular respiration is a complex process involving many steps. Generally, these steps are grouped into 3 main phases: glycolysis, the citric acid cycle, and the electron transport chain. The process of glycolysis can be summarised as follows:\n\n[figure1]\n\nDuring glycolysis, there are steps at which ATP is created from ADP and phosphate and steps at which ATP is broken down. Which of the following options can be ruled out based on the information provided?\nA: The creation and breakdown of ATP occur at different locations within the cell.\nB: The order of reactions is irrelevant, net ATP produced is the same.\nC: The breakdown of ATP releases energy that is required for glycolysis to proceed.\nD: Glycolysis involves energy investment and payoff phases which are both necessary for respiration.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCellular respiration is a complex process involving many steps. Generally, these steps are grouped into 3 main phases: glycolysis, the citric acid cycle, and the electron transport chain. The process of glycolysis can be summarised as follows:\n\n[figure1]\n\nDuring glycolysis, there are steps at which ATP is created from ADP and phosphate and steps at which ATP is broken down. Which of the following options can be ruled out based on the information provided?\n\nA: The creation and breakdown of ATP occur at different locations within the cell.\nB: The order of reactions is irrelevant, net ATP produced is the same.\nC: The breakdown of ATP releases energy that is required for glycolysis to proceed.\nD: Glycolysis involves energy investment and payoff phases which are both necessary for respiration.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-34.jpg?height=1204&width=832&top_left_y=532&top_left_x=286"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_444",
        "problem": "TM4 为侵染耻垢分枝杆菌的双链 DNA 噬菌体。耻垢分枝杆菌的 $\\operatorname{stpK} 7$ 基因是维持 TM4 噬菌体的吸附能力并抑制细胞死亡的关键基因。以 TM4、敲除 $\\operatorname{stpK} 7$ 和未敲除 $\\operatorname{stpK} 7$的耻垢分枝杆菌为实验材料, 按照赫尔希和蔡斯的噬菌体侵染大肠杆菌实验流程, 进行下表中的相关实验，实验结果分析错误的是（ ）\n\n| 选 | ${ }^{32} \\mathrm{P} 、{ }^{35} \\mathrm{~S}$ 标记情况和 TM4 侵染情况 | 实验结果分析 |\n| :---: | :---: | :---: |\n| A | 用 ${ }^{35} \\mathrm{~S}$ 标记的 TM4 侵染敲除 stpK7 <br> 的耻垢分枝杆菌 | 放射性集中在上清液中 |\n| B | 用未标记的 TM4 侵染 ${ }^{32} \\mathrm{P}$ 标记的未 <br> 敲除 $\\operatorname{stpK} 7$ 的耻垢分枝杆菌 | 子代 TM4 的两条 DNA 单链均带有 ${ }^{32} \\mathrm{P}$ 标记 |\n| C | 用 ${ }^{32} \\mathrm{P}$ 标记的 TM4 分别侵染未敲除 <br> stpK7 和敲除 stpK7 的耻垢分枝杆 <br> 菌 | 沉淀中放射性强度敲除 $\\operatorname{stpK7}$ 的耻垢分枝杆 <br> 菌组低于未敲除 $\\operatorname{stpK} 7$ 组 |\n\n\n| $\\mathrm{D}$ | 用未标记的 TM4 侵染 ${ }^{35} \\mathrm{~S}$ 标记的未 <br> 敲除 $\\operatorname{stpK} 7$ 和敲除 $\\operatorname{stpK} 7$ 的耻垢分 <br> 枝杆菌 | 两组子代 TM4 放射性强度有明显差别 |\n| :--- | :--- | :--- |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nTM4 为侵染耻垢分枝杆菌的双链 DNA 噬菌体。耻垢分枝杆菌的 $\\operatorname{stpK} 7$ 基因是维持 TM4 噬菌体的吸附能力并抑制细胞死亡的关键基因。以 TM4、敲除 $\\operatorname{stpK} 7$ 和未敲除 $\\operatorname{stpK} 7$的耻垢分枝杆菌为实验材料, 按照赫尔希和蔡斯的噬菌体侵染大肠杆菌实验流程, 进行下表中的相关实验，实验结果分析错误的是（ ）\n\n| 选 | ${ }^{32} \\mathrm{P} 、{ }^{35} \\mathrm{~S}$ 标记情况和 TM4 侵染情况 | 实验结果分析 |\n| :---: | :---: | :---: |\n| A | 用 ${ }^{35} \\mathrm{~S}$ 标记的 TM4 侵染敲除 stpK7 <br> 的耻垢分枝杆菌 | 放射性集中在上清液中 |\n| B | 用未标记的 TM4 侵染 ${ }^{32} \\mathrm{P}$ 标记的未 <br> 敲除 $\\operatorname{stpK} 7$ 的耻垢分枝杆菌 | 子代 TM4 的两条 DNA 单链均带有 ${ }^{32} \\mathrm{P}$ 标记 |\n| C | 用 ${ }^{32} \\mathrm{P}$ 标记的 TM4 分别侵染未敲除 <br> stpK7 和敲除 stpK7 的耻垢分枝杆 <br> 菌 | 沉淀中放射性强度敲除 $\\operatorname{stpK7}$ 的耻垢分枝杆 <br> 菌组低于未敲除 $\\operatorname{stpK} 7$ 组 |\n\n\n| $\\mathrm{D}$ | 用未标记的 TM4 侵染 ${ }^{35} \\mathrm{~S}$ 标记的未 <br> 敲除 $\\operatorname{stpK} 7$ 和敲除 $\\operatorname{stpK} 7$ 的耻垢分 <br> 枝杆菌 | 两组子代 TM4 放射性强度有明显差别 |\n| :--- | :--- | :--- |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_583",
        "problem": "南瓜是雌雄同株异花, 以长蔓种居多, 长蔓种在生长发育过程中所需空间较大, 单位面积产量不高。科研人员在研究中发现了一株突变的短蔓种, 并通过培育得到自交不分离短蔓种, 将长蔓种和此短蔓种做正反交实验, 结果如下表。下列叙述错误的是\n\n| 亲本 |  | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :--- | :--- | :--- | :--- |\n| 长蔓种O | 短蔓种へ | 短蔓种 36 <br> 株 | 短蔓种 363 株, 长曼种 28 <br> 株 |\n| 长曼种へ | 短曼种O | 短蔓种 54 <br> 株 | 短蔓种 445 株, 长蔓种 34 <br> 株 |\nA: 控制短蔓的基因受常染色体上两对等位基因控制, 且独立遗传\nB: 短蔓种产生的配子中, 存在致死现象\nC: 在育种中需要经过“去雄-套袋-传粉-套袋”操作\nD: 通过单倍体育种技术可获得自交不分离短蔓种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n南瓜是雌雄同株异花, 以长蔓种居多, 长蔓种在生长发育过程中所需空间较大, 单位面积产量不高。科研人员在研究中发现了一株突变的短蔓种, 并通过培育得到自交不分离短蔓种, 将长蔓种和此短蔓种做正反交实验, 结果如下表。下列叙述错误的是\n\n| 亲本 |  | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :--- | :--- | :--- | :--- |\n| 长蔓种O | 短蔓种へ | 短蔓种 36 <br> 株 | 短蔓种 363 株, 长曼种 28 <br> 株 |\n| 长曼种へ | 短曼种O | 短蔓种 54 <br> 株 | 短蔓种 445 株, 长蔓种 34 <br> 株 |\n\nA: 控制短蔓的基因受常染色体上两对等位基因控制, 且独立遗传\nB: 短蔓种产生的配子中, 存在致死现象\nC: 在育种中需要经过“去雄-套袋-传粉-套袋”操作\nD: 通过单倍体育种技术可获得自交不分离短蔓种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1000",
        "problem": "[figure1]\n\nThe cross section (b) indicated here is a:\nA: Root\nB: Stem\nC: Leaf petiole\nD: Flower bud\nE: None of the above\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nThe cross section (b) indicated here is a:\n\nA: Root\nB: Stem\nC: Leaf petiole\nD: Flower bud\nE: None of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_c11579df664240d6158ag-05.jpg?height=442&width=480&top_left_y=1655&top_left_x=1077"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_387",
        "problem": "DNA 标记是 DNA 或基因中一些稳定的特征性碱基序列, 通过相应 DNA 标记 PCR 扩增产物的电泳图谱可进行遗传病致病基因的定位。下图为某家庭中甲、乙两种遗传病相关基因（包含正常基因和致病基因）的电泳图，该家庭中父母均不患乙病，儿子同时患甲、乙两种病，不考虑 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体同源区段。下列叙述错误的是（）\n\n[图1]\nA: PCR 扩增需要知道基因中全部的特征性碱基序列\nB: 甲病基因位于常染色体上, 乙病基因位于 X 染色体上\nC: 若该家庭再生一个女儿, 患病的概率为 $100 \\%$ D．遗传咨询和产前诊断能在一定程度上预防甲、乙病\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nDNA 标记是 DNA 或基因中一些稳定的特征性碱基序列, 通过相应 DNA 标记 PCR 扩增产物的电泳图谱可进行遗传病致病基因的定位。下图为某家庭中甲、乙两种遗传病相关基因（包含正常基因和致病基因）的电泳图，该家庭中父母均不患乙病，儿子同时患甲、乙两种病，不考虑 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体同源区段。下列叙述错误的是（）\n\n[图1]\n\nA: PCR 扩增需要知道基因中全部的特征性碱基序列\nB: 甲病基因位于常染色体上, 乙病基因位于 X 染色体上\nC: 若该家庭再生一个女儿, 患病的概率为 $100 \\%$ D．遗传咨询和产前诊断能在一定程度上预防甲、乙病\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-60.jpg?height=466&width=648&top_left_y=1543&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_966",
        "problem": "When the base composition of DNA from bacterium Fusobacterium novum was determined, $32 \\%$ of the bases were found to be guanine. What is the pyrimidine content?\nA: $18 \\%$\nB: $32 \\%$\nC: $50 \\%$\nD: $64 \\%$\nE: $72 \\%$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen the base composition of DNA from bacterium Fusobacterium novum was determined, $32 \\%$ of the bases were found to be guanine. What is the pyrimidine content?\n\nA: $18 \\%$\nB: $32 \\%$\nC: $50 \\%$\nD: $64 \\%$\nE: $72 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_771",
        "problem": "科研人员在种植野生型玉米的田间, 发现两株隐性矮秆突变体 $\\mathrm{X}$ 和 $\\mathrm{Y}(\\mathrm{X}$ 和 $\\mathrm{Y}$ 均为单基因突变), 为探究两种突变体是否是同一位点基因突变导致, 让两种突变体杂交后, $F_{1}$ 再自交 (不考虑染色体互换), 观察并统计 $F_{2}$ 的表型及比例。下列分析错误的是\nA: 若两突变基因为同一位点基因突变导致, 则 $F_{2}$ 均为矮秆\nB: 若两突变基因是非同源染色体不同位点的基因, 则 $F_{2}$ 为高秆:矮秆 $=9: 7$\nC: 若两突变基因是一对同源染色体不同位点的基因, 则 $F_{2}$ 为高秆:矮秆 $=3: 1$\nD: 若两突变基因的遗传遵循自由组合定律，则 $\\mathrm{F}_{2}$ 矮秆植株中纯合子占 $3 / 7$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n科研人员在种植野生型玉米的田间, 发现两株隐性矮秆突变体 $\\mathrm{X}$ 和 $\\mathrm{Y}(\\mathrm{X}$ 和 $\\mathrm{Y}$ 均为单基因突变), 为探究两种突变体是否是同一位点基因突变导致, 让两种突变体杂交后, $F_{1}$ 再自交 (不考虑染色体互换), 观察并统计 $F_{2}$ 的表型及比例。下列分析错误的是\n\nA: 若两突变基因为同一位点基因突变导致, 则 $F_{2}$ 均为矮秆\nB: 若两突变基因是非同源染色体不同位点的基因, 则 $F_{2}$ 为高秆:矮秆 $=9: 7$\nC: 若两突变基因是一对同源染色体不同位点的基因, 则 $F_{2}$ 为高秆:矮秆 $=3: 1$\nD: 若两突变基因的遗传遵循自由组合定律，则 $\\mathrm{F}_{2}$ 矮秆植株中纯合子占 $3 / 7$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1327",
        "problem": "Which one of the following assumptions is not necessary in order to estimate the size of an isolated mouse population, using he capture/mark/recapture method?\nA: Marked and unmarked mice are randomly distributed in the population\nB: Recaptured mice are a random sample of the population\nC: There is no difference in the survival of marked and unmarked mice\nD: Marked an unmarked mice have the same reproductive success\nE: Marked and unmarked mice are equally likely to be recaptured\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following assumptions is not necessary in order to estimate the size of an isolated mouse population, using he capture/mark/recapture method?\n\nA: Marked and unmarked mice are randomly distributed in the population\nB: Recaptured mice are a random sample of the population\nC: There is no difference in the survival of marked and unmarked mice\nD: Marked an unmarked mice have the same reproductive success\nE: Marked and unmarked mice are equally likely to be recaptured\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_692",
        "problem": "某植物的叶色同时受 $\\mathrm{E} 、 \\mathrm{e}$ 与 $\\mathrm{F} 、 \\mathrm{f}$ 两对基因控制, 两对基因都位于常染色体上。绿叶植株基因型为 E_ff, 紫叶植株基因型为 $e_{-} \\mathrm{F}_{-}$。将某绿叶植株与紫叶植株作为亲本进行杂交, 所得红叶植株自交得 $F_{2}, F_{2}$ 的表型及比例为红叶: 紫叶: 绿叶: 黄叶=7:3:1:1,不考虑突变。下列有关说法错误的是（ ）\nA: $\\mathrm{F}_{1}$ 产生的基因型为 $\\mathrm{eF}$ 的雌配子或雄配子致死\nB: 这两对基因的遗传遵循自由组合定律\nC: $F_{2}$ 中红叶植株的基因型有 4 种，且比例为 $3: 2: 1: 1$\nD: 取 $\\mathrm{F}_{2}$ 绿叶植株与紫叶植株进行正反交, 所得子代的表型比例不同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某植物的叶色同时受 $\\mathrm{E} 、 \\mathrm{e}$ 与 $\\mathrm{F} 、 \\mathrm{f}$ 两对基因控制, 两对基因都位于常染色体上。绿叶植株基因型为 E_ff, 紫叶植株基因型为 $e_{-} \\mathrm{F}_{-}$。将某绿叶植株与紫叶植株作为亲本进行杂交, 所得红叶植株自交得 $F_{2}, F_{2}$ 的表型及比例为红叶: 紫叶: 绿叶: 黄叶=7:3:1:1,不考虑突变。下列有关说法错误的是（ ）\n\nA: $\\mathrm{F}_{1}$ 产生的基因型为 $\\mathrm{eF}$ 的雌配子或雄配子致死\nB: 这两对基因的遗传遵循自由组合定律\nC: $F_{2}$ 中红叶植株的基因型有 4 种，且比例为 $3: 2: 1: 1$\nD: 取 $\\mathrm{F}_{2}$ 绿叶植株与紫叶植株进行正反交, 所得子代的表型比例不同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1416",
        "problem": "The pangolin is an endangered mammal, with 4 species native to Africa and another 4 species native to the Asian continent. Pangolins are protected by CITES, the Convention on International Trade in Endangered Species of Wild Fauna and Flora. CITES aims to ensure that the international trade in wild animals and plants does not threaten their survival.\n\nCITES records of intercepted pangolin trades (1977-2014) are summarised below. Red sectors in the pie graphs indicate that a country was the origin of an exported pangolin, blue indicates that a country was the destination for an imported pangolin. The width of the arrow and diameter of pie graphs indicate the relative size of the trade.\n\n[figure1]\n\nWhich of these statements is NOT supported by the data?\nA: New Zealand (NZ) is involved in the illegal trade of pangolins\nB: The USA (US) is the main destination for imported pangolins\nC: More pangolins are exported from Asia than Africa\nD: Italy (IT) exports more pangolins than the USA (US)\nE: Asian countries are involved in both the importation and exportation of pangolins\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe pangolin is an endangered mammal, with 4 species native to Africa and another 4 species native to the Asian continent. Pangolins are protected by CITES, the Convention on International Trade in Endangered Species of Wild Fauna and Flora. CITES aims to ensure that the international trade in wild animals and plants does not threaten their survival.\n\nCITES records of intercepted pangolin trades (1977-2014) are summarised below. Red sectors in the pie graphs indicate that a country was the origin of an exported pangolin, blue indicates that a country was the destination for an imported pangolin. The width of the arrow and diameter of pie graphs indicate the relative size of the trade.\n\n[figure1]\n\nWhich of these statements is NOT supported by the data?\n\nA: New Zealand (NZ) is involved in the illegal trade of pangolins\nB: The USA (US) is the main destination for imported pangolins\nC: More pangolins are exported from Asia than Africa\nD: Italy (IT) exports more pangolins than the USA (US)\nE: Asian countries are involved in both the importation and exportation of pangolins\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-35.jpg?height=1070&width=1056&top_left_y=630&top_left_x=500"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1428",
        "problem": "[figure1]\n\nA researcher is trying to synthesise a protein with tertiary structure. They have been able to produce a copy of the amino acid sequence, but have been unsuccessful in creating even one properly folded protein. What is the most likely reason for this?\nA: The experiment's temperature is too low.\nB: The solution is deficient in critical amino acids.\nC: The researchers have isolated the wrong segment of DNA\nD: The experiment's $\\mathrm{pH}$ does not mimic the correct cellular conditions.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nA researcher is trying to synthesise a protein with tertiary structure. They have been able to produce a copy of the amino acid sequence, but have been unsuccessful in creating even one properly folded protein. What is the most likely reason for this?\n\nA: The experiment's temperature is too low.\nB: The solution is deficient in critical amino acids.\nC: The researchers have isolated the wrong segment of DNA\nD: The experiment's $\\mathrm{pH}$ does not mimic the correct cellular conditions.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-36.jpg?height=560&width=1394&top_left_y=388&top_left_x=248"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_68",
        "problem": "'Coefficient of relatedness' (or 'genetic relatedness') refers to the probability of two related individuals inheriting a particular allele of a single gene from their common ancestor.\n\n[figure1]\n\nIn this family tree of diploid individuals, which of the following 'coefficient of relatedness' is not true?\n\nCoefficient of relatedness of\nA: A being $1 / 2$\nB: B being $1 / 2$\nC: C being $1 / 4$\nD: D being 1\nE: E being 1/4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n'Coefficient of relatedness' (or 'genetic relatedness') refers to the probability of two related individuals inheriting a particular allele of a single gene from their common ancestor.\n\n[figure1]\n\nIn this family tree of diploid individuals, which of the following 'coefficient of relatedness' is not true?\n\nCoefficient of relatedness of\n\nA: A being $1 / 2$\nB: B being $1 / 2$\nC: C being $1 / 4$\nD: D being 1\nE: E being 1/4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-50.jpg?height=508&width=1142&top_left_y=474&top_left_x=523"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1429",
        "problem": "The brain is a vitally important organ that has evolved a multitude of protective barriers, including the skull, cerebrospinal fluid, and the meninges.\n\nAnother protective element is the blood-brain barrier, which separates the blood in the circulatory system from the cerebrospinal fluid of the central nervous system. The bloodbrain barrier was discovered in the late 19th century, when Paul Ehrlich, a German physician, injected a blue dye into the bloodstream of a mouse. All of the tissues in the mouse turned blue, with the exception of the brain and spinal cord. The effectiveness of the blood-brain barrier depends on the cells that line the interior of blood vessels. In capillaries that form the blood-brain barrier, these cells are packed more closely together than in other capillaries.\n\nWhich of the following statements is true?\nA: The blue dye was composed of large hydrophilic molecules which were unable to diffuse across the blood-brain barrier\nB: The cells of the brain and spinal cord do not have a specific receptor for the dye to bind; therefore, it did not cause a colour change in these tissues.\nC: The molecules of blue dye were transported through the blood bound to large proteins, which were unable to diffuse across the blood-brain barrier.\nD: Cells that make up the capillary wall lack a phospholipid bilayer; therefore, the dye was unable to diffuse across the blood-brain barrier.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe brain is a vitally important organ that has evolved a multitude of protective barriers, including the skull, cerebrospinal fluid, and the meninges.\n\nAnother protective element is the blood-brain barrier, which separates the blood in the circulatory system from the cerebrospinal fluid of the central nervous system. The bloodbrain barrier was discovered in the late 19th century, when Paul Ehrlich, a German physician, injected a blue dye into the bloodstream of a mouse. All of the tissues in the mouse turned blue, with the exception of the brain and spinal cord. The effectiveness of the blood-brain barrier depends on the cells that line the interior of blood vessels. In capillaries that form the blood-brain barrier, these cells are packed more closely together than in other capillaries.\n\nWhich of the following statements is true?\n\nA: The blue dye was composed of large hydrophilic molecules which were unable to diffuse across the blood-brain barrier\nB: The cells of the brain and spinal cord do not have a specific receptor for the dye to bind; therefore, it did not cause a colour change in these tissues.\nC: The molecules of blue dye were transported through the blood bound to large proteins, which were unable to diffuse across the blood-brain barrier.\nD: Cells that make up the capillary wall lack a phospholipid bilayer; therefore, the dye was unable to diffuse across the blood-brain barrier.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_445",
        "problem": "基因 Bax 和 Bcl-2 分别促进和抑制细胞凋亡。研究人员利用 siRNA 干扰技术降低 TRPM7 基因表达，研究其对细胞调亡的影响，结果如图所示。下列叙述正确的是（）\n[图1]\nA: TRPM7 基因可能通过促进 Bax 基因的表达来抑制细胞凋亡\nB: TRPM7 基因可能通过抑制 Bcl-2 基因的表达来抑制细胞调亡\nC: 可通过抑制癌细胞中 TRPM7 基因的表达来治疗相关癌症\nD: 可通过促进癌细胞中 TRPM7 基因的表达来治疗相关癌症\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n基因 Bax 和 Bcl-2 分别促进和抑制细胞凋亡。研究人员利用 siRNA 干扰技术降低 TRPM7 基因表达，研究其对细胞调亡的影响，结果如图所示。下列叙述正确的是（）\n[图1]\n\nA: TRPM7 基因可能通过促进 Bax 基因的表达来抑制细胞凋亡\nB: TRPM7 基因可能通过抑制 Bcl-2 基因的表达来抑制细胞调亡\nC: 可通过抑制癌细胞中 TRPM7 基因的表达来治疗相关癌症\nD: 可通过促进癌细胞中 TRPM7 基因的表达来治疗相关癌症\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-32.jpg?height=424&width=988&top_left_y=156&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1494",
        "problem": "The animals below are found in the same food web, but fit into different food chains. One chain takes place close to the ocean surface, and one begins with dead animals resting on the seabed.\n\nWhich animals go into the seabed food chain? The order does not matter.\nA: Scavenging crab, octopus, hammerhead shark\nB: Phytoplankton / algae, zooplankton, shrimp, whale, small fish\nC: Phytoplankton / algae, zooplankton, scavenging crab, octopus, small fish\nD: Octopus, shrimp, scavenging shark\nE: Shrimp, small fish, octopus\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe animals below are found in the same food web, but fit into different food chains. One chain takes place close to the ocean surface, and one begins with dead animals resting on the seabed.\n\nWhich animals go into the seabed food chain? The order does not matter.\n\nA: Scavenging crab, octopus, hammerhead shark\nB: Phytoplankton / algae, zooplankton, shrimp, whale, small fish\nC: Phytoplankton / algae, zooplankton, scavenging crab, octopus, small fish\nD: Octopus, shrimp, scavenging shark\nE: Shrimp, small fish, octopus\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_904",
        "problem": "某二倍体动物, 其控制毛色的等位基因 $G 、 g$ 只位于 $\\mathrm{X}$ 染色体上, 仅 $\\mathrm{G}$ 表达时为黑\n\n色, 仅 $\\mathrm{g}$ 表达时为白色, 二者均不表达时也为白色。受表观遗传的影响, $\\mathrm{G} 、 \\mathrm{~g}$ 来自父本时才表达, 来自母本时不表达。某雄性与杂合子雌性杂交, 获得 4 只基因型互不相同的 $F_{1}$, 亲本和 $F_{1}$ 组成的群体中, 白色个体所占比例不可能是 ( )\nA: $1 / 6$\nB: $1 / 2$\nC: $2 / 3$\nD: $5 / 6$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二倍体动物, 其控制毛色的等位基因 $G 、 g$ 只位于 $\\mathrm{X}$ 染色体上, 仅 $\\mathrm{G}$ 表达时为黑\n\n色, 仅 $\\mathrm{g}$ 表达时为白色, 二者均不表达时也为白色。受表观遗传的影响, $\\mathrm{G} 、 \\mathrm{~g}$ 来自父本时才表达, 来自母本时不表达。某雄性与杂合子雌性杂交, 获得 4 只基因型互不相同的 $F_{1}$, 亲本和 $F_{1}$ 组成的群体中, 白色个体所占比例不可能是 ( )\n\nA: $1 / 6$\nB: $1 / 2$\nC: $2 / 3$\nD: $5 / 6$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1155",
        "problem": "The following diagram is what a nurse saw on a patient's ECG trace, which measures the electrical impulse passed by the heart in one heartbeat. The first arrow on the left signifies the beginning of the beat and the second arrow on the right signifies the end of the beat.\n\n[figure1]\n\nHow many beats per minute did the nurse find in this patient?\nA: 100 beats per minute.\nB: 75 beats per minute.\nC: 60 beats per minute.\nD: 54 beats per minute.\nE: 48 beats per minute.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following diagram is what a nurse saw on a patient's ECG trace, which measures the electrical impulse passed by the heart in one heartbeat. The first arrow on the left signifies the beginning of the beat and the second arrow on the right signifies the end of the beat.\n\n[figure1]\n\nHow many beats per minute did the nurse find in this patient?\n\nA: 100 beats per minute.\nB: 75 beats per minute.\nC: 60 beats per minute.\nD: 54 beats per minute.\nE: 48 beats per minute.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-07.jpg?height=403&width=783&top_left_y=2197&top_left_x=131"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_888",
        "problem": "下图表示 NAT10 蛋白介导的 mRNA 乙酰化修饰参与癌症进展的机制, 相关叙述错误的是 ( )\n\n[图1]\nA: 过程(1)中以核糖核苷酸为原料\nB: 过程(2)中的 mRNA 乙酰化修饰，可以提高 mRNA 的稳定性\nC: 过程(1)和(3)的碱基互补配对方式不完全相同\nD: 胃癌细胞易发生转化与转移是由于 NAT10 蛋白的表达水平低于正常细胞\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图表示 NAT10 蛋白介导的 mRNA 乙酰化修饰参与癌症进展的机制, 相关叙述错误的是 ( )\n\n[图1]\n\nA: 过程(1)中以核糖核苷酸为原料\nB: 过程(2)中的 mRNA 乙酰化修饰，可以提高 mRNA 的稳定性\nC: 过程(1)和(3)的碱基互补配对方式不完全相同\nD: 胃癌细胞易发生转化与转移是由于 NAT10 蛋白的表达水平低于正常细胞\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-31.jpg?height=512&width=1151&top_left_y=2194&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_158",
        "problem": "The figure below shows the change in the abundance pattern of three trophic levels in a lake when it was polluted by city sewage. Ground-feeding carps increase in frequency because they benefit directly from additional mineral nutrients.\n\nTrophic level\n\n[figure1]\n\n## After eutrophication\n\nIncrease\n\nDecrease\n\nIncrease\n\nWhich ecological control methods could improve the water quality of the lake?\n\n<Ecological control methods>\n\nMechanism\n\nControl methods\n\nI. Top-down control: Attempt to introduce predatory fish on carp.\n\nII. Top-down control: Attempt to reduce the nutrients in the river entering the lake.\n\nIII. Bottom-up control: Attempt to inhibit recycling of nutrients accumulated in the substrate of the lake.\n\nIV. Bottom-up control: Attempt to reduce primary producers as well as consumers by introducing more carp.\nA: Only I and II\nB: Only I and III\nC: Only I and IV\nD: Only II and III\nE: Only II and IV\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe figure below shows the change in the abundance pattern of three trophic levels in a lake when it was polluted by city sewage. Ground-feeding carps increase in frequency because they benefit directly from additional mineral nutrients.\n\nTrophic level\n\n[figure1]\n\n## After eutrophication\n\nIncrease\n\nDecrease\n\nIncrease\n\nWhich ecological control methods could improve the water quality of the lake?\n\n<Ecological control methods>\n\nMechanism\n\nControl methods\n\nI. Top-down control: Attempt to introduce predatory fish on carp.\n\nII. Top-down control: Attempt to reduce the nutrients in the river entering the lake.\n\nIII. Bottom-up control: Attempt to inhibit recycling of nutrients accumulated in the substrate of the lake.\n\nIV. Bottom-up control: Attempt to reduce primary producers as well as consumers by introducing more carp.\n\nA: Only I and II\nB: Only I and III\nC: Only I and IV\nD: Only II and III\nE: Only II and IV\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-51.jpg?height=392&width=428&top_left_y=889&top_left_x=654"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_334",
        "problem": "某二倍体植物的红花对白花为完全显性，这一对相对性状由等位基因 B、b 控制。现一纯合红花植株和一白花植株杂交, $F_{1}$ 中偶然发现一株基因型为 Bbb 的三体红花植株(等位基因 $\\mathrm{B} 、 \\mathrm{~b}$ 所在的同源染色体多了一条)。该三体红花植株在减数分裂时联会的三条染色体中的任意两条移向一极, 另一条移向另一极, 产生的配子均可育。下列相关叙述, 错误的是 ( )\nA: 该三体红花植株的产生可能与亲本白花植株的减数分裂 $I$ 过程有关, 也可能与减数分裂II过程有关\nB: 该三体红花植株经减数分裂产生的配子的基因型有四种\nC: 取该三体红花植株的花药进行离体培养, 获得的幼苗经秋水仙素处理后可得到开红花的纯合体\nD: 该三体红花植株与 $F_{1}$ 中其他红花植株杂交, 后代中出现红花植株的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二倍体植物的红花对白花为完全显性，这一对相对性状由等位基因 B、b 控制。现一纯合红花植株和一白花植株杂交, $F_{1}$ 中偶然发现一株基因型为 Bbb 的三体红花植株(等位基因 $\\mathrm{B} 、 \\mathrm{~b}$ 所在的同源染色体多了一条)。该三体红花植株在减数分裂时联会的三条染色体中的任意两条移向一极, 另一条移向另一极, 产生的配子均可育。下列相关叙述, 错误的是 ( )\n\nA: 该三体红花植株的产生可能与亲本白花植株的减数分裂 $I$ 过程有关, 也可能与减数分裂II过程有关\nB: 该三体红花植株经减数分裂产生的配子的基因型有四种\nC: 取该三体红花植株的花药进行离体培养, 获得的幼苗经秋水仙素处理后可得到开红花的纯合体\nD: 该三体红花植株与 $F_{1}$ 中其他红花植株杂交, 后代中出现红花植株的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_302",
        "problem": "大肠杆菌乳糖操纵子包括 lacZ、lacY、lacA 三个结构基因（编码参与乳糖代谢的酶，\n\n其中酶 a 能够水解乳糖)，以及操纵基因、启动子和调节基因。培养基中无乳糖存在时，调节基因表达的阻遏蛋白和操纵基因结合，导致 RNA 聚合酶不能与启动子结合，使结构基因无法转录; 乳糖存在时，结构基因才能正常表达，调节过程如下图所示。下列说法错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 结构基因转录时只能以 $\\beta$ 链为模板, 表达出来的酶 $\\mathrm{a}$ 会使结构基因的表达受到抑制\nB: 过程(1)的碱基配对方式与(2)不完全相同, 参与过程(2)的氨基酸有的可被多种 tRNA 转运\nC: 若调节基因被甲基化修饰，可能导致结构基因持续表达，造成大肠杆菌物质和能量的浪费\nD: 据图可知, 乳糖能够调节大肠杆菌中基因的选择性表达, 该过程发生细胞的分化\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n大肠杆菌乳糖操纵子包括 lacZ、lacY、lacA 三个结构基因（编码参与乳糖代谢的酶，\n\n其中酶 a 能够水解乳糖)，以及操纵基因、启动子和调节基因。培养基中无乳糖存在时，调节基因表达的阻遏蛋白和操纵基因结合，导致 RNA 聚合酶不能与启动子结合，使结构基因无法转录; 乳糖存在时，结构基因才能正常表达，调节过程如下图所示。下列说法错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 结构基因转录时只能以 $\\beta$ 链为模板, 表达出来的酶 $\\mathrm{a}$ 会使结构基因的表达受到抑制\nB: 过程(1)的碱基配对方式与(2)不完全相同, 参与过程(2)的氨基酸有的可被多种 tRNA 转运\nC: 若调节基因被甲基化修饰，可能导致结构基因持续表达，造成大肠杆菌物质和能量的浪费\nD: 据图可知, 乳糖能够调节大肠杆菌中基因的选择性表达, 该过程发生细胞的分化\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-86.jpg?height=391&width=880&top_left_y=150&top_left_x=334",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-86.jpg?height=343&width=828&top_left_y=565&top_left_x=388"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1511",
        "problem": "The complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nWhich pathway causes the complement system to target everything, even an entirely new pathogen?\nA: Pathway A\nB: Pathway B\nC: Pathway C\nD: Pathway D\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nWhich pathway causes the complement system to target everything, even an entirely new pathogen?\n\nA: Pathway A\nB: Pathway B\nC: Pathway C\nD: Pathway D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=790&width=1714&top_left_y=486&top_left_x=228",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=930&width=1806&top_left_y=1335&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_789",
        "problem": "已知甲、乙两种病独立遗传，分别由一对等位基因控制。甲病为“卵子死亡”症，患者因卵子萎缩、退化而不育，已知 $\\mathrm{I}_{1} 、 \\mathrm{II}_{4}$ 含甲病基因， $\\mathrm{I}_{2} 、 \\mathrm{II}_{2}$ 不含甲病基因。如图是某家系关于甲病和乙病的系谱图。下列有关说法正确的是（）\n\n[图1]\nA: 甲病的遗传方式是伴 X 染色体显性遗传\nB: $\\mathrm{II}_{3}$ 与含甲病基因的男子结婚, 可以生出正常孩子\nC: 若与 $\\mathrm{II}_{4}$ 基因型相同的个体和乙病患者结婚生儿子均患乙病, 则乙病为 XY 染色体同源区段隐性遗传病\nD: 若乙病为常染色体隐性遗传病, 则 $\\mathrm{III}_{5}$ 与乙病患者结婚, 生的孩子患病概率为 $1 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知甲、乙两种病独立遗传，分别由一对等位基因控制。甲病为“卵子死亡”症，患者因卵子萎缩、退化而不育，已知 $\\mathrm{I}_{1} 、 \\mathrm{II}_{4}$ 含甲病基因， $\\mathrm{I}_{2} 、 \\mathrm{II}_{2}$ 不含甲病基因。如图是某家系关于甲病和乙病的系谱图。下列有关说法正确的是（）\n\n[图1]\n\nA: 甲病的遗传方式是伴 X 染色体显性遗传\nB: $\\mathrm{II}_{3}$ 与含甲病基因的男子结婚, 可以生出正常孩子\nC: 若与 $\\mathrm{II}_{4}$ 基因型相同的个体和乙病患者结婚生儿子均患乙病, 则乙病为 XY 染色体同源区段隐性遗传病\nD: 若乙病为常染色体隐性遗传病, 则 $\\mathrm{III}_{5}$ 与乙病患者结婚, 生的孩子患病概率为 $1 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-11.jpg?height=480&width=1348&top_left_y=1822&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_77",
        "problem": "In the photic zone of freshwater and marine environments, where light penetrates, cyanobacteria are found in the upper part of the zone, and purple and green bacteria are in the lower part of the zone. Which of the following statements best explains the vertical distribution of the photosynthetic bacteria?\nA: Green and purple bacteria are anaerobic, while cyanobacteria are aerobic.\nB: Green and purple bacteria are better able to use light wavelengths that cyanobacteria do not use as efficiently.\nC: Habitat isolation develops due to competition for nutrient and oxygen.\nD: Cyanobacteria are better able to use oxygen as an electron donor for photosynthesis.\nE: It is the result of adaptation to lower temperatures in purple and green bacteria.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the photic zone of freshwater and marine environments, where light penetrates, cyanobacteria are found in the upper part of the zone, and purple and green bacteria are in the lower part of the zone. Which of the following statements best explains the vertical distribution of the photosynthetic bacteria?\n\nA: Green and purple bacteria are anaerobic, while cyanobacteria are aerobic.\nB: Green and purple bacteria are better able to use light wavelengths that cyanobacteria do not use as efficiently.\nC: Habitat isolation develops due to competition for nutrient and oxygen.\nD: Cyanobacteria are better able to use oxygen as an electron donor for photosynthesis.\nE: It is the result of adaptation to lower temperatures in purple and green bacteria.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_947",
        "problem": "细胞中氨基酸有 2 个来源: 一是从细胞外摄取, 二是细胞内利用氨基酸合成酶自己合成。当细胞缺乏氨基酸时, 某种 RNA 无法结合氨基酸, 空载的某种 RNA 与核糖体结合后引发 RclA 利用 GDP 和 ATP 合成 ppGpp（如图 1), ppGpp 是细胞内的一种信号分子，可提高 $\\mathrm{A}$ 类基因或降低 $\\mathrm{B}$ 类基因的转录水平，也可直接影响翻译过程（如图 2)。下列叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 图 1 中的某种 RNA 为 tRNA, 它通过碱基互补配对识别并结合氨基酸\nB: 图 2 所示为翻译过程, mRNA 的启动子在靠近右侧的部位\nC: 推测 $\\mathrm{A}$ 类基因属于促进消耗氨基酸的基因，B 类基因属于促进产生氨基酸的基因\nD: ppGpp 调节机制属于负反馈调节, 可能 RNA 聚合酶有 ppGpp 作用靶点\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n细胞中氨基酸有 2 个来源: 一是从细胞外摄取, 二是细胞内利用氨基酸合成酶自己合成。当细胞缺乏氨基酸时, 某种 RNA 无法结合氨基酸, 空载的某种 RNA 与核糖体结合后引发 RclA 利用 GDP 和 ATP 合成 ppGpp（如图 1), ppGpp 是细胞内的一种信号分子，可提高 $\\mathrm{A}$ 类基因或降低 $\\mathrm{B}$ 类基因的转录水平，也可直接影响翻译过程（如图 2)。下列叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 图 1 中的某种 RNA 为 tRNA, 它通过碱基互补配对识别并结合氨基酸\nB: 图 2 所示为翻译过程, mRNA 的启动子在靠近右侧的部位\nC: 推测 $\\mathrm{A}$ 类基因属于促进消耗氨基酸的基因，B 类基因属于促进产生氨基酸的基因\nD: ppGpp 调节机制属于负反馈调节, 可能 RNA 聚合酶有 ppGpp 作用靶点\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-22.jpg?height=540&width=511&top_left_y=1306&top_left_x=361",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-22.jpg?height=557&width=442&top_left_y=1343&top_left_x=887"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_359",
        "problem": "我国科学家研究发现, 真核细胞的核基因表达调节机制——含子 miRNA 通过下调转录因子来调控基因的表达, 图甲和图乙是发生在细胞核内的正向和反向调节机制模式图。请据所学知识分析, 下列叙述错误的是 ( )\n\n[图1]\n\n甲内含子miRNA通过下调转录抑制因子正向调节宿主基因的表达模式图\n\n[图2]\n\n乙内含子miRNA通过下调转录激活因子反向调节宿主基因的表达模式图\nA: 内含子 miRNA 是在转录水平来调控基因的表达\nB: 内含子 miRNA 调控基因的表达存在反馈调节机制\nC: 形成主基因 mRNA 和成熟 miRNA 需解旋酶和限制酶的作用\nD: 该研究为某些疾病如肿瘤提供多个靶点联合治疗\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n我国科学家研究发现, 真核细胞的核基因表达调节机制——含子 miRNA 通过下调转录因子来调控基因的表达, 图甲和图乙是发生在细胞核内的正向和反向调节机制模式图。请据所学知识分析, 下列叙述错误的是 ( )\n\n[图1]\n\n甲内含子miRNA通过下调转录抑制因子正向调节宿主基因的表达模式图\n\n[图2]\n\n乙内含子miRNA通过下调转录激活因子反向调节宿主基因的表达模式图\n\nA: 内含子 miRNA 是在转录水平来调控基因的表达\nB: 内含子 miRNA 调控基因的表达存在反馈调节机制\nC: 形成主基因 mRNA 和成熟 miRNA 需解旋酶和限制酶的作用\nD: 该研究为某些疾病如肿瘤提供多个靶点联合治疗\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-046.jpg?height=535&width=696&top_left_y=1937&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-046.jpg?height=542&width=694&top_left_y=1942&top_left_x=1092"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_92",
        "problem": "About 50 years ago, Charles Yanofsky studied the sequence of the tryptophan synthetase of E. coli. The wild type protein (1) has a glycine in position 38. Yanofsky isolated two inactive trp mutants: 2 and 3. Mutant 2 had Arg instead of Gly at position 38, and mutant 3 had Glu at position 38. Mutants 2 and 3 were plated on minimal medium (without tryptophan). Colonies appearing correspond to spontaneous mutations that restored tryptophan synthetase function. The amino acid at position 38 was identified as described in figure A. Assume that each amino acid replacement results from a single base-pair change.\n\n[figure1]\n\n[figure2]\nA: Mutant 2 results from a base transition at the first position of codon 38 .\nB: Strains 7 and 8 likely have the same sequence as the wild type strain.\nC: The codon 38 in strain 10 is $5^{\\prime}$ GTA3'\nD: The codon 38 in strain 6 is 5'AGC3'.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAbout 50 years ago, Charles Yanofsky studied the sequence of the tryptophan synthetase of E. coli. The wild type protein (1) has a glycine in position 38. Yanofsky isolated two inactive trp mutants: 2 and 3. Mutant 2 had Arg instead of Gly at position 38, and mutant 3 had Glu at position 38. Mutants 2 and 3 were plated on minimal medium (without tryptophan). Colonies appearing correspond to spontaneous mutations that restored tryptophan synthetase function. The amino acid at position 38 was identified as described in figure A. Assume that each amino acid replacement results from a single base-pair change.\n\n[figure1]\n\n[figure2]\n\nA: Mutant 2 results from a base transition at the first position of codon 38 .\nB: Strains 7 and 8 likely have the same sequence as the wild type strain.\nC: The codon 38 in strain 10 is $5^{\\prime}$ GTA3'\nD: The codon 38 in strain 6 is 5'AGC3'.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-088.jpg?height=503&width=965&top_left_y=939&top_left_x=591",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-088.jpg?height=1168&width=1045&top_left_y=1483&top_left_x=563",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-090.jpg?height=534&width=1077&top_left_y=1292&top_left_x=615"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_768",
        "problem": "秀丽隐杆线虫从卵发育到成虫大约需要 3 天, 利用秀丽隐杆线虫的突变体 (基因型为 dpy-5, 表型为体型短粗）验证基因分离定律的实验交配方案如下图所示, 其中“\"̛”代表雌雄同体, “+”代表野生型基因。雌雄同体个体能自体受精或与雄虫交配, 但雌雄同体的不同个体之间不能交配, 且具有突变性状的雄虫交配能力弱。下列相关叙述正确的是 ( )\n\n[图1]\nA: 秀丽隐杆线虫从卵发育到成虫过程中没有细胞的凋亡, 是遗传学的良好实验材料\nB: 本实验若采用野生型雌雄同体个体与基因型为 dpy- 5 的雄虫杂交的反交设计,效果相同\nC: 若 $F_{1}$ 中出现雌雄同体个体与雄虫的数目比例接近于 1: 1, 则可验证分离定律\nD: 对 $\\mathrm{F}_{2}$ 计数统计后, 若体型正常: 体型短粗 $=598: 210$, 则可验证分离定律\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n秀丽隐杆线虫从卵发育到成虫大约需要 3 天, 利用秀丽隐杆线虫的突变体 (基因型为 dpy-5, 表型为体型短粗）验证基因分离定律的实验交配方案如下图所示, 其中“\"̛”代表雌雄同体, “+”代表野生型基因。雌雄同体个体能自体受精或与雄虫交配, 但雌雄同体的不同个体之间不能交配, 且具有突变性状的雄虫交配能力弱。下列相关叙述正确的是 ( )\n\n[图1]\n\nA: 秀丽隐杆线虫从卵发育到成虫过程中没有细胞的凋亡, 是遗传学的良好实验材料\nB: 本实验若采用野生型雌雄同体个体与基因型为 dpy- 5 的雄虫杂交的反交设计,效果相同\nC: 若 $F_{1}$ 中出现雌雄同体个体与雄虫的数目比例接近于 1: 1, 则可验证分离定律\nD: 对 $\\mathrm{F}_{2}$ 计数统计后, 若体型正常: 体型短粗 $=598: 210$, 则可验证分离定律\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-67.jpg?height=731&width=880&top_left_y=571&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_390",
        "problem": "蜜蜂中, 雌蜂是雌雄配子结合产生的二倍体, 雄蜂是由未受精的卵直接发育而来的。某对蜜蜂所产生子代的基因型为：雌蜂是 $A A D D 、 A A D d 、 A a D D 、 A a D d ；$ 雄蜂是 $A D 、 A d 、 a D 、 a d$ ．这对蜜蜂的基因型是（）\nA: AADd 和 ad\nB: AaDd 和 $\\mathrm{Ad}$\nC: AaDd 和 $A D$\nD: Aadd 和 AD\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n蜜蜂中, 雌蜂是雌雄配子结合产生的二倍体, 雄蜂是由未受精的卵直接发育而来的。某对蜜蜂所产生子代的基因型为：雌蜂是 $A A D D 、 A A D d 、 A a D D 、 A a D d ；$ 雄蜂是 $A D 、 A d 、 a D 、 a d$ ．这对蜜蜂的基因型是（）\n\nA: AADd 和 ad\nB: AaDd 和 $\\mathrm{Ad}$\nC: AaDd 和 $A D$\nD: Aadd 和 AD\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_190",
        "problem": "Electrophoresis patterns of PCR products of several chromosome pairs (pairs shown on bottom) in five different individual (1-5) in figure below allow evaluation of presence or absence of chromosome number abnormality. Monosomy of autosomal chromosomes is known to be lethal. Relative height of the peaks in each box reflects the copy number ratio of the two chromosomes in that box. The horizontal axis shows migration, and the vertical axis shows fluorescence intensity.\n\n[figure1]\nA: Three individuals show trisomy.\nB: Two individuals show abnormal monosomy.\nC: Two individuals have normal karyotypes.\nD: PCR products related to different chromosomes should have different sizes to allow copy number evaluation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nElectrophoresis patterns of PCR products of several chromosome pairs (pairs shown on bottom) in five different individual (1-5) in figure below allow evaluation of presence or absence of chromosome number abnormality. Monosomy of autosomal chromosomes is known to be lethal. Relative height of the peaks in each box reflects the copy number ratio of the two chromosomes in that box. The horizontal axis shows migration, and the vertical axis shows fluorescence intensity.\n\n[figure1]\n\nA: Three individuals show trisomy.\nB: Two individuals show abnormal monosomy.\nC: Two individuals have normal karyotypes.\nD: PCR products related to different chromosomes should have different sizes to allow copy number evaluation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-09.jpg?height=1025&width=1380&top_left_y=624&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1333",
        "problem": "Part of the citric acid cycle is represented below.\n\na-ketoglutarate $\\rightarrow$ succinate $\\rightarrow$ fumarate $\\rightarrow$ malate\n\nIf the enzyme responsible for the conversion of succinate to fumarate became inactive, which one of the following would NOToccur?\nA: some accumulation of succinate\nB: a halt in the production of malate\nC: gradual disappearance of fumarate\nD: continued conversion of $\\alpha$-ketoglutarate\nE: an increase in the concentration of $\\alpha$-ketoglutarate\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPart of the citric acid cycle is represented below.\n\na-ketoglutarate $\\rightarrow$ succinate $\\rightarrow$ fumarate $\\rightarrow$ malate\n\nIf the enzyme responsible for the conversion of succinate to fumarate became inactive, which one of the following would NOToccur?\n\nA: some accumulation of succinate\nB: a halt in the production of malate\nC: gradual disappearance of fumarate\nD: continued conversion of $\\alpha$-ketoglutarate\nE: an increase in the concentration of $\\alpha$-ketoglutarate\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_264",
        "problem": "Laboratory experiments were conducted to examine the effects of temperature on interspecific competition between two stream salmonid fishes, Salvelinus malma and $S$. leucomaenis, with largely allopatric altitudinal distribution. Three combinations of species population, including allopatric populations of S. malma and S. leucomaenis, and sympatric populations of both species, were treated with low temperature $\\left(6^{\\circ} \\mathrm{C}\\right)$ and high temperature $\\left(12^{\\circ} \\mathrm{C}\\right)$, in which thriving allopatric populations of $S$. malma $\\left(6^{\\circ} \\mathrm{C}\\right)$ and $S$. leucomaenis $\\left(12^{\\circ} \\mathrm{C}\\right)$ are commonly found.\n\nLow temperature High temperature\n\n[figure1]\n\nFigure Q. 44\nA: Competition of these two species is likely to have been affected by temperature and altitude\nB: S. malma may be distributed at higher altitudinal ranges than is $S$. leucomaenis\nC: S. leucomaenis is likely to be more low-temperature stress-resistant than S. malma\nD: S. malma has a narrower fundamental niche than does S. leucomaenis\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nLaboratory experiments were conducted to examine the effects of temperature on interspecific competition between two stream salmonid fishes, Salvelinus malma and $S$. leucomaenis, with largely allopatric altitudinal distribution. Three combinations of species population, including allopatric populations of S. malma and S. leucomaenis, and sympatric populations of both species, were treated with low temperature $\\left(6^{\\circ} \\mathrm{C}\\right)$ and high temperature $\\left(12^{\\circ} \\mathrm{C}\\right)$, in which thriving allopatric populations of $S$. malma $\\left(6^{\\circ} \\mathrm{C}\\right)$ and $S$. leucomaenis $\\left(12^{\\circ} \\mathrm{C}\\right)$ are commonly found.\n\nLow temperature High temperature\n\n[figure1]\n\nFigure Q. 44\n\nA: Competition of these two species is likely to have been affected by temperature and altitude\nB: S. malma may be distributed at higher altitudinal ranges than is $S$. leucomaenis\nC: S. leucomaenis is likely to be more low-temperature stress-resistant than S. malma\nD: S. malma has a narrower fundamental niche than does S. leucomaenis\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-096.jpg?height=1199&width=1082&top_left_y=885&top_left_x=521"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1020",
        "problem": "Earth's four major terrestrial biomes are\nA: Forest, taiga, grassland, desert\nB: Forest, grassland, tundra, desert.\nC: Forest, grassland, savanna, desert\nD: Savanna, forest, grassland, tundra\nE: Savanna, taiga, grassland, estuary\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEarth's four major terrestrial biomes are\n\nA: Forest, taiga, grassland, desert\nB: Forest, grassland, tundra, desert.\nC: Forest, grassland, savanna, desert\nD: Savanna, forest, grassland, tundra\nE: Savanna, taiga, grassland, estuary\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_905",
        "problem": "甲病、乙病和丙病是人群中的三种单基因遗传病, 控制甲病和乙病的等位基因均位于常染色体上，控制丙病的等位基因位于 X 染色体上。假设一对夫妇婚后生育甲病小孩的概率为 $\\mathrm{a}$, 生育乙病小孩的概率为 $\\mathrm{b}$. 生育丙病小孩的概率为 $\\mathrm{c}$. 不考虑其他疾病,下列叙述正确的是 （ ）\nA: 他们所生的男孩同时患甲病和乙病的概率为 $\\mathrm{a} \\cdot \\mathrm{b}$\nB: 他们生一个同时患甲病和丙病的女孩的概率为 $1 / 2 \\mathrm{a} \\cdot \\mathrm{c}$\nC: 他们所生的小孩患乙病但不患丙病的概率为 b$\\cdot$(1-c)\nD: 他们所生的小孩只患丙病的概率为 $(1-\\mathrm{a}) \\cdot(1-\\mathrm{b}) \\cdot \\mathrm{c}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n甲病、乙病和丙病是人群中的三种单基因遗传病, 控制甲病和乙病的等位基因均位于常染色体上，控制丙病的等位基因位于 X 染色体上。假设一对夫妇婚后生育甲病小孩的概率为 $\\mathrm{a}$, 生育乙病小孩的概率为 $\\mathrm{b}$. 生育丙病小孩的概率为 $\\mathrm{c}$. 不考虑其他疾病,下列叙述正确的是 （ ）\n\nA: 他们所生的男孩同时患甲病和乙病的概率为 $\\mathrm{a} \\cdot \\mathrm{b}$\nB: 他们生一个同时患甲病和丙病的女孩的概率为 $1 / 2 \\mathrm{a} \\cdot \\mathrm{c}$\nC: 他们所生的小孩患乙病但不患丙病的概率为 b$\\cdot$(1-c)\nD: 他们所生的小孩只患丙病的概率为 $(1-\\mathrm{a}) \\cdot(1-\\mathrm{b}) \\cdot \\mathrm{c}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_817",
        "problem": "果蝇 $(2 n=8)$ 体内两条 $\\mathrm{X}$ 染色体有时可融合成一条 $\\mathrm{X}$ 染色体, 称为“并连 $\\mathrm{X}$ ”（记作“ $X^{\\wedge} X^{\\prime}$ ”), 其形成过程如图所示。一只含有“并连 $X^{\\prime}$ 的雌蝇（ $X^{\\wedge} X Y ）$ 和一只正常雄蝇杂交, 其子代的基因型与亲代相同，子代连续交配也是如此，该种品系称为“并连 $\\mathrm{X}$ 保持系”。下列叙述正确的是（ ）\n\n[图1]\nA: 含有“并连 $X^{\\prime}$ ”的雌蝇（ $\\left.X^{\\wedge} X Y\\right)$ 在减数分裂过程中可形成 3 个四分体\nB: 子代中仅染色体组成为 YY 的果蝇无法发育为新个体\nC: 在“并连 X 保持系”中, 亲本雄蝇的 Y 染色体传向子代雄蝇\nD: 利用“并连 X 保持系”, 可“监控”雄蝇 X 染色体上的新发突变\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇 $(2 n=8)$ 体内两条 $\\mathrm{X}$ 染色体有时可融合成一条 $\\mathrm{X}$ 染色体, 称为“并连 $\\mathrm{X}$ ”（记作“ $X^{\\wedge} X^{\\prime}$ ”), 其形成过程如图所示。一只含有“并连 $X^{\\prime}$ 的雌蝇（ $X^{\\wedge} X Y ）$ 和一只正常雄蝇杂交, 其子代的基因型与亲代相同，子代连续交配也是如此，该种品系称为“并连 $\\mathrm{X}$ 保持系”。下列叙述正确的是（ ）\n\n[图1]\n\nA: 含有“并连 $X^{\\prime}$ ”的雌蝇（ $\\left.X^{\\wedge} X Y\\right)$ 在减数分裂过程中可形成 3 个四分体\nB: 子代中仅染色体组成为 YY 的果蝇无法发育为新个体\nC: 在“并连 X 保持系”中, 亲本雄蝇的 Y 染色体传向子代雄蝇\nD: 利用“并连 X 保持系”, 可“监控”雄蝇 X 染色体上的新发突变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-078.jpg?height=283&width=354&top_left_y=1800&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_929",
        "problem": "现用某野生植物甲 $(A A B B)$ 、乙 (aabb) 两品系作亲本杂交得 $F_{1}, F_{1}$ 的测交结果如下表。下列推测或分析不正确的是（ ）\n\n| 品系 |  | 测交后代基因型种类及比值 |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 父本 | 母本 | $\\mathrm{AaBb}$ | $\\mathrm{Aabb}$ | $\\mathrm{aaBb}$ | $\\mathrm{aabb}$ |\n| $\\mathrm{F}_{1}$ | 乙 | 1 | 2 | 2 | 2 |\n| 乙 | $\\mathrm{F}_{1}$ | 1 | 1 | 1 | 1 |\nA: $F_{1}$ 自交得到的 $F_{2}$ 有 9 种基因型\nB: $\\mathrm{F}_{1}$ 产生的基因型为 $\\mathrm{AB}$ 的花粉 $50 \\%$ 不能萌发而不能受精\nC: $F_{1}$ 自交后代 $F_{2}$ 中重组类型的比例是 $3 / 7$\nD: 正反交结果不同，说明这两对基因的遗传不遵循自由组合定律\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n现用某野生植物甲 $(A A B B)$ 、乙 (aabb) 两品系作亲本杂交得 $F_{1}, F_{1}$ 的测交结果如下表。下列推测或分析不正确的是（ ）\n\n| 品系 |  | 测交后代基因型种类及比值 |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 父本 | 母本 | $\\mathrm{AaBb}$ | $\\mathrm{Aabb}$ | $\\mathrm{aaBb}$ | $\\mathrm{aabb}$ |\n| $\\mathrm{F}_{1}$ | 乙 | 1 | 2 | 2 | 2 |\n| 乙 | $\\mathrm{F}_{1}$ | 1 | 1 | 1 | 1 |\n\nA: $F_{1}$ 自交得到的 $F_{2}$ 有 9 种基因型\nB: $\\mathrm{F}_{1}$ 产生的基因型为 $\\mathrm{AB}$ 的花粉 $50 \\%$ 不能萌发而不能受精\nC: $F_{1}$ 自交后代 $F_{2}$ 中重组类型的比例是 $3 / 7$\nD: 正反交结果不同，说明这两对基因的遗传不遵循自由组合定律\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_200",
        "problem": "Strigolactone (SL) is a plant hormone that controls shoot branching. In Arabidopsis thaliana, several SL-related mutants, such as max1, $\\max 2$, and $\\max 4$, which have loss-of-function mutations in the genes MAX1, MAX2, and MAX4, respectively, have been isolated. While MAX2 encodes a key component of the SL receptor complex, MAXI and MAX4 each encode an enzyme for SL biosynthesis (Figure 1); MAX4 for the production of the SL precursor carlactone (CL), and MAXI for the conversion of CL into SL. Grafting experiments using these mutants and the wild type (WT) were conducted, and the number of shoot branches were counted (Figure $2 \\& 3$ ). In this experiment, neither mRNAs nor proteins of the $M A X$ genes were found to move across the grafting junction.\n\nGrafting of seedlings\n\n[figure1]\n\nStrigolactone (SL)\n\nFigure 1.\n\nBiosynthetic pathway of strigolactone\n\n[figure2]\n\nFigure 2.\n\nSchematic illustration of grafting experiments\n\n[figure3]\n\nFigure 3. Number of shoot branches in the grafted plants\nA: The MAX2 gene mainly functions in the root.\nB: SL is synthesized both in the root and shoot.\nC: CL, the substrate of MAX1, is transported between the root and shoot in either direction.\nD: If a shoot scion of max4 is grafted on a rootstock of max2, shoot branching will be normal.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nStrigolactone (SL) is a plant hormone that controls shoot branching. In Arabidopsis thaliana, several SL-related mutants, such as max1, $\\max 2$, and $\\max 4$, which have loss-of-function mutations in the genes MAX1, MAX2, and MAX4, respectively, have been isolated. While MAX2 encodes a key component of the SL receptor complex, MAXI and MAX4 each encode an enzyme for SL biosynthesis (Figure 1); MAX4 for the production of the SL precursor carlactone (CL), and MAXI for the conversion of CL into SL. Grafting experiments using these mutants and the wild type (WT) were conducted, and the number of shoot branches were counted (Figure $2 \\& 3$ ). In this experiment, neither mRNAs nor proteins of the $M A X$ genes were found to move across the grafting junction.\n\nGrafting of seedlings\n\n[figure1]\n\nStrigolactone (SL)\n\nFigure 1.\n\nBiosynthetic pathway of strigolactone\n\n[figure2]\n\nFigure 2.\n\nSchematic illustration of grafting experiments\n\n[figure3]\n\nFigure 3. Number of shoot branches in the grafted plants\n\nA: The MAX2 gene mainly functions in the root.\nB: SL is synthesized both in the root and shoot.\nC: CL, the substrate of MAX1, is transported between the root and shoot in either direction.\nD: If a shoot scion of max4 is grafted on a rootstock of max2, shoot branching will be normal.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-43.jpg?height=382&width=291&top_left_y=1114&top_left_x=331",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-43.jpg?height=440&width=1077&top_left_y=1159&top_left_x=795",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-43.jpg?height=568&width=1314&top_left_y=1926&top_left_x=294"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1032",
        "problem": "When found on the extracellular side of the cell, which of the following classes of membrane molecules is a marker for the phagocytosis of apoptotic cells?\nA: Glycolipid\nB: Sphingomyelin\nC: Phosphatidylethanolamine\nD: Phosphatidylserine\nE: Phosphatidylcholine\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen found on the extracellular side of the cell, which of the following classes of membrane molecules is a marker for the phagocytosis of apoptotic cells?\n\nA: Glycolipid\nB: Sphingomyelin\nC: Phosphatidylethanolamine\nD: Phosphatidylserine\nE: Phosphatidylcholine\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_207",
        "problem": "To understand the effect of desiccation on herbaceous plants and their responses, a student conducted a study on three Ranunculus species in their natural habitats, including $R$. bulbosus in dry medow, $R$. lanuginosus in humid medow, and $R$. acris in both habitats. He measured leaf water potential and hydraulic conductance of these species in response to dehydration (Fig.Q62). Xylem staining experiment on $R$. acris in dry habitat was used to estimate loss of conductivity due to embolism. An estimated $50 \\%$ loss of xylem hydraulic conductivity due to embolism occurred at $-2 \\mathrm{MPa}$ or less. Previously, leaf hydraulic vulnerability studies found $50 \\%$ reduction in leaf hydraulic conductance between -1 and $-1.8 \\mathrm{MPa}$ in parennial grasses and at $-1.8 \\mathrm{MPa}$ in woody plant species.\n\n[figure1]\n\nFig.Q62. Leaf hydaulic conductance of Ranunculus species/populations in response to dehydration. Solid and dashed vertical lines indicate, respectively, fitted $50 \\%$ and $88 \\%$ leaf hydraulic conductance losses.\nA: All species were very vulnerable to water stress. In species with narrow ecological amplitude, the drought-exposed $R$. bulbosus was less vulnerable to desiccation than the humid habitat $R$. lanuginosus.\nB: Herbaceous species would be more vulnerable to water stress than woody species and perennial grasses, but also would show interspecific and intraspecific adjustments in hydraulic vulnerability based on the water availability of their respective habitats.\nC: The leaf hydraulics method employs hydraulic conductance including both xylary and extraxylary pathways.\nD: The effect of drought in these plant species is found to be a loss of leaf hydraulic conductance at moderate water potential based on extraxylary pathways rather than embolism formation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nTo understand the effect of desiccation on herbaceous plants and their responses, a student conducted a study on three Ranunculus species in their natural habitats, including $R$. bulbosus in dry medow, $R$. lanuginosus in humid medow, and $R$. acris in both habitats. He measured leaf water potential and hydraulic conductance of these species in response to dehydration (Fig.Q62). Xylem staining experiment on $R$. acris in dry habitat was used to estimate loss of conductivity due to embolism. An estimated $50 \\%$ loss of xylem hydraulic conductivity due to embolism occurred at $-2 \\mathrm{MPa}$ or less. Previously, leaf hydraulic vulnerability studies found $50 \\%$ reduction in leaf hydraulic conductance between -1 and $-1.8 \\mathrm{MPa}$ in parennial grasses and at $-1.8 \\mathrm{MPa}$ in woody plant species.\n\n[figure1]\n\nFig.Q62. Leaf hydaulic conductance of Ranunculus species/populations in response to dehydration. Solid and dashed vertical lines indicate, respectively, fitted $50 \\%$ and $88 \\%$ leaf hydraulic conductance losses.\n\nA: All species were very vulnerable to water stress. In species with narrow ecological amplitude, the drought-exposed $R$. bulbosus was less vulnerable to desiccation than the humid habitat $R$. lanuginosus.\nB: Herbaceous species would be more vulnerable to water stress than woody species and perennial grasses, but also would show interspecific and intraspecific adjustments in hydraulic vulnerability based on the water availability of their respective habitats.\nC: The leaf hydraulics method employs hydraulic conductance including both xylary and extraxylary pathways.\nD: The effect of drought in these plant species is found to be a loss of leaf hydraulic conductance at moderate water potential based on extraxylary pathways rather than embolism formation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-028.jpg?height=1345&width=944&top_left_y=1071&top_left_x=565"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_38",
        "problem": "The following figure shows nitrogen cycling in an ecosystem. Numbers $\\mathrm{I} \\sim \\mathrm{V}$ represent different chemical conversion steps in the cycle.\n\n[figure1]\n\nWhich process ( $\\mathrm{I} \\sim \\mathrm{V}$ ) is correctly paired with the organismal group performing that step?\nA: I- photoautotrophs.\nB: II- bacteria symbiotic with plants.\nC: III- anaerobic bacteria living in conditions such as wetland ecosystem.\nD: IV- eukaryotic organisms.\nE: V - nitrogen fixing bacteria such as Rhizobium or Cyanobacteria.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following figure shows nitrogen cycling in an ecosystem. Numbers $\\mathrm{I} \\sim \\mathrm{V}$ represent different chemical conversion steps in the cycle.\n\n[figure1]\n\nWhich process ( $\\mathrm{I} \\sim \\mathrm{V}$ ) is correctly paired with the organismal group performing that step?\n\nA: I- photoautotrophs.\nB: II- bacteria symbiotic with plants.\nC: III- anaerobic bacteria living in conditions such as wetland ecosystem.\nD: IV- eukaryotic organisms.\nE: V - nitrogen fixing bacteria such as Rhizobium or Cyanobacteria.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-54.jpg?height=617&width=760&top_left_y=537&top_left_x=662"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_156",
        "problem": "Correlation between genome size in polyploid and diploid lineages and life history and occupation of alpine habitat for the genus Veronica (Plantaginaceae) were investigated (Meudt et al., 2015). The genus primarily originated in northern hemisphere. The phylogenetic tree below includes variables of interest including:\n\n- 1C-value (amount of DNA contained in a haploid genome)\n- life history (annual: bar; perennial: no bar)\n- habitat (non-alpine: no bar; alpine: bar)\n- ploidy $(2 \\times$ to $18 \\times)$.\n\n[figure1]\nA: All the Southern hemisphere species are polyploid.\nB: Polyploidy is always accompanied by genomic upsizing.\nC: Transitions from annual to perennial life history can be accompanied by migration to the higher elevations or cooler climates.\nD: Annual species are never found in alpine habitats.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nCorrelation between genome size in polyploid and diploid lineages and life history and occupation of alpine habitat for the genus Veronica (Plantaginaceae) were investigated (Meudt et al., 2015). The genus primarily originated in northern hemisphere. The phylogenetic tree below includes variables of interest including:\n\n- 1C-value (amount of DNA contained in a haploid genome)\n- life history (annual: bar; perennial: no bar)\n- habitat (non-alpine: no bar; alpine: bar)\n- ploidy $(2 \\times$ to $18 \\times)$.\n\n[figure1]\n\nA: All the Southern hemisphere species are polyploid.\nB: Polyploidy is always accompanied by genomic upsizing.\nC: Transitions from annual to perennial life history can be accompanied by migration to the higher elevations or cooler climates.\nD: Annual species are never found in alpine habitats.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-38.jpg?height=1339&width=1448&top_left_y=795&top_left_x=310"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1510",
        "problem": "Peaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nIn actuality, nectarines are the result of a recessive mutation in a single gene. A farmer grows only peaches. She crosses her peaches and gives the seeds to a neighbour to set-up his own farm. However, $9 \\%$ of the plants on his new farm turn into nectarines.\n\nWhat is the frequency of the mutation on the neighbour's new farm?\nA: 0.03\nB: 0.09\nC: 0.3\nD: 0.9\nE: 0.45\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPeaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nIn actuality, nectarines are the result of a recessive mutation in a single gene. A farmer grows only peaches. She crosses her peaches and gives the seeds to a neighbour to set-up his own farm. However, $9 \\%$ of the plants on his new farm turn into nectarines.\n\nWhat is the frequency of the mutation on the neighbour's new farm?\n\nA: 0.03\nB: 0.09\nC: 0.3\nD: 0.9\nE: 0.45\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-45.jpg?height=782&width=1231&top_left_y=474&top_left_x=241"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_462",
        "problem": "番茄茎的颜色受一对等位基因控制, 且紫茎对绿茎为完全显性。现将某番茄群体中的紫茎植株与绿茎植株杂交, 子一代中紫茎植株和绿茎植株的比值为 5:1, 如果将亲本紫茎植株自交， $F_{1}$ 中紫茎植株和绿茎植株的比值为（）\nA: $7: 2$\nB: $5: 3$\nC: 9:7\nD: 11:1\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n番茄茎的颜色受一对等位基因控制, 且紫茎对绿茎为完全显性。现将某番茄群体中的紫茎植株与绿茎植株杂交, 子一代中紫茎植株和绿茎植株的比值为 5:1, 如果将亲本紫茎植株自交， $F_{1}$ 中紫茎植株和绿茎植株的比值为（）\n\nA: $7: 2$\nB: $5: 3$\nC: 9:7\nD: 11:1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1226",
        "problem": "The following table shows various inhalational anaesthetic agents with comparison between pungency, potency, onset and blood-gas solubility. Potency is measured by minimal alveolar concentration (MAC), which is the concentration of vapour required to produce immobility on surgical stimulus in $50 \\%$ of patients. A high MAC indicates a low potency. Onset is the time it takes for the effects of a drug to be observed after administration.\n\n| Anaesthetic <br> Agent | Pungency | Potency <br> $($ MAC) | Onset | Blood-gas <br> Solubility |\n| :--- | :--- | :--- | :--- | :--- |\n| Nitrous oxide | Non-pungent | $101 \\%$ | Rapid | Low |\n| Halothane | Non-pungent | $0.86 \\%$ | Slow | Intermediate |\n| Isoflurane | Pungent | $1.1 \\%$ | Medium | Intermediate |\n| Sevoflurane | Non-pungent | $1.7 \\%$ | Rapid | Low |\n| Desflurane | Pungent | $6.0 \\%$ | Rapid | Low |\n\nWhich of the following statements is true?\nA: Nitrous oxide has a high potency and rapid onset.\nB: Sevoflurane and Desflurane are both pungent agents with rapid onsets.\nC: Halothane has a high potency and low blood-gas solubility.\nD: Halothane and Isoflurane are both highly potent with intermediate blood-gas solubility.\nE: Desflurane has a very low potency and rapid onset.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following table shows various inhalational anaesthetic agents with comparison between pungency, potency, onset and blood-gas solubility. Potency is measured by minimal alveolar concentration (MAC), which is the concentration of vapour required to produce immobility on surgical stimulus in $50 \\%$ of patients. A high MAC indicates a low potency. Onset is the time it takes for the effects of a drug to be observed after administration.\n\n| Anaesthetic <br> Agent | Pungency | Potency <br> $($ MAC) | Onset | Blood-gas <br> Solubility |\n| :--- | :--- | :--- | :--- | :--- |\n| Nitrous oxide | Non-pungent | $101 \\%$ | Rapid | Low |\n| Halothane | Non-pungent | $0.86 \\%$ | Slow | Intermediate |\n| Isoflurane | Pungent | $1.1 \\%$ | Medium | Intermediate |\n| Sevoflurane | Non-pungent | $1.7 \\%$ | Rapid | Low |\n| Desflurane | Pungent | $6.0 \\%$ | Rapid | Low |\n\nWhich of the following statements is true?\n\nA: Nitrous oxide has a high potency and rapid onset.\nB: Sevoflurane and Desflurane are both pungent agents with rapid onsets.\nC: Halothane has a high potency and low blood-gas solubility.\nD: Halothane and Isoflurane are both highly potent with intermediate blood-gas solubility.\nE: Desflurane has a very low potency and rapid onset.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_478",
        "problem": "某小组对某种动物进行有丝分裂和减数分裂细胞中染色体的数目进行观察, 该种动物体细胞中的染色体数为 $2 \\mathrm{n}$ 。下图中的细胞类型是依据不同时期细胞中染色体数和核 DNA 分子数的数量关系而划分的。下列有关叙述错误的是（ ）\n\n[图1]\n\n## 细胞类型\nA: 一定含有同源染色体的细胞类型有 $\\mathrm{a} 、 \\mathrm{~b}$\nB: $b$ 转变为 $a 、 d$ 转变为 $c$ 是因为着丝粒分裂\nC: b 类型的细胞进行有丝分裂, c 类型的细胞进行减数分裂\nD: 细胞 e 可能为精细胞, 精细胞需要经过变形才能成为精子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某小组对某种动物进行有丝分裂和减数分裂细胞中染色体的数目进行观察, 该种动物体细胞中的染色体数为 $2 \\mathrm{n}$ 。下图中的细胞类型是依据不同时期细胞中染色体数和核 DNA 分子数的数量关系而划分的。下列有关叙述错误的是（ ）\n\n[图1]\n\n## 细胞类型\n\nA: 一定含有同源染色体的细胞类型有 $\\mathrm{a} 、 \\mathrm{~b}$\nB: $b$ 转变为 $a 、 d$ 转变为 $c$ 是因为着丝粒分裂\nC: b 类型的细胞进行有丝分裂, c 类型的细胞进行减数分裂\nD: 细胞 e 可能为精细胞, 精细胞需要经过变形才能成为精子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-15.jpg?height=559&width=1079&top_left_y=1231&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_255",
        "problem": "Long analyzed DNA samples from three families using six loci of short tandem repeat (STR) on six different autosomal chromosomes. Each STR locus has two alleles which are usually different from each other (the number in the table below e.g 3/5). In the first family, the father is Hung, the mother is Huong and their son is Dung. In the second family, the father is Nhan and his two sons are Tin and Nghia. In the third family the father is Phu, and his son is Quy. Long also included a DNA sample from Dat who is unrelated to any of the three families. Tubes were randomly encoded with number but the key was lost except for Huong's sample.\n\nSTR loci\n\n|  | Locus 1 | Locus 2 | Locus 3 | Locus 4 | Locus 5 | Locus 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Huong | $3 / 5$ | $2 / 2$ | $5 / 6$ | $3 / 3$ | $1 / 2$ | $2 / 6$ |\n| 669 | $3 / 5$ | $2 / 7$ | $4 / 9$ | $5 / 8$ | $3 / 7$ | $4 / 5$ |\n| 297 | $1 / 5$ | $3 / 3$ | $3 / 3$ | $6 / 9$ | $2 / 7$ | $4 / 8$ |\n| 653 | $1 / 5$ | $2 / 2$ | $6 / 6$ | $3 / 7$ | $2 / 9$ | $4 / 5$ |\n| 735 | $5 / 7$ | $2 / 4$ | $5 / 5$ | $3 / 4$ | $2 / 2$ | $1 / 2$ |\n| 130 | $5 / 7$ | $7 / 7$ | $5 / 9$ | $3 / 8$ | $2 / 7$ | $4 / 5$ |\n| 860 | $1 / 6$ | $2 / 3$ | $3 / 5$ | $6 / 7$ | $1 / 7$ | $2 / 8$ |\n| 938 | $3 / 7$ | $4 / 5$ | $5 / 6$ | $4 / 4$ | $2 / 3$ | $1 / 2$ |\n| 264 | $3 / 7$ | $7 / 7$ | $1 / 4$ | $5 / 9$ | $7 / 9$ | $3 / 4$ |\nA: Sample 735 is Dat's DNA\nB: Sample 669 is Nhan's DNA\nC: Sample 938 is Hung's DNA\nD: Sample 297 can be Phu's DNA\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nLong analyzed DNA samples from three families using six loci of short tandem repeat (STR) on six different autosomal chromosomes. Each STR locus has two alleles which are usually different from each other (the number in the table below e.g 3/5). In the first family, the father is Hung, the mother is Huong and their son is Dung. In the second family, the father is Nhan and his two sons are Tin and Nghia. In the third family the father is Phu, and his son is Quy. Long also included a DNA sample from Dat who is unrelated to any of the three families. Tubes were randomly encoded with number but the key was lost except for Huong's sample.\n\nSTR loci\n\n|  | Locus 1 | Locus 2 | Locus 3 | Locus 4 | Locus 5 | Locus 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| Huong | $3 / 5$ | $2 / 2$ | $5 / 6$ | $3 / 3$ | $1 / 2$ | $2 / 6$ |\n| 669 | $3 / 5$ | $2 / 7$ | $4 / 9$ | $5 / 8$ | $3 / 7$ | $4 / 5$ |\n| 297 | $1 / 5$ | $3 / 3$ | $3 / 3$ | $6 / 9$ | $2 / 7$ | $4 / 8$ |\n| 653 | $1 / 5$ | $2 / 2$ | $6 / 6$ | $3 / 7$ | $2 / 9$ | $4 / 5$ |\n| 735 | $5 / 7$ | $2 / 4$ | $5 / 5$ | $3 / 4$ | $2 / 2$ | $1 / 2$ |\n| 130 | $5 / 7$ | $7 / 7$ | $5 / 9$ | $3 / 8$ | $2 / 7$ | $4 / 5$ |\n| 860 | $1 / 6$ | $2 / 3$ | $3 / 5$ | $6 / 7$ | $1 / 7$ | $2 / 8$ |\n| 938 | $3 / 7$ | $4 / 5$ | $5 / 6$ | $4 / 4$ | $2 / 3$ | $1 / 2$ |\n| 264 | $3 / 7$ | $7 / 7$ | $1 / 4$ | $5 / 9$ | $7 / 9$ | $3 / 4$ |\n\nA: Sample 735 is Dat's DNA\nB: Sample 669 is Nhan's DNA\nC: Sample 938 is Hung's DNA\nD: Sample 297 can be Phu's DNA\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1366",
        "problem": "[figure1]\n\nLabel B, H and L are unique to cell 2 . These labels represent what organelles\nA: mitochondria, vacuole, cell membrane\nB: chloroplast, vacuole, cell wall\nC: endoplasmic reticulum, nucleus, cell wall\nD: chloroplast, vacuole, cell membrane\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nLabel B, H and L are unique to cell 2 . These labels represent what organelles\n\nA: mitochondria, vacuole, cell membrane\nB: chloroplast, vacuole, cell wall\nC: endoplasmic reticulum, nucleus, cell wall\nD: chloroplast, vacuole, cell membrane\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-03.jpg?height=614&width=920&top_left_y=304&top_left_x=548"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_307",
        "problem": "我国科研人员对水稻的无融合生殖 (不发生雌、雄配子的融合而产生种子的一种繁殖过程）进行研究，解决了普通杂交水稻需每年制种的问题。水稻无融合生殖受两对基因控制: 含基因 $\\mathrm{E}$ 的植株形成雌配子时, 减数分裂 I 时同源染色体移向同一极, 使雌配子染色体数目加倍; 含基因 $\\mathrm{F}$ 的植株产生的雌配子不经受精直接发育成植株。雄配子的发育不受基因 $\\mathrm{E} 、 \\mathrm{~F}$ 的影响。图为部分水稻品系杂交的示意图。下列叙述错误的是\n\n[图1]\nA: 普通杂交水稻自交后会出现性状分离\nB: 图中?表示的基因型为 eeff\nC: 子代中II号个体自交后代的基因型有 3 种\nD: 子代中 IV 号个体的杂种优势可稳定遗传\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n我国科研人员对水稻的无融合生殖 (不发生雌、雄配子的融合而产生种子的一种繁殖过程）进行研究，解决了普通杂交水稻需每年制种的问题。水稻无融合生殖受两对基因控制: 含基因 $\\mathrm{E}$ 的植株形成雌配子时, 减数分裂 I 时同源染色体移向同一极, 使雌配子染色体数目加倍; 含基因 $\\mathrm{F}$ 的植株产生的雌配子不经受精直接发育成植株。雄配子的发育不受基因 $\\mathrm{E} 、 \\mathrm{~F}$ 的影响。图为部分水稻品系杂交的示意图。下列叙述错误的是\n\n[图1]\n\nA: 普通杂交水稻自交后会出现性状分离\nB: 图中?表示的基因型为 eeff\nC: 子代中II号个体自交后代的基因型有 3 种\nD: 子代中 IV 号个体的杂种优势可稳定遗传\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-76.jpg?height=246&width=666&top_left_y=982&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_375",
        "problem": "踠豆子叶的颜色黄色 (Y) 对绿色 (y) 为显性, 种子的圆粒 $(\\mathrm{R})$ 对皱粒 $(\\mathrm{r})$ 为显性, 两对基因独立遗传。某人用黄色圆粒和绿色圆粒的豌豆进行杂交, 发现 $F_{1}$ 出现 4 种类型, 对性状的统计结果如图所示, 下列说法不正确的是 ( )\n\n[图1]\nA: 亲本的基因型是 $Y y R r$ 和 yyRr\nB: $F_{1}$ 表现型不同于亲本的个体在子代中的比例为 $1 / 4$\nC: 若用 $F_{1}$ 中的一株黄色圆粒豌豆与绿色皱粒踠豆杂交, 得到的 $F_{2}$ 的表现型及比例为黄色圆粒：绿色圆粒：黄色皱粒：绿色皱粒=2: 2: 1: 1\nD: 若将 $F_{1}$ 中的黄色圆粒踠豆自交，得到的 $F_{2}$ 的表现型及比例为黄色圆粒: 绿色圆粒：黄色皱粒：绿色皱粒=15：5：3：1\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n踠豆子叶的颜色黄色 (Y) 对绿色 (y) 为显性, 种子的圆粒 $(\\mathrm{R})$ 对皱粒 $(\\mathrm{r})$ 为显性, 两对基因独立遗传。某人用黄色圆粒和绿色圆粒的豌豆进行杂交, 发现 $F_{1}$ 出现 4 种类型, 对性状的统计结果如图所示, 下列说法不正确的是 ( )\n\n[图1]\n\nA: 亲本的基因型是 $Y y R r$ 和 yyRr\nB: $F_{1}$ 表现型不同于亲本的个体在子代中的比例为 $1 / 4$\nC: 若用 $F_{1}$ 中的一株黄色圆粒豌豆与绿色皱粒踠豆杂交, 得到的 $F_{2}$ 的表现型及比例为黄色圆粒：绿色圆粒：黄色皱粒：绿色皱粒=2: 2: 1: 1\nD: 若将 $F_{1}$ 中的黄色圆粒踠豆自交，得到的 $F_{2}$ 的表现型及比例为黄色圆粒: 绿色圆粒：黄色皱粒：绿色皱粒=15：5：3：1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-095.jpg?height=557&width=862&top_left_y=178&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_423",
        "problem": "摩尔根等人用纯合体灰身长翅果蝇和黑身残翅果蝇交配, 子一代 $\\left(\\mathrm{F}_{1}\\right)$ 都是灰身长\n\n翅。已知灰身和黑身是由等位基因 $\\mathrm{B}$ 和控制，长翅和残翅是由等位基因 $\\mathrm{V}$ 和 $\\mathrm{v}$ 控制，下列叙述正确的是（ ）\nA: 若两对等位基因位于两对常染色体上, $\\mathrm{F}_{1}$ 雌雄果蝇相互交配, 随机取 16 只果蝇,则其中有 9 只是灰身长翅\nB: 由于 $F_{1}$ 雌雄果蝇的表现型相同, 故两对等位基因均位于常染色体上\nC: 若两对等位基因位于常染色体, 则 $F_{1}$ 的测交后代有四种表现型且比例为 1：1： $1: 1$\nD: 若 $F_{1}$ 测交后代中灰身长翅：灰身残翅：黑身长翅：黑身残翅 $=4: 1: 1: 4$, 则可能 BV 在一条染色体上, bv 在另一条同源染色体上\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n摩尔根等人用纯合体灰身长翅果蝇和黑身残翅果蝇交配, 子一代 $\\left(\\mathrm{F}_{1}\\right)$ 都是灰身长\n\n翅。已知灰身和黑身是由等位基因 $\\mathrm{B}$ 和控制，长翅和残翅是由等位基因 $\\mathrm{V}$ 和 $\\mathrm{v}$ 控制，下列叙述正确的是（ ）\n\nA: 若两对等位基因位于两对常染色体上, $\\mathrm{F}_{1}$ 雌雄果蝇相互交配, 随机取 16 只果蝇,则其中有 9 只是灰身长翅\nB: 由于 $F_{1}$ 雌雄果蝇的表现型相同, 故两对等位基因均位于常染色体上\nC: 若两对等位基因位于常染色体, 则 $F_{1}$ 的测交后代有四种表现型且比例为 1：1： $1: 1$\nD: 若 $F_{1}$ 测交后代中灰身长翅：灰身残翅：黑身长翅：黑身残翅 $=4: 1: 1: 4$, 则可能 BV 在一条染色体上, bv 在另一条同源染色体上\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1084",
        "problem": "The process of photosynthesis needs carbon dioxide to diffuse from the atmosphere into the leaf and into the carboxylation site of RUBISCO. This happens through a series of steps. Each step in this diffusion pathway imposes resistance to $\\mathrm{CO}_{2}$ diffusion. This is depicted in the diagram. Various resistances are listed below.\n\ni. Boundary layer resistance\n\nii. Stomatal resistance\n\niii. Liquid phase resistance\n\niv. Intercellular air space resistance\n\n[figure1]\n\nWhich of the following shows the correct combination?\nA: $P$ - i; Q - iii; R - ii; S - iv\nB: $P$ - i; Q - ii; R - iii; S - iv\nC: P - iii; Q - iv; R - ii; S - i\nD: P-iv; Q - iii; R - ii; S - i\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe process of photosynthesis needs carbon dioxide to diffuse from the atmosphere into the leaf and into the carboxylation site of RUBISCO. This happens through a series of steps. Each step in this diffusion pathway imposes resistance to $\\mathrm{CO}_{2}$ diffusion. This is depicted in the diagram. Various resistances are listed below.\n\ni. Boundary layer resistance\n\nii. Stomatal resistance\n\niii. Liquid phase resistance\n\niv. Intercellular air space resistance\n\n[figure1]\n\nWhich of the following shows the correct combination?\n\nA: $P$ - i; Q - iii; R - ii; S - iv\nB: $P$ - i; Q - ii; R - iii; S - iv\nC: P - iii; Q - iv; R - ii; S - i\nD: P-iv; Q - iii; R - ii; S - i\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-04.jpg?height=889&width=873&top_left_y=1751&top_left_x=1103"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1017",
        "problem": "Of the cellular compartments listed below, which has the lowest $\\mathrm{pH}$ ?\nA: Mitochondrial matrix\nB: Endoplasmic reticulum\nC: Lysosome\nD: Cytoplasm\nE: Nucleus\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOf the cellular compartments listed below, which has the lowest $\\mathrm{pH}$ ?\n\nA: Mitochondrial matrix\nB: Endoplasmic reticulum\nC: Lysosome\nD: Cytoplasm\nE: Nucleus\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1358",
        "problem": "Auxin (IAA) is a hormone produced primarily in shoot apical meristems and young leaves. Root apical meristems also produce some auxin, although the root depends on the shoot for much of its auxin. Amongst its many functions in plant growth, auxin plays an important role in gravitropism.\n\nFigure A shows the anatomy of a root. Figure B shows the relationship between auxin concentration and growth of roots and shoots.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\nWhich of the following statements is true?\nA: The concentration of auxin in the roots is inversely proportional to the concentration of auxin the shoots.\nB: Maximal shoot growth occurs at auxin concentrations of less than $1 \\times 10^{3}$ parts per billion.\nC: Maximal shoot growth occurs at auxin concentrations of less than $1 \\times 10^{-7}$ parts per billion.\nD: At a given concentration, auxin may not stimulate growth of both shoots and roots.\nE: At a given concentration, auxin may inhibit the growth of both shoots and roots.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAuxin (IAA) is a hormone produced primarily in shoot apical meristems and young leaves. Root apical meristems also produce some auxin, although the root depends on the shoot for much of its auxin. Amongst its many functions in plant growth, auxin plays an important role in gravitropism.\n\nFigure A shows the anatomy of a root. Figure B shows the relationship between auxin concentration and growth of roots and shoots.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\nWhich of the following statements is true?\n\nA: The concentration of auxin in the roots is inversely proportional to the concentration of auxin the shoots.\nB: Maximal shoot growth occurs at auxin concentrations of less than $1 \\times 10^{3}$ parts per billion.\nC: Maximal shoot growth occurs at auxin concentrations of less than $1 \\times 10^{-7}$ parts per billion.\nD: At a given concentration, auxin may not stimulate growth of both shoots and roots.\nE: At a given concentration, auxin may inhibit the growth of both shoots and roots.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-17.jpg?height=731&width=599&top_left_y=700&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-17.jpg?height=671&width=683&top_left_y=767&top_left_x=1052"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_763",
        "problem": "实验中下列关于化学试剂或相关操作的叙述正确的的是（ ）\nA: 酒精在“新冠病毒预防”和“检测生物组织中的脂肪”中的作用相同\nB: 盐酸在“观察根尖分生区细胞有丝分裂”和“探究 $\\mathrm{pH}$ 对酶活性的影响”中的作用相同\nC: $\\mathrm{CuSO}_{4}$ 在“检测生物组织中的还原糖”和“检测生物组织中的蛋白质\"中的作用相同\nD: 显微镜在“观察植物细胞的质壁分离和复原”和“观察蝗虫精母细胞减数分裂装片”的用法不同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n实验中下列关于化学试剂或相关操作的叙述正确的的是（ ）\n\nA: 酒精在“新冠病毒预防”和“检测生物组织中的脂肪”中的作用相同\nB: 盐酸在“观察根尖分生区细胞有丝分裂”和“探究 $\\mathrm{pH}$ 对酶活性的影响”中的作用相同\nC: $\\mathrm{CuSO}_{4}$ 在“检测生物组织中的还原糖”和“检测生物组织中的蛋白质\"中的作用相同\nD: 显微镜在“观察植物细胞的质壁分离和复原”和“观察蝗虫精母细胞减数分裂装片”的用法不同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1399",
        "problem": "The graph shows the birth rate and death rate for a population of kangaroos over a 100year period.\n\n[figure1]\n\nAssuming equal emigration and immigration, from 1900 to 2000 , the population has:\nA: Increased\nB: Decreased\nC: Stayed the same\nD: Increased until 1930, then decreased\nE: Decreased until 1930, then stayed the same\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows the birth rate and death rate for a population of kangaroos over a 100year period.\n\n[figure1]\n\nAssuming equal emigration and immigration, from 1900 to 2000 , the population has:\n\nA: Increased\nB: Decreased\nC: Stayed the same\nD: Increased until 1930, then decreased\nE: Decreased until 1930, then stayed the same\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-29.jpg?height=851&width=1322&top_left_y=1025&top_left_x=310"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_540",
        "problem": "某种鸟的羽毛有覆羽和单羽是一对相对性状（基因为 A、a； B、b 基因位于常染试卷第 24 页，共 99 页\n色体上, bb 可使部分应表现为单羽的个体表现为覆羽, 其余基因型无此影响。从群体中随机选取三对组合进行杂交实验，如下表所示。下列说法错误的是（）\n\n| 杂交组合 | 子代 |\n| :--- | :--- |\n| 组合一: 单羽雄鸟×覆羽雌鸟 | 覆羽雄鸟: 单羽雌鸟 $=1: 1$ |\n| 组合二: 覆羽雄鸟×单羽雌鸟 | 全为覆羽 |\n| 组合三: 覆羽雄鸟×覆羽雌鸟 | 出现单羽雄鸟 |\nA: 控制羽毛性状的基因位于 $\\mathrm{Z}$ 染色体上，覆羽由 $\\mathrm{Z}$ 染色体上的 $\\mathrm{A}$ 基因控制\nB: 组合一亲本基因型可能是 $b b Z^{a} Z^{a} 、 B B Z^{A} W$\nC: 组合二子代覆羽雄鸟基因型是 $Z^{A} Z^{a}$ 或 $Z^{A} Z^{A}$\nD: 组合三子代单羽雄鸟基因型可能为 bbZa $Z^{a}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种鸟的羽毛有覆羽和单羽是一对相对性状（基因为 A、a； B、b 基因位于常染试卷第 24 页，共 99 页\n色体上, bb 可使部分应表现为单羽的个体表现为覆羽, 其余基因型无此影响。从群体中随机选取三对组合进行杂交实验，如下表所示。下列说法错误的是（）\n\n| 杂交组合 | 子代 |\n| :--- | :--- |\n| 组合一: 单羽雄鸟×覆羽雌鸟 | 覆羽雄鸟: 单羽雌鸟 $=1: 1$ |\n| 组合二: 覆羽雄鸟×单羽雌鸟 | 全为覆羽 |\n| 组合三: 覆羽雄鸟×覆羽雌鸟 | 出现单羽雄鸟 |\n\nA: 控制羽毛性状的基因位于 $\\mathrm{Z}$ 染色体上，覆羽由 $\\mathrm{Z}$ 染色体上的 $\\mathrm{A}$ 基因控制\nB: 组合一亲本基因型可能是 $b b Z^{a} Z^{a} 、 B B Z^{A} W$\nC: 组合二子代覆羽雄鸟基因型是 $Z^{A} Z^{a}$ 或 $Z^{A} Z^{A}$\nD: 组合三子代单羽雄鸟基因型可能为 bbZa $Z^{a}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_467",
        "problem": "晼豆子叶的黄色对绿色为显性, 种子的圆粒对皱粒为显性, 并且两对性状独立遗传。以 1 株黄色圆粒和 1 株绿色皱粒的踠豆作为亲本, 杂交得到 $\\mathrm{F}_{1}$, 其自交得到的 $\\mathrm{F}_{2}$ 中黄色圆粒: 黄色皱粒: 绿色圆粒: 绿色皱粒=9: 3 :\n\n15: 5, 则黄色圆粒的亲本产生的配子种类有\nA: 1 种\nB: 2 种\nC: 3 种\nD: 4 种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n晼豆子叶的黄色对绿色为显性, 种子的圆粒对皱粒为显性, 并且两对性状独立遗传。以 1 株黄色圆粒和 1 株绿色皱粒的踠豆作为亲本, 杂交得到 $\\mathrm{F}_{1}$, 其自交得到的 $\\mathrm{F}_{2}$ 中黄色圆粒: 黄色皱粒: 绿色圆粒: 绿色皱粒=9: 3 :\n\n15: 5, 则黄色圆粒的亲本产生的配子种类有\n\nA: 1 种\nB: 2 种\nC: 3 种\nD: 4 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1388",
        "problem": "Review the following map and answer the 5 subsequent questions.\n\n[figure1]\n\nEcologists have been studying the small archipelago shown in the map above. They have been comparing the ecology of the mainland with those on the nearby islands and have come up with a few discoveries. The area has not been known to have been inhabited by humans, so it remains very natural. The mainland contains a low mountain range approximately $50 \\mathrm{~km}$ inland from the coast (highest peak approximately $400 \\mathrm{~m}$ above sea level) and the prevailing winds are westerlies.\n\nThe ecologists studying the biodiversity on the islands notice that the geckos on island 3 look darker than those on island 1 , even though genetic analysis shows that they are the same species.\n\nWhat is most likely to produce this phenomenon?\nA: Radiant heat differs on each island\nB: The islands differ in their gecko prey availability\nC: Predation pressure\nD: Genetic bottleneck\nE: Low reproduction rates\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nReview the following map and answer the 5 subsequent questions.\n\n[figure1]\n\nEcologists have been studying the small archipelago shown in the map above. They have been comparing the ecology of the mainland with those on the nearby islands and have come up with a few discoveries. The area has not been known to have been inhabited by humans, so it remains very natural. The mainland contains a low mountain range approximately $50 \\mathrm{~km}$ inland from the coast (highest peak approximately $400 \\mathrm{~m}$ above sea level) and the prevailing winds are westerlies.\n\nThe ecologists studying the biodiversity on the islands notice that the geckos on island 3 look darker than those on island 1 , even though genetic analysis shows that they are the same species.\n\nWhat is most likely to produce this phenomenon?\n\nA: Radiant heat differs on each island\nB: The islands differ in their gecko prey availability\nC: Predation pressure\nD: Genetic bottleneck\nE: Low reproduction rates\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-39.jpg?height=925&width=1479&top_left_y=451&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_198",
        "problem": "An experiment is designed to test this hypothesis: the number of yellowjacket wasps at a feeding site visually affects the feeding-site choices of workers collecting nectar.. Four feeders with zero, one, two or eight individual decoys are prepared, as shown in the figure below. One nectar dish is placed in the middle of each feeder. You then observe the feeding-site choice made by each worker.\n\n[figure1]\n\nfeeder 1\n\n[figure2]\n\nfeeder 2\n\n[figure3]\n\nfeeder 3\n\n[figure4]\n\nfeeder 4\n\nWhich of the followings should not be included in this experimental design?\nA: Using live individuals as decoys.\nB: Placing the four feeders with nectar dishes randomly and alternatively.\nC: Using the nectar solutions of equal concentration among the feeders.\nD: Preventing other species from visiting the feeders.\nE: Preventing successive visits by the same worker.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn experiment is designed to test this hypothesis: the number of yellowjacket wasps at a feeding site visually affects the feeding-site choices of workers collecting nectar.. Four feeders with zero, one, two or eight individual decoys are prepared, as shown in the figure below. One nectar dish is placed in the middle of each feeder. You then observe the feeding-site choice made by each worker.\n\n[figure1]\n\nfeeder 1\n\n[figure2]\n\nfeeder 2\n\n[figure3]\n\nfeeder 3\n\n[figure4]\n\nfeeder 4\n\nWhich of the followings should not be included in this experimental design?\n\nA: Using live individuals as decoys.\nB: Placing the four feeders with nectar dishes randomly and alternatively.\nC: Using the nectar solutions of equal concentration among the feeders.\nD: Preventing other species from visiting the feeders.\nE: Preventing successive visits by the same worker.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-39.jpg?height=323&width=388&top_left_y=815&top_left_x=366",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-39.jpg?height=319&width=342&top_left_y=820&top_left_x=797",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-39.jpg?height=317&width=346&top_left_y=821&top_left_x=1186",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-39.jpg?height=319&width=343&top_left_y=820&top_left_x=1573"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_691",
        "problem": "育种专家为获得基因型为 AA 的高产小麦品种, 以基因型为 Aa 的小麦为亲本, 通过逐代自交，且逐代淘汰基因型为 aa 的个体的方法进行育种。下列说法正确的是（）\nA: 育种专家逐代淘汰基因型为 aa 的个体属于自然选择的过程\nB: 基因型为 $\\mathrm{Aa}$ 的小麦自交后代中出现基因型为 $\\mathrm{aa}$ 的个体是基因重组的结果\nC: 该育种过程所得 $F_{2}$ 中，经筛选后基因型为 $A A$ 的个体占 $3 / 5$\nD: 若对基因型为 $\\mathrm{Aa}$ 的小麦进行花药离体培养之后, 再用秋水仙素处理萌发的种子或幼苗，可快速获得基因型为 $\\mathrm{AA}$ 的高产小麦品种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n育种专家为获得基因型为 AA 的高产小麦品种, 以基因型为 Aa 的小麦为亲本, 通过逐代自交，且逐代淘汰基因型为 aa 的个体的方法进行育种。下列说法正确的是（）\n\nA: 育种专家逐代淘汰基因型为 aa 的个体属于自然选择的过程\nB: 基因型为 $\\mathrm{Aa}$ 的小麦自交后代中出现基因型为 $\\mathrm{aa}$ 的个体是基因重组的结果\nC: 该育种过程所得 $F_{2}$ 中，经筛选后基因型为 $A A$ 的个体占 $3 / 5$\nD: 若对基因型为 $\\mathrm{Aa}$ 的小麦进行花药离体培养之后, 再用秋水仙素处理萌发的种子或幼苗，可快速获得基因型为 $\\mathrm{AA}$ 的高产小麦品种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1463",
        "problem": "The following graph depicts the results of a study where the number of tadpoles within a particular habitat (finite resources) were experimentally varied.\n\nNote that the numbers next to the lines are the numbers of individuals in each treatment.\n\n[figure1]\n\nAs the number of tadpoles in the habitat increases, the mean body mass at week three:\nA: increases\nB: decreases\nC: stays the same\nD: cannot tell from this diagram\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph depicts the results of a study where the number of tadpoles within a particular habitat (finite resources) were experimentally varied.\n\nNote that the numbers next to the lines are the numbers of individuals in each treatment.\n\n[figure1]\n\nAs the number of tadpoles in the habitat increases, the mean body mass at week three:\n\nA: increases\nB: decreases\nC: stays the same\nD: cannot tell from this diagram\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-07.jpg?height=602&width=1205&top_left_y=1561&top_left_x=266"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_78",
        "problem": "The figure below depicts the relation between substrate concentration and enzyme reaction rate.\n\n[figure1]\nA: At saturating levels of substrate, the rate of an enzyme catalysed reaction is proportional to the enzyme concentration.\nB: If enough substrate is added, the $V_{\\max }$ of an enzyme catalysed reaction in the presence of a non-competitive inhibitor can be the same as $V_{\\max }$ of the reaction in absence of the inhibitor.\nC: The rate of an enzyme catalysed reaction in the presence of a rate-limiting concentration of substrate decreases with time.\nD: The sigmoidal shape of the $\\mathrm{V}$ versus [S] curve obtained with allosteric enzymes indicates that the affinity of the enzyme for substrate decreases as the substrate\nE: The affinity of an allosteric enzyme for substrate varies with enzyme concentration.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe figure below depicts the relation between substrate concentration and enzyme reaction rate.\n\n[figure1]\n\nA: At saturating levels of substrate, the rate of an enzyme catalysed reaction is proportional to the enzyme concentration.\nB: If enough substrate is added, the $V_{\\max }$ of an enzyme catalysed reaction in the presence of a non-competitive inhibitor can be the same as $V_{\\max }$ of the reaction in absence of the inhibitor.\nC: The rate of an enzyme catalysed reaction in the presence of a rate-limiting concentration of substrate decreases with time.\nD: The sigmoidal shape of the $\\mathrm{V}$ versus [S] curve obtained with allosteric enzymes indicates that the affinity of the enzyme for substrate decreases as the substrate\nE: The affinity of an allosteric enzyme for substrate varies with enzyme concentration.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-23.jpg?height=554&width=880&top_left_y=448&top_left_x=588"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1430",
        "problem": "Turmeric is a flowering plant, Curcuma longa of the ginger family, Zingiberaceae. It has a very long history of medicinal use, dating back nearly 4000 years. In the last 25 years, over 3000 new publications have been released regarding the health effects of turmeric. The primary compound in turmeric is curcumin, which has been purported to have antiinflammatory and antioxidant effects. Scientists conducted a study investigating the effect of varying concentrations and length of exposure of curcumin on adipocytes from a human cell line. Figure A shows the effect of curcumin concentration on the viability of a population of adipocytes.\n\n[figure1]\n\nFurther investigation showed that the effect on viability was due to changes in the regulation of apoptosis. Caspases are known to play essential roles in apoptosis and are an important regulator of this process is the Bax protein, which causes the release of cytochrome $c$.\n\nWhich organelle releases cytochrome $c$ ?\nA: Mitochondrion\nB: Nucleolus\nC: Smooth endoplasmic reticulum\nD: Golgi apparatus\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTurmeric is a flowering plant, Curcuma longa of the ginger family, Zingiberaceae. It has a very long history of medicinal use, dating back nearly 4000 years. In the last 25 years, over 3000 new publications have been released regarding the health effects of turmeric. The primary compound in turmeric is curcumin, which has been purported to have antiinflammatory and antioxidant effects. Scientists conducted a study investigating the effect of varying concentrations and length of exposure of curcumin on adipocytes from a human cell line. Figure A shows the effect of curcumin concentration on the viability of a population of adipocytes.\n\n[figure1]\n\nFurther investigation showed that the effect on viability was due to changes in the regulation of apoptosis. Caspases are known to play essential roles in apoptosis and are an important regulator of this process is the Bax protein, which causes the release of cytochrome $c$.\n\nWhich organelle releases cytochrome $c$ ?\n\nA: Mitochondrion\nB: Nucleolus\nC: Smooth endoplasmic reticulum\nD: Golgi apparatus\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-08.jpg?height=560&width=797&top_left_y=774&top_left_x=618"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_746",
        "problem": "某自花传粉植物, 有紫花和白花性状, 受细胞核基因控制。选择某紫色植株自交,所得子代数量足够多, 统计发现 $F_{1}$ 中开白花植株的比例为 $7 / 16$, 其余均开紫花 (不考虑基因突变和互换）。相关分析错误的是（ ）\nA: 若受两对等位基因控制, 对亲本植株进行测交, 则子代中白花植株的比例为 $3 / 4$\nB: 若受两对等位基因控制, $F_{1}$ 的紫花植株进行自交, 后代中有 $11 / 36$ 的植株开白花\nC: 若受一对等位基因控制, 可能是杂合子植株产生的某种配子中有 $6 / 7$ 不参与受精\nD: 若受一对等位基因控制, $F_{1}$ 的紫花植株进行自交, 后代中有 $2 / 9$ 的植株开白花\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某自花传粉植物, 有紫花和白花性状, 受细胞核基因控制。选择某紫色植株自交,所得子代数量足够多, 统计发现 $F_{1}$ 中开白花植株的比例为 $7 / 16$, 其余均开紫花 (不考虑基因突变和互换）。相关分析错误的是（ ）\n\nA: 若受两对等位基因控制, 对亲本植株进行测交, 则子代中白花植株的比例为 $3 / 4$\nB: 若受两对等位基因控制, $F_{1}$ 的紫花植株进行自交, 后代中有 $11 / 36$ 的植株开白花\nC: 若受一对等位基因控制, 可能是杂合子植株产生的某种配子中有 $6 / 7$ 不参与受精\nD: 若受一对等位基因控制, $F_{1}$ 的紫花植株进行自交, 后代中有 $2 / 9$ 的植株开白花\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1027",
        "problem": "Which of the following statements is incorrect regarding hemophilia, a maternally inherited sex linked trait?\nA: A son of a normal non-carrier mother, will not be affected by hemophilia even if the father has the disease.\nB: A daughter of the same parents will be a heterozygous carrier.\nC: Males are considered to be hemizygous.\nD: If a female has a hemophilic gene on one of her chromosomes, she will not show signs of the disease.\nE: A son will be protected from the disease if neither parent shows any symptoms.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following statements is incorrect regarding hemophilia, a maternally inherited sex linked trait?\n\nA: A son of a normal non-carrier mother, will not be affected by hemophilia even if the father has the disease.\nB: A daughter of the same parents will be a heterozygous carrier.\nC: Males are considered to be hemizygous.\nD: If a female has a hemophilic gene on one of her chromosomes, she will not show signs of the disease.\nE: A son will be protected from the disease if neither parent shows any symptoms.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_530",
        "problem": "某植物的花色有红色和白色两种, 该对相对性状可能由一对等位基因（A/a）控制,也可能由两对等位基因（A/a 和 $\\mathrm{B} / \\mathrm{b}$ ）控制。让红花植株甲进行自花传粉，所得子代中红花:白花 $=15: 1$ 。下列推测不支持“15:1”的是（）\nA: 植株甲的基因型为 $\\mathrm{Aa}$ 其产生的含 a 基因的雌雄配子可育率都为 $1 / 3$\nB: 植株甲的基因型为 $\\mathrm{AaBb}$ 其产生的可育雌雄配子各有 4 种且比例相同\nC: 植株甲的基因型为 $\\mathrm{Aa}$ 其产生的含 a 基因的雌配子可育率为 $1 / 7$\nD: 植株甲的基因型为 $\\mathrm{AaBb}$ ，其产生的可育雌雄配子均只有 $\\mathrm{AB} 、 \\mathrm{ab}$ 且比例为 $2: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某植物的花色有红色和白色两种, 该对相对性状可能由一对等位基因（A/a）控制,也可能由两对等位基因（A/a 和 $\\mathrm{B} / \\mathrm{b}$ ）控制。让红花植株甲进行自花传粉，所得子代中红花:白花 $=15: 1$ 。下列推测不支持“15:1”的是（）\n\nA: 植株甲的基因型为 $\\mathrm{Aa}$ 其产生的含 a 基因的雌雄配子可育率都为 $1 / 3$\nB: 植株甲的基因型为 $\\mathrm{AaBb}$ 其产生的可育雌雄配子各有 4 种且比例相同\nC: 植株甲的基因型为 $\\mathrm{Aa}$ 其产生的含 a 基因的雌配子可育率为 $1 / 7$\nD: 植株甲的基因型为 $\\mathrm{AaBb}$ ，其产生的可育雌雄配子均只有 $\\mathrm{AB} 、 \\mathrm{ab}$ 且比例为 $2: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_605",
        "problem": "进行有性生殖的某种植物含有 $\\mathrm{n}$ 对独立遗传的等位基因, 每对基因只控制一种性状,相应基因可以依次用 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b} 、 \\mathrm{C} / \\mathrm{c} \\ldots .$. 表示，已知植株 $\\mathrm{A}$ 的 $\\mathrm{n}$ 对基因均杂合（杂合子表现为显性性状）。下列说法正确的是（）\nA: 植株 A 测交, 子代中纯合子和杂合子比例相等\nB: 仅考虑两对基因, 若两亲本杂交子代的表型之比为 $1: 1: 1: 1$, 则亲本之一肯定为隐性纯合子\nC: 某植株 $\\mathrm{n}$ 对基因均杂合，不考虑变异的情况下，其测交子代中杂合子所占的比例为 $1 / 2^{\\mathrm{n}}$\nD: 某植株 $\\mathrm{n}$ 对基因均杂合，不考虑变异的情况下，其测交子代中单杂合子（仅一对基因杂合）的比例为 $\\mathrm{n} / 2^{\\mathrm{n}}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n进行有性生殖的某种植物含有 $\\mathrm{n}$ 对独立遗传的等位基因, 每对基因只控制一种性状,相应基因可以依次用 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b} 、 \\mathrm{C} / \\mathrm{c} \\ldots .$. 表示，已知植株 $\\mathrm{A}$ 的 $\\mathrm{n}$ 对基因均杂合（杂合子表现为显性性状）。下列说法正确的是（）\n\nA: 植株 A 测交, 子代中纯合子和杂合子比例相等\nB: 仅考虑两对基因, 若两亲本杂交子代的表型之比为 $1: 1: 1: 1$, 则亲本之一肯定为隐性纯合子\nC: 某植株 $\\mathrm{n}$ 对基因均杂合，不考虑变异的情况下，其测交子代中杂合子所占的比例为 $1 / 2^{\\mathrm{n}}$\nD: 某植株 $\\mathrm{n}$ 对基因均杂合，不考虑变异的情况下，其测交子代中单杂合子（仅一对基因杂合）的比例为 $\\mathrm{n} / 2^{\\mathrm{n}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_865",
        "problem": "用转基因技术将抗虫基因 $\\mathrm{A}$ 和抗除草剂基因 $\\mathrm{B}$ 成功导入某植株细胞 $(2 \\mathrm{n}=40)$ 的染色体组中, 随后将该细胞培养成植株 $\\mathrm{Y}$ 。植株 $\\mathrm{Y}$ 自交, 子代中既不抗虫也不抗除草剂的植株约占 $1 / 16$ 。取植株 Y 某部位的一个细胞放在适宜条件下培养, 产生 4 个子细胞。用苂光分子标记追踪基因 A 和 B(基因 A 和 B 均能被苂光标记, 且培养过程中不发生突变)。下列叙述不正确的是\nA: 若 4 个子细胞分别含 $2 、 2 、 2 、 2$ 个苂光点, 则该细胞含有同源染色体\nB: 若某后期细胞有 3 个荧光点, 其产生原理与植株 Y 获得抗虫基因和抗除草剂基因的原理相同\nC: 若 4 个子细胞分别含 $1 、 1 、 1 、 1$ 个荧光点, 则取材部位可能为根尖分生区\nD: 若 4 个子细胞分别含 $2 、 1 、 1 、 0$ 个苂光点, 则细胞分裂过程中出现过四分体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n用转基因技术将抗虫基因 $\\mathrm{A}$ 和抗除草剂基因 $\\mathrm{B}$ 成功导入某植株细胞 $(2 \\mathrm{n}=40)$ 的染色体组中, 随后将该细胞培养成植株 $\\mathrm{Y}$ 。植株 $\\mathrm{Y}$ 自交, 子代中既不抗虫也不抗除草剂的植株约占 $1 / 16$ 。取植株 Y 某部位的一个细胞放在适宜条件下培养, 产生 4 个子细胞。用苂光分子标记追踪基因 A 和 B(基因 A 和 B 均能被苂光标记, 且培养过程中不发生突变)。下列叙述不正确的是\n\nA: 若 4 个子细胞分别含 $2 、 2 、 2 、 2$ 个苂光点, 则该细胞含有同源染色体\nB: 若某后期细胞有 3 个荧光点, 其产生原理与植株 Y 获得抗虫基因和抗除草剂基因的原理相同\nC: 若 4 个子细胞分别含 $1 、 1 、 1 、 1$ 个荧光点, 则取材部位可能为根尖分生区\nD: 若 4 个子细胞分别含 $2 、 1 、 1 、 0$ 个苂光点, 则细胞分裂过程中出现过四分体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_229",
        "problem": "0\nA: A particular type of colon cancer can be caused by recessive alleles even though its inheritable pattern appears similar to that of a dominant trait.\nB: In one patient, normal cells have only one mutated $p 53$ allele but cancer cells have two identical mutated $p 53$ alleles. Then it can be concluded that the second mutated $p 53$ allele is formed by gene conversion.\nC: Some cancers have been effectively treated with drugs that cause demethylation. Then it can be concluded that genes causing those cancers are more likely to be oncogenes.\nD: Chromosome inversions can produce novel oncogenes.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\n0\n\nA: A particular type of colon cancer can be caused by recessive alleles even though its inheritable pattern appears similar to that of a dominant trait.\nB: In one patient, normal cells have only one mutated $p 53$ allele but cancer cells have two identical mutated $p 53$ alleles. Then it can be concluded that the second mutated $p 53$ allele is formed by gene conversion.\nC: Some cancers have been effectively treated with drugs that cause demethylation. Then it can be concluded that genes causing those cancers are more likely to be oncogenes.\nD: Chromosome inversions can produce novel oncogenes.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1449",
        "problem": "The picture below shows a white-light microscopy image of Paramecium, a unicellular organism which is commonly found in freshwater environments.\n\nGiven that the scale bar represents $120 \\mu \\mathrm{m}$, approximately how many identical organisms can fit length-to-length across a petri dish with a $100 \\mathrm{~mm}$ diameter?\n\n[figure1]\nA: 17\nB: 30\nC: 58\nD: 80\nE: 108\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe picture below shows a white-light microscopy image of Paramecium, a unicellular organism which is commonly found in freshwater environments.\n\nGiven that the scale bar represents $120 \\mu \\mathrm{m}$, approximately how many identical organisms can fit length-to-length across a petri dish with a $100 \\mathrm{~mm}$ diameter?\n\n[figure1]\n\nA: 17\nB: 30\nC: 58\nD: 80\nE: 108\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-23.jpg?height=597&width=648&top_left_y=564&top_left_x=698"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_579",
        "problem": "下图甲表示某基因型为 HH 的二倍体动物细胞在不同时期三种结构或物质的数量情况，图乙表示该个体正在发生的细胞分裂过程，下列叙述正确的是（）\n\n[图1]\n\n图甲\n\n[图2]\n\n图乙\nA: 图甲中 $\\mathrm{a}$ 可表示染色体, $\\mathrm{b}$ 表示染色单体, $\\mathrm{c}$ 表示核 DNA\nB: 图甲中处于III时期的细胞含两个染色体组\nC: 图乙 B 细胞内同时出现 $\\mathrm{H}$ 和 $\\mathrm{h}$ 一定是基因突变造成的\nD: 由图乙 B 的细胞分裂特点可知, 该动物一定是雄性动物\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图甲表示某基因型为 HH 的二倍体动物细胞在不同时期三种结构或物质的数量情况，图乙表示该个体正在发生的细胞分裂过程，下列叙述正确的是（）\n\n[图1]\n\n图甲\n\n[图2]\n\n图乙\n\nA: 图甲中 $\\mathrm{a}$ 可表示染色体, $\\mathrm{b}$ 表示染色单体, $\\mathrm{c}$ 表示核 DNA\nB: 图甲中处于III时期的细胞含两个染色体组\nC: 图乙 B 细胞内同时出现 $\\mathrm{H}$ 和 $\\mathrm{h}$ 一定是基因突变造成的\nD: 由图乙 B 的细胞分裂特点可知, 该动物一定是雄性动物\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-08.jpg?height=423&width=920&top_left_y=488&top_left_x=334",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-08.jpg?height=477&width=728&top_left_y=1029&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_350",
        "problem": "有甲、乙两种人类遗传病, 相关基因分别用 $\\mathrm{A} 、 \\mathrm{a}$ 和 $\\mathrm{B} 、 \\mathrm{~b}$ 表示, 下图为某家族系谱图, 系谱中 6 号个体无致病基因。相关叙述正确的是（ ）\n\n[图1]\nA: 甲病为常染色体显性遗传病\nB: 5 号个体基因型为 $\\mathrm{AaX}^{\\mathrm{B}} X^{\\mathrm{B}}$ 或 $\\mathrm{Aa} X^{\\mathrm{B}} X^{\\mathrm{b}}$\nC: 7 号个体基因型为 $a \\mathrm{aX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{B}}$\nD: 若 8 与 9 婚配, 生出患病孩子的概率为 $25 / 32$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n有甲、乙两种人类遗传病, 相关基因分别用 $\\mathrm{A} 、 \\mathrm{a}$ 和 $\\mathrm{B} 、 \\mathrm{~b}$ 表示, 下图为某家族系谱图, 系谱中 6 号个体无致病基因。相关叙述正确的是（ ）\n\n[图1]\n\nA: 甲病为常染色体显性遗传病\nB: 5 号个体基因型为 $\\mathrm{AaX}^{\\mathrm{B}} X^{\\mathrm{B}}$ 或 $\\mathrm{Aa} X^{\\mathrm{B}} X^{\\mathrm{b}}$\nC: 7 号个体基因型为 $a \\mathrm{aX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{B}}$\nD: 若 8 与 9 婚配, 生出患病孩子的概率为 $25 / 32$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-087.jpg?height=365&width=571&top_left_y=1845&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_479",
        "problem": "图为甲、乙两个家族的遗传系谱图, 已知甲、乙两个家族个体不含对方的致病基因, 6 号不携带致病基因（不考虑基因突变）。假设甲病由 $\\mathrm{A} 、 \\mathrm{a}$ 基因控制，乙病由 $\\mathrm{B} 、 \\mathrm{~b}$ 基因控制，女性群体中这两种病的发病率均为 $1 / 10000$ 。下列叙述正确的是（）\n\n[图1]\n\n甲家族\n\n[图2]\n\n乙家族\n\n$O \\square$ 正常男性、女性\n\n- 患甲病男性、女性\n\n患乙病男性、女性\nA: 乙病的遗传方式为常染色体显性遗传\nB: 选择患者家系调查乙病发病率时, 调查结果应接近 $1 / 10000$\nC: 甲家族的 8 号与人群中一男性婚配, 生育后代患甲病的概率为 $101 / 800$\nD: 若 7 号与 18 号婚配, 生育了一位乙病患儿, 则后代男孩中两病兼患的概率为 $1 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图为甲、乙两个家族的遗传系谱图, 已知甲、乙两个家族个体不含对方的致病基因, 6 号不携带致病基因（不考虑基因突变）。假设甲病由 $\\mathrm{A} 、 \\mathrm{a}$ 基因控制，乙病由 $\\mathrm{B} 、 \\mathrm{~b}$ 基因控制，女性群体中这两种病的发病率均为 $1 / 10000$ 。下列叙述正确的是（）\n\n[图1]\n\n甲家族\n\n[图2]\n\n乙家族\n\n$O \\square$ 正常男性、女性\n\n- 患甲病男性、女性\n\n患乙病男性、女性\n\nA: 乙病的遗传方式为常染色体显性遗传\nB: 选择患者家系调查乙病发病率时, 调查结果应接近 $1 / 10000$\nC: 甲家族的 8 号与人群中一男性婚配, 生育后代患甲病的概率为 $101 / 800$\nD: 若 7 号与 18 号婚配, 生育了一位乙病患儿, 则后代男孩中两病兼患的概率为 $1 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-064.jpg?height=320&width=391&top_left_y=157&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-064.jpg?height=362&width=400&top_left_y=156&top_left_x=768"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_409",
        "problem": "某种二倍体高等植物的性别决定类型为 XY 型。该植物有宽叶和窄叶两种叶形, 宽叶对窄叶为显性。控制这对相对性状的基因 (B/b) 位于 $\\mathrm{X}$ 染色体上，含有基因 $\\mathrm{b}$ 的花粉不育。下列叙述错误的是 $(\\quad)$\nA: 窄叶性状只能出现在雄株中, 不可能出现在雌株中\nB: 宽叶雌株与宽叶雄株杂交, 子代中可能出现窄叶雄株\nC: 宽叶雌株与窄叶雄株杂交, 子代中既有雌株又有雄株\nD: 若亲本杂交后子代雄株均为宽叶, 则亲本雌株是纯合子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种二倍体高等植物的性别决定类型为 XY 型。该植物有宽叶和窄叶两种叶形, 宽叶对窄叶为显性。控制这对相对性状的基因 (B/b) 位于 $\\mathrm{X}$ 染色体上，含有基因 $\\mathrm{b}$ 的花粉不育。下列叙述错误的是 $(\\quad)$\n\nA: 窄叶性状只能出现在雄株中, 不可能出现在雌株中\nB: 宽叶雌株与宽叶雄株杂交, 子代中可能出现窄叶雄株\nC: 宽叶雌株与窄叶雄株杂交, 子代中既有雌株又有雄株\nD: 若亲本杂交后子代雄株均为宽叶, 则亲本雌株是纯合子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1376",
        "problem": "John takes $1 \\mathrm{~g}$ of soil and adds $9 \\mathrm{~mL}$ of water (making total volume $10 \\mathrm{~mL}$ ). Assume all the bacteria in the soil end up evenly distributed in this suspension. Two serial 10 -fold dilutions of this soil suspension are performed and $0.2 \\mathrm{~mL}$ of the last dilution is plated on an agar plate. After a two-day incubation period, 250 colony forming units (CFU) were counted on the agar plate. What is the concentration of bacteria in the initial soil sample?\nA: $2.50 \\times 106 \\mathrm{CFU} / \\mathrm{gram}$ of soil\nB: $1.25 \\times 105 \\mathrm{CFU} / \\mathrm{gram}$ of soil\nC: $1.25 \\times 106 \\mathrm{CFU} / \\mathrm{gram}$ of soil\nD: $1.25 \\times 107 \\mathrm{CFU} / \\mathrm{gram}$ of soil\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nJohn takes $1 \\mathrm{~g}$ of soil and adds $9 \\mathrm{~mL}$ of water (making total volume $10 \\mathrm{~mL}$ ). Assume all the bacteria in the soil end up evenly distributed in this suspension. Two serial 10 -fold dilutions of this soil suspension are performed and $0.2 \\mathrm{~mL}$ of the last dilution is plated on an agar plate. After a two-day incubation period, 250 colony forming units (CFU) were counted on the agar plate. What is the concentration of bacteria in the initial soil sample?\n\nA: $2.50 \\times 106 \\mathrm{CFU} / \\mathrm{gram}$ of soil\nB: $1.25 \\times 105 \\mathrm{CFU} / \\mathrm{gram}$ of soil\nC: $1.25 \\times 106 \\mathrm{CFU} / \\mathrm{gram}$ of soil\nD: $1.25 \\times 107 \\mathrm{CFU} / \\mathrm{gram}$ of soil\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1151",
        "problem": "An $70 \\mathrm{~kg}$ adult male has approximately $42 \\mathrm{~L}$ of water in his body. Of this volume, $55 \\%$ of the water is found within the cells and $45 \\%$ of water is found outside the cells. For the water found outside the cells, $83.3 \\%$ is found outside the blood vessels.\n\nWhat is the approximate amount of water found within the blood vessels?\nA: $5.26 \\mathrm{~kg}$\nB: $6.43 \\mathrm{~kg}$\nC: $3.16 \\mathrm{~L}$\nD: $3.86 \\mathrm{~L}$\nE: $16.86 \\mathrm{~L}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn $70 \\mathrm{~kg}$ adult male has approximately $42 \\mathrm{~L}$ of water in his body. Of this volume, $55 \\%$ of the water is found within the cells and $45 \\%$ of water is found outside the cells. For the water found outside the cells, $83.3 \\%$ is found outside the blood vessels.\n\nWhat is the approximate amount of water found within the blood vessels?\n\nA: $5.26 \\mathrm{~kg}$\nB: $6.43 \\mathrm{~kg}$\nC: $3.16 \\mathrm{~L}$\nD: $3.86 \\mathrm{~L}$\nE: $16.86 \\mathrm{~L}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1165",
        "problem": "## DEAR ENEMIES \\& NASTY NEIGHBORS\n\n\"Dear Enemy, I curse you, and hope that something slightly unpleasant happens to you, like an onion falling on your head.\"\n\n## Blackadder I, The Archbishop\n\nMany species show territorial behaviour (actively defending an area containing resources) allowing the species to monopolise resources such as food or mates. Territoriality is costly in terms of energy and time so some species reduce these costs by being less aggressive towards their neighbours than towards unfamiliar strangers, the so called \"dear enemy\" hypothesis. The converse are \"nasty neighbours\" in which species are more, not less aggressive towards their neighbours.\n\nDr Brunton's research group, at the Institute of Natural Sciences at Massey University, has been studying the behaviour of the New Zealand bellbird,\n\n[figure1]\n(Anthornis melanura) on Tiritiri Matangi Island. Both sexes sing prolifically and are known to use vocalisations to recognise individuals. Male and female bellbirds counter-sing in response to the vocalisations of their neighbours in a territorial behaviour that may lead to chasing of an individual bird intruding on a territory.\n\nThe researchers used speakers to play the song of neighbouring females or stranger females at different breeding stages. They recorded the responses of male and female bellbirds, including counter-singing and whether the birds approached the speaker. An asterisk indicates significant differences between the response to neighbours and strangers. Due to small sample sizes, the responses to the neighbours versus strangers were not tested during the incubation stage. Control 1 was silence, control 2 played back the\n\n[figure2]\nsong of a different species.\n[figure3]\n\nb) Counter-singing\n[figure4]\n\nUsing the information and the graphs above answer Questions 54 - 58 on the following page.This research provides evidence for which of the following statements?\nA: The female bellbird responses support the \"dear enemy\" hypothesis.\nB: The female bellbird responses support the \"nasty neighbours\" hypothesis.\nC: The male bellbird responses support the \"dear enemy\" hypothesis.\nD: The male bellbird responses support the \"nasty neighbours\" hypothesis.\nE: Both male and female bellbird responses support \"dear enemy\" hypothesis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## DEAR ENEMIES \\& NASTY NEIGHBORS\n\n\"Dear Enemy, I curse you, and hope that something slightly unpleasant happens to you, like an onion falling on your head.\"\n\n## Blackadder I, The Archbishop\n\nMany species show territorial behaviour (actively defending an area containing resources) allowing the species to monopolise resources such as food or mates. Territoriality is costly in terms of energy and time so some species reduce these costs by being less aggressive towards their neighbours than towards unfamiliar strangers, the so called \"dear enemy\" hypothesis. The converse are \"nasty neighbours\" in which species are more, not less aggressive towards their neighbours.\n\nDr Brunton's research group, at the Institute of Natural Sciences at Massey University, has been studying the behaviour of the New Zealand bellbird,\n\n[figure1]\n(Anthornis melanura) on Tiritiri Matangi Island. Both sexes sing prolifically and are known to use vocalisations to recognise individuals. Male and female bellbirds counter-sing in response to the vocalisations of their neighbours in a territorial behaviour that may lead to chasing of an individual bird intruding on a territory.\n\nThe researchers used speakers to play the song of neighbouring females or stranger females at different breeding stages. They recorded the responses of male and female bellbirds, including counter-singing and whether the birds approached the speaker. An asterisk indicates significant differences between the response to neighbours and strangers. Due to small sample sizes, the responses to the neighbours versus strangers were not tested during the incubation stage. Control 1 was silence, control 2 played back the\n\n[figure2]\nsong of a different species.\n[figure3]\n\nb) Counter-singing\n[figure4]\n\nUsing the information and the graphs above answer Questions 54 - 58 on the following page.\n\nproblem:\nThis research provides evidence for which of the following statements?\n\nA: The female bellbird responses support the \"dear enemy\" hypothesis.\nB: The female bellbird responses support the \"nasty neighbours\" hypothesis.\nC: The male bellbird responses support the \"dear enemy\" hypothesis.\nD: The male bellbird responses support the \"nasty neighbours\" hypothesis.\nE: Both male and female bellbird responses support \"dear enemy\" hypothesis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=357&width=571&top_left_y=524&top_left_x=1368",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=485&width=311&top_left_y=931&top_left_x=1495",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=486&width=1734&top_left_y=1493&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=534&width=1744&top_left_y=2042&top_left_x=108"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1047",
        "problem": "In many bird and mammal species, more often males have secondary sexual characters like the antlers of the deer or colored plumage in birds and/or perform elaborate courtship displays in order to attract females for copulations. Why do you think females are often said to be the 'choosy' sex?\nA: Sperm production by the male is more expensive than egg production by the females, so they can invest more time in mate choice.\nB: The embryo usually develops inside females until birth or egg laying. This involves a great amount of energy and causes females to be choosy.\nC: Males produce much more sperm than females produce eggs. The limited gamete production by females makes them more choosy than males.\nD: A, B and C\nE: B and C\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn many bird and mammal species, more often males have secondary sexual characters like the antlers of the deer or colored plumage in birds and/or perform elaborate courtship displays in order to attract females for copulations. Why do you think females are often said to be the 'choosy' sex?\n\nA: Sperm production by the male is more expensive than egg production by the females, so they can invest more time in mate choice.\nB: The embryo usually develops inside females until birth or egg laying. This involves a great amount of energy and causes females to be choosy.\nC: Males produce much more sperm than females produce eggs. The limited gamete production by females makes them more choosy than males.\nD: A, B and C\nE: B and C\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1157",
        "problem": "Evolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).Assume that once flight is lost it cannot be regained. If both the ancestor of all palaeognaths (at $72.3 \\mathrm{Ma}$ ) and the ancestor of elephant birds and kiwi (at 50.0 Ma) could fly, what is the minimum number of times flightlessness must have arisen among the palaeognaths?\nA: 3\nB: 4\nC: 5\nD: 6\nE: 7\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nEvolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).\n\nproblem:\nAssume that once flight is lost it cannot be regained. If both the ancestor of all palaeognaths (at $72.3 \\mathrm{Ma}$ ) and the ancestor of elephant birds and kiwi (at 50.0 Ma) could fly, what is the minimum number of times flightlessness must have arisen among the palaeognaths?\n\nA: 3\nB: 4\nC: 5\nD: 6\nE: 7\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-28.jpg?height=1868&width=1627&top_left_y=748&top_left_x=206"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_641",
        "problem": "鹅(ZW 型)的我掌颜色有黑色、花色和黄色三种, 该性状由两对等位基因 $\\mathrm{M} / \\mathrm{m}$ 和 $\\mathrm{T} / \\mathrm{t}$共同控制。已知 $\\mathrm{M}$ 基因控制黑色素的合成，无 $\\mathrm{M}$ 基因表现为黄掌。 $\\mathrm{T}$ 基因使鹅掌中的黑色素均匀分布, 表现为全黑性状: 不含 $\\mathrm{T}$ 基因的个体，黑色素随机分布，出现花掌(黑黄相间)。现让两只黑掌我交配, $\\mathrm{F}_{1}$ 表型及比例如下表所示, 已知不考虑 $\\mathrm{Z} 、 \\mathrm{~W}$ 染色体上的同源区段。据表分析，下列叙述错误的是（）\n\n| 性别 | 黑掌 | 花掌 | 黄掌 |\n| :---: | :---: | :---: | :---: |\n| 雄性 | 6 | 0 | 2 |\n| 雌性 | 3 | 3 | 2 |\nA: $\\mathrm{M} / \\mathrm{m}$ 和 $\\mathrm{T} / \\mathrm{t}$ 基因分别位于常染色体和 $\\mathrm{Z}$ 染色\nB: 亲代雌雄个体的基因型分别是 $M m Z^{T} \\mathrm{~W} 、 \\mathrm{MmZ}^{\\mathrm{T}} \\mathrm{Z}^{\\mathrm{t}}$\nC: $F_{1}$ 个体中花掌鹅和黄掌鹅的基因型均有 2 种\nD: 将 $F_{1}$ 花掌鹅与纯种黄掌鹅交配可验证花掌鹅的基因型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n鹅(ZW 型)的我掌颜色有黑色、花色和黄色三种, 该性状由两对等位基因 $\\mathrm{M} / \\mathrm{m}$ 和 $\\mathrm{T} / \\mathrm{t}$共同控制。已知 $\\mathrm{M}$ 基因控制黑色素的合成，无 $\\mathrm{M}$ 基因表现为黄掌。 $\\mathrm{T}$ 基因使鹅掌中的黑色素均匀分布, 表现为全黑性状: 不含 $\\mathrm{T}$ 基因的个体，黑色素随机分布，出现花掌(黑黄相间)。现让两只黑掌我交配, $\\mathrm{F}_{1}$ 表型及比例如下表所示, 已知不考虑 $\\mathrm{Z} 、 \\mathrm{~W}$ 染色体上的同源区段。据表分析，下列叙述错误的是（）\n\n| 性别 | 黑掌 | 花掌 | 黄掌 |\n| :---: | :---: | :---: | :---: |\n| 雄性 | 6 | 0 | 2 |\n| 雌性 | 3 | 3 | 2 |\n\nA: $\\mathrm{M} / \\mathrm{m}$ 和 $\\mathrm{T} / \\mathrm{t}$ 基因分别位于常染色体和 $\\mathrm{Z}$ 染色\nB: 亲代雌雄个体的基因型分别是 $M m Z^{T} \\mathrm{~W} 、 \\mathrm{MmZ}^{\\mathrm{T}} \\mathrm{Z}^{\\mathrm{t}}$\nC: $F_{1}$ 个体中花掌鹅和黄掌鹅的基因型均有 2 种\nD: 将 $F_{1}$ 花掌鹅与纯种黄掌鹅交配可验证花掌鹅的基因型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_36",
        "problem": "The tryptophan operon (trp operon) of $E$. coli is transcriptionally regulated by a repressor that is activated by the binding of tryptophan. The active form repressor binds to the operator sequence located between the promoter and the transcription initiating point and blocks the RNA polymerase. There is another expression control system called the attenuator linked to transcription and translation in the trp operon.\n\nBetween the operator sequence and the $\\operatorname{trpE}$ gene, which is the first structural gene of the trp operon, there are four sequences of about 15 bases called Region 1-4 (Figure 1). Region 1 and Region 2, and Region 3 and Region 4 have complementary sequences, respectively. When these regions are transcribed as mRNA, they are paired with each other and form stem-loop structures (Figure 2). Furthermore, the sequences of Region 2 and Region 3 are also complementary, so a stem loop structure can be formed.\n\n[figure1]\n\nRegion 1/Region 2, Region 3/Region 4; complementary\n\nRegion 2/Region 3; complementary\n\nLeader sequence (Region 1) encodes short peptide containing two tryptophan residues (Trp)\n\n## Figure 1\n\nA short peptide of 14 amino acids containing two tryptophan codons called a leader peptide is encoded in Region 1 (Figure 1).\n\nIf the trp operon mRNA is not translated at the same time as it is transcribed by RNA polymerase, Region 1 and Region 2 of mRNA, and Region 3 and Region 4 pair with each other to form stem loop structures, respectively. In this case, a consecutive $U$ bases is located immediately after Region 4. Since the form in which $U$ bases continue immediately after the stem loop structure functions as a transcription termination signal in the\nprocaryote, the RNA polymerase is released and the transcription is terminated (Figure 2).\n[figure2]\n\nFigure 2\n\nWhen the translation of the leader sequence occurs at the same time as the transcription of the mRNA, the ribosome can translate the mRNA with the stem loop structure, but the transcription also ends by forming the stem loop structure of Region 3 and Region 4.\n[figure3]\n\nFigure 3\n\nWhen tryptophan is deficient, it takes time to translate the Trp codons in the leader sequence, and the ribosome temporarily stays in Region 1 . The mRNA transcribed during that time will be paired with Region 2 and Region 3 to form a stem loop structure. In this case, since Region 4 does not pair, a transcription termination signal is not formed, and RNA polymerase continues transcription of the trpEDCBA operon encoding the downstream Trp biosynthetic enzymes (Figure 3).\nA: The transcription rate is much faster than the translation rate in the $E$ coli cell.\nB: In a mutant strain of $E$. coli lacking the trp operator sequence, transcription-truncated mRNA is generated when tryptophan is present in the medium.\nC: In the mutant strain in which the tryptophan codons in the leader peptide is deleted, the growth is delayed when tryptophan is deficient in the medium.\nD: The tryptophan concentration in the cells increases in the mutant strain in which 10 tryptophan codons are present in the leader peptide.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe tryptophan operon (trp operon) of $E$. coli is transcriptionally regulated by a repressor that is activated by the binding of tryptophan. The active form repressor binds to the operator sequence located between the promoter and the transcription initiating point and blocks the RNA polymerase. There is another expression control system called the attenuator linked to transcription and translation in the trp operon.\n\nBetween the operator sequence and the $\\operatorname{trpE}$ gene, which is the first structural gene of the trp operon, there are four sequences of about 15 bases called Region 1-4 (Figure 1). Region 1 and Region 2, and Region 3 and Region 4 have complementary sequences, respectively. When these regions are transcribed as mRNA, they are paired with each other and form stem-loop structures (Figure 2). Furthermore, the sequences of Region 2 and Region 3 are also complementary, so a stem loop structure can be formed.\n\n[figure1]\n\nRegion 1/Region 2, Region 3/Region 4; complementary\n\nRegion 2/Region 3; complementary\n\nLeader sequence (Region 1) encodes short peptide containing two tryptophan residues (Trp)\n\n## Figure 1\n\nA short peptide of 14 amino acids containing two tryptophan codons called a leader peptide is encoded in Region 1 (Figure 1).\n\nIf the trp operon mRNA is not translated at the same time as it is transcribed by RNA polymerase, Region 1 and Region 2 of mRNA, and Region 3 and Region 4 pair with each other to form stem loop structures, respectively. In this case, a consecutive $U$ bases is located immediately after Region 4. Since the form in which $U$ bases continue immediately after the stem loop structure functions as a transcription termination signal in the\nprocaryote, the RNA polymerase is released and the transcription is terminated (Figure 2).\n[figure2]\n\nFigure 2\n\nWhen the translation of the leader sequence occurs at the same time as the transcription of the mRNA, the ribosome can translate the mRNA with the stem loop structure, but the transcription also ends by forming the stem loop structure of Region 3 and Region 4.\n[figure3]\n\nFigure 3\n\nWhen tryptophan is deficient, it takes time to translate the Trp codons in the leader sequence, and the ribosome temporarily stays in Region 1 . The mRNA transcribed during that time will be paired with Region 2 and Region 3 to form a stem loop structure. In this case, since Region 4 does not pair, a transcription termination signal is not formed, and RNA polymerase continues transcription of the trpEDCBA operon encoding the downstream Trp biosynthetic enzymes (Figure 3).\n\nA: The transcription rate is much faster than the translation rate in the $E$ coli cell.\nB: In a mutant strain of $E$. coli lacking the trp operator sequence, transcription-truncated mRNA is generated when tryptophan is present in the medium.\nC: In the mutant strain in which the tryptophan codons in the leader peptide is deleted, the growth is delayed when tryptophan is deficient in the medium.\nD: The tryptophan concentration in the cells increases in the mutant strain in which 10 tryptophan codons are present in the leader peptide.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-24.jpg?height=768&width=1288&top_left_y=1084&top_left_x=218",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-25.jpg?height=690&width=1210&top_left_y=361&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-25.jpg?height=1226&width=1182&top_left_y=1355&top_left_x=330"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_139",
        "problem": "Figures $\\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$ show the sagittal sections (the plane which divides body into left and right parts) of a frog embryo from the surface to the depth, respectively. Figures a-d are cross sections of the same embryo shown in Figure Z.\n[figure1]\nA: Figure \"a\" is the cross section from region \" 1 \".\nB: Figure \"b\" is the cross section from region \" 3 \".\nC: Figure \" $c$ \" is the cross section from region \" 4 \".\nD: Figure \" $\\mathrm{d}$ \" is the cross section from region \" 2 \".\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFigures $\\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$ show the sagittal sections (the plane which divides body into left and right parts) of a frog embryo from the surface to the depth, respectively. Figures a-d are cross sections of the same embryo shown in Figure Z.\n[figure1]\n\nA: Figure \"a\" is the cross section from region \" 1 \".\nB: Figure \"b\" is the cross section from region \" 3 \".\nC: Figure \" $c$ \" is the cross section from region \" 4 \".\nD: Figure \" $\\mathrm{d}$ \" is the cross section from region \" 2 \".\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-25.jpg?height=1110&width=1538&top_left_y=563&top_left_x=250"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1557",
        "problem": "The main purpose of flower pigments is to create contrasting patterns which guide pollinators to pollen. However, scientists wanted to investigate whether pigments also protect pollen. Some flowers have anthers exposed in a dish of petals, whereas some flowers have anthers enveloped under petals.\n\n[figure1]\n\nThe change in pigmentation of these types of flowers was measured across the world over time. The change in temperature due to global warming, and the change in UV exposure due to depletion and recovery of the ozone layer varies between regions. The correlation between these variables is plotted below.\n\n[figure2]\n\nHumans have stopped depleting the ozone layer, but the emission of $\\mathrm{CO}_{2}$ is accelerating. What effect will this have on flower pigmentation in future?\nA: Pigmentation will increase in flowers with enveloped anthers.\nB: Pigmentation will increase in flowers with exposed anthers.\nC: Pigmentation will increase in both kinds of flower.\nD: Pigmentation will decrease in both kinds of flower.\nE: Pigmentation changes will improve the attractiveness of flowers to pollinators.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe main purpose of flower pigments is to create contrasting patterns which guide pollinators to pollen. However, scientists wanted to investigate whether pigments also protect pollen. Some flowers have anthers exposed in a dish of petals, whereas some flowers have anthers enveloped under petals.\n\n[figure1]\n\nThe change in pigmentation of these types of flowers was measured across the world over time. The change in temperature due to global warming, and the change in UV exposure due to depletion and recovery of the ozone layer varies between regions. The correlation between these variables is plotted below.\n\n[figure2]\n\nHumans have stopped depleting the ozone layer, but the emission of $\\mathrm{CO}_{2}$ is accelerating. What effect will this have on flower pigmentation in future?\n\nA: Pigmentation will increase in flowers with enveloped anthers.\nB: Pigmentation will increase in flowers with exposed anthers.\nC: Pigmentation will increase in both kinds of flower.\nD: Pigmentation will decrease in both kinds of flower.\nE: Pigmentation changes will improve the attractiveness of flowers to pollinators.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-34.jpg?height=616&width=1176&top_left_y=523&top_left_x=240",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-34.jpg?height=854&width=1625&top_left_y=1338&top_left_x=227"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_276",
        "problem": "The following summaries describe recently published research results.\n\nResearch 1. Wu and Li (1985): The comparative analysis of homologous genes between human and mouse genomes suggests that the evolutionary rate of homologous genes was higher in the mouse lineage than in the human lineage.\n\nResearch 2. Smith and Donohue (2008): The plant families Caprifoliaceae, Asclepiadaceae, and Lamiaceae are composed of both herbaceous and arborescent species. The comparative analysis of homologous genes between the herbaceous and the arborescent species within a single plant family suggest that the evolutionary rates of homologous genes in herbaceous lineages were faster than that of arborescent lineages in all three plant families.\n\nResearch 3. Gilman et al. (2009): The comparative analysis of 130 homologous mitochondrial genes between a sister species pair of vertebrates from the temperate region and from the tropical region indicate that the base substitution rates of homologous genes from the tropical region are 1.7 times faster than that of the temperate region.\n\nBased on these studies, which of the following statements best describes the common evolutionary processes in plant and animal genes?\nA: The evolutionary rates of genes are accelerated in short-lived animals and plants.\nB: The evolutionary rates of genes are accelerated in higher animals and plants.\nC: The evolutionary rates of genes are accelerated in animals and plants which lived in higher temperature regions.\nD: Direct comparisons of homologous genes between animals and plants show that plants evolve faster than animals.\nE: The fast evolutionary rates of mitochondria genes make them ideal for phylogenetic comparison between distant lineages.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following summaries describe recently published research results.\n\nResearch 1. Wu and Li (1985): The comparative analysis of homologous genes between human and mouse genomes suggests that the evolutionary rate of homologous genes was higher in the mouse lineage than in the human lineage.\n\nResearch 2. Smith and Donohue (2008): The plant families Caprifoliaceae, Asclepiadaceae, and Lamiaceae are composed of both herbaceous and arborescent species. The comparative analysis of homologous genes between the herbaceous and the arborescent species within a single plant family suggest that the evolutionary rates of homologous genes in herbaceous lineages were faster than that of arborescent lineages in all three plant families.\n\nResearch 3. Gilman et al. (2009): The comparative analysis of 130 homologous mitochondrial genes between a sister species pair of vertebrates from the temperate region and from the tropical region indicate that the base substitution rates of homologous genes from the tropical region are 1.7 times faster than that of the temperate region.\n\nBased on these studies, which of the following statements best describes the common evolutionary processes in plant and animal genes?\n\nA: The evolutionary rates of genes are accelerated in short-lived animals and plants.\nB: The evolutionary rates of genes are accelerated in higher animals and plants.\nC: The evolutionary rates of genes are accelerated in animals and plants which lived in higher temperature regions.\nD: Direct comparisons of homologous genes between animals and plants show that plants evolve faster than animals.\nE: The fast evolutionary rates of mitochondria genes make them ideal for phylogenetic comparison between distant lineages.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1303",
        "problem": "Spines and thorns on plants look similar, and both provide protection from herbivores. However, not all plants with spines or thorns have descended from a recent common ancestor. Spines are modified leaves, and thorns are modified stems.\n\nWhich of the following statements best describes how this information provides evidence for evolution by natural selection?\nA: It shows that different organisms sometimes look alike.\nB: It shows that herbivores are the strongest selection force on organisms.\nC: It shows that a variety of structures can be effective in protecting an organism from herbivores.\nD: It shows that environmental pressures can cause unrelated species to change in similar ways.\nE: It shows that spines and thorns provide the best protection from herbivores.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSpines and thorns on plants look similar, and both provide protection from herbivores. However, not all plants with spines or thorns have descended from a recent common ancestor. Spines are modified leaves, and thorns are modified stems.\n\nWhich of the following statements best describes how this information provides evidence for evolution by natural selection?\n\nA: It shows that different organisms sometimes look alike.\nB: It shows that herbivores are the strongest selection force on organisms.\nC: It shows that a variety of structures can be effective in protecting an organism from herbivores.\nD: It shows that environmental pressures can cause unrelated species to change in similar ways.\nE: It shows that spines and thorns provide the best protection from herbivores.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1508",
        "problem": "For the three different diseases, pedigree analysis was performed on family trees.\n\n[figure1]\n\nWhich is true?\nA: 0.05\nB: 0.1\nC: 0.25\nD: 0.5\nE: 1\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor the three different diseases, pedigree analysis was performed on family trees.\n\n[figure1]\n\nWhich is true?\n\nA: 0.05\nB: 0.1\nC: 0.25\nD: 0.5\nE: 1\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-43.jpg?height=1110&width=1636&top_left_y=500&top_left_x=240"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_86",
        "problem": "ErbB-2 is a receptor found on the plasma membrane of mammalian cells which can move from the plasma membrane to the nucleus. Because most proteins that shuttle between the cytoplasm and the nucleus are soluble and not integral membrane bound proteins, the mechanism by which ErbB-2 undergoes transport to the nucleus is of particular interest. Three experiments described below were done to shed light on the underlying mechanisms.\n\nExperiment 1: siRNA knock down of importin $\\beta 1$ expression in target cells. The cells were transfected with importin $\\beta 1$ siRNA (Imp), non-functional siRNA control (N.S.), or buffer only (-). Proteins from the cytoplasmic and nuclear fraction of cell lysates were analysed by Western blotting using importin $\\beta 1$, and ErbB-2, $\\alpha$ tubulin and histone H3 antibodies as indicated.\n\n[figure1]\n\nExperiment 2: Mutant cells lacking ErbB-2 gene were transfected with wild-type ErbB-2 (WT), ErbB-2 mutant containing a deficient nuclear localization signal ( $\\triangle \\mathrm{NLS})$, or vector control (-). Proteins from the cytoplasmic fraction, nuclear fraction were then analysed by Western blotting with ErbB-2, importin $\\beta 1, \\alpha$ tubulin and histone $\\mathrm{H} 3$ antibodies as indicated.\n\n[figure2]\n\nExperiment 3: Cell lysates from the cells were immune-precipitated with anti-ErbB-2 or mouse IgG (mIgG). The precipitated immune-complexes and the cell lysates were then analysed by Western blotting with, clathrin, importin $\\beta 1$, and ErbB-2 antibodies.\n\n[figure3]\nA: The data in Figure (a) suggest that ErbB-2 requires importin $\\beta 1$ in order to enter the nucleus.\nB: It is predicted that the antibody against importin $\\beta 1$ does not precipitate the ErbB-2( $\\triangle$ NLS).\nC: The data presented suggest that localization of histones to the nucleus is mediated by a mechanism distinct from that used to shuttle ErbB-2.\nD: As true for other membrane bound receptors, ErbB2 enters the cytoplasm by endocytosis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nErbB-2 is a receptor found on the plasma membrane of mammalian cells which can move from the plasma membrane to the nucleus. Because most proteins that shuttle between the cytoplasm and the nucleus are soluble and not integral membrane bound proteins, the mechanism by which ErbB-2 undergoes transport to the nucleus is of particular interest. Three experiments described below were done to shed light on the underlying mechanisms.\n\nExperiment 1: siRNA knock down of importin $\\beta 1$ expression in target cells. The cells were transfected with importin $\\beta 1$ siRNA (Imp), non-functional siRNA control (N.S.), or buffer only (-). Proteins from the cytoplasmic and nuclear fraction of cell lysates were analysed by Western blotting using importin $\\beta 1$, and ErbB-2, $\\alpha$ tubulin and histone H3 antibodies as indicated.\n\n[figure1]\n\nExperiment 2: Mutant cells lacking ErbB-2 gene were transfected with wild-type ErbB-2 (WT), ErbB-2 mutant containing a deficient nuclear localization signal ( $\\triangle \\mathrm{NLS})$, or vector control (-). Proteins from the cytoplasmic fraction, nuclear fraction were then analysed by Western blotting with ErbB-2, importin $\\beta 1, \\alpha$ tubulin and histone $\\mathrm{H} 3$ antibodies as indicated.\n\n[figure2]\n\nExperiment 3: Cell lysates from the cells were immune-precipitated with anti-ErbB-2 or mouse IgG (mIgG). The precipitated immune-complexes and the cell lysates were then analysed by Western blotting with, clathrin, importin $\\beta 1$, and ErbB-2 antibodies.\n\n[figure3]\n\nA: The data in Figure (a) suggest that ErbB-2 requires importin $\\beta 1$ in order to enter the nucleus.\nB: It is predicted that the antibody against importin $\\beta 1$ does not precipitate the ErbB-2( $\\triangle$ NLS).\nC: The data presented suggest that localization of histones to the nucleus is mediated by a mechanism distinct from that used to shuttle ErbB-2.\nD: As true for other membrane bound receptors, ErbB2 enters the cytoplasm by endocytosis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-14.jpg?height=623&width=1085&top_left_y=885&top_left_x=497",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-14.jpg?height=628&width=1108&top_left_y=1845&top_left_x=477",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-15.jpg?height=586&width=626&top_left_y=455&top_left_x=772"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_410",
        "problem": "三体细胞 (染色体比正常体细胞多 1 条) 在有丝分裂时, 三条染色体中的一条随机丢失, 可产生染色体数目正常的体细胞, 这种现象称为“三体自救”。某家庭中的男性为\n正常个体, 女性患有进行性肌营养不良（伴 X 染色体隐性遗传病）, 这对夫妇生下一个染色体正常的患病女孩。下列说法错误的是（）\nA: 该母亲在减数分裂产生卵细胞的过程中, 一定是减数分裂II后期发生异常\nB: 该患病女孩的两条 X 染色体上的进行性肌营养不良致病基因可能均来自母方\nC: 该女孩患病的原因可能是来自父方 $\\mathrm{X}$ 染色体上的正常基因突变成致病基因\nD: 若发育为该女孩的受精卵发生“三体自救”，则丢失的染色体可能来自父方\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n三体细胞 (染色体比正常体细胞多 1 条) 在有丝分裂时, 三条染色体中的一条随机丢失, 可产生染色体数目正常的体细胞, 这种现象称为“三体自救”。某家庭中的男性为\n正常个体, 女性患有进行性肌营养不良（伴 X 染色体隐性遗传病）, 这对夫妇生下一个染色体正常的患病女孩。下列说法错误的是（）\n\nA: 该母亲在减数分裂产生卵细胞的过程中, 一定是减数分裂II后期发生异常\nB: 该患病女孩的两条 X 染色体上的进行性肌营养不良致病基因可能均来自母方\nC: 该女孩患病的原因可能是来自父方 $\\mathrm{X}$ 染色体上的正常基因突变成致病基因\nD: 若发育为该女孩的受精卵发生“三体自救”，则丢失的染色体可能来自父方\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_574",
        "problem": "亨廷顿舞蹈症(HD)是由于编码亨廷顿蛋白的基因(IT15 基因) 序列中的三个核苷酸 (CAG)发生多次重复所致, 某 HD 家庭的系谱图及对该家庭中部分个体相关基因分析的电泳图如图所示。下列叙述正确的是\n\n[图1]\nA: 结合系谱图和电泳图推测该病为常染色体隐性遗传病\nB: $\\mathrm{III}_{3}$ 与正常女性婚配, 所生两个孩子只有一个患 $\\mathrm{HD}$ 的概率为 $1 / 4$\nC: 该致病基因不可能存在于 $\\mathrm{X} 、 \\mathrm{Y}$ 的同源区段\nD: HD 患者体内编码亨廷顿蛋白的基因中嘌呤的比例比正常基因高\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n亨廷顿舞蹈症(HD)是由于编码亨廷顿蛋白的基因(IT15 基因) 序列中的三个核苷酸 (CAG)发生多次重复所致, 某 HD 家庭的系谱图及对该家庭中部分个体相关基因分析的电泳图如图所示。下列叙述正确的是\n\n[图1]\n\nA: 结合系谱图和电泳图推测该病为常染色体隐性遗传病\nB: $\\mathrm{III}_{3}$ 与正常女性婚配, 所生两个孩子只有一个患 $\\mathrm{HD}$ 的概率为 $1 / 4$\nC: 该致病基因不可能存在于 $\\mathrm{X} 、 \\mathrm{Y}$ 的同源区段\nD: HD 患者体内编码亨廷顿蛋白的基因中嘌呤的比例比正常基因高\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-72.jpg?height=277&width=1264&top_left_y=801&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_152",
        "problem": "The analysis of DNA and protein sequences nowadays is widely used in constructing phylogenetic trees.\nA: The number of differences in DNA or RNA nucleotide sequences or in amino acid sequences in two organisms reflects how much time has passed since the groups branched from a common ancestor.\nB: If two polypeptides from two organisms sharing a most recent common ancestor differ from each other in only one amino acid, then one substitution had occurred during the evolution of the two polypeptides.\nC: rRNA sequence analysis is useful not only for phylogenetic analysis of organisms belonging to different domains but also for phylogenetic relationships among species within genera.\nD: Pseudogenes often used for constructing phylogenetic trees.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe analysis of DNA and protein sequences nowadays is widely used in constructing phylogenetic trees.\n\nA: The number of differences in DNA or RNA nucleotide sequences or in amino acid sequences in two organisms reflects how much time has passed since the groups branched from a common ancestor.\nB: If two polypeptides from two organisms sharing a most recent common ancestor differ from each other in only one amino acid, then one substitution had occurred during the evolution of the two polypeptides.\nC: rRNA sequence analysis is useful not only for phylogenetic analysis of organisms belonging to different domains but also for phylogenetic relationships among species within genera.\nD: Pseudogenes often used for constructing phylogenetic trees.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_509",
        "problem": "果蝇的 $\\mathrm{X}$ 连锁隐性致死基因的突变率为 $0.1 \\%$, 即每 1000 个配子中，有 1 个 $\\mathrm{X}$ 染色体有。隐性致死突变。果蝇的第II、第III染色体的隐性致死突变率均为 $0.5 \\%$ 。第IV染色体为点状，突变率极低，忽略不计。下列说法正确的是\nA: 果蝇染色体 DNA 中只要发生碱基的替换、增添或缺失，即为基因突变\nB: 果蝇的所有染色体都可以发生隐性致死突变, 说明基因突变具有不定向性\nC: 仅考虑 X 染色体, 正常果蝇群体的后代雄性个体中隐性致死的概率为 $0.1 \\%$\nD: 果蝇群体中总的隐性致死突变率为 $1.1 \\%$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的 $\\mathrm{X}$ 连锁隐性致死基因的突变率为 $0.1 \\%$, 即每 1000 个配子中，有 1 个 $\\mathrm{X}$ 染色体有。隐性致死突变。果蝇的第II、第III染色体的隐性致死突变率均为 $0.5 \\%$ 。第IV染色体为点状，突变率极低，忽略不计。下列说法正确的是\n\nA: 果蝇染色体 DNA 中只要发生碱基的替换、增添或缺失，即为基因突变\nB: 果蝇的所有染色体都可以发生隐性致死突变, 说明基因突变具有不定向性\nC: 仅考虑 X 染色体, 正常果蝇群体的后代雄性个体中隐性致死的概率为 $0.1 \\%$\nD: 果蝇群体中总的隐性致死突变率为 $1.1 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1003",
        "problem": "An individual with Turner Syndrome would have which of the following sex chromosomes arrangements?\nA: XO\nB: XXYY\nC: XXXXY\nD: YO\nE: XXX\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn individual with Turner Syndrome would have which of the following sex chromosomes arrangements?\n\nA: XO\nB: XXYY\nC: XXXXY\nD: YO\nE: XXX\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_441",
        "problem": "致死基因的存在可影响后代性状分离比。现有基因型为 $\\mathrm{AaBb}$ 的个体，两对等位基因独立遗传，但具有某种基因型的配子或个体致死，若该个体自交，下列说法不正确的是 $(\\quad)$\nA: 后代分离比为 $5: 3: 3: 1$, 则可推测原因可能是基因型为 $A B$ 的雄配子或雌配子致死\nB: 后代分离比为 7: 3: 1: 1, 则可推测原因可能是基因型为 $\\mathrm{Ab}$ 的雄配子或雌配子致死\nC: 后代分离比为 9: 3: 3, 则可推测原因可能是基因型为 $a b$ 的雄配子或雌配子致死\nD: 后代分离比为 $4: 2: 2: 1$, 则可推测原因可能是 A 基因和 B 基因显性纯合致死\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n致死基因的存在可影响后代性状分离比。现有基因型为 $\\mathrm{AaBb}$ 的个体，两对等位基因独立遗传，但具有某种基因型的配子或个体致死，若该个体自交，下列说法不正确的是 $(\\quad)$\n\nA: 后代分离比为 $5: 3: 3: 1$, 则可推测原因可能是基因型为 $A B$ 的雄配子或雌配子致死\nB: 后代分离比为 7: 3: 1: 1, 则可推测原因可能是基因型为 $\\mathrm{Ab}$ 的雄配子或雌配子致死\nC: 后代分离比为 9: 3: 3, 则可推测原因可能是基因型为 $a b$ 的雄配子或雌配子致死\nD: 后代分离比为 $4: 2: 2: 1$, 则可推测原因可能是 A 基因和 B 基因显性纯合致死\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_945",
        "problem": "血友病是一种伴 $\\mathrm{X}$ 隐性遗传病, 经研究发现其致病基因 $\\mathrm{d}$ 有两种突变形式, 记作 $\\mathrm{dA}$\n\n与 $\\mathrm{dB}$ 。如图表示某血友病家族系谱图, $\\mathrm{II}_{1}$ 同时还患有克氏综合征 (性染色体组成为 XXY)。不考虑新的突变和染色体片段交换, 下列分析正确的是（）\n\nI\n\nII\n\n[图1]\n\n[图2]\n\n$\\mathrm{AB} \\quad \\mathrm{A}$正常男性、女姓\n\n患伴X染色体隐性遗传病男性、女性\n\n$\\mathrm{Ad}$ 基因突变形式是 $\\mathrm{dA}$\n\n$B \\mathrm{~d}$ 基因突变形式是 $\\mathrm{dB}$\n\n$\\mathrm{AB} \\mathrm{d}$ 基因突变形式兼有 $\\mathrm{d} A$ 和 $d B$\nA: $\\mathrm{II}_{1}$ 性染色体异常, 是因为 $\\mathrm{I}_{2}$ 减数分裂 $I$ 时同源染色体不分离\nB: $\\mathrm{II}_{2}$ 与正常女性婚配, 所生子女患有该伴 X 染色体隐性遗传病的概率是 $1 / 2$\nC: $\\mathrm{II}_{3}$ 的一个次级卵母细胞与 $\\mathrm{I}_{1}$ 的一个次级精母细胞中含有的 $\\mathrm{dA}$ 数目相同\nD: $\\mathrm{II}_{4}$ 与基因型与 $\\mathrm{II}_{2}$ 相同的个体结婚, 所生子女患血友病的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n血友病是一种伴 $\\mathrm{X}$ 隐性遗传病, 经研究发现其致病基因 $\\mathrm{d}$ 有两种突变形式, 记作 $\\mathrm{dA}$\n\n与 $\\mathrm{dB}$ 。如图表示某血友病家族系谱图, $\\mathrm{II}_{1}$ 同时还患有克氏综合征 (性染色体组成为 XXY)。不考虑新的突变和染色体片段交换, 下列分析正确的是（）\n\nI\n\nII\n\n[图1]\n\n[图2]\n\n$\\mathrm{AB} \\quad \\mathrm{A}$正常男性、女姓\n\n患伴X染色体隐性遗传病男性、女性\n\n$\\mathrm{Ad}$ 基因突变形式是 $\\mathrm{dA}$\n\n$B \\mathrm{~d}$ 基因突变形式是 $\\mathrm{dB}$\n\n$\\mathrm{AB} \\mathrm{d}$ 基因突变形式兼有 $\\mathrm{d} A$ 和 $d B$\n\nA: $\\mathrm{II}_{1}$ 性染色体异常, 是因为 $\\mathrm{I}_{2}$ 减数分裂 $I$ 时同源染色体不分离\nB: $\\mathrm{II}_{2}$ 与正常女性婚配, 所生子女患有该伴 X 染色体隐性遗传病的概率是 $1 / 2$\nC: $\\mathrm{II}_{3}$ 的一个次级卵母细胞与 $\\mathrm{I}_{1}$ 的一个次级精母细胞中含有的 $\\mathrm{dA}$ 数目相同\nD: $\\mathrm{II}_{4}$ 与基因型与 $\\mathrm{II}_{2}$ 相同的个体结婚, 所生子女患血友病的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-015.jpg?height=183&width=488&top_left_y=1947&top_left_x=470",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-015.jpg?height=129&width=249&top_left_y=1780&top_left_x=595"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_575",
        "problem": "果蝇的眼型和体色分别受两对等位基因 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 控制 (若在性染色体上, 不考虑 $\\mathrm{X}$与 $\\mathrm{Y}$ 同源区段)。让灰体正常眼雌果蝇群体与黄体正常眼雄果蝇群体随机交配, 所得 $\\mathrm{F}_{1}$全为灰体。让 $F_{1}$ 雌雄果蝇随机交配得到 $F_{2}, F_{2}$ 结果如下表, 下列说法错误的是 ( )\n\n| $F$ | 表现型 | 灰体正常 | 黄体正常 | 灰体正常 | 灰体棒 | 黄体正常 | 黄体棒 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n\n\n| 2 |  | 眼雌性 | 眼雌性 | 眼雄性 | 眼雄性 | 眼雄性 | 眼雄性 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 数量 | 438 | 146 | 657 | 219 | 219 | 73 |\nA: 基因 A 与基因 B 的本质区别是碱基（对）的数量、排列顺序不同\nB: F1 雌果蝇的基因型为 $B b X^{A} X^{A}$ 和 $B b X^{A} X^{a}$\nC: $F_{2}$ 中雌性无棒眼的原因可能是含 $X^{\\mathrm{a}}$ 的雄配子致死\nD: $F_{2}$ 雌雄性个体的基因型分别有 4 种和 6 种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的眼型和体色分别受两对等位基因 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 控制 (若在性染色体上, 不考虑 $\\mathrm{X}$与 $\\mathrm{Y}$ 同源区段)。让灰体正常眼雌果蝇群体与黄体正常眼雄果蝇群体随机交配, 所得 $\\mathrm{F}_{1}$全为灰体。让 $F_{1}$ 雌雄果蝇随机交配得到 $F_{2}, F_{2}$ 结果如下表, 下列说法错误的是 ( )\n\n| $F$ | 表现型 | 灰体正常 | 黄体正常 | 灰体正常 | 灰体棒 | 黄体正常 | 黄体棒 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n\n\n| 2 |  | 眼雌性 | 眼雌性 | 眼雄性 | 眼雄性 | 眼雄性 | 眼雄性 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 数量 | 438 | 146 | 657 | 219 | 219 | 73 |\n\nA: 基因 A 与基因 B 的本质区别是碱基（对）的数量、排列顺序不同\nB: F1 雌果蝇的基因型为 $B b X^{A} X^{A}$ 和 $B b X^{A} X^{a}$\nC: $F_{2}$ 中雌性无棒眼的原因可能是含 $X^{\\mathrm{a}}$ 的雄配子致死\nD: $F_{2}$ 雌雄性个体的基因型分别有 4 种和 6 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1121",
        "problem": "An experiment to understand the relationship between a herbivorous crab Mithrax forceps and the coral Oculina arbuscula was undertaken in a shallow water coastal ecosystem. Observation on predation of crab, growth of algae, growth and mortality of corals were made. The following graphs indicate the results obtained during the experiment.\n[figure1]\n\nA few statements based on the results obtained are made.\n\ni. The presence of crabs has a negative influence on algal growth and positive influence on coral growth.\n\nii. Presence of M. forceps is obligatory for the survival of $O$.arbuscula.\n\niii. The coral species plays an important role in preventing predation of $M$. forceps.\n\niv. The algae outcompete O.arbuscula in the absence of M. Forceps.\n\nWhich of these statements are true?\nA: i, ii and iii only\nB: i and iv only\nC: i, iii and iv only\nD: iii and iv only\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn experiment to understand the relationship between a herbivorous crab Mithrax forceps and the coral Oculina arbuscula was undertaken in a shallow water coastal ecosystem. Observation on predation of crab, growth of algae, growth and mortality of corals were made. The following graphs indicate the results obtained during the experiment.\n[figure1]\n\nA few statements based on the results obtained are made.\n\ni. The presence of crabs has a negative influence on algal growth and positive influence on coral growth.\n\nii. Presence of M. forceps is obligatory for the survival of $O$.arbuscula.\n\niii. The coral species plays an important role in preventing predation of $M$. forceps.\n\niv. The algae outcompete O.arbuscula in the absence of M. Forceps.\n\nWhich of these statements are true?\n\nA: i, ii and iii only\nB: i and iv only\nC: i, iii and iv only\nD: iii and iv only\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-11.jpg?height=1516&width=1632&top_left_y=1041&top_left_x=206"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_480",
        "problem": "如图 1 表示某二倍体生物卵原细胞中的 4 条染色体, 若此细胞中的一条 1 号染色体与一条 X 染色体连接在一起, 形成如图 2 所示的变异。已知染色体的同源区段发生联会后彼此分离进入不同的配子中。下列有关叙述正确的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 图 2 所示变异类型是染色体结构变异, 导致基因重组\nB: 图 2 细胞在减数分裂 I 过程中不能发生等位基因的分离\nC: 若图 2 细胞进行减数分裂, 染色体数正常的子细胞的概率为 $1 / 2$\nD: 若图 2 细胞进行有丝分裂, 染色体数正常的子细胞的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图 1 表示某二倍体生物卵原细胞中的 4 条染色体, 若此细胞中的一条 1 号染色体与一条 X 染色体连接在一起, 形成如图 2 所示的变异。已知染色体的同源区段发生联会后彼此分离进入不同的配子中。下列有关叙述正确的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 图 2 所示变异类型是染色体结构变异, 导致基因重组\nB: 图 2 细胞在减数分裂 I 过程中不能发生等位基因的分离\nC: 若图 2 细胞进行减数分裂, 染色体数正常的子细胞的概率为 $1 / 2$\nD: 若图 2 细胞进行有丝分裂, 染色体数正常的子细胞的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-78.jpg?height=277&width=283&top_left_y=501&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-78.jpg?height=275&width=280&top_left_y=505&top_left_x=705"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_389",
        "problem": "某种鼠的毛色由常染色体上的两对等位基因（A、a 和 B、b）控制, 其中 $\\mathrm{A}$ 基因控制褐色素的合成, B 基因控制黑色素的合成，两种色素均不合成时毛色呈白色。当 $\\mathrm{A} 、$ $\\mathrm{B}$ 基因同时存在时, 二者的转录产物会形成双链结构。用纯合的褐色和黑色亲本杂交,\n$F_{1}$ 全为白色, $F_{1}$ 雌雄个体相互交配得到 $F_{2}$, 不考虑交叉互换, 下列分析正确的是 ( )\nA: $F_{1}$ 全为白色的原因是 $A 、 B$ 两基因不能正常转录\nB: $F_{1}$ 与双隐性的个体测交, 后代中黑色个体占 $1 / 4$\nC: 若 $F_{2}$ 中褐色个体的比例接近 $1 / 4$, 则白色个体接近 $1 / 2$\nD: 若 $F_{2}$ 中有一定比例的白色个体，则白色个体中纯合子占 $1 / 9$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种鼠的毛色由常染色体上的两对等位基因（A、a 和 B、b）控制, 其中 $\\mathrm{A}$ 基因控制褐色素的合成, B 基因控制黑色素的合成，两种色素均不合成时毛色呈白色。当 $\\mathrm{A} 、$ $\\mathrm{B}$ 基因同时存在时, 二者的转录产物会形成双链结构。用纯合的褐色和黑色亲本杂交,\n$F_{1}$ 全为白色, $F_{1}$ 雌雄个体相互交配得到 $F_{2}$, 不考虑交叉互换, 下列分析正确的是 ( )\n\nA: $F_{1}$ 全为白色的原因是 $A 、 B$ 两基因不能正常转录\nB: $F_{1}$ 与双隐性的个体测交, 后代中黑色个体占 $1 / 4$\nC: 若 $F_{2}$ 中褐色个体的比例接近 $1 / 4$, 则白色个体接近 $1 / 2$\nD: 若 $F_{2}$ 中有一定比例的白色个体，则白色个体中纯合子占 $1 / 9$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1315",
        "problem": "Many laboratory procedures involve the use of dilutions. If a solution has a 1/10 dilution the number represents 1 part of the sample added to 9 parts of diluent. The dilution factor equals the final volume divided by the sample volume. A serial dilution is any dilution in which the concentration decreases by the same quantity in each successive step. Serial dilutions are multiplicative.\n\nBacterial counts in a stream water sample can be determined by placing a known volume of the stream water into a liquefied agar medium that is then poured into a petri dish. The agar solidifies and bacterial colonies grow within the agar. These colonies can then be accurately counted as the bacteria are equally distributed through the agar. In practice the number of bacteria is usually so great that a serial dilution must be made first so that the number of colonies can be counted. Note that the amount put on the plate is also a dilution as colonies are normally reported as per $\\mathrm{ml}$. Plates with 30-300 colonies are used for the calculation as plates with greater than 300 and less than 30 have a high degree of error. Air contaminants can contribute significantly to a really low count and a high count can be confounded by error in counting too many small colonies.\n\n[figure1]\n\nCalculate the number of bacterial cells (E. coli) per $\\mathrm{ml}$ in the original water sample.\nA: $1.28 \\times 10^{7}$\nB: $1.28 \\times 10^{8}$\nC: $1.28 \\times 10^{9}$\nD: $1.20 \\times 10^{8}$\nE: $1.20 \\times 10^{9}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMany laboratory procedures involve the use of dilutions. If a solution has a 1/10 dilution the number represents 1 part of the sample added to 9 parts of diluent. The dilution factor equals the final volume divided by the sample volume. A serial dilution is any dilution in which the concentration decreases by the same quantity in each successive step. Serial dilutions are multiplicative.\n\nBacterial counts in a stream water sample can be determined by placing a known volume of the stream water into a liquefied agar medium that is then poured into a petri dish. The agar solidifies and bacterial colonies grow within the agar. These colonies can then be accurately counted as the bacteria are equally distributed through the agar. In practice the number of bacteria is usually so great that a serial dilution must be made first so that the number of colonies can be counted. Note that the amount put on the plate is also a dilution as colonies are normally reported as per $\\mathrm{ml}$. Plates with 30-300 colonies are used for the calculation as plates with greater than 300 and less than 30 have a high degree of error. Air contaminants can contribute significantly to a really low count and a high count can be confounded by error in counting too many small colonies.\n\n[figure1]\n\nCalculate the number of bacterial cells (E. coli) per $\\mathrm{ml}$ in the original water sample.\n\nA: $1.28 \\times 10^{7}$\nB: $1.28 \\times 10^{8}$\nC: $1.28 \\times 10^{9}$\nD: $1.20 \\times 10^{8}$\nE: $1.20 \\times 10^{9}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-06.jpg?height=497&width=968&top_left_y=1987&top_left_x=133"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_106",
        "problem": "A protein can be integrated into a membrane through a polypeptide sequence or via a lipid anchor. The attachment of eukaryotic proteins to the outer leaflet of the plasma membrane occurs only via Glycosylphosphatidylinositol (GPI) anchors. The biosynthesis of GPI glycolipid is a multistep process relies on many proteins, including GPI transamidase. In Arabidopsis plants, AtGPI8 gene encodes the enzyme GPI transamidase. To study the role of this gene in plant development, scientists constructed a mutant (atgpi8-l) plant line. He observed phenotypes of both wild type (WT) and mutant plants.\n[figure1]\n\nFig.Q63-1. Cotyledonous epidermis of wild type (A) and atgpi8-l (B) plants\n[figure2]\n\nFig.Q63-2. Growth phenotypes of wild type and atgpi8-1 plants. (A) Seedlings, (B) cotyledons and (C) first two leaves of seedlings. (D) 30-day-old and (E) 60-day-old plants. F- G: Inflorescences. H- L: Morphometric analysis of wildtype (gray bars) and atgpi8-1 (white bars) seedlings and mature plants. All values in $\\mathrm{H}$ to $\\mathrm{O}$ are statistically significant.\nA: The early post germination growth of cotyledons and first two leaves are not affected by the mutation. However, root growth, hypocotyl elongation and stomata differentiation are strongly affected by the mutation.\nB: The data suggest that GPI anchoring promotes the growth of leaves in vegetative plants; however, it inhibits axillary shoot formation.\nC: The atgpi8-1 mutation leads to reduced internode and pedicel elongation. However, the height of atgpi8-I plants is only moderately reduced likely because the number of internodes is increased.\nD: The results indicate that $A t G P I 8$ gene promotes early transition to flowering, flower formation and seed production.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA protein can be integrated into a membrane through a polypeptide sequence or via a lipid anchor. The attachment of eukaryotic proteins to the outer leaflet of the plasma membrane occurs only via Glycosylphosphatidylinositol (GPI) anchors. The biosynthesis of GPI glycolipid is a multistep process relies on many proteins, including GPI transamidase. In Arabidopsis plants, AtGPI8 gene encodes the enzyme GPI transamidase. To study the role of this gene in plant development, scientists constructed a mutant (atgpi8-l) plant line. He observed phenotypes of both wild type (WT) and mutant plants.\n[figure1]\n\nFig.Q63-1. Cotyledonous epidermis of wild type (A) and atgpi8-l (B) plants\n[figure2]\n\nFig.Q63-2. Growth phenotypes of wild type and atgpi8-1 plants. (A) Seedlings, (B) cotyledons and (C) first two leaves of seedlings. (D) 30-day-old and (E) 60-day-old plants. F- G: Inflorescences. H- L: Morphometric analysis of wildtype (gray bars) and atgpi8-1 (white bars) seedlings and mature plants. All values in $\\mathrm{H}$ to $\\mathrm{O}$ are statistically significant.\n\nA: The early post germination growth of cotyledons and first two leaves are not affected by the mutation. However, root growth, hypocotyl elongation and stomata differentiation are strongly affected by the mutation.\nB: The data suggest that GPI anchoring promotes the growth of leaves in vegetative plants; however, it inhibits axillary shoot formation.\nC: The atgpi8-1 mutation leads to reduced internode and pedicel elongation. However, the height of atgpi8-I plants is only moderately reduced likely because the number of internodes is increased.\nD: The results indicate that $A t G P I 8$ gene promotes early transition to flowering, flower formation and seed production.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-031.jpg?height=686&width=1414&top_left_y=916&top_left_x=318",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-032.jpg?height=1738&width=1626&top_left_y=164&top_left_x=221"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_10",
        "problem": "Spermatozoa motility is essential for the fertilization of an oocyte. In most animals, including fish, spermatozoa are immotile in the male reproductive organ (testis or sperm duct). Spermatozoa motility is triggered after being ejaculated into the female reproductive tract (in animals with internal fertilization) or after it is released into the aquatic environment (in animals with external fertilization). Ionic composition and osmolality of freshwater, seawater, and seminal plasma of pike (Esox lucius), sturgeon (Acipenser ruthenus), and cod (Gadus morhua) are shown in the table. Pike and sturgeon spawn in freshwater, and cod spawns in seawater.\n\n|  | Freshwater | Seawater | Pike | Sturgeon | Cod |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| Sodium $\\mathrm{Na}^{+}(\\mathrm{mM})$ | 0.26 | 469 | 75 | 20 | 197 |\n| Chloride $\\mathrm{Cl}^{-}(\\mathrm{mM})$ | 0.22 | 546 | 112 | 6 | 179 |\n| Potassium K (mM) | 0.07 | 10 | 82 | 1 | 6 |\n| Calcium $\\mathrm{Ca}^{2+}(\\mathrm{mM})$ | 0.38 | 10 | 2 | 0.2 | 3 |\n| Osmolality (mOsmol/kg) | $<1-5$ | 1000 | 302 | 51 | 385 |\n\n[figure1]\n\nA) Effect of osmolality on sperm motility initiation in pike (open markers) and cod (filled markers). $\\mathrm{NaCl}$ and Sucrose were used to make activation medium for pike sperm. $\\mathrm{NaCl}$ and artificial sea salt were used to make activation medium for cod sperm.\n\nB) Effect of osmolality on sperm motility initiation in sturgeon. $\\mathrm{NaCl}$ and Sucrose were used to make activation medium for sturgeon sperm.\n\nC) Effect of potassium ions $\\left(\\mathrm{K}^{+}\\right)$on sperm motility initiation and sperm velocity in sturgeon. $\\mathrm{KCl}(\\mathrm{mM})$ was added into activation medium made by $\\mathrm{NaCl}$ or sucrose with osomolality $40 \\mathrm{mOsmol} / \\mathrm{kg}$.\nA: Sperm motility in pike and cod become triggered in a hypo-osmotic and hyperosmotic environment, respectively.\nB: Under physiological condition, osmolality is the main factor that inhibits sperm motility initiation in sturgeon.\nC: Sperm motility in sturgeon become triggered after discharged into a hypoosmotic environment with lower $\\mathrm{K}^{+}$ions.\nD: Environmental osmolality is a key signal to trigger sperm motility initiation after release from reproductive system into aquatic environment in marine fish\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nSpermatozoa motility is essential for the fertilization of an oocyte. In most animals, including fish, spermatozoa are immotile in the male reproductive organ (testis or sperm duct). Spermatozoa motility is triggered after being ejaculated into the female reproductive tract (in animals with internal fertilization) or after it is released into the aquatic environment (in animals with external fertilization). Ionic composition and osmolality of freshwater, seawater, and seminal plasma of pike (Esox lucius), sturgeon (Acipenser ruthenus), and cod (Gadus morhua) are shown in the table. Pike and sturgeon spawn in freshwater, and cod spawns in seawater.\n\n|  | Freshwater | Seawater | Pike | Sturgeon | Cod |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| Sodium $\\mathrm{Na}^{+}(\\mathrm{mM})$ | 0.26 | 469 | 75 | 20 | 197 |\n| Chloride $\\mathrm{Cl}^{-}(\\mathrm{mM})$ | 0.22 | 546 | 112 | 6 | 179 |\n| Potassium K (mM) | 0.07 | 10 | 82 | 1 | 6 |\n| Calcium $\\mathrm{Ca}^{2+}(\\mathrm{mM})$ | 0.38 | 10 | 2 | 0.2 | 3 |\n| Osmolality (mOsmol/kg) | $<1-5$ | 1000 | 302 | 51 | 385 |\n\n[figure1]\n\nA) Effect of osmolality on sperm motility initiation in pike (open markers) and cod (filled markers). $\\mathrm{NaCl}$ and Sucrose were used to make activation medium for pike sperm. $\\mathrm{NaCl}$ and artificial sea salt were used to make activation medium for cod sperm.\n\nB) Effect of osmolality on sperm motility initiation in sturgeon. $\\mathrm{NaCl}$ and Sucrose were used to make activation medium for sturgeon sperm.\n\nC) Effect of potassium ions $\\left(\\mathrm{K}^{+}\\right)$on sperm motility initiation and sperm velocity in sturgeon. $\\mathrm{KCl}(\\mathrm{mM})$ was added into activation medium made by $\\mathrm{NaCl}$ or sucrose with osomolality $40 \\mathrm{mOsmol} / \\mathrm{kg}$.\n\nA: Sperm motility in pike and cod become triggered in a hypo-osmotic and hyperosmotic environment, respectively.\nB: Under physiological condition, osmolality is the main factor that inhibits sperm motility initiation in sturgeon.\nC: Sperm motility in sturgeon become triggered after discharged into a hypoosmotic environment with lower $\\mathrm{K}^{+}$ions.\nD: Environmental osmolality is a key signal to trigger sperm motility initiation after release from reproductive system into aquatic environment in marine fish\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-23.jpg?height=798&width=1422&top_left_y=1004&top_left_x=316"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_600",
        "problem": "细菌利用黏肽合成细胞壁的反应需要黏肽转肽酶的催化, 青霉素的结构和黏肽的末端结构类似, 能与黍肽转肽酶的活性中心稳定结合, 是主要在细菌繁殖期起杀菌作用的一类抗生素。在 20 世纪, 我国青霉素的临床用量为 $20 \\sim 40$ 万单位/次; 近年来, 由于细菌的耐药性逐渐增强, 目前青霉素的临床用量为 160 320 万单位/次。下列相关叙述正确的是 ( )\nA: 青霉素能与黏肽竞争酶活性位点, 抑制细菌细胞壁的合成, 造成其细胞壁缺损\nB: 青霉素可用于治疗支原体肺炎, 但有引发患者发生过敏反应的风险\nC: 注入体液中的青霉素能消灭患者体内的多种病原体, 该过程属于非特异性免疫\nD: 青霉素能诱导细菌产生耐药性突变, 经逐年诱导, 敏感型细菌的耐药性增强\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n细菌利用黏肽合成细胞壁的反应需要黏肽转肽酶的催化, 青霉素的结构和黏肽的末端结构类似, 能与黍肽转肽酶的活性中心稳定结合, 是主要在细菌繁殖期起杀菌作用的一类抗生素。在 20 世纪, 我国青霉素的临床用量为 $20 \\sim 40$ 万单位/次; 近年来, 由于细菌的耐药性逐渐增强, 目前青霉素的临床用量为 160 320 万单位/次。下列相关叙述正确的是 ( )\n\nA: 青霉素能与黏肽竞争酶活性位点, 抑制细菌细胞壁的合成, 造成其细胞壁缺损\nB: 青霉素可用于治疗支原体肺炎, 但有引发患者发生过敏反应的风险\nC: 注入体液中的青霉素能消灭患者体内的多种病原体, 该过程属于非特异性免疫\nD: 青霉素能诱导细菌产生耐药性突变, 经逐年诱导, 敏感型细菌的耐药性增强\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1390",
        "problem": "The left ventricle of the heart pumps blood around the body. The amount of blood pumped out of the heart during one contraction is called the cardiac output. The cardiac output varies according to end diastolic volume (see diagram below). The end diastolic volume, also known as the preload, is the volume of blood in the left ventricle before contraction. The cardiac output is also influenced by contractility, which is the ability of the heart muscle to contract.\n\n[figure1]\n\nWhich of the following would increase cardiac output?\n\nI. Increased contractility\n\nII. Increased preload\n\nIII. Weakened heart muscle\n\nIV. Decreased blood volume\n\nV. Fight or flight response\nA: I and II\nB: I, II and III\nC: I, II and IV\nD: I, II, and V\nE: I, II, III and IV\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe left ventricle of the heart pumps blood around the body. The amount of blood pumped out of the heart during one contraction is called the cardiac output. The cardiac output varies according to end diastolic volume (see diagram below). The end diastolic volume, also known as the preload, is the volume of blood in the left ventricle before contraction. The cardiac output is also influenced by contractility, which is the ability of the heart muscle to contract.\n\n[figure1]\n\nWhich of the following would increase cardiac output?\n\nI. Increased contractility\n\nII. Increased preload\n\nIII. Weakened heart muscle\n\nIV. Decreased blood volume\n\nV. Fight or flight response\n\nA: I and II\nB: I, II and III\nC: I, II and IV\nD: I, II, and V\nE: I, II, III and IV\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-19.jpg?height=939&width=1245&top_left_y=661&top_left_x=405"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_856",
        "problem": "美国生物学家摩尔根偶然在一群红眼果蝇中发现了一只白眼雄果蝇, 并利用该果蝇与野生型红眼雌果蝇杂交证明了基因在染色体上。实验过程如下，下列说法错误的是\n\n实验一: 白眼雄果蝇 $\\times$ 野生型红眼雌果蝇 $\\rightarrow F_{1}$ 雌雄全为红眼。 $\\mathrm{F}_{1}$ 雌雄果蝇交配 $\\rightarrow \\mathrm{F}_{2}$ 红眼:白眼 $=3: 1$, 且 $F_{2}$ 中的白眼全为雄性。\n\n实验二: 白眼雄果蝇 $\\times F_{1}$ 红眼雌果蝇 $\\rightarrow F_{3}$ 红眼雌: 红眼雄: 白眼雌: 白眼雄 $=1: 1: 1$ : 1\n\n实验三: $F_{3}$ 白雌果蝇 $\\times$ 野生型红眼雄果蝇 $\\rightarrow F_{4}$ 雌性全为红眼, 雄性全为白眼\nA: 摩尔根偶然发现的一只白眼雄果蝇出现的根本原因可能是基因突变\nB: 根据实验一结果判断果蝇的眼色遗传符合分离定律, 且红眼对白眼为显性\nC: 仅根据实验一和实验二的结果就能判断控制眼色的基因只位于 X 染色体上\nD: 用 $F_{1}$ 红眼雌果蝇与野生型红眼雄果蝇杂交能证明控制眼色的基因只位于 $X$ 染色体上\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n美国生物学家摩尔根偶然在一群红眼果蝇中发现了一只白眼雄果蝇, 并利用该果蝇与野生型红眼雌果蝇杂交证明了基因在染色体上。实验过程如下，下列说法错误的是\n\n实验一: 白眼雄果蝇 $\\times$ 野生型红眼雌果蝇 $\\rightarrow F_{1}$ 雌雄全为红眼。 $\\mathrm{F}_{1}$ 雌雄果蝇交配 $\\rightarrow \\mathrm{F}_{2}$ 红眼:白眼 $=3: 1$, 且 $F_{2}$ 中的白眼全为雄性。\n\n实验二: 白眼雄果蝇 $\\times F_{1}$ 红眼雌果蝇 $\\rightarrow F_{3}$ 红眼雌: 红眼雄: 白眼雌: 白眼雄 $=1: 1: 1$ : 1\n\n实验三: $F_{3}$ 白雌果蝇 $\\times$ 野生型红眼雄果蝇 $\\rightarrow F_{4}$ 雌性全为红眼, 雄性全为白眼\n\nA: 摩尔根偶然发现的一只白眼雄果蝇出现的根本原因可能是基因突变\nB: 根据实验一结果判断果蝇的眼色遗传符合分离定律, 且红眼对白眼为显性\nC: 仅根据实验一和实验二的结果就能判断控制眼色的基因只位于 X 染色体上\nD: 用 $F_{1}$ 红眼雌果蝇与野生型红眼雄果蝇杂交能证明控制眼色的基因只位于 $X$ 染色体上\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_471",
        "problem": "若马的毛色受常染色体上一对等位基因控制, 棕色马与白色马交配, $F_{1}$ 均为淡棕色马, $F_{1}$ 随机交配, $F_{2}$ 中棕色马：淡棕色马：白色马=1：2：1。下列叙述正确的是（ ）\nA: 马的毛色性状中, 棕色对白色为完全显性\nB: $F_{2}$ 中出现棕色、淡棕色和白色是基因重组的结果\nC: $F_{2}$ 中相同毛色的雌雄马交配，其子代中雌性棕色马所占的比例为 $3 / 8$\nD: $F_{2}$ 中淡棕色马与棕色马交配, 其子代基因型的比例与表现型的比例相同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n若马的毛色受常染色体上一对等位基因控制, 棕色马与白色马交配, $F_{1}$ 均为淡棕色马, $F_{1}$ 随机交配, $F_{2}$ 中棕色马：淡棕色马：白色马=1：2：1。下列叙述正确的是（ ）\n\nA: 马的毛色性状中, 棕色对白色为完全显性\nB: $F_{2}$ 中出现棕色、淡棕色和白色是基因重组的结果\nC: $F_{2}$ 中相同毛色的雌雄马交配，其子代中雌性棕色马所占的比例为 $3 / 8$\nD: $F_{2}$ 中淡棕色马与棕色马交配, 其子代基因型的比例与表现型的比例相同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_304",
        "problem": "如图为甲、乙两种单基因遗传病的遗传家系图, 甲病由等位基因 $\\mathrm{A} / \\mathrm{a}$ 控制, 乙病由等位基因 $\\mathrm{B} / \\mathrm{b}$ 控制， $\\mathrm{II}_{4}$ 不携带甲病致病基因，人群中乙病的发病率为 $1 / 400$ 。不考虑基因突变和染色体畸变。下列叙述正确的是（）\n[图1]\n\n[图2]\n\n[图3]\n\n[图4]\nA: 人群中患甲病的女性多于男性\nB: $\\mathrm{III}_{4}$ 与 $\\mathrm{I}_{1}$ 基因型相同的概率为 $1 / 2$\nC: 若 $\\mathrm{III}_{2}$ 和 $\\mathrm{III}_{3}$ 婚配, 生育的子女中只患甲病的概率是 $1 / 12$\nD:  [图5]\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图为甲、乙两种单基因遗传病的遗传家系图, 甲病由等位基因 $\\mathrm{A} / \\mathrm{a}$ 控制, 乙病由等位基因 $\\mathrm{B} / \\mathrm{b}$ 控制， $\\mathrm{II}_{4}$ 不携带甲病致病基因，人群中乙病的发病率为 $1 / 400$ 。不考虑基因突变和染色体畸变。下列叙述正确的是（）\n[图1]\n\n[图2]\n\n[图3]\n\n[图4]\n\nA: 人群中患甲病的女性多于男性\nB: $\\mathrm{III}_{4}$ 与 $\\mathrm{I}_{1}$ 基因型相同的概率为 $1 / 2$\nC: 若 $\\mathrm{III}_{2}$ 和 $\\mathrm{III}_{3}$ 婚配, 生育的子女中只患甲病的概率是 $1 / 12$\nD:  [图5]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-068.jpg?height=442&width=774&top_left_y=196&top_left_x=380",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-068.jpg?height=85&width=460&top_left_y=246&top_left_x=1232",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-068.jpg?height=68&width=297&top_left_y=366&top_left_x=1236",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-068.jpg?height=89&width=494&top_left_y=458&top_left_x=1238",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-068.jpg?height=49&width=1088&top_left_y=958&top_left_x=404",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-068.jpg?height=48&width=1373&top_left_y=2009&top_left_x=347",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-068.jpg?height=54&width=1380&top_left_y=2337&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_922",
        "problem": "现建立“动物精原细胞 $(2 n=4)$ 有丝分裂和减数分裂过程”模型。将染色体 DNA 都被 ${ }^{32} \\mathrm{P}$ 标记的 1 个精原细胞置于不含 ${ }^{32} \\mathrm{P}$ 的培养液中正常培养, 分裂为 2 个子细胞, 其中 1 个子细胞发育为下图细胞(1)。细胞(1)和(2)的染色体组成如图所示, $\\mathrm{H} / \\mathrm{h} 、 \\mathrm{R} / \\mathrm{r}$ 是其中的两对基因。下列叙述错误的是（ ）\n\n[图1]\nA: 细胞 (1)中含 ${ }^{32} \\mathrm{P}$ 的核 DNA 分子数是 4 或 5\nB: 细胞 1)的形成过程中发生了基因突变和基因重组\nC: 细胞(4) (7)中含 ${ }^{32} \\mathrm{P}$ 的核 DNA 分子数可能分别是 2、1、1、0\nD: 该模型中动物精原细胞的基因型为 $\\mathrm{HhRr}$ 或 $\\mathrm{Hhrr}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n现建立“动物精原细胞 $(2 n=4)$ 有丝分裂和减数分裂过程”模型。将染色体 DNA 都被 ${ }^{32} \\mathrm{P}$ 标记的 1 个精原细胞置于不含 ${ }^{32} \\mathrm{P}$ 的培养液中正常培养, 分裂为 2 个子细胞, 其中 1 个子细胞发育为下图细胞(1)。细胞(1)和(2)的染色体组成如图所示, $\\mathrm{H} / \\mathrm{h} 、 \\mathrm{R} / \\mathrm{r}$ 是其中的两对基因。下列叙述错误的是（ ）\n\n[图1]\n\nA: 细胞 (1)中含 ${ }^{32} \\mathrm{P}$ 的核 DNA 分子数是 4 或 5\nB: 细胞 1)的形成过程中发生了基因突变和基因重组\nC: 细胞(4) (7)中含 ${ }^{32} \\mathrm{P}$ 的核 DNA 分子数可能分别是 2、1、1、0\nD: 该模型中动物精原细胞的基因型为 $\\mathrm{HhRr}$ 或 $\\mathrm{Hhrr}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-54.jpg?height=520&width=1176&top_left_y=168&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1040",
        "problem": "Suppose you are walking through a deciduous forest. You notice a dead tree that had been cut down and cut into logs that were tagged and labeled for removal from the forest three years prior. You and a friend try to move one of the smaller logs in order to build a camp fire. You find that the small log is heavy and difficult to move. The mass of the $\\log$ is actually made-up mostly of:\nA: Water from the soil\nB: Macronutrients from the soil\nC: Micronutrients from the soil\nD: An invisible gas from the air\nE: Membrane-bound cell organelles of the prokaryotes\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSuppose you are walking through a deciduous forest. You notice a dead tree that had been cut down and cut into logs that were tagged and labeled for removal from the forest three years prior. You and a friend try to move one of the smaller logs in order to build a camp fire. You find that the small log is heavy and difficult to move. The mass of the $\\log$ is actually made-up mostly of:\n\nA: Water from the soil\nB: Macronutrients from the soil\nC: Micronutrients from the soil\nD: An invisible gas from the air\nE: Membrane-bound cell organelles of the prokaryotes\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_934",
        "problem": "图 1 表示 $14 / 21$ 染色体发生易位的过程, 图 2 表示 $14 / 21$ 染色体易位患者进行减数分裂时相关染色体的联会情况, 减数分裂时联会的任意两条染色体分开移向细胞的两极,另一条随机移向细胞的一极, 且这种方式产生的配子均能存活, 不考虑其他变异。下列叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\nA: 形成 $14 / 21$ 易位染色体的过程可能会使某些基因功能失活\nB: 该患者发生了染色体结构变异和数目变异\nC: 上述染色体结构变异只能发生在减数分裂过程中\nD: 该患者与正常个体婚配生育的子女理论上有 6 种染色体组成\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 1 表示 $14 / 21$ 染色体发生易位的过程, 图 2 表示 $14 / 21$ 染色体易位患者进行减数分裂时相关染色体的联会情况, 减数分裂时联会的任意两条染色体分开移向细胞的两极,另一条随机移向细胞的一极, 且这种方式产生的配子均能存活, 不考虑其他变异。下列叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\nA: 形成 $14 / 21$ 易位染色体的过程可能会使某些基因功能失活\nB: 该患者发生了染色体结构变异和数目变异\nC: 上述染色体结构变异只能发生在减数分裂过程中\nD: 该患者与正常个体婚配生育的子女理论上有 6 种染色体组成\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-005.jpg?height=386&width=506&top_left_y=224&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-005.jpg?height=500&width=743&top_left_y=178&top_left_x=859"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_724",
        "problem": "已知正常小鼠体细胞中染色体数目为 20 对。中科院动物研究所研究团队发现\n\nCyclinB3（细胞周期蛋白）在雌鼠卵细胞形成过程中发挥了独特作用, 当 CyclinB3 缺失时雌鼠不能产生后代。此外, 基因组分析发现 CyclinB3 缺失的卵母细胞仍能正常进行减数第二次分裂。研究者对 CyclinB3 缺失雌鼠和正常雌鼠卵细胞的形成过程对比观察并绘制了下图。下列说法正确的是（）\n\n[图1]\nA: 正常雌鼠中期II染色体数为 40 条, CyclinB3 缺失雌鼠中期II染色体数为 80 条\nB: 若 CyclinB3 缺失雌鼠可正常受精, 则受精后形成的受精卵中染色体数为 60\nC: 推测细胞周期蛋白 CyclinB3 的功能是抑制同源染色体的分离\nD: 减数分裂和有丝分裂对同种生物前后代染色体数目保持恒定起到了决定性作用\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知正常小鼠体细胞中染色体数目为 20 对。中科院动物研究所研究团队发现\n\nCyclinB3（细胞周期蛋白）在雌鼠卵细胞形成过程中发挥了独特作用, 当 CyclinB3 缺失时雌鼠不能产生后代。此外, 基因组分析发现 CyclinB3 缺失的卵母细胞仍能正常进行减数第二次分裂。研究者对 CyclinB3 缺失雌鼠和正常雌鼠卵细胞的形成过程对比观察并绘制了下图。下列说法正确的是（）\n\n[图1]\n\nA: 正常雌鼠中期II染色体数为 40 条, CyclinB3 缺失雌鼠中期II染色体数为 80 条\nB: 若 CyclinB3 缺失雌鼠可正常受精, 则受精后形成的受精卵中染色体数为 60\nC: 推测细胞周期蛋白 CyclinB3 的功能是抑制同源染色体的分离\nD: 减数分裂和有丝分裂对同种生物前后代染色体数目保持恒定起到了决定性作用\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-043.jpg?height=355&width=871&top_left_y=168&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_470",
        "problem": "小鼠的毛色野生型 (A) 对突变型 (a) 为完全显性, 在生殖细胞的发育过程中, 原有的甲基化均会被清除, 再生成的所有雌配子中控制毛色的基因均不会被甲基化, 所有雄配子中控制毛色的基因均会被甲基化修饰而使该基因在后代中不能表达。两只突变型小鼠杂交得到 $F_{1}, F_{1}$ 中小鼠自由交配得到 $F_{2}$, 下列说法错误的是 ( )\nA: 小鼠的毛色遗传现象属于表观遗传\nB: 若 $F_{1}$ 只有一种表型, 则亲本基因型均为 aa\nC: 推测 $F_{2}$ 中野生型最多占比为 $1 / 2$\nD: 若 $F_{2}$ 逐代自由交配, 后代表型比例不变\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n小鼠的毛色野生型 (A) 对突变型 (a) 为完全显性, 在生殖细胞的发育过程中, 原有的甲基化均会被清除, 再生成的所有雌配子中控制毛色的基因均不会被甲基化, 所有雄配子中控制毛色的基因均会被甲基化修饰而使该基因在后代中不能表达。两只突变型小鼠杂交得到 $F_{1}, F_{1}$ 中小鼠自由交配得到 $F_{2}$, 下列说法错误的是 ( )\n\nA: 小鼠的毛色遗传现象属于表观遗传\nB: 若 $F_{1}$ 只有一种表型, 则亲本基因型均为 aa\nC: 推测 $F_{2}$ 中野生型最多占比为 $1 / 2$\nD: 若 $F_{2}$ 逐代自由交配, 后代表型比例不变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_308",
        "problem": "某种植物有花的颜色（红、白）、果的形状（扁、圆）等相对性状，相关基因用 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b} 、 \\mathrm{C} / \\mathrm{c} \\ldots . .$. 表示, 某科研人员利用红花扁果纯合子和白花圆果纯合子进行杂交得 $F_{1}, F_{1}$ 自交得 $F_{2}$, 结果如下表所示, 下列分析错误的是 ( )\n\n| 表现型 | 红花扁果 | 红花圆果 | 白花扁果 | 白花圆果 |\n| :--- | :--- | :--- | :--- | :--- |\n| 比例 | 27 | 9 | 21 | 7 |\nA: 根据表格可知, 该种植物花的颜色的遗传受 2 对等位基因控制\nB: $F_{1}$ 的基因型是 $\\mathrm{AaBbCc}$ ，其测交后代中表现型与之相同的概率是 $1 / 8$\nC: $F_{2}$ 红花圆果中杂合子占 $8 / 9$\nD: 可以通过自交判断某红花圆果植株的基因型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种植物有花的颜色（红、白）、果的形状（扁、圆）等相对性状，相关基因用 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b} 、 \\mathrm{C} / \\mathrm{c} \\ldots . .$. 表示, 某科研人员利用红花扁果纯合子和白花圆果纯合子进行杂交得 $F_{1}, F_{1}$ 自交得 $F_{2}$, 结果如下表所示, 下列分析错误的是 ( )\n\n| 表现型 | 红花扁果 | 红花圆果 | 白花扁果 | 白花圆果 |\n| :--- | :--- | :--- | :--- | :--- |\n| 比例 | 27 | 9 | 21 | 7 |\n\nA: 根据表格可知, 该种植物花的颜色的遗传受 2 对等位基因控制\nB: $F_{1}$ 的基因型是 $\\mathrm{AaBbCc}$ ，其测交后代中表现型与之相同的概率是 $1 / 8$\nC: $F_{2}$ 红花圆果中杂合子占 $8 / 9$\nD: 可以通过自交判断某红花圆果植株的基因型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_927",
        "problem": "将一个基因型为 $\\mathrm{AaX} X^{\\mathrm{B}} \\mathrm{Y}$ 的精原细胞所有核 DNA 双链均用 ${ }^{15} \\mathrm{~N}$ 标记后, 置于只含 ${ }^{14} \\mathrm{~N}$的培养基中培养, 经一次有丝分裂后, 又分别完成减数分裂, 产生了一个基因型为 $\\mathrm{aX}^{\\mathrm{B}} \\mathrm{Y}$的异常精细胞。若无其他染色体变异和互换发生, 下列说法错误的是（）\nA: 可利用 ${ }^{15} \\mathrm{~N}$ 的放射性追踪初级精母细胞中染色体的位置移动\nB: 初级精母细胞中含 ${ }^{15} \\mathrm{~N}$ 标记的核 DNA 分子占 $1 / 2$\nC: 次级精母细胞中含有 0 条或 1 条或 2 条 X 染色体\nD: 异常精细胞可能由减数分裂 I 时同源染色体 XY 未分离导致\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将一个基因型为 $\\mathrm{AaX} X^{\\mathrm{B}} \\mathrm{Y}$ 的精原细胞所有核 DNA 双链均用 ${ }^{15} \\mathrm{~N}$ 标记后, 置于只含 ${ }^{14} \\mathrm{~N}$的培养基中培养, 经一次有丝分裂后, 又分别完成减数分裂, 产生了一个基因型为 $\\mathrm{aX}^{\\mathrm{B}} \\mathrm{Y}$的异常精细胞。若无其他染色体变异和互换发生, 下列说法错误的是（）\n\nA: 可利用 ${ }^{15} \\mathrm{~N}$ 的放射性追踪初级精母细胞中染色体的位置移动\nB: 初级精母细胞中含 ${ }^{15} \\mathrm{~N}$ 标记的核 DNA 分子占 $1 / 2$\nC: 次级精母细胞中含有 0 条或 1 条或 2 条 X 染色体\nD: 异常精细胞可能由减数分裂 I 时同源染色体 XY 未分离导致\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_241",
        "problem": "One of the basic ways to illustrate trophic relationship in an ecosystem is to use pyramids of energy and biomass, which are the stack of rectangles. Each represents the amount of energy or biomass within one trophic level. We know a proportion of energy and biomass is lost from the ecosystem when transferring from one to the next level. Therefore, the size of each rectangle decreases when we move from one level to the level above it.\n\n[figure1]\n\nEnergy\n\n[figure2]\n\nEnergy\n\n[figure3]\n\nBiomass\n\n[figure4]\n\nBiomass\n\nExamples of pyramids schemes in different ecosystems. As you can see biomass pyramid can be inverted in some ecosystems.\nA: Inverted biomass pyramids are caused by low density of heterotrophs at any given time.\nB: We can't have regular energy pyramid when biomass pyramid is inverted, thus the energy pyramid for the inverted biomass must be inverted as well.\nC: Biomass of autotroph are usually greater than biomass of heterotroph in terrestrial ecosystems.\nD: Given that phytoplankton has more rapid turnover, the biomass pyramid is more likely to be inverted.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nOne of the basic ways to illustrate trophic relationship in an ecosystem is to use pyramids of energy and biomass, which are the stack of rectangles. Each represents the amount of energy or biomass within one trophic level. We know a proportion of energy and biomass is lost from the ecosystem when transferring from one to the next level. Therefore, the size of each rectangle decreases when we move from one level to the level above it.\n\n[figure1]\n\nEnergy\n\n[figure2]\n\nEnergy\n\n[figure3]\n\nBiomass\n\n[figure4]\n\nBiomass\n\nExamples of pyramids schemes in different ecosystems. As you can see biomass pyramid can be inverted in some ecosystems.\n\nA: Inverted biomass pyramids are caused by low density of heterotrophs at any given time.\nB: We can't have regular energy pyramid when biomass pyramid is inverted, thus the energy pyramid for the inverted biomass must be inverted as well.\nC: Biomass of autotroph are usually greater than biomass of heterotroph in terrestrial ecosystems.\nD: Given that phytoplankton has more rapid turnover, the biomass pyramid is more likely to be inverted.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-46.jpg?height=354&width=491&top_left_y=631&top_left_x=517",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-46.jpg?height=368&width=491&top_left_y=1095&top_left_x=520",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-46.jpg?height=352&width=528&top_left_y=635&top_left_x=1024",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-46.jpg?height=357&width=332&top_left_y=1101&top_left_x=1116"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1172",
        "problem": "Which one of the following is an example of divergent evolution?\nA: eye of locust and blackbird\nB: forelimb of pigeon and dolphin\nC: wings of cockroach and bat\nD: skeleton of tortoise and lobster\nE: scales of shark and trout\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following is an example of divergent evolution?\n\nA: eye of locust and blackbird\nB: forelimb of pigeon and dolphin\nC: wings of cockroach and bat\nD: skeleton of tortoise and lobster\nE: scales of shark and trout\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_695",
        "problem": "玉米是常见的遗传实验材料, 玉米花序与授粉方式模式如图, 下列说法错误的是\n\n[图1]\nA: 自然状态下, 玉米能进行自交和自由交配\nB: 可通过I和II两种受粉方式来判断甜和非甜的显隐性\nC: 用玉米做杂交实验中，常对未成熟的雌花序采取套袋的方式来避免自交\nD: 若在传粉时节正好遇上连续阴雨天气，可采用生长素调节剂处理的方法来提高产量\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n玉米是常见的遗传实验材料, 玉米花序与授粉方式模式如图, 下列说法错误的是\n\n[图1]\n\nA: 自然状态下, 玉米能进行自交和自由交配\nB: 可通过I和II两种受粉方式来判断甜和非甜的显隐性\nC: 用玉米做杂交实验中，常对未成熟的雌花序采取套袋的方式来避免自交\nD: 若在传粉时节正好遇上连续阴雨天气，可采用生长素调节剂处理的方法来提高产量\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-83.jpg?height=359&width=600&top_left_y=503&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_987",
        "problem": "Which of the following statement(s) is/are INCORRECT:\n1. Cartilage heals slower than skin because cartilage is a deeper tissue\n2. The inside lining of the intestine has a large surface area because of the presence of cilia\n3. Adipose is a type of connective tissue because that is where fat is stored\nA: \\#1 and \\#2 are incorrect\nB: $\\# 2$ and $\\# 3$ are incorrect\nC: \\#1 and \\#3 are incorrect\nD: All are correct statements\nE: All are incorrect statements\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following statement(s) is/are INCORRECT:\n1. Cartilage heals slower than skin because cartilage is a deeper tissue\n2. The inside lining of the intestine has a large surface area because of the presence of cilia\n3. Adipose is a type of connective tissue because that is where fat is stored\n\nA: \\#1 and \\#2 are incorrect\nB: $\\# 2$ and $\\# 3$ are incorrect\nC: \\#1 and \\#3 are incorrect\nD: All are correct statements\nE: All are incorrect statements\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_955",
        "problem": "Which of the following would you expect to happen if a bacterial cell lacked restriction enzymes?\nA: Bacteriophages could easily infect and lyse the cell\nB: Incomplete plasmids would be the cell's products\nC: Replication would not be possible\nD: The bacteria would be an obligate parasite\nE: Both C and D\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following would you expect to happen if a bacterial cell lacked restriction enzymes?\n\nA: Bacteriophages could easily infect and lyse the cell\nB: Incomplete plasmids would be the cell's products\nC: Replication would not be possible\nD: The bacteria would be an obligate parasite\nE: Both C and D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_726",
        "problem": "水稻为两性花、花小, 杂交育种时操作难度较大, 因此找到合适的雄性不育系是杂交育种的关键。如图表示我国科学家利用光/温敏雄性不育系水稻留种及获得 $\\mathrm{F}_{1}$ 杂交种的过程。但是即使是在高温或长日照下，光/温敏雄性不育系仍有 $5 \\% ~ 10 \\%$ 的自交结实率，导致制备的杂交种中混有纯合子。为解决这一问题，杂交制种时，常借助其他性状如紫叶和绿叶（绿叶对紫叶为显性）进行篮选，以淘汰纯合子。下列相关叙述错误的是\n\n[图1]\nA: 雄性不育这一性状是基因型和环境条件共同决定的\nB: 相较于普通的雄性不育系, 光/温敏雄性不育系留种更方便\nC: 杂交时只收获雄性不育系植株上的种子\nD: 杂交制种时, 选用雄性可育系纯合绿叶稻与光/温敏雄性不育系纯合紫叶稻杂交, 并在子代的秧苗期内剔除绿叶秧苗即可淘汰纯合子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n水稻为两性花、花小, 杂交育种时操作难度较大, 因此找到合适的雄性不育系是杂交育种的关键。如图表示我国科学家利用光/温敏雄性不育系水稻留种及获得 $\\mathrm{F}_{1}$ 杂交种的过程。但是即使是在高温或长日照下，光/温敏雄性不育系仍有 $5 \\% ~ 10 \\%$ 的自交结实率，导致制备的杂交种中混有纯合子。为解决这一问题，杂交制种时，常借助其他性状如紫叶和绿叶（绿叶对紫叶为显性）进行篮选，以淘汰纯合子。下列相关叙述错误的是\n\n[图1]\n\nA: 雄性不育这一性状是基因型和环境条件共同决定的\nB: 相较于普通的雄性不育系, 光/温敏雄性不育系留种更方便\nC: 杂交时只收获雄性不育系植株上的种子\nD: 杂交制种时, 选用雄性可育系纯合绿叶稻与光/温敏雄性不育系纯合紫叶稻杂交, 并在子代的秧苗期内剔除绿叶秧苗即可淘汰纯合子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-051.jpg?height=483&width=902&top_left_y=1866&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_35",
        "problem": "Cardiac output (CO) is the volume of blood pumped by the heart in one minute. Cardiac output is affected by the stroke volume (SV) and the heart rate (HR).\n\nCardiac output can be measured indirectly using the Fick's equation: $\\mathrm{CO}=\\mathrm{Q} /(\\mathrm{A}-\\mathrm{V})$, where $\\mathrm{Q}$ is the rate of oxygen consumption $(\\mathrm{mL} / \\mathrm{min}), A-V$ is the difference between oxygen concentration in the arterial blood (A) and in the venous blood (V).\n\nThe data below were measured from a healthy person before and during physical exercise.\n\n| Parameters | Before <br> Exercise | During <br> Exercise |\n| :--- | :---: | :---: |\n| Rate of oxygen <br> consumption (Q) | $250 \\mathrm{~mL} / \\mathrm{min}$ | $1500 \\mathrm{~mL} / \\mathrm{min}$ |\n| Oxygen difference <br> (A-V) | $50 \\mathrm{~mL} / \\mathrm{L}$ <br> blood | $150 \\mathrm{~mL} / \\mathrm{L}$ <br> blood |\n| Heart rate (HR) | 60 beats $/ \\mathrm{min}$ | 120 beats $/ \\mathrm{min}$ |\nA: Cardiac output increased by two-fold during exercise.\nB: Stroke volume during exercise was higher than that before exercise.\nC: Physical exercise caused a reduction in hemoglobin's affinity for oxygen, resulting in a three-fold increase in the amount of oxygen released to tissues.\nD: From the data above, it can be concluded that change in cardiac output during exercise is caused by the changes in both heart rate and stroke volume.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nCardiac output (CO) is the volume of blood pumped by the heart in one minute. Cardiac output is affected by the stroke volume (SV) and the heart rate (HR).\n\nCardiac output can be measured indirectly using the Fick's equation: $\\mathrm{CO}=\\mathrm{Q} /(\\mathrm{A}-\\mathrm{V})$, where $\\mathrm{Q}$ is the rate of oxygen consumption $(\\mathrm{mL} / \\mathrm{min}), A-V$ is the difference between oxygen concentration in the arterial blood (A) and in the venous blood (V).\n\nThe data below were measured from a healthy person before and during physical exercise.\n\n| Parameters | Before <br> Exercise | During <br> Exercise |\n| :--- | :---: | :---: |\n| Rate of oxygen <br> consumption (Q) | $250 \\mathrm{~mL} / \\mathrm{min}$ | $1500 \\mathrm{~mL} / \\mathrm{min}$ |\n| Oxygen difference <br> (A-V) | $50 \\mathrm{~mL} / \\mathrm{L}$ <br> blood | $150 \\mathrm{~mL} / \\mathrm{L}$ <br> blood |\n| Heart rate (HR) | 60 beats $/ \\mathrm{min}$ | 120 beats $/ \\mathrm{min}$ |\n\nA: Cardiac output increased by two-fold during exercise.\nB: Stroke volume during exercise was higher than that before exercise.\nC: Physical exercise caused a reduction in hemoglobin's affinity for oxygen, resulting in a three-fold increase in the amount of oxygen released to tissues.\nD: From the data above, it can be concluded that change in cardiac output during exercise is caused by the changes in both heart rate and stroke volume.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_416",
        "problem": "我国研究人员发现 $P$ 基因表达的蛋白质能促进细胞凋亡。肝癌细胞和正常肝细胞中的 P 基因存在差异, 如图所示。下列叙述正确的是（ ）\n\n## $\\mathrm{P}$ 基因的启动子\n\n## 高度甲 <br> 正常肝细胞的P基因 <br> 基化 <br> 启动子沉默\n\nHintir\n\n## 肝癌细胞的P基因\nA: 肝细胞中的 P 基因可能属于抑癌基因\nB: 肝细胞癌变的原因是 $\\mathrm{P}$ 基因启动子甲基化, 使得 $\\mathrm{P}$ 基因大量表达\nC: 甲基化会引发 P 基因碱基排列顺序发生变化，进而影响 P 基因的表达\nD: P 基因启动子的甲基化是不可逆的, 因此肝癌细胞一旦形成就不可逆转\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n我国研究人员发现 $P$ 基因表达的蛋白质能促进细胞凋亡。肝癌细胞和正常肝细胞中的 P 基因存在差异, 如图所示。下列叙述正确的是（ ）\n\n## $\\mathrm{P}$ 基因的启动子\n\n## 高度甲 <br> 正常肝细胞的P基因 <br> 基化 <br> 启动子沉默\n\nHintir\n\n## 肝癌细胞的P基因\n\nA: 肝细胞中的 P 基因可能属于抑癌基因\nB: 肝细胞癌变的原因是 $\\mathrm{P}$ 基因启动子甲基化, 使得 $\\mathrm{P}$ 基因大量表达\nC: 甲基化会引发 P 基因碱基排列顺序发生变化，进而影响 P 基因的表达\nD: P 基因启动子的甲基化是不可逆的, 因此肝癌细胞一旦形成就不可逆转\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1092",
        "problem": "The nuclear membrane disappears during cell division. After completion of cell division, it re-appears during the interphase. Which of the following contributes towards formation of the nuclear membrane?\nA: Spindle fibre proteins\nB: Cytoskeletal elements\nC: Endoplasmic reticulum\nD: Golgi bodies\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe nuclear membrane disappears during cell division. After completion of cell division, it re-appears during the interphase. Which of the following contributes towards formation of the nuclear membrane?\n\nA: Spindle fibre proteins\nB: Cytoskeletal elements\nC: Endoplasmic reticulum\nD: Golgi bodies\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1137",
        "problem": "When an animal population is being estimated by the capture-recapture technique, which one of the following would lead to the size of the population being underestimated?\nA: a higher predation of marked than of unmarked animals\nB: a greater attraction of marked than of unmarked animals to traps\nC: an immigration of animals into the area between samplings\nD: a high but proportionately equal mortality of both marked and unmarked animals\nE: emigration of a higher proportion of marked than unmarked individuals from the sampling area\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen an animal population is being estimated by the capture-recapture technique, which one of the following would lead to the size of the population being underestimated?\n\nA: a higher predation of marked than of unmarked animals\nB: a greater attraction of marked than of unmarked animals to traps\nC: an immigration of animals into the area between samplings\nD: a high but proportionately equal mortality of both marked and unmarked animals\nE: emigration of a higher proportion of marked than unmarked individuals from the sampling area\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1141",
        "problem": "The graph shows the effect of exercise on the rate of oxygen uptake.\n\n[figure1]\n\nThe total additional volume of oxygen used due to the exercise is approximately\nA: $12 \\mathrm{dm}^{3}$\nB: $15 \\mathrm{dm}^{3}$\nC: $18 \\mathrm{dm}^{3}$\nD: $20 \\mathrm{dm}^{3}$\nE: $24 \\mathrm{dm}^{3}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows the effect of exercise on the rate of oxygen uptake.\n\n[figure1]\n\nThe total additional volume of oxygen used due to the exercise is approximately\n\nA: $12 \\mathrm{dm}^{3}$\nB: $15 \\mathrm{dm}^{3}$\nC: $18 \\mathrm{dm}^{3}$\nD: $20 \\mathrm{dm}^{3}$\nE: $24 \\mathrm{dm}^{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-11.jpg?height=545&width=1259&top_left_y=1846&top_left_x=404"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_135",
        "problem": "| Metabolic <br> substrate | Concentration <br> $[\\mathrm{mM}]$ | Output power <br> $[\\mathrm{W}]$ | Expected speed <br> $[\\mathrm{m} / \\mathrm{s}]$ | Exercise duration <br> $[\\mathrm{s}]$ |\n| :---: | :---: | :---: | :---: | :---: |\n| ATP | 8 | 6400 | 27 | $2-4$ |\n| CP | 26 | 6000 | 25 | $10-17$ |\n| Glycogen | 90 | 1640 | 6.7 | $>6000$ |\n| Fat | $7-25$ | 1100 | 4.6 |  |\n\nTable 1. Types of metabolic substrate and its concentration as an energy source in human muscle cells. The predicted values of output power produced by the muscle tissue, the expected speed at which the athlete ran with that power, and the duration of exercise are shown when only the respective energy sources were used. $\\mathrm{CP}$ indicates creatine phosphate.\nA: Athletes running a 100-meter sprint are supposed to run using ATP originally stored in muscle cells during the former half. During the last half, ATP produced by respiration is used.\nB: It is possible that marathon runners continue exercising using muscle tissue without ATP.\nC: A crucial point for middle-distance runners of $1,500 \\mathrm{~m}$ is switching smoothly from running with $\\mathrm{CP}$ to that with ATP produced by aerobic breathing. 78\nD: Similar to bird migration, stored fat is one of the major energy sources for long-distance runners, although it has some metabolic delay for conversion into ATP.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\n| Metabolic <br> substrate | Concentration <br> $[\\mathrm{mM}]$ | Output power <br> $[\\mathrm{W}]$ | Expected speed <br> $[\\mathrm{m} / \\mathrm{s}]$ | Exercise duration <br> $[\\mathrm{s}]$ |\n| :---: | :---: | :---: | :---: | :---: |\n| ATP | 8 | 6400 | 27 | $2-4$ |\n| CP | 26 | 6000 | 25 | $10-17$ |\n| Glycogen | 90 | 1640 | 6.7 | $>6000$ |\n| Fat | $7-25$ | 1100 | 4.6 |  |\n\nTable 1. Types of metabolic substrate and its concentration as an energy source in human muscle cells. The predicted values of output power produced by the muscle tissue, the expected speed at which the athlete ran with that power, and the duration of exercise are shown when only the respective energy sources were used. $\\mathrm{CP}$ indicates creatine phosphate.\n\nA: Athletes running a 100-meter sprint are supposed to run using ATP originally stored in muscle cells during the former half. During the last half, ATP produced by respiration is used.\nB: It is possible that marathon runners continue exercising using muscle tissue without ATP.\nC: A crucial point for middle-distance runners of $1,500 \\mathrm{~m}$ is switching smoothly from running with $\\mathrm{CP}$ to that with ATP produced by aerobic breathing. 78\nD: Similar to bird migration, stored fat is one of the major energy sources for long-distance runners, although it has some metabolic delay for conversion into ATP.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_243",
        "problem": "Nitrogen is an important nutrient for all plants. While there is an abundance of nitrogen $\\left(\\mathrm{N}_{2}\\right)$ in the atmosphere, most plants are unable to convert $\\mathrm{N}_{2}$ into a usable form. Fixation of nitrogen gas into ammonia is an ability restricted to nitrogen-fixing bacteria. The figure below shows prerequisite step for nitrogen fixation through symbiosis between these bacteria and some plant species.\n\n[figure1]\nA: Between two major inorganic nitrogen forms, $\\mathrm{NH}_{4}{ }^{+}$and $\\mathrm{NO}_{3}{ }^{-}$, the former is mobile in the plant, while the latter is not.\nB: Immediately after recognition of plant by bacteria and recognition of bacteria by host plant, a calcium spike occurs in the root cells.\nC: The plant hormone cytokinin is needed for initiation of nodule formation.\nD: Release of bacterial exopolysaccharides is the necessary and sufficient condition for a functional symbiosis between rhizobia and its appropriate host.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nNitrogen is an important nutrient for all plants. While there is an abundance of nitrogen $\\left(\\mathrm{N}_{2}\\right)$ in the atmosphere, most plants are unable to convert $\\mathrm{N}_{2}$ into a usable form. Fixation of nitrogen gas into ammonia is an ability restricted to nitrogen-fixing bacteria. The figure below shows prerequisite step for nitrogen fixation through symbiosis between these bacteria and some plant species.\n\n[figure1]\n\nA: Between two major inorganic nitrogen forms, $\\mathrm{NH}_{4}{ }^{+}$and $\\mathrm{NO}_{3}{ }^{-}$, the former is mobile in the plant, while the latter is not.\nB: Immediately after recognition of plant by bacteria and recognition of bacteria by host plant, a calcium spike occurs in the root cells.\nC: The plant hormone cytokinin is needed for initiation of nodule formation.\nD: Release of bacterial exopolysaccharides is the necessary and sufficient condition for a functional symbiosis between rhizobia and its appropriate host.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-33.jpg?height=788&width=1313&top_left_y=577&top_left_x=380"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_326",
        "problem": "人体内肾上腺素与激素 $\\mathrm{X}$ 参与血糖调节的部分调节机理如图。下列说法正确的是\n\n[图1]\nA: 下丘脑的一定区域通过兴奋交感神经直接作用于肝脏, 能促进肝糖原的分解\nB: 激素 X 与肾上腺素具有协同作用, 调节激素 X 分泌的信号分子有神经递质等\nC: 某病人出现高血糖一定是其体内出现 $\\mathrm{G}_{2}$ 蛋白偶联受体的抗体所致\nD: 若激素 X 受体的控制基因发生突变, 患者体内激素 X 相对正常人降低\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人体内肾上腺素与激素 $\\mathrm{X}$ 参与血糖调节的部分调节机理如图。下列说法正确的是\n\n[图1]\n\nA: 下丘脑的一定区域通过兴奋交感神经直接作用于肝脏, 能促进肝糖原的分解\nB: 激素 X 与肾上腺素具有协同作用, 调节激素 X 分泌的信号分子有神经递质等\nC: 某病人出现高血糖一定是其体内出现 $\\mathrm{G}_{2}$ 蛋白偶联受体的抗体所致\nD: 若激素 X 受体的控制基因发生突变, 患者体内激素 X 相对正常人降低\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-036.jpg?height=748&width=1087&top_left_y=151&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1552",
        "problem": "The complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nWhich pathway amplifies the activity of the complement system to increase its speed and destructive power?\nA: Pathway A\nB: Pathway B\nC: Pathway C\nD: Pathway D\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nWhich pathway amplifies the activity of the complement system to increase its speed and destructive power?\n\nA: Pathway A\nB: Pathway B\nC: Pathway C\nD: Pathway D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=790&width=1714&top_left_y=486&top_left_x=228",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=930&width=1806&top_left_y=1335&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_455",
        "problem": "下图是甲、乙两种独立遗传的单基因遗传病的系谱图, 已知其中一种属于伴 X 染色体遗传病。据调查发现, 人群中乙病的发病率为 $1 / 10000$ 。\n\nI\n\nII\n\n[图1]\n\n$\\square \\bigcirc$ 正常男女甲病男乙病女\n\n下列叙述错误的是（ ）\nA: 通过遗传系谱图分析可知, 乙病为常染色体隐性遗传\nB: 羊膜腔穿刺技术可用来诊断胎儿是否患甲病或乙病\nC: 若已知 $\\mathrm{II}_{3}$ 含有乙病致病基因, 则 $\\mathrm{II}_{3}$ 和 $^{2} \\mathrm{II}_{4}$ 生出两病兼患男孩的概率为 $1 / 48$\nD: 若 $I_{5}$ 与人群中一个表型正常的男子婚配, 后代中女孩患乙病的概率为 $1 / 202$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图是甲、乙两种独立遗传的单基因遗传病的系谱图, 已知其中一种属于伴 X 染色体遗传病。据调查发现, 人群中乙病的发病率为 $1 / 10000$ 。\n\nI\n\nII\n\n[图1]\n\n$\\square \\bigcirc$ 正常男女甲病男乙病女\n\n下列叙述错误的是（ ）\n\nA: 通过遗传系谱图分析可知, 乙病为常染色体隐性遗传\nB: 羊膜腔穿刺技术可用来诊断胎儿是否患甲病或乙病\nC: 若已知 $\\mathrm{II}_{3}$ 含有乙病致病基因, 则 $\\mathrm{II}_{3}$ 和 $^{2} \\mathrm{II}_{4}$ 生出两病兼患男孩的概率为 $1 / 48$\nD: 若 $I_{5}$ 与人群中一个表型正常的男子婚配, 后代中女孩患乙病的概率为 $1 / 202$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-37.jpg?height=403&width=716&top_left_y=1552&top_left_x=430"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1354",
        "problem": "Evolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).At what minimum date is flightlessness thought to have arisen in the emu lineage?\nA: $55 \\mathrm{Ma}$\nB: $50 \\mathrm{Ma}$\nC: $25 \\mathrm{Ma}$\nD: $20 \\mathrm{Ma}$\nE: $15 \\mathrm{Ma}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nEvolutionary relationships or phylogeny can be represented by branching lines that end at groups of organisms on diagrams called cladograms. At the point of branching, a common ancestor is believed to have existed. Clades or groupings of organisms, are characterized by synapomorphies, characters present in the last common ancestor.\n\nMitchell et al. sequenced the mitochondrial genomes of two elephant birds and used these to infer relationships within the palaeognaths. These data are presented in the cladogram below. Divergence dates are given in the blue numbers above branches with the blue bars representing the $95 \\%$ probability around that date. Blue arrows mark the minimum date for the evolution of flightlessness in lineages for which fossil evidence is available. The scale is given in millions of years before the present. Silhouettes indicate the relative size of representative taxa. Species diversity for each major clade is presented in parentheses, with extinct groups shown in red. The dagger symbol $(\\dagger)$ indicates that the number of species is uncertain.\n\n[figure1]\n\nSource: Mitchell et al. Science 344, 898 (2014).\n\nproblem:\nAt what minimum date is flightlessness thought to have arisen in the emu lineage?\n\nA: $55 \\mathrm{Ma}$\nB: $50 \\mathrm{Ma}$\nC: $25 \\mathrm{Ma}$\nD: $20 \\mathrm{Ma}$\nE: $15 \\mathrm{Ma}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-28.jpg?height=1868&width=1627&top_left_y=748&top_left_x=206"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1462",
        "problem": "[figure1]\n\nFig 2A: Regulation of PFK activity\n\n[figure2]\n\nFig 2B: Graph between the relative velocity and fructose- 6 phosphate\n\nFigure 2A and 2B: Enzyme activity under different conditions. (PFK = Phosphofructokinase)\n\nFigure 2A and Figure 2B show that when [fructose 6-phosphate] is high the enzyme:\nA: exhibits product inhibition.\nB: exhibits non-competitive inhibition.\nC: is stabilised in the high activity state.\nD: is stabilised in the low activity state.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nFig 2A: Regulation of PFK activity\n\n[figure2]\n\nFig 2B: Graph between the relative velocity and fructose- 6 phosphate\n\nFigure 2A and 2B: Enzyme activity under different conditions. (PFK = Phosphofructokinase)\n\nFigure 2A and Figure 2B show that when [fructose 6-phosphate] is high the enzyme:\n\nA: exhibits product inhibition.\nB: exhibits non-competitive inhibition.\nC: is stabilised in the high activity state.\nD: is stabilised in the low activity state.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-33.jpg?height=508&width=671&top_left_y=186&top_left_x=201",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-33.jpg?height=548&width=574&top_left_y=157&top_left_x=1072"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1523",
        "problem": "Blue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nAre satellite images better at estimating the change in population or the total population?\nA: Change in population\nB: Total population.\nC: Both\nD: Neither\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBlue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nAre satellite images better at estimating the change in population or the total population?\n\nA: Change in population\nB: Total population.\nC: Both\nD: Neither\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-36.jpg?height=948&width=1673&top_left_y=657&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_215",
        "problem": "A set of experiments on the regulation of hormone secretion and effects of various drugs on the activities of endocrine glands were conducted on rats. Rats were divided to different groups and each group was injected with a hormone or a drug. Some physiological parameters were collected and analysed.\nA: The group of rats injected with the drug that reduces the sensitivity of hypothalamus to cortisol resulted in higher plasma levels of both glucose and insulin than those in the group of rats injected with the drug that reduces the sensitivity of adrenocorticotropic hormone (ACTH) receptors.\nB: The group of rats injected with the drug that increases the sensitivity of hypothalamus to thyroxine resulted in higher metabolic rate and body temperature than those in the group of rats injected with the drug that increases the sensitivity of target cells to thyroid releasing hormone (TRH).\nC: The group of rats injected with propylthiouracil (which blocks thyroid hormone synthesis) resulted in smaller thyroid gland and body weight than those in the group of rats injected with placebo.\nD: The group of rats injected with thyroid stimulating hormone (TSH) had smaller pituitary gland and bigger adrenal glands compared with the group of rats injected with corticotrophin-releasing hormone (CRH).\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA set of experiments on the regulation of hormone secretion and effects of various drugs on the activities of endocrine glands were conducted on rats. Rats were divided to different groups and each group was injected with a hormone or a drug. Some physiological parameters were collected and analysed.\n\nA: The group of rats injected with the drug that reduces the sensitivity of hypothalamus to cortisol resulted in higher plasma levels of both glucose and insulin than those in the group of rats injected with the drug that reduces the sensitivity of adrenocorticotropic hormone (ACTH) receptors.\nB: The group of rats injected with the drug that increases the sensitivity of hypothalamus to thyroxine resulted in higher metabolic rate and body temperature than those in the group of rats injected with the drug that increases the sensitivity of target cells to thyroid releasing hormone (TRH).\nC: The group of rats injected with propylthiouracil (which blocks thyroid hormone synthesis) resulted in smaller thyroid gland and body weight than those in the group of rats injected with placebo.\nD: The group of rats injected with thyroid stimulating hormone (TSH) had smaller pituitary gland and bigger adrenal glands compared with the group of rats injected with corticotrophin-releasing hormone (CRH).\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_194",
        "problem": "The following table describes the rate of blood flow to different parts of the body including cardiac muscle, brain, skin, and intestines at rest and during strenuous exercise.\n\n| Part of the body | Rate of blood flow/cm $\\mathbf{3}^{3} / \\mathbf{m i n}$ |  |\n| :---: | :--- | :--- |\n|  | At rest | During exercise |\n| I | 250 | 1200 |\n| II | 500 | 500 |\n| III | 500 | 1000 |\n| IV | 2500 | 90 |\nA: At rest, ATP of the cells of part I comes mainly from oxidation of fatty acid.\nB: The activity of insulin receptors in the cells of part II is increased during exercise, enhancing glucose uptake.\nC: The increase in blood flow to part III during exercise helps to regulate the body temperature.\nD: Epinephrine decreases blood flow to part IV via $\\beta$-receptor.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe following table describes the rate of blood flow to different parts of the body including cardiac muscle, brain, skin, and intestines at rest and during strenuous exercise.\n\n| Part of the body | Rate of blood flow/cm $\\mathbf{3}^{3} / \\mathbf{m i n}$ |  |\n| :---: | :--- | :--- |\n|  | At rest | During exercise |\n| I | 250 | 1200 |\n| II | 500 | 500 |\n| III | 500 | 1000 |\n| IV | 2500 | 90 |\n\nA: At rest, ATP of the cells of part I comes mainly from oxidation of fatty acid.\nB: The activity of insulin receptors in the cells of part II is increased during exercise, enhancing glucose uptake.\nC: The increase in blood flow to part III during exercise helps to regulate the body temperature.\nD: Epinephrine decreases blood flow to part IV via $\\beta$-receptor.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_69",
        "problem": "The figure shows muscle fibers, muscle spindle, and their nerve innervations of biceps of human arm.\n\n[figure1]\n\n$a$ : afferent nerves innervating muscle fibers of the spindle\n\n$b$ : efferent nerve innervating muscle fibers outside of spindle\n\n$c$ : efferent nerve innervating muscle fibers of the spindle\n\n$d$ : muscle spindle\n\n$e$ : nerve endings of $(a)$\n\n$f$ : muscle fibers outside of spindle\n\nNerve (a) is sensitive to the stretch of muscle fibers outside of the spindle when muscle fibers within the spindle are relaxed. Choose a case when the afferent signals in nerve (a) increase?\nA: Signals in (b) are increased.\nB: Signals in (c) are decreased.\nC: Triceps are contracted.\nD: $(f)$ are contracted.\nE: The lenght of $(d)$ remained constant.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe figure shows muscle fibers, muscle spindle, and their nerve innervations of biceps of human arm.\n\n[figure1]\n\n$a$ : afferent nerves innervating muscle fibers of the spindle\n\n$b$ : efferent nerve innervating muscle fibers outside of spindle\n\n$c$ : efferent nerve innervating muscle fibers of the spindle\n\n$d$ : muscle spindle\n\n$e$ : nerve endings of $(a)$\n\n$f$ : muscle fibers outside of spindle\n\nNerve (a) is sensitive to the stretch of muscle fibers outside of the spindle when muscle fibers within the spindle are relaxed. Choose a case when the afferent signals in nerve (a) increase?\n\nA: Signals in (b) are increased.\nB: Signals in (c) are decreased.\nC: Triceps are contracted.\nD: $(f)$ are contracted.\nE: The lenght of $(d)$ remained constant.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-24.jpg?height=622&width=774&top_left_y=474&top_left_x=707"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_842",
        "problem": "玉米是二倍体植物, 玉米果穗上的每一个籽粒都是受精后发育而来。科学家发现了单基因突变（BB 突变成 $\\mathrm{Bb}$ ）的甲品系玉米，其自交后的果穗上约 $1 / 4$ 籽粒严重干瘜且无发芽能力。研究人员采用相关技术克隆了 $\\mathrm{B}$ 基因, 并将 $\\mathrm{B}$ 基因导入到甲品系中, 获得了转入单个 $\\mathrm{B}$ 基因的转基因玉米。假定转入的 $\\mathrm{B}$ 基因已插入 $\\mathrm{b}$ 基因所在染色体的非同源染色体上, 且能正常表达, 让转基因玉米自交得到的子代表型及比例是（ ）\nA: 正常籽粒:干疮籽粒 $\\approx 15: 1$\nB: 正常籽粒:干瘜籽粒 $\\approx 3: 1$\nC: 正常籽粒:干瘜籽粒 $\\approx 7: 1$\nD: 正常籽粒:干瘜籽粒 $\\approx 9: 7$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n玉米是二倍体植物, 玉米果穗上的每一个籽粒都是受精后发育而来。科学家发现了单基因突变（BB 突变成 $\\mathrm{Bb}$ ）的甲品系玉米，其自交后的果穗上约 $1 / 4$ 籽粒严重干瘜且无发芽能力。研究人员采用相关技术克隆了 $\\mathrm{B}$ 基因, 并将 $\\mathrm{B}$ 基因导入到甲品系中, 获得了转入单个 $\\mathrm{B}$ 基因的转基因玉米。假定转入的 $\\mathrm{B}$ 基因已插入 $\\mathrm{b}$ 基因所在染色体的非同源染色体上, 且能正常表达, 让转基因玉米自交得到的子代表型及比例是（ ）\n\nA: 正常籽粒:干疮籽粒 $\\approx 15: 1$\nB: 正常籽粒:干瘜籽粒 $\\approx 3: 1$\nC: 正常籽粒:干瘜籽粒 $\\approx 7: 1$\nD: 正常籽粒:干瘜籽粒 $\\approx 9: 7$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1120",
        "problem": "There are several types of enzyme catalyzed reactions. In one type of enzyme catalyzed reaction, in addition to the catalytic site to which the substrate $(X)$ binds, the enzyme also has a site to which some other substance $(Y)$ can bind. When $Y$ binds to such an enzyme, the enzyme can still bind to the substrate but cannot convert it to the product. Which of the following will occur in such a case?\n\ni. The affinity of the enzyme for the substrate will reduce.\n\nii. Vmax of the reaction will decrease.\n\niii. Y will alter the conformation of $X$.\n\niv. The conformation of the catalytic site will be altered by binding of $Y$.\n\nv. The effect of $Y$ can be overcome by increasing the concentration of $X$.\nA: Only i, iii and v\nB: Only ii, iv and $v$\nC: Only i, ii and v\nD: Only ii and iv\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThere are several types of enzyme catalyzed reactions. In one type of enzyme catalyzed reaction, in addition to the catalytic site to which the substrate $(X)$ binds, the enzyme also has a site to which some other substance $(Y)$ can bind. When $Y$ binds to such an enzyme, the enzyme can still bind to the substrate but cannot convert it to the product. Which of the following will occur in such a case?\n\ni. The affinity of the enzyme for the substrate will reduce.\n\nii. Vmax of the reaction will decrease.\n\niii. Y will alter the conformation of $X$.\n\niv. The conformation of the catalytic site will be altered by binding of $Y$.\n\nv. The effect of $Y$ can be overcome by increasing the concentration of $X$.\n\nA: Only i, iii and v\nB: Only ii, iv and $v$\nC: Only i, ii and v\nD: Only ii and iv\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_502",
        "problem": "图 1 表示某二倍体小鼠细胞正常分裂过程中某物质数量变化曲线的一部分。研究发现, 细胞中染色体的正确排列、分离与粘连蛋白有关, 粘连蛋白的水解是着丝粒分裂的原因，如图 2 所示。下列叙述正确的是（）\n\n[图1]\nA: 水解粘连蛋白的酶发挥作用的同时会发生核膜、核仁的消失\nB: 若图 1 纵坐标表示同源染色体对数, 则该曲线可能表示减数分裂\nC: 若图 1 纵坐标表示染色体数量, 则曲线 BD 段可能不会发生等位基因的分离\nD: 若图 1 纵坐标表示染色体组数, 则曲线 $\\mathrm{CD}$ 段的染色体数等于 $\\mathrm{AB}$ 段的染色单体数\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 1 表示某二倍体小鼠细胞正常分裂过程中某物质数量变化曲线的一部分。研究发现, 细胞中染色体的正确排列、分离与粘连蛋白有关, 粘连蛋白的水解是着丝粒分裂的原因，如图 2 所示。下列叙述正确的是（）\n\n[图1]\n\nA: 水解粘连蛋白的酶发挥作用的同时会发生核膜、核仁的消失\nB: 若图 1 纵坐标表示同源染色体对数, 则该曲线可能表示减数分裂\nC: 若图 1 纵坐标表示染色体数量, 则曲线 BD 段可能不会发生等位基因的分离\nD: 若图 1 纵坐标表示染色体组数, 则曲线 $\\mathrm{CD}$ 段的染色体数等于 $\\mathrm{AB}$ 段的染色单体数\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-26.jpg?height=360&width=946&top_left_y=814&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_309",
        "problem": "某科研人员在某地区野生型果蝇（正常眼色）种群中发现两只突变型褐眼雌果蝇,分别记为果蝇 $\\mathrm{A}$ 和果蝇 $\\mathrm{B}$ 。为研究果蝇 $\\mathrm{A}$ 和果蝇 $\\mathrm{B}$ 的突变是否为同一突变类型, 进行了如下实验（突变基因均能独立控制褐色素的合成而表现褐眼）。据此分析正确的是\n\n| 组别 | 亲本 | 子代表现型及比例 |\n| :---: | :---: | :---: |\n| 实验一 | $\\mathrm{A} \\times$ 纯合正常雄果蝇 | 40 正常 (甲): 38 褐眼 (q): 42 正常 (○) |\n| 实验二 | B×纯合正常雄果蝇 | 62 正常 (中): 62 褐眼 (中): 65 正常 (へ): 63 褐 <br> 眼 (の) |\n| 实验三 | 实验二的子代褐眼 <br> 雌、雄果蝇互相交配 | 25 正常 (甲): 49 褐眼 (甲): 23 正常 (へ): 47 褐 <br> 眼 (の) |\nA: 果蝇 A 发生了隐性突变, 突变基因位于 X 染色体上\nB: 果蝇 A 发生了显性突变, 突变基因位于常染色体上\nC: 果蝇 B 发生了显性突变, 突变基因位于 X 染色体上\nD: 让果蝇 $\\mathrm{A}$ 与实验二中 $\\mathrm{F}_{1}$ 代褐眼雄果蝇杂交, 其后代出现褐眼果蝇的概率是 $2 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某科研人员在某地区野生型果蝇（正常眼色）种群中发现两只突变型褐眼雌果蝇,分别记为果蝇 $\\mathrm{A}$ 和果蝇 $\\mathrm{B}$ 。为研究果蝇 $\\mathrm{A}$ 和果蝇 $\\mathrm{B}$ 的突变是否为同一突变类型, 进行了如下实验（突变基因均能独立控制褐色素的合成而表现褐眼）。据此分析正确的是\n\n| 组别 | 亲本 | 子代表现型及比例 |\n| :---: | :---: | :---: |\n| 实验一 | $\\mathrm{A} \\times$ 纯合正常雄果蝇 | 40 正常 (甲): 38 褐眼 (q): 42 正常 (○) |\n| 实验二 | B×纯合正常雄果蝇 | 62 正常 (中): 62 褐眼 (中): 65 正常 (へ): 63 褐 <br> 眼 (の) |\n| 实验三 | 实验二的子代褐眼 <br> 雌、雄果蝇互相交配 | 25 正常 (甲): 49 褐眼 (甲): 23 正常 (へ): 47 褐 <br> 眼 (の) |\n\nA: 果蝇 A 发生了隐性突变, 突变基因位于 X 染色体上\nB: 果蝇 A 发生了显性突变, 突变基因位于常染色体上\nC: 果蝇 B 发生了显性突变, 突变基因位于 X 染色体上\nD: 让果蝇 $\\mathrm{A}$ 与实验二中 $\\mathrm{F}_{1}$ 代褐眼雄果蝇杂交, 其后代出现褐眼果蝇的概率是 $2 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_538",
        "problem": "已知某种细胞有 4 条染色体, 且两对等位基因分别位于两对同源染色体上。某同学用示意图表示这种细胞在正常减数分裂过程中可能产生的细胞。其中表示错误的是（）\nA: [图1]\nB: [图2]\nC: [图3]\nD: [图4]\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知某种细胞有 4 条染色体, 且两对等位基因分别位于两对同源染色体上。某同学用示意图表示这种细胞在正常减数分裂过程中可能产生的细胞。其中表示错误的是（）\n\nA: [图1]\nB: [图2]\nC: [图3]\nD: [图4]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-14.jpg?height=360&width=208&top_left_y=366&top_left_x=276",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-14.jpg?height=352&width=186&top_left_y=370&top_left_x=615",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-14.jpg?height=344&width=374&top_left_y=374&top_left_x=884",
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-14.jpg?height=336&width=370&top_left_y=381&top_left_x=1300"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_370",
        "problem": "在群体遗传学中, 赖特把小的群体中不同基因型个体生育的子代数有所变动而引起基因频率随机波动的现象称为遗传漂变。下图表示种群个体数 (N) 分别是 $25 、 250 、 2500$的 A 基因频率的变迁。下列有关叙述正确的是（ ）\n\n[图1]\nA: 一般来说, 种群越小遗传漂变就越显著, 遗传漂变产生新的可遗传变异, 从而引起生物进化\nB: 基因突变、基因重组、遗传漂变、迁移都会影响图中种群的 A 基因频率\nC: 自然选择是引起遗传漂变的主要原因, 且遗传漂变对种群基因频率的影响具有随机性\nD: 若群体随机交配, 第 125 代时 $\\mathrm{N}$ 为 250 的群体中 $\\mathrm{Aa}$ 基因型频率比 $\\mathrm{N}$ 为 2500 的群体的小\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在群体遗传学中, 赖特把小的群体中不同基因型个体生育的子代数有所变动而引起基因频率随机波动的现象称为遗传漂变。下图表示种群个体数 (N) 分别是 $25 、 250 、 2500$的 A 基因频率的变迁。下列有关叙述正确的是（ ）\n\n[图1]\n\nA: 一般来说, 种群越小遗传漂变就越显著, 遗传漂变产生新的可遗传变异, 从而引起生物进化\nB: 基因突变、基因重组、遗传漂变、迁移都会影响图中种群的 A 基因频率\nC: 自然选择是引起遗传漂变的主要原因, 且遗传漂变对种群基因频率的影响具有随机性\nD: 若群体随机交配, 第 125 代时 $\\mathrm{N}$ 为 250 的群体中 $\\mathrm{Aa}$ 基因型频率比 $\\mathrm{N}$ 为 2500 的群体的小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-81.jpg?height=362&width=825&top_left_y=156&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_784",
        "problem": "人的发形有直发、波浪发和卷发三种，其遗传受一对等位基因 A/a 控制，且基因型为 $\\mathrm{Aa}$ 个体为波浪发; 秃发和非秃发受非同源染色体上的另一对等位基因 $\\mathrm{B} / \\mathrm{b}$ 控制, 且男性中基因型为 $\\mathrm{BB}$ 个体表现为非秃发，基因型为 $\\mathrm{Bb}$ 和 $\\mathrm{bb}$ 个体均表现为秃发，女性中基因型为 $\\mathrm{BB}$ 和 $\\mathrm{Bb}$ 个体均表现为非秃发， $\\mathrm{bb}$ 个体表现为秃发。现有多对基因型波浪发男女，其随机婚配后所生子女中发形一定会出现（假设子女数量足够多）（）\nA: 男性有 6 种表现型\nB: 女性有 3 种表现型\nC: 男性有 $1 / 4$ 纯合子\nD: 女性有 $1 / 2$ 秃发\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人的发形有直发、波浪发和卷发三种，其遗传受一对等位基因 A/a 控制，且基因型为 $\\mathrm{Aa}$ 个体为波浪发; 秃发和非秃发受非同源染色体上的另一对等位基因 $\\mathrm{B} / \\mathrm{b}$ 控制, 且男性中基因型为 $\\mathrm{BB}$ 个体表现为非秃发，基因型为 $\\mathrm{Bb}$ 和 $\\mathrm{bb}$ 个体均表现为秃发，女性中基因型为 $\\mathrm{BB}$ 和 $\\mathrm{Bb}$ 个体均表现为非秃发， $\\mathrm{bb}$ 个体表现为秃发。现有多对基因型波浪发男女，其随机婚配后所生子女中发形一定会出现（假设子女数量足够多）（）\n\nA: 男性有 6 种表现型\nB: 女性有 3 种表现型\nC: 男性有 $1 / 4$ 纯合子\nD: 女性有 $1 / 2$ 秃发\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_267",
        "problem": "In the African clawed frog, Xenopus laevis, the mode of cell division shifts from cleavage to somatic cell division, which has interphase, at the 12th cleavage after fertilization. This is called the mid-blastula transition (MBT).\n\nMicroinjection of mRNA of genes that are required for nuclear membrane formation at one-cell stage results in the increase of the nuclear size, but cell size does not change compared with a control embryo. In this experiment, MBT occurs earlier than the 12th cleavage (Figure 1, left). Conversely, when the nuclear size is artificially reduced, the cell size does not again change but MBT occurs later than the 12th cleavage (Figure 1, right). Note: These treatments do not alter the time required for each cleavage.\n\n[figure1]\n\nFigure 1\nA: This experiment indicates that that MBT occurs when the volume ratio of nucleus/cytoplasm is high.\nB: When MBT occurs before the 12th cleavage stage, the duration from the fertilization to the 12th cleavage stage is reduced.\nC: The timing of MBT depends on the number of divisions after fertilization.\nD: These results indicate that MBT occurs when the amount of histone per nucleus is greater than a certain value (Note: No manipulations performed in this experiment affect amount of histone).\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the African clawed frog, Xenopus laevis, the mode of cell division shifts from cleavage to somatic cell division, which has interphase, at the 12th cleavage after fertilization. This is called the mid-blastula transition (MBT).\n\nMicroinjection of mRNA of genes that are required for nuclear membrane formation at one-cell stage results in the increase of the nuclear size, but cell size does not change compared with a control embryo. In this experiment, MBT occurs earlier than the 12th cleavage (Figure 1, left). Conversely, when the nuclear size is artificially reduced, the cell size does not again change but MBT occurs later than the 12th cleavage (Figure 1, right). Note: These treatments do not alter the time required for each cleavage.\n\n[figure1]\n\nFigure 1\n\nA: This experiment indicates that that MBT occurs when the volume ratio of nucleus/cytoplasm is high.\nB: When MBT occurs before the 12th cleavage stage, the duration from the fertilization to the 12th cleavage stage is reduced.\nC: The timing of MBT depends on the number of divisions after fertilization.\nD: These results indicate that MBT occurs when the amount of histone per nucleus is greater than a certain value (Note: No manipulations performed in this experiment affect amount of histone).\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-34.jpg?height=651&width=1082&top_left_y=1062&top_left_x=401"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_739",
        "problem": "自然界中，同源染色体上不同对基因间的重组情况在雌雄个体间的表现可能不同。家虫中, 血液黄色 $(\\mathrm{Y})$ 对白色 $(\\mathrm{y})$ 为显性, 幼蚕皮斑普通斑 $(\\mathrm{P})$ 对素白斑 $(\\mathrm{p})$ 为显性。让纯合普通斑黄血蚕与素白斑白血蚕杂交得 $\\mathrm{F}_{1}$, 并进行以下杂交实验:\n\n杂交 1: $F_{1}\\left(\\right.$ O ) ×素白斑白血 $\\left(O^{\\lambda}\\right) \\rightarrow F_{2}$ 普通斑黄血: 素白斑白血 $=1: 1$;\n\n杂交 2: $F_{1}\\left(O^{\\prime}\\right) \\times$ 素白斑白血（） $\\rightarrow F_{2}$ 普通斑黄血: 普通斑白血: 素白斑黄血: 素白斑白血=3:1: 1: 3。\n以上的杂交子代中每种表现型的雌雄数量相当。下列叙述正确的是\nA: 杂交 1 与杂交 2 实质上互为正反交实验, 而它们不都是测交实验\nB: 因为杂交 1 与杂交 2 结果不同, 所以 $\\mathrm{Y} / \\mathrm{y}$ 和 $\\mathrm{P} / \\mathrm{p}$ 两对基因中至少有一对位于性染色体上\nC: 杂交 1 说明 $\\mathrm{Y} / \\mathrm{y}$ 和 $\\mathrm{P} / \\mathrm{p}$ 两对基因没有重组, 而杂交 2 说明 $\\mathrm{Y} / \\mathrm{y}$ 和 $\\mathrm{P} / \\mathrm{p}$ 符合自由组合关系\nD: 若选取 $F_{1}(甲) \\times F_{1}(O)$ 进行杂交, 可预测子代中普通斑黄血所占的比例为 11/16\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n自然界中，同源染色体上不同对基因间的重组情况在雌雄个体间的表现可能不同。家虫中, 血液黄色 $(\\mathrm{Y})$ 对白色 $(\\mathrm{y})$ 为显性, 幼蚕皮斑普通斑 $(\\mathrm{P})$ 对素白斑 $(\\mathrm{p})$ 为显性。让纯合普通斑黄血蚕与素白斑白血蚕杂交得 $\\mathrm{F}_{1}$, 并进行以下杂交实验:\n\n杂交 1: $F_{1}\\left(\\right.$ O ) ×素白斑白血 $\\left(O^{\\lambda}\\right) \\rightarrow F_{2}$ 普通斑黄血: 素白斑白血 $=1: 1$;\n\n杂交 2: $F_{1}\\left(O^{\\prime}\\right) \\times$ 素白斑白血（） $\\rightarrow F_{2}$ 普通斑黄血: 普通斑白血: 素白斑黄血: 素白斑白血=3:1: 1: 3。\n以上的杂交子代中每种表现型的雌雄数量相当。下列叙述正确的是\n\nA: 杂交 1 与杂交 2 实质上互为正反交实验, 而它们不都是测交实验\nB: 因为杂交 1 与杂交 2 结果不同, 所以 $\\mathrm{Y} / \\mathrm{y}$ 和 $\\mathrm{P} / \\mathrm{p}$ 两对基因中至少有一对位于性染色体上\nC: 杂交 1 说明 $\\mathrm{Y} / \\mathrm{y}$ 和 $\\mathrm{P} / \\mathrm{p}$ 两对基因没有重组, 而杂交 2 说明 $\\mathrm{Y} / \\mathrm{y}$ 和 $\\mathrm{P} / \\mathrm{p}$ 符合自由组合关系\nD: 若选取 $F_{1}(甲) \\times F_{1}(O)$ 进行杂交, 可预测子代中普通斑黄血所占的比例为 11/16\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1035",
        "problem": "Which of the following membranes is/are responsible for respiration in developing birds and reptiles?\nA: Amnion\nB: Allantois\nC: Chorion\nD: $A$ and $C$\nE: B and C\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following membranes is/are responsible for respiration in developing birds and reptiles?\n\nA: Amnion\nB: Allantois\nC: Chorion\nD: $A$ and $C$\nE: B and C\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_912",
        "problem": "有时, 两条 $\\mathrm{X}$ 染色体可融合成一条 $\\mathrm{X}$ 染色体，称为并联 $\\mathrm{X}$ (记作“ $\\mathrm{X}^{\\wedge} \\mathrm{X}$ ”), 其形成过程如图所示。一只含有并联 $\\mathrm{X}$ 的雌果蝇 $\\left(X^{\\wedge} X Y\\right)$ 和一只正常雄果蝇杂交, 子代的基因型与亲代完全相同, 子代连续交配也是如此, 因而称为并联 $\\mathrm{X}$ 保持系。下列叙述正确的是 ( )\n\n[图1]\n\n并联X\nA: 形成 $\\mathrm{X}^{\\wedge} \\mathrm{X}$ 的过程中发生了染色体变异\nB: 染色体组成为 $X^{\\wedge} X X 、 Y Y$ 的果蝇胚胎致死\nC: 在并联 X 保持系中, 亲本雄果蝇的 X 染色体传向子代雌果蝇\nD: 利用该保持系，可“监控”和“记录”雄果蝇 X 染色体上的新发突变\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n有时, 两条 $\\mathrm{X}$ 染色体可融合成一条 $\\mathrm{X}$ 染色体，称为并联 $\\mathrm{X}$ (记作“ $\\mathrm{X}^{\\wedge} \\mathrm{X}$ ”), 其形成过程如图所示。一只含有并联 $\\mathrm{X}$ 的雌果蝇 $\\left(X^{\\wedge} X Y\\right)$ 和一只正常雄果蝇杂交, 子代的基因型与亲代完全相同, 子代连续交配也是如此, 因而称为并联 $\\mathrm{X}$ 保持系。下列叙述正确的是 ( )\n\n[图1]\n\n并联X\n\nA: 形成 $\\mathrm{X}^{\\wedge} \\mathrm{X}$ 的过程中发生了染色体变异\nB: 染色体组成为 $X^{\\wedge} X X 、 Y Y$ 的果蝇胚胎致死\nC: 在并联 X 保持系中, 亲本雄果蝇的 X 染色体传向子代雌果蝇\nD: 利用该保持系，可“监控”和“记录”雄果蝇 X 染色体上的新发突变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-097.jpg?height=286&width=414&top_left_y=1796&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_171",
        "problem": "Plants conduct sugars and other substances throughout their body by means of phloem. Phloem is a compound tissue, including different cell types (e.g. sieve-tubes elements (in angiosperms), companion cells and parenchymal cells.). In a study, DNA genomic samples prepared from sieve-tubes and companion cells of an angiosperm and investigated for DNA corresponding to X. Gene-specific primers were used to amplify gene $\\mathrm{X}$ among whole genome of cells by PCR.\n\n[figure1]\n\nNote: Each band represents a unique DNA molecule, asterisks denotes bands corresponding to functional genes.\nA: Results of sieve-tube element genome analysis suggest that the plant is heterozygote in the gene $\\mathrm{X}$.\nB: The results obtained for the sieve-tube cells can be explained by gene $\\mathrm{X}$ being present both in the mitochondrial and chloroplast DNA.\nC: Band A most likely corresponds to an organelle-related DNA sequences in the nuclear genome.\nD: Band B most likely corresponds to a pseudogene in nuclear genome.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPlants conduct sugars and other substances throughout their body by means of phloem. Phloem is a compound tissue, including different cell types (e.g. sieve-tubes elements (in angiosperms), companion cells and parenchymal cells.). In a study, DNA genomic samples prepared from sieve-tubes and companion cells of an angiosperm and investigated for DNA corresponding to X. Gene-specific primers were used to amplify gene $\\mathrm{X}$ among whole genome of cells by PCR.\n\n[figure1]\n\nNote: Each band represents a unique DNA molecule, asterisks denotes bands corresponding to functional genes.\n\nA: Results of sieve-tube element genome analysis suggest that the plant is heterozygote in the gene $\\mathrm{X}$.\nB: The results obtained for the sieve-tube cells can be explained by gene $\\mathrm{X}$ being present both in the mitochondrial and chloroplast DNA.\nC: Band A most likely corresponds to an organelle-related DNA sequences in the nuclear genome.\nD: Band B most likely corresponds to a pseudogene in nuclear genome.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-50.jpg?height=737&width=734&top_left_y=634&top_left_x=661"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_760",
        "problem": "假设在特定环境中, 某种动物基因型为 $\\mathrm{BB}$ 和 $\\mathrm{Bb}$ 的受精卵均可发育成个体，基因型为 $\\mathrm{bb}$ 的受精卵全部死亡。现有基因型均为 $\\mathrm{Bb}$ 的该动物 1000 对 (每对含有 1 个父本和 1 个母本)，在这种环境中，若每对亲本只形成一个受精卵，则理论上该群体的子一代中 $\\mathrm{BB} 、 \\mathrm{Bb} 、 \\mathrm{bb}$ 个体的数目依次为 $(\\quad)$\nA: $250 、 500 、 0$\nB: $250 、 500 、 250$\nC: $500 、 250 、 0$\nD: $750 、 250 、 0$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n假设在特定环境中, 某种动物基因型为 $\\mathrm{BB}$ 和 $\\mathrm{Bb}$ 的受精卵均可发育成个体，基因型为 $\\mathrm{bb}$ 的受精卵全部死亡。现有基因型均为 $\\mathrm{Bb}$ 的该动物 1000 对 (每对含有 1 个父本和 1 个母本)，在这种环境中，若每对亲本只形成一个受精卵，则理论上该群体的子一代中 $\\mathrm{BB} 、 \\mathrm{Bb} 、 \\mathrm{bb}$ 个体的数目依次为 $(\\quad)$\n\nA: $250 、 500 、 0$\nB: $250 、 500 、 250$\nC: $500 、 250 、 0$\nD: $750 、 250 、 0$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1517",
        "problem": "moebas are found in many different environments. An amoeba found living in a lake formed by a melting glacier was placed in a bucket of water taken from a river estuary. An amoeba taken from the ocean was also placed in the bucket.\n\n[figure1]\n\n\nWhat happened to the amoeba taken from the glacier?\nA: It expanded.\nB: It shrank.\nC: It stayed the same.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nmoebas are found in many different environments. An amoeba found living in a lake formed by a melting glacier was placed in a bucket of water taken from a river estuary. An amoeba taken from the ocean was also placed in the bucket.\n\n[figure1]\n\n\nWhat happened to the amoeba taken from the glacier?\n\nA: It expanded.\nB: It shrank.\nC: It stayed the same.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-03.jpg?height=751&width=1099&top_left_y=752&top_left_x=244"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_935",
        "problem": "$\\mathrm{S}$ 品系家虫是人工构建的家蚕品系, 构建过程如下图。图中 $\\mathrm{a} 、 \\mathrm{~b}$ 为隐性致死基因,呈连锁状态的基因间无互换。Os 基因控制油蚕, 蚕体透明, 可用于家蚕的笕选。 $\\mathrm{S}$ 品系家蚕可作种蚕和普通蚕 (非油蚕) 杂交, 用于家蚕的制种。通常, 雄蚕食桑叶少且蚕丝质量好, 因而蚕农喜欢只养雄蚕。下列说法正确的是（）\n\n[图1]\nA:  构建 $\\mathrm{S}$ 品系家蚕过程中只发生了基因突变\nB: S 品系家蚕的基因型有 3 种\nC: $\\mathrm{S}$ 品系雄蚕作种蚕时, 子代全为雄蚕\nD: $\\mathrm{S}$ 品系雌蚕作种蚕时, 子代应去除油蝅\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{S}$ 品系家虫是人工构建的家蚕品系, 构建过程如下图。图中 $\\mathrm{a} 、 \\mathrm{~b}$ 为隐性致死基因,呈连锁状态的基因间无互换。Os 基因控制油蚕, 蚕体透明, 可用于家蚕的笕选。 $\\mathrm{S}$ 品系家蚕可作种蚕和普通蚕 (非油蚕) 杂交, 用于家蚕的制种。通常, 雄蚕食桑叶少且蚕丝质量好, 因而蚕农喜欢只养雄蚕。下列说法正确的是（）\n\n[图1]\n\nA:  构建 $\\mathrm{S}$ 品系家蚕过程中只发生了基因突变\nB: S 品系家蚕的基因型有 3 种\nC: $\\mathrm{S}$ 品系雄蚕作种蚕时, 子代全为雄蚕\nD: $\\mathrm{S}$ 品系雌蚕作种蚕时, 子代应去除油蝅\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-079.jpg?height=366&width=1174&top_left_y=485&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1184",
        "problem": "The gastrocnemius (calf) muscle of a frog and the sciatic nerve supplying it can be kept alive in a special salt solution (Ringer's solution) and stimulated to contract by a single electric shock delivered to the nerve supplying the muscle. The response is called a twitch, and can be recorded by attaching the tendon of the muscle to a lever attached to a pen that makes a mark on a revolving drum (called a kymograph). The recording below is of a single twitch after stimulation of the nerve supplying the muscle.\n\n[figure1]\n\nThe distance (d) is NOT affected by the\nA: Velocity of the nerve impulse.\nB: Temperature of the Ringer's solution.\nC: Strength of the stimulus.\nD: Speed of rotation of the kymograph drum.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe gastrocnemius (calf) muscle of a frog and the sciatic nerve supplying it can be kept alive in a special salt solution (Ringer's solution) and stimulated to contract by a single electric shock delivered to the nerve supplying the muscle. The response is called a twitch, and can be recorded by attaching the tendon of the muscle to a lever attached to a pen that makes a mark on a revolving drum (called a kymograph). The recording below is of a single twitch after stimulation of the nerve supplying the muscle.\n\n[figure1]\n\nThe distance (d) is NOT affected by the\n\nA: Velocity of the nerve impulse.\nB: Temperature of the Ringer's solution.\nC: Strength of the stimulus.\nD: Speed of rotation of the kymograph drum.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-16.jpg?height=374&width=708&top_left_y=2326&top_left_x=114"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1124",
        "problem": "Breeding in birds is mostly of the non-cooperative type, in which the male and female rear the young ones. As they mature into adults, they leave the nest in search of a new nesting site. However, in cooperative breeding, individuals other than the genetic\nparents help raise the young. They are called \"helpers\". Very often these helpers are male offspring from the previous clutch. The average brood size of a pair with helpers has always been found to be larger than the one without helpers.\n\nFor a population currently without co-operative breeding, analyse the following situations and choose which of them will favour co-operative breeding.\nA: High rate of adult and juvenile survival.\nB: Surplus of mature individuals relative to available territory.\nC: A large difference in territory quality with some with very poor resources.\nD: Female-biased skewed sex ratio in the population.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBreeding in birds is mostly of the non-cooperative type, in which the male and female rear the young ones. As they mature into adults, they leave the nest in search of a new nesting site. However, in cooperative breeding, individuals other than the genetic\nparents help raise the young. They are called \"helpers\". Very often these helpers are male offspring from the previous clutch. The average brood size of a pair with helpers has always been found to be larger than the one without helpers.\n\nFor a population currently without co-operative breeding, analyse the following situations and choose which of them will favour co-operative breeding.\n\nA: High rate of adult and juvenile survival.\nB: Surplus of mature individuals relative to available territory.\nC: A large difference in territory quality with some with very poor resources.\nD: Female-biased skewed sex ratio in the population.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_6",
        "problem": "Brown adipose tissue (BAT) activation through $\\beta$-adrenergic signalling pathway is associated with expression of thermogenic genes and the process of thermogenesis. Glycogen synthase kinase-3 (GSK3) acts as a regulator of $\\beta$-adrenergic signalling in brown adipocytes. ISO is also a chemical stimulator of BAT activation; its effect on GSK3 is shown by the Western blot below (p-GSK3 is phosphorylated form of GSK3). SB216763 is an inhibitor of GSK3. Effects of these two agents on the expression of thermogenic genes ( $F g f 21, U c p 1, D i o 2$, and Ppargcla) are shown in the Figure below.\n\n[figure1]\nA: GSK3 acts as a negative regulator of $\\beta$-adrenergic signalling in brown adipocytes.\nB: Phosphorylation of GSK3 causes decreased expression of Fgf21.\nC: SB216763 may prevent diet-induced obesity.\nD: Use of SB216763 and ISO together cause much higher increase in the number of Ppargcla mRNA transcripts as compared to $F g f 21$ mRNA transcripts\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBrown adipose tissue (BAT) activation through $\\beta$-adrenergic signalling pathway is associated with expression of thermogenic genes and the process of thermogenesis. Glycogen synthase kinase-3 (GSK3) acts as a regulator of $\\beta$-adrenergic signalling in brown adipocytes. ISO is also a chemical stimulator of BAT activation; its effect on GSK3 is shown by the Western blot below (p-GSK3 is phosphorylated form of GSK3). SB216763 is an inhibitor of GSK3. Effects of these two agents on the expression of thermogenic genes ( $F g f 21, U c p 1, D i o 2$, and Ppargcla) are shown in the Figure below.\n\n[figure1]\n\nA: GSK3 acts as a negative regulator of $\\beta$-adrenergic signalling in brown adipocytes.\nB: Phosphorylation of GSK3 causes decreased expression of Fgf21.\nC: SB216763 may prevent diet-induced obesity.\nD: Use of SB216763 and ISO together cause much higher increase in the number of Ppargcla mRNA transcripts as compared to $F g f 21$ mRNA transcripts\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-21.jpg?height=802&width=1442&top_left_y=730&top_left_x=318"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_517",
        "problem": "在某一动物种群内基因型为 $\\mathrm{AA}$ 的个体（雌雄各半）占 20\\%, 基因型为 $\\mathrm{Aa}$ 的个体 （雌雄各半）占 $80 \\%$, 若它们随机交配繁殖一代, 经调查, 后代中基因型为 $\\mathrm{Aa}$ 的个体占 $9 / 20$, 原因可能是（）\nA: 基因型为 aa 的个体不能成活\nB: 后代中基因型为 $\\mathrm{Aa}$ 的个体迁出 50\\%\nC: 基因型为 $\\mathrm{AA}$ 的雄性个体无受精能力\nD: 含 $\\mathrm{a}$ 基因的雄配子仅有 50\\%具有受精能力\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在某一动物种群内基因型为 $\\mathrm{AA}$ 的个体（雌雄各半）占 20\\%, 基因型为 $\\mathrm{Aa}$ 的个体 （雌雄各半）占 $80 \\%$, 若它们随机交配繁殖一代, 经调查, 后代中基因型为 $\\mathrm{Aa}$ 的个体占 $9 / 20$, 原因可能是（）\n\nA: 基因型为 aa 的个体不能成活\nB: 后代中基因型为 $\\mathrm{Aa}$ 的个体迁出 50\\%\nC: 基因型为 $\\mathrm{AA}$ 的雄性个体无受精能力\nD: 含 $\\mathrm{a}$ 基因的雄配子仅有 50\\%具有受精能力\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_849",
        "problem": "雄蚕 $(\\mathrm{ZZ})$ 与雌蚕 $(\\mathrm{ZW})$ 相比具有食桑少、出丝率高和丝质优等特点。研究人员通过诱变和一系列杂交实验创建了家蚕性连锁平衡致死系。雄性的 1 对性染色体上分别带有 2 个不等位的隐性纯合致死基因 $\\mathrm{a}$ 和 $\\mathrm{b}$ ，会导致家蚕在胚胎期致死，不能成功孵化。用性连锁平衡致死系与普通家蚕杂交，过程如图(不考虑基因突变和染色体变异)。下列叙述错误的是（ ）\n[图1]\n全部是雄蟇, 鉴别到个别雌奋\n\n[图2]\nA: 普通家蚕 $\\mathrm{W}$ 染色体上不含致死基因的等位基因\nB: $\\mathrm{a}$ 和 $\\mathrm{b}$ 基因同时存在时才导致雌蚕胚胎期致死\nC: 平衡致死系产生雄配子时发生染色体互换导致11现象\nD: 创建平衡致死系时发生了染色体互换可能导致(2)现象\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n雄蚕 $(\\mathrm{ZZ})$ 与雌蚕 $(\\mathrm{ZW})$ 相比具有食桑少、出丝率高和丝质优等特点。研究人员通过诱变和一系列杂交实验创建了家蚕性连锁平衡致死系。雄性的 1 对性染色体上分别带有 2 个不等位的隐性纯合致死基因 $\\mathrm{a}$ 和 $\\mathrm{b}$ ，会导致家蚕在胚胎期致死，不能成功孵化。用性连锁平衡致死系与普通家蚕杂交，过程如图(不考虑基因突变和染色体变异)。下列叙述错误的是（ ）\n[图1]\n全部是雄蟇, 鉴别到个别雌奋\n\n[图2]\n\nA: 普通家蚕 $\\mathrm{W}$ 染色体上不含致死基因的等位基因\nB: $\\mathrm{a}$ 和 $\\mathrm{b}$ 基因同时存在时才导致雌蚕胚胎期致死\nC: 平衡致死系产生雄配子时发生染色体互换导致11现象\nD: 创建平衡致死系时发生了染色体互换可能导致(2)现象\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-40.jpg?height=504&width=810&top_left_y=2030&top_left_x=432",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-40.jpg?height=155&width=234&top_left_y=2304&top_left_x=1162"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_902",
        "problem": "对下列示意图所表示的生物学意义的描述, 准确的是\n[图1]\nA: 甲图中生物自交后产生基因型为 Aadd 个体的概率为 $1 / 4$\nB: 若只研究细胞中每对同源染色体上的一对基因, 则乙图细胞分裂完成后, 可能同时产生 2 种、 3 种或 4 种不同基因型的配子\nC: 丙图所示家系中男性患者明显多于女性患者, 该病最有可能是伴 X 隐性遗传病\nD: 丁图表示某果蝇染色体组成, 该果蝇只能产生 $\\mathrm{AX}^{\\mathrm{w}} 、 \\mathrm{aX}^{\\mathrm{w}}$ 两种基因型的配子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n对下列示意图所表示的生物学意义的描述, 准确的是\n[图1]\n\nA: 甲图中生物自交后产生基因型为 Aadd 个体的概率为 $1 / 4$\nB: 若只研究细胞中每对同源染色体上的一对基因, 则乙图细胞分裂完成后, 可能同时产生 2 种、 3 种或 4 种不同基因型的配子\nC: 丙图所示家系中男性患者明显多于女性患者, 该病最有可能是伴 X 隐性遗传病\nD: 丁图表示某果蝇染色体组成, 该果蝇只能产生 $\\mathrm{AX}^{\\mathrm{w}} 、 \\mathrm{aX}^{\\mathrm{w}}$ 两种基因型的配子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-17.jpg?height=292&width=1284&top_left_y=248&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1122",
        "problem": "Following are microscopic sections of quadriceps muscles from two athletes. The cells are stained to show the activity of myosin ATPase.\n\nMark the most appropriate description from the following.\n[figure1]\nA: $\\mathrm{P}$ is likely to contain more mitochondria than $\\mathrm{Q}$.\nB: $\\mathrm{Q}$ is likely to have large stores of glycogen than $\\mathrm{P}$.\nC: $P$ is likely to represent muscle of a high jumper and sprinter.\nD: Creatine phosphate is the major source of energy in the muscle depicted in Q but not in P.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFollowing are microscopic sections of quadriceps muscles from two athletes. The cells are stained to show the activity of myosin ATPase.\n\nMark the most appropriate description from the following.\n[figure1]\n\nA: $\\mathrm{P}$ is likely to contain more mitochondria than $\\mathrm{Q}$.\nB: $\\mathrm{Q}$ is likely to have large stores of glycogen than $\\mathrm{P}$.\nC: $P$ is likely to represent muscle of a high jumper and sprinter.\nD: Creatine phosphate is the major source of energy in the muscle depicted in Q but not in P.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-07.jpg?height=694&width=654&top_left_y=1105&top_left_x=711"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1432",
        "problem": "Rhizophores are leafless branches that arise from the stem and grow downward, producing roots at its tip when it reaches the soil. Rhizophores are commonly known as aerial roots. Scientists measured length and height of rhizophores of mangrove plant (Rhizophora mangle, Fig A). They also made cross sections of rhizophores and observed their anatomical characteristics. The rhizophore cross sections were classified in orders according to the number of arches away from the main stem. The results are shown in Fig B and Fig $\\mathrm{C}$.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\n[figure3]\n\nFig: Rhizophores of Rhizophora mangle plants.\n\nA: Rhizophore height and length measurement.\n\nB: Change in height (empty square) and in length/height proportion (full circle) in five sequential orders of rhizophores.\n\nC: Relative proportions of bark (including aerenchyma), xylem and pith along the length of individual first-order rhizophores (left), and at the base of rhizophores of sequential orders (right).\n\nWhat mechanism makes water rise from the root system to the leaves\nA: forces created in the cells of the root\nB: increased respiratory activity in root cells\nC: osmotic force in the shoot system\nD: tension in the cell sap due to transpiration\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRhizophores are leafless branches that arise from the stem and grow downward, producing roots at its tip when it reaches the soil. Rhizophores are commonly known as aerial roots. Scientists measured length and height of rhizophores of mangrove plant (Rhizophora mangle, Fig A). They also made cross sections of rhizophores and observed their anatomical characteristics. The rhizophore cross sections were classified in orders according to the number of arches away from the main stem. The results are shown in Fig B and Fig $\\mathrm{C}$.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\n[figure3]\n\nFig: Rhizophores of Rhizophora mangle plants.\n\nA: Rhizophore height and length measurement.\n\nB: Change in height (empty square) and in length/height proportion (full circle) in five sequential orders of rhizophores.\n\nC: Relative proportions of bark (including aerenchyma), xylem and pith along the length of individual first-order rhizophores (left), and at the base of rhizophores of sequential orders (right).\n\nWhat mechanism makes water rise from the root system to the leaves\n\nA: forces created in the cells of the root\nB: increased respiratory activity in root cells\nC: osmotic force in the shoot system\nD: tension in the cell sap due to transpiration\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-09.jpg?height=454&width=508&top_left_y=681&top_left_x=266",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-09.jpg?height=448&width=377&top_left_y=690&top_left_x=814",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-09.jpg?height=496&width=552&top_left_y=640&top_left_x=1226"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1351",
        "problem": "$E$ and $T$ represent alleles of two linked genes in maize. Plants of the genotype EETT were crossed with plants of the genotype eett. The resulting F1 plants were backcrossed with eett plants to produce a generation of which 36391 plants showed either both dominant or both recessive characters, while 9141 plants showed only one of the dominant characters. Which one of the following most closely represents the cross-over value between the two genes?\nA: $0.25 \\%$\nB: $5 \\%$\nC: $30 \\%$\nD: $25 \\%$\nE: $20 \\%$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n$E$ and $T$ represent alleles of two linked genes in maize. Plants of the genotype EETT were crossed with plants of the genotype eett. The resulting F1 plants were backcrossed with eett plants to produce a generation of which 36391 plants showed either both dominant or both recessive characters, while 9141 plants showed only one of the dominant characters. Which one of the following most closely represents the cross-over value between the two genes?\n\nA: $0.25 \\%$\nB: $5 \\%$\nC: $30 \\%$\nD: $25 \\%$\nE: $20 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_234",
        "problem": "Antifreeze glycoproteins (AFGPs) possess the ability to inhibit the formation of ice and are therefore essential to the survival of many marine teleost fishes that routinely encounter sub-zero temperatures. A typical AFGP consists of repeating tripeptide units, the alanyl-threonyl-alanyl (Ala-Thr-Ala) $\\mathrm{n}_{\\mathrm{n}}$ unit connected to a disaccharide through a glycosidic bond at the second hydroxyl group of the threonine residue. To identify chemical groups which affect antifreeze activities of this glycoprotein, scientists synthesized numerous AFGP analogues by modifying both the structure of the sugar moieties and the peptide by replacing three groups $R_{1}, R_{2}, R_{3}$ as shown in Fig.Q. 54 with different chemical groups and recorded the antifreeze activity.\n\n[figure1]\n\nFig.Q.54. The structure of a typical AFGP\n\nThe results of the study are shown in the following table.\n\n| $\\mathbf{R}_{\\mathbf{1}}$ | $\\mathbf{R}_{\\mathbf{2}}$ | $\\mathbf{R}_{\\mathbf{3}}$ | Antifreeze activity |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{HO}$ | $\\mathrm{CH}_{3}$ | Galactosyl | No |\n| $N$-Acetyl | $\\mathrm{CH}_{3}$ | Galactosyl | Yes |\n| $N$-Acetyl | $\\mathrm{H}$ | Galactosyl | No |\n| $N$-Acetyl | $\\mathrm{CH}_{3}$ | $\\mathrm{H}$ | Yes |\n| $O$-Acetyl | $\\mathrm{CH}_{3}$ | $\\mathrm{H}$ | No |\n| $N$-Acetyl | $\\mathrm{CH}_{3}$ | Galactosyl-Galactosyl | No |\nA: A disaccharide bound to the threonine residue is required for antifreeze activity.\nB: A mutant that has threonine residues replaced with serine residues significantly reduces antifeeze activities.\nC: $N$-acetyl group at the $\\mathrm{C}$-2 position is required for antifreeze activity.\nD: Different numbers of repetitive motifs in AFGP genes amongst closely related species might have been caused by DNA polymerase inaccuracy.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAntifreeze glycoproteins (AFGPs) possess the ability to inhibit the formation of ice and are therefore essential to the survival of many marine teleost fishes that routinely encounter sub-zero temperatures. A typical AFGP consists of repeating tripeptide units, the alanyl-threonyl-alanyl (Ala-Thr-Ala) $\\mathrm{n}_{\\mathrm{n}}$ unit connected to a disaccharide through a glycosidic bond at the second hydroxyl group of the threonine residue. To identify chemical groups which affect antifreeze activities of this glycoprotein, scientists synthesized numerous AFGP analogues by modifying both the structure of the sugar moieties and the peptide by replacing three groups $R_{1}, R_{2}, R_{3}$ as shown in Fig.Q. 54 with different chemical groups and recorded the antifreeze activity.\n\n[figure1]\n\nFig.Q.54. The structure of a typical AFGP\n\nThe results of the study are shown in the following table.\n\n| $\\mathbf{R}_{\\mathbf{1}}$ | $\\mathbf{R}_{\\mathbf{2}}$ | $\\mathbf{R}_{\\mathbf{3}}$ | Antifreeze activity |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{HO}$ | $\\mathrm{CH}_{3}$ | Galactosyl | No |\n| $N$-Acetyl | $\\mathrm{CH}_{3}$ | Galactosyl | Yes |\n| $N$-Acetyl | $\\mathrm{H}$ | Galactosyl | No |\n| $N$-Acetyl | $\\mathrm{CH}_{3}$ | $\\mathrm{H}$ | Yes |\n| $O$-Acetyl | $\\mathrm{CH}_{3}$ | $\\mathrm{H}$ | No |\n| $N$-Acetyl | $\\mathrm{CH}_{3}$ | Galactosyl-Galactosyl | No |\n\nA: A disaccharide bound to the threonine residue is required for antifreeze activity.\nB: A mutant that has threonine residues replaced with serine residues significantly reduces antifeeze activities.\nC: $N$-acetyl group at the $\\mathrm{C}$-2 position is required for antifreeze activity.\nD: Different numbers of repetitive motifs in AFGP genes amongst closely related species might have been caused by DNA polymerase inaccuracy.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-011.jpg?height=530&width=777&top_left_y=973&top_left_x=659"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_271",
        "problem": "Nakamura TM, Morin GB, Chapman KB, Weinrich SL, Andrews WH, Lingner J, Harley CB \\& Cech TR (1997) Telomerase catalytic subunit homologs from fission yeast and human. Science 277, 955-959.\n\nReoxygenation after a period of lack of oxygen causes cardiomyocyte damage. One of the most potential indices evaluating myocardial functions is mitochondrial membrane potential, which is labeled by a cell permeant dye (positively-charged, grey color) readily accumulating in active mitochondria due to their relative negative charge. The figure below illustrates hypoxia/reoxygenation (HR)-treated single myocyte model (1) with or without pre-hypoxic treatment of drug A. Myocyte images were captured at time points $(a, b, c)$.\n\n[figure1]\n\n(1)\n\n[figure2]\n\n(2)\n\n[figure3]\n\n(3) $\\simeq$ Drug A\n\n$\\Longrightarrow$ Low 02\n\n$\\varpi$ Normal $\\mathrm{O2}$\n\n$-\\mathrm{HR}$\n\n...... Drug A\n\npretreatment\n\nC\nprotreatment\n\nFig.Q.53.\nA: As seen in Fig.Q.53.(2)a, cardiomyocytes are a type of striated muscle cells.\nB: Hypoxia induces acidic $\\mathrm{pH}$ in myocardial mitochondria.\nC: Drug A pretreatment is good for cell because it prevents the collapse of mitochondrial membrane potential in HR.\nD: Captured images in drug A pretreatment group are presented in (2) and captured images in HR treatment without pretreatment of drug A are presented in (3).\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nNakamura TM, Morin GB, Chapman KB, Weinrich SL, Andrews WH, Lingner J, Harley CB \\& Cech TR (1997) Telomerase catalytic subunit homologs from fission yeast and human. Science 277, 955-959.\n\nReoxygenation after a period of lack of oxygen causes cardiomyocyte damage. One of the most potential indices evaluating myocardial functions is mitochondrial membrane potential, which is labeled by a cell permeant dye (positively-charged, grey color) readily accumulating in active mitochondria due to their relative negative charge. The figure below illustrates hypoxia/reoxygenation (HR)-treated single myocyte model (1) with or without pre-hypoxic treatment of drug A. Myocyte images were captured at time points $(a, b, c)$.\n\n[figure1]\n\n(1)\n\n[figure2]\n\n(2)\n\n[figure3]\n\n(3) $\\simeq$ Drug A\n\n$\\Longrightarrow$ Low 02\n\n$\\varpi$ Normal $\\mathrm{O2}$\n\n$-\\mathrm{HR}$\n\n...... Drug A\n\npretreatment\n\nC\nprotreatment\n\nFig.Q.53.\n\nA: As seen in Fig.Q.53.(2)a, cardiomyocytes are a type of striated muscle cells.\nB: Hypoxia induces acidic $\\mathrm{pH}$ in myocardial mitochondria.\nC: Drug A pretreatment is good for cell because it prevents the collapse of mitochondrial membrane potential in HR.\nD: Captured images in drug A pretreatment group are presented in (2) and captured images in HR treatment without pretreatment of drug A are presented in (3).\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-009.jpg?height=401&width=455&top_left_y=824&top_left_x=228",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-009.jpg?height=356&width=344&top_left_y=835&top_left_x=682",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-009.jpg?height=344&width=313&top_left_y=841&top_left_x=1016"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_184",
        "problem": "The following figure shows two histological sections of an organ from an angiosperm plant. The tissue section on the left was prepared from the medial part of the organ, whereas the section on the right was from the lateral part.\n[figure1]\nA: The sections were prepared from a monocotyledonous plant.\nB: The sections represent a modified stem.\nC: Transpiration occurs in this organ.\nD: The vascular bundles are of bicolateral type, meaning that phloem is present on both sides of xylem.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following figure shows two histological sections of an organ from an angiosperm plant. The tissue section on the left was prepared from the medial part of the organ, whereas the section on the right was from the lateral part.\n[figure1]\n\nA: The sections were prepared from a monocotyledonous plant.\nB: The sections represent a modified stem.\nC: Transpiration occurs in this organ.\nD: The vascular bundles are of bicolateral type, meaning that phloem is present on both sides of xylem.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-40.jpg?height=540&width=1470&top_left_y=552&top_left_x=293"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_251",
        "problem": "A GFP (green fluorescent protein) - tagged form of protein P was expressed in fibroblast cells. The subcellular distribution of protein $\\mathrm{P}$ can be observed using fluorescence microscopy. To determine the precise movement mechanism of protein $\\mathrm{P}$ in the cells, fluorescence recovery after photobleaching (FRAP) was performed. As shown below, protein $\\mathrm{P}$ is expressed in the nucleus (ROI 1) and in the cytoplasm (ROI 2). Protein P in the ROI 1 area was photobleached using a laser beam. Photobleaching causes an irreversible loss of flouorescence. Changes in the fluorescence intensity of protein P in ROI 1 and ROI 2 following photobleaching are shown in the graph and figures below.\n[figure1]\n\n[figure2]\n\n$0 \\mathrm{sec}$\n\n[figure3]\n\n$40 \\mathrm{sec}$\n\n[figure4]\n\n$80 \\mathrm{sec}$\n\n[figure5]\n\n$120 \\mathrm{sec}$\n\n[figure6]\n\n$160 \\mathrm{sec}$\n\nWhich of the following is the best explanation for the distribution and movement of protein P?\nA: $\\mathrm{P}$ is a nuclear membrane protein.\nB: P is imported to the nucleus through a nuclear pore.\nC: P binds to the nuclear pore complex.\nD: P is imported to the nucleus via vesicular trafficking through Golgi and ER.\nE: P is capable of moving from the nucleus to the cytoplasm.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA GFP (green fluorescent protein) - tagged form of protein P was expressed in fibroblast cells. The subcellular distribution of protein $\\mathrm{P}$ can be observed using fluorescence microscopy. To determine the precise movement mechanism of protein $\\mathrm{P}$ in the cells, fluorescence recovery after photobleaching (FRAP) was performed. As shown below, protein $\\mathrm{P}$ is expressed in the nucleus (ROI 1) and in the cytoplasm (ROI 2). Protein P in the ROI 1 area was photobleached using a laser beam. Photobleaching causes an irreversible loss of flouorescence. Changes in the fluorescence intensity of protein P in ROI 1 and ROI 2 following photobleaching are shown in the graph and figures below.\n[figure1]\n\n[figure2]\n\n$0 \\mathrm{sec}$\n\n[figure3]\n\n$40 \\mathrm{sec}$\n\n[figure4]\n\n$80 \\mathrm{sec}$\n\n[figure5]\n\n$120 \\mathrm{sec}$\n\n[figure6]\n\n$160 \\mathrm{sec}$\n\nWhich of the following is the best explanation for the distribution and movement of protein P?\n\nA: $\\mathrm{P}$ is a nuclear membrane protein.\nB: P is imported to the nucleus through a nuclear pore.\nC: P binds to the nuclear pore complex.\nD: P is imported to the nucleus via vesicular trafficking through Golgi and ER.\nE: P is capable of moving from the nucleus to the cytoplasm.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-08.jpg?height=554&width=1266&top_left_y=971&top_left_x=514",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-08.jpg?height=288&width=255&top_left_y=1558&top_left_x=495",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-08.jpg?height=279&width=251&top_left_y=1568&top_left_x=754",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-08.jpg?height=294&width=254&top_left_y=1555&top_left_x=1004",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-08.jpg?height=288&width=260&top_left_y=1558&top_left_x=1269",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-08.jpg?height=288&width=255&top_left_y=1558&top_left_x=1523"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_562",
        "problem": "科学家曾提出 DNA 复制方式的三种假说 全保留复制、半保留复制和分散复制（图甲)。针对假说, 科学家以大肠杆菌为实验材料, 进行了如下实验（图乙): 下列相关叙述正确的是（ ）\n[图1]\n\n转移到含 ${ }^{14} \\mathrm{NH}_{4} \\mathrm{Cl}$ 的培养液中\n\n[图2]\nA: 实验使用放射性同位素 ${ }^{15} \\mathrm{~N}$ 标记 DNA 后再通过离心来观察 DNA 的分布情况\nB: 第一代细菌 DNA 离心后, 试管中出现 1 条中带, 说明 DNA 复制方式一定是半保留复制\nC: 如果将第二代细菌提取出的 DNA 解旋后再离心, 三种假说所得结果均会出现 1 条轻带和 1 条重带\nD: 若复制方式是半保留复制, 继续培养至第三代, 细菌 DNA 离心后试管中会出现 1 条中带和 1 条轻带\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n科学家曾提出 DNA 复制方式的三种假说 全保留复制、半保留复制和分散复制（图甲)。针对假说, 科学家以大肠杆菌为实验材料, 进行了如下实验（图乙): 下列相关叙述正确的是（ ）\n[图1]\n\n转移到含 ${ }^{14} \\mathrm{NH}_{4} \\mathrm{Cl}$ 的培养液中\n\n[图2]\n\nA: 实验使用放射性同位素 ${ }^{15} \\mathrm{~N}$ 标记 DNA 后再通过离心来观察 DNA 的分布情况\nB: 第一代细菌 DNA 离心后, 试管中出现 1 条中带, 说明 DNA 复制方式一定是半保留复制\nC: 如果将第二代细菌提取出的 DNA 解旋后再离心, 三种假说所得结果均会出现 1 条轻带和 1 条重带\nD: 若复制方式是半保留复制, 继续培养至第三代, 细菌 DNA 离心后试管中会出现 1 条中带和 1 条轻带\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-59.jpg?height=764&width=1302&top_left_y=912&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-59.jpg?height=574&width=962&top_left_y=1683&top_left_x=678"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1093",
        "problem": "In a lung cancer cell line, it is observed that a single point mutation (cysteine to phenylalanine) makes a protein defective in phosphorylation function.\n\nThe loss of function is most likely due to:\nA: increase in solubility of the protein in aqueous environment.\nB: alteration of protein conformation essential for protein-protein interaction.\nC: increase in the net charge of the protein leading to loss of active site.\nD: decrease in hydrophobic interactions of the protein with other proteins.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn a lung cancer cell line, it is observed that a single point mutation (cysteine to phenylalanine) makes a protein defective in phosphorylation function.\n\nThe loss of function is most likely due to:\n\nA: increase in solubility of the protein in aqueous environment.\nB: alteration of protein conformation essential for protein-protein interaction.\nC: increase in the net charge of the protein leading to loss of active site.\nD: decrease in hydrophobic interactions of the protein with other proteins.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_48",
        "problem": "The following data were collected by a researcher over seven successive years from a population of a fish in a pond. He calculated population growth rate $(r)$ for each year using this formula\n\n$$\nr=\\ln N_{t+1}-\\ln N_{t}\n$$\n\nNote: $\\mathrm{N}=$ the population size, $\\mathrm{K}=$ the carrying capacity.\n\n| Year | Number of Fish | Population growth rate (r) |\n| :---: | :---: | :---: |\n| $\\mathbf{2 0 0 2}$ | 2 | 2.77 |\n| $\\mathbf{2 0 0 3}$ | 32 | 1.02 |\n| $\\mathbf{2 0 0 4}$ | 89 | -0.75 |\n| $\\mathbf{2 0 0 5}$ | 42 | 0.39 |\n| $\\mathbf{2 0 0 6}$ | 62 | 0.15 |\n| $\\mathbf{2 0 0 7}$ | 72 | -0.15 |\n| $\\mathbf{2 0 0 8}$ | 62 | - |\nA: The fish population grows exponentially as $\\mathrm{r}$ is bigger than $\\mathbf{1}$ in all years.\nB: When fecundity is less than mortality we can assume $\\mathrm{N}<\\mathrm{K}$, which means that the population is growing.\nC: The fraction of the resources that were used by the fish in each year can be calculated by $\\mathrm{N} / \\mathrm{K}$.\nD: The pond carrying capacity $(K)$ is between 62 and 72 .\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe following data were collected by a researcher over seven successive years from a population of a fish in a pond. He calculated population growth rate $(r)$ for each year using this formula\n\n$$\nr=\\ln N_{t+1}-\\ln N_{t}\n$$\n\nNote: $\\mathrm{N}=$ the population size, $\\mathrm{K}=$ the carrying capacity.\n\n| Year | Number of Fish | Population growth rate (r) |\n| :---: | :---: | :---: |\n| $\\mathbf{2 0 0 2}$ | 2 | 2.77 |\n| $\\mathbf{2 0 0 3}$ | 32 | 1.02 |\n| $\\mathbf{2 0 0 4}$ | 89 | -0.75 |\n| $\\mathbf{2 0 0 5}$ | 42 | 0.39 |\n| $\\mathbf{2 0 0 6}$ | 62 | 0.15 |\n| $\\mathbf{2 0 0 7}$ | 72 | -0.15 |\n| $\\mathbf{2 0 0 8}$ | 62 | - |\n\nA: The fish population grows exponentially as $\\mathrm{r}$ is bigger than $\\mathbf{1}$ in all years.\nB: When fecundity is less than mortality we can assume $\\mathrm{N}<\\mathrm{K}$, which means that the population is growing.\nC: The fraction of the resources that were used by the fish in each year can be calculated by $\\mathrm{N} / \\mathrm{K}$.\nD: The pond carrying capacity $(K)$ is between 62 and 72 .\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_285",
        "problem": "Eusocial honeybees have a specific system of sex determination. Females are diploid (2n) and develop from fertilized eggs; males are haploid (n) and develop from unfertilized eggs. Assuming that the queen copulated with a single male, which of the following is/are most likely true for this social group?\n\nI. The males have mothers but not fathers.\n\nII. A female should foster her brothers to increase her inclusive fitness rather than trying to increase her direct reproduction.\n\nIII. It is advantageous to females' (workers') fitness if the queen produces sons and daughters in equal proportions.\n\nIV. A female should remove the eggs of other females (workers) from the nest to increase her fitness.\nA: Only I and II\nB: Only I and III\nC: Only I and IV\nD: Only II and III\nE: Only III and IV\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEusocial honeybees have a specific system of sex determination. Females are diploid (2n) and develop from fertilized eggs; males are haploid (n) and develop from unfertilized eggs. Assuming that the queen copulated with a single male, which of the following is/are most likely true for this social group?\n\nI. The males have mothers but not fathers.\n\nII. A female should foster her brothers to increase her inclusive fitness rather than trying to increase her direct reproduction.\n\nIII. It is advantageous to females' (workers') fitness if the queen produces sons and daughters in equal proportions.\n\nIV. A female should remove the eggs of other females (workers) from the nest to increase her fitness.\n\nA: Only I and II\nB: Only I and III\nC: Only I and IV\nD: Only II and III\nE: Only III and IV\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_141",
        "problem": "Thyroid hormones are transported in blood by proteins. The TBG (thyroid hormone binding globulin) is the main thyroid hormone transporting protein. Many factors affect TBG concentration such as oestrogen and OCP (oral contraceptive pill), etc. OCP increases TBG concentration. T3RU (T3 resin uptake) assay is a method of quantification of unbound TBG in the blood. It indirectly measures the capacity of patients TBG to bind radioactive labelled T3. The less patients thyroid hormone level is the more radioactive labelled T3 will bound to TBG. Thus, since radioactive T3 is only detectable in unbound form, the T3RU assay result will be lower.\nA: In primary hypothyroidism, the thyroid function test would be like below: TSH increased T4 decreased T3RU decreased\nB: A person who uses OCP and is euthyroid (normal functioning of thyroid) the thyroid function test would be like below: TSH normal T4 decreased T3RU decreased\nC: In primary hyperthyroidism the thyroid function test would be like below: TSH increased T4 Increased T3RU increased\nD: In secondary hypothyroidism (pituitary dysfunction) the thyroid function test would be like below: TSH decreased T4 decreased T3RU decreased\nE: In tertiary hypothyroidism (hypothalamic dysfunction) the thyroid function test would be like below: TSH normal T4 decreased T3RU decreased\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThyroid hormones are transported in blood by proteins. The TBG (thyroid hormone binding globulin) is the main thyroid hormone transporting protein. Many factors affect TBG concentration such as oestrogen and OCP (oral contraceptive pill), etc. OCP increases TBG concentration. T3RU (T3 resin uptake) assay is a method of quantification of unbound TBG in the blood. It indirectly measures the capacity of patients TBG to bind radioactive labelled T3. The less patients thyroid hormone level is the more radioactive labelled T3 will bound to TBG. Thus, since radioactive T3 is only detectable in unbound form, the T3RU assay result will be lower.\n\nA: In primary hypothyroidism, the thyroid function test would be like below: TSH increased T4 decreased T3RU decreased\nB: A person who uses OCP and is euthyroid (normal functioning of thyroid) the thyroid function test would be like below: TSH normal T4 decreased T3RU decreased\nC: In primary hyperthyroidism the thyroid function test would be like below: TSH increased T4 Increased T3RU increased\nD: In secondary hypothyroidism (pituitary dysfunction) the thyroid function test would be like below: TSH decreased T4 decreased T3RU decreased\nE: In tertiary hypothyroidism (hypothalamic dysfunction) the thyroid function test would be like below: TSH normal T4 decreased T3RU decreased\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_816",
        "problem": "亚硝酸盐可使 DNA 的某些碱基脱去氨基而发生变化: C 转变为 $\\mathrm{U}$ ( $\\mathrm{U}$ 与 $\\mathrm{A}$ 配对), $A$ 转变为 $I$ ( $I$ 为次黄嘌呤, 与 $C$ 配对)。已知某双链 DNA 的两条链分别是(1)链和(2)链, (1)链的一段碱基序列为 $5^{\\prime}$-AGTCG-3', 此 DNA 片段经亚硝酸盐作用后, 其中一条链中的 $\\mathrm{A} 、 \\mathrm{C}$ 发生了脱氨基作用, 经过两轮复制后, 子代 DNA 片段之一是 -GGTTG-/-CCAAC-。下列说法错误的是（）\nA: 该 DNA 片段在未经亚硝酸盐作用时，核苷酸之间含有 8 个磷酸二酯键、 13 个氢键\nB: 经亚硝酸盐作用后, (2)链中碱基 A 和碱基 C 发生了脱氨基作用\nC: 经亚硝酸盐作用后, DNA 片段经两轮复制产生的异常 DNA 片段有 2 个\nD: 经过两轮复制共消耗游离的胞嘧啶脱氧核苷酸 10 个\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n亚硝酸盐可使 DNA 的某些碱基脱去氨基而发生变化: C 转变为 $\\mathrm{U}$ ( $\\mathrm{U}$ 与 $\\mathrm{A}$ 配对), $A$ 转变为 $I$ ( $I$ 为次黄嘌呤, 与 $C$ 配对)。已知某双链 DNA 的两条链分别是(1)链和(2)链, (1)链的一段碱基序列为 $5^{\\prime}$-AGTCG-3', 此 DNA 片段经亚硝酸盐作用后, 其中一条链中的 $\\mathrm{A} 、 \\mathrm{C}$ 发生了脱氨基作用, 经过两轮复制后, 子代 DNA 片段之一是 -GGTTG-/-CCAAC-。下列说法错误的是（）\n\nA: 该 DNA 片段在未经亚硝酸盐作用时，核苷酸之间含有 8 个磷酸二酯键、 13 个氢键\nB: 经亚硝酸盐作用后, (2)链中碱基 A 和碱基 C 发生了脱氨基作用\nC: 经亚硝酸盐作用后, DNA 片段经两轮复制产生的异常 DNA 片段有 2 个\nD: 经过两轮复制共消耗游离的胞嘧啶脱氧核苷酸 10 个\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_669",
        "problem": "某育种工作者在一次杂交实验时, 偶然发现了一个罕见现象: 选取的高茎 (AA)\n踠豆植株与矮茎 (aa) 㱧豆植株杂交, 得到的 $F_{1}$ 全为高茎; 其中有一株 $F_{1}$ 植株自交得到的 $\\mathrm{F}_{2}$ 出现了高茎: 矮茎 $=35: 1$ 的性状分离比, 分析此现象可能是由于环境骤变如降温影响，以下说法正确的是（）\nA: 这一 $F_{1}$ 植株基因型为 AAaa, 由于低温抑制了着丝粒的分裂导致染色体数目加倍\nB: 这一 $F_{1}$ 植株自交出现 35:1 的性状分离比是因为发生了基因突变\nC: 这一 $F_{1}$ 植株自交, 产生的 $F_{2}$ 基因型有 5 种, 比例为 $1: 8: 18: 8: 1$\nD: 这一 $F_{1}$ 植株产生的含有隐性基因的配子所占比例为 $1 / 6$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某育种工作者在一次杂交实验时, 偶然发现了一个罕见现象: 选取的高茎 (AA)\n踠豆植株与矮茎 (aa) 㱧豆植株杂交, 得到的 $F_{1}$ 全为高茎; 其中有一株 $F_{1}$ 植株自交得到的 $\\mathrm{F}_{2}$ 出现了高茎: 矮茎 $=35: 1$ 的性状分离比, 分析此现象可能是由于环境骤变如降温影响，以下说法正确的是（）\n\nA: 这一 $F_{1}$ 植株基因型为 AAaa, 由于低温抑制了着丝粒的分裂导致染色体数目加倍\nB: 这一 $F_{1}$ 植株自交出现 35:1 的性状分离比是因为发生了基因突变\nC: 这一 $F_{1}$ 植株自交, 产生的 $F_{2}$ 基因型有 5 种, 比例为 $1: 8: 18: 8: 1$\nD: 这一 $F_{1}$ 植株产生的含有隐性基因的配子所占比例为 $1 / 6$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_709",
        "problem": "脑源性神经营养因子（BDNF）是由两条肽链构成的蛋白质, 能够促进和维持中枢神经系统正常的生长发育。若 BDNF 基因表达受阻, 则会导致精神异常。图示为 BDNF 基因的表达及调控过程，下列有关叙述正确的是（ ）\n\n[图1]\nA: 图中 miRNA-195 基因是具有遗传效应的 RNA 片段\nB: 同一个体不同组织细胞的同一条染色体 DNA 进行甲过程时, 发挥作用的启动子不完全相同\nC: 图中 $\\mathrm{A}$ 侧为 $\\mathrm{mRNA}$ 的 5 '端, 多个核糖体结合在同一 mRNA 上, 有利于提高翻译效率\nD: 若 miRNA-195 基因的一条链中 $(\\mathrm{A}+\\mathrm{G}) /(\\mathrm{T}+\\mathrm{C})$ 为 1.25 , 则 miRNA-195 中 $(\\mathrm{U}+\\mathrm{C}) /(\\mathrm{A}+\\mathrm{G})$ 为 0.8\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n脑源性神经营养因子（BDNF）是由两条肽链构成的蛋白质, 能够促进和维持中枢神经系统正常的生长发育。若 BDNF 基因表达受阻, 则会导致精神异常。图示为 BDNF 基因的表达及调控过程，下列有关叙述正确的是（ ）\n\n[图1]\n\nA: 图中 miRNA-195 基因是具有遗传效应的 RNA 片段\nB: 同一个体不同组织细胞的同一条染色体 DNA 进行甲过程时, 发挥作用的启动子不完全相同\nC: 图中 $\\mathrm{A}$ 侧为 $\\mathrm{mRNA}$ 的 5 '端, 多个核糖体结合在同一 mRNA 上, 有利于提高翻译效率\nD: 若 miRNA-195 基因的一条链中 $(\\mathrm{A}+\\mathrm{G}) /(\\mathrm{T}+\\mathrm{C})$ 为 1.25 , 则 miRNA-195 中 $(\\mathrm{U}+\\mathrm{C}) /(\\mathrm{A}+\\mathrm{G})$ 为 0.8\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-38.jpg?height=363&width=1105&top_left_y=892&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_900",
        "problem": "眼白化病是由酪氨酸酶缺乏或功能减退引起的单基因（A/a) 遗传病, 患者眼部黑色素的生成减少。某家族眼白化病的遗传情况如图 1 所示, 对该家族部分成员的基因 $\\mathrm{A} / \\mathrm{a}$酶切后电泳的结果如图 2 所示。下列分析错误的是（ ）\n[图1]\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n眼白化病是由酪氨酸酶缺乏或功能减退引起的单基因（A/a) 遗传病, 患者眼部黑色素的生成减少。某家族眼白化病的遗传情况如图 1 所示, 对该家族部分成员的基因 $\\mathrm{A} / \\mathrm{a}$酶切后电泳的结果如图 2 所示。下列分析错误的是（ ）\n[图1]\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-39.jpg?height=777&width=1060&top_left_y=1642&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1265",
        "problem": "Which one of the following would not increase the rate of water uptake from soil to root?\nA: an increase in the concentration of soluble metabolites in root cells\nB: a decrease in the concentration of mineral ions in the soil\nC: an increase in root surface area\nD: a decrease in pressure potential of the root xylem sap\nE: a decrease in the concentration gradient of soluble metabolites across the root cortex\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following would not increase the rate of water uptake from soil to root?\n\nA: an increase in the concentration of soluble metabolites in root cells\nB: a decrease in the concentration of mineral ions in the soil\nC: an increase in root surface area\nD: a decrease in pressure potential of the root xylem sap\nE: a decrease in the concentration gradient of soluble metabolites across the root cortex\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1421",
        "problem": "What would cause increased capillary hydrostatic pressure?\nA: Hypotension\nB: Increased fluid volume\nC: Decreased fluid volume\nD: Increased diameter of the capillary\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat would cause increased capillary hydrostatic pressure?\n\nA: Hypotension\nB: Increased fluid volume\nC: Decreased fluid volume\nD: Increased diameter of the capillary\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_690",
        "problem": "“筮菌体侵染大肠杆菌”的实验示意图如图所示。已知在以下实验条件中，该噬菌体在大肠杆菌中每 20 分钟复制一代, 不考虑大肠杆菌裂解, 下列叙述正确的是（ ）\n\n[图1]\nA: 该实验证明了 DNA 的复制方式为半保留复制\nB: 离心前应充分搅拌使大肠杆菌裂解, 释放出子代噬菌体\nC: 提取 A 组试管 III 沉淀中的子代噬菌体 DNA, 仅少量 DNA 含有 ${ }^{32} \\mathrm{P}$\nD: B 组试管 III 上清液中的放射性强度与接种后的培养时间成正比\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n“筮菌体侵染大肠杆菌”的实验示意图如图所示。已知在以下实验条件中，该噬菌体在大肠杆菌中每 20 分钟复制一代, 不考虑大肠杆菌裂解, 下列叙述正确的是（ ）\n\n[图1]\n\nA: 该实验证明了 DNA 的复制方式为半保留复制\nB: 离心前应充分搅拌使大肠杆菌裂解, 释放出子代噬菌体\nC: 提取 A 组试管 III 沉淀中的子代噬菌体 DNA, 仅少量 DNA 含有 ${ }^{32} \\mathrm{P}$\nD: B 组试管 III 上清液中的放射性强度与接种后的培养时间成正比\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-65.jpg?height=311&width=1348&top_left_y=164&top_left_x=320"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1474",
        "problem": "[figure1]\n\nFigure 1. 3D visualisation of the protein backbone structure for an enzyme involved in glycolysis. Please note some of the following questions relate to the circles and rectangles positioned on this visualisation.\n\nThe reaction that this enzyme catalyses is:\n\n[figure2]\n\nThe enzyme is composed of four subunits. There are certain sites on the enzyme that can bind to other molecules, indicated by the rectangles and ovals. Binding sites at the front of the molecule are indicated by solid lines. Binding sites at the back of the molecule is indicated by dotted lines. Oval sites can bind to fructose-6-phosphate. The sites indicated with rectangles bind fructose-2,6-bisphosphate. Interestingly, ATP can bind to both sites (ovals and rectangles), but ADP can only bind to rectangular sites.\n\nATP, ADP, and fructose-6-phosphate are all negative molecules.\n\nIn addition, please note that the term allosteric relates to enzymes that change the shape of the binding site of a molecule.\n\nEnzymes catalyse biochemical reactions by altering which of the following quantities associated with the reaction?\nA: The enthalpy of formation, $\\Delta \\mathrm{H}$\nB: The equilibrium constant, Keq\nC: The change in Gibb's free energy, $\\Delta \\mathrm{G}$\nD: The activation energy, Ea\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nFigure 1. 3D visualisation of the protein backbone structure for an enzyme involved in glycolysis. Please note some of the following questions relate to the circles and rectangles positioned on this visualisation.\n\nThe reaction that this enzyme catalyses is:\n\n[figure2]\n\nThe enzyme is composed of four subunits. There are certain sites on the enzyme that can bind to other molecules, indicated by the rectangles and ovals. Binding sites at the front of the molecule are indicated by solid lines. Binding sites at the back of the molecule is indicated by dotted lines. Oval sites can bind to fructose-6-phosphate. The sites indicated with rectangles bind fructose-2,6-bisphosphate. Interestingly, ATP can bind to both sites (ovals and rectangles), but ADP can only bind to rectangular sites.\n\nATP, ADP, and fructose-6-phosphate are all negative molecules.\n\nIn addition, please note that the term allosteric relates to enzymes that change the shape of the binding site of a molecule.\n\nEnzymes catalyse biochemical reactions by altering which of the following quantities associated with the reaction?\n\nA: The enthalpy of formation, $\\Delta \\mathrm{H}$\nB: The equilibrium constant, Keq\nC: The change in Gibb's free energy, $\\Delta \\mathrm{G}$\nD: The activation energy, Ea\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-31.jpg?height=442&width=400&top_left_y=276&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-31.jpg?height=280&width=965&top_left_y=1105&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_459",
        "problem": "用 ${ }^{32} \\mathrm{P}$ 充分标记蚕豆 $(2 \\mathrm{~N}=12)$ 根尖细胞的核 $\\mathrm{DNA}$ 分子后, 将该细胞转入不含 ${ }^{32} \\mathrm{P}$的培养基中培养完成一个细胞周期，下列叙述不正确的是（ ）\nA: 分裂间期, 细胞利用未标记的脱氧核苷酸为原料进行 DNA 的复制\nB: 分裂中期, 每条染色体都被标记, 但都只有一个 DNA 分子被标记\nC: 分裂后期, 移向细胞两极的 DNA 都被标记, 标记 DNA 分子共 24 个\nD: 分裂结束, 每个细胞中的每条染色体上的 DNA 都只有一条链被标记\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n用 ${ }^{32} \\mathrm{P}$ 充分标记蚕豆 $(2 \\mathrm{~N}=12)$ 根尖细胞的核 $\\mathrm{DNA}$ 分子后, 将该细胞转入不含 ${ }^{32} \\mathrm{P}$的培养基中培养完成一个细胞周期，下列叙述不正确的是（ ）\n\nA: 分裂间期, 细胞利用未标记的脱氧核苷酸为原料进行 DNA 的复制\nB: 分裂中期, 每条染色体都被标记, 但都只有一个 DNA 分子被标记\nC: 分裂后期, 移向细胞两极的 DNA 都被标记, 标记 DNA 分子共 24 个\nD: 分裂结束, 每个细胞中的每条染色体上的 DNA 都只有一条链被标记\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_469",
        "problem": "某雄性哺乳动物的基因型为 Dd, 图甲是该动物体内不同时期的细胞中染色体数与\n\n核 DNA 分子数的关系图。若某细胞在形成细胞(7)的过程中, D 和 $\\mathrm{d}$ 所在染色体出现\n\n了如图乙所示的变化, 即当染色体的端粒断裂后, 姐妹染色单体会在断裂处发生融合,融合的染色体在细胞分裂后期由于纺锤丝的牵引, 形成染色体桥并在两个着丝粒之间随机断裂，形成的两条子染色体移到两极。下列说法错误的是（ ）\n\n[图1]\n\n核DNA分子数/个\n\n甲\n\n[图2]\nA: 图甲中细胞(3)(4)(5)肯定含有两个染色体组\nB: 图乙中染色体桥形成发生在有丝分裂\nC: 图乙中姐妹染色单体上出现等位基因只可能是基因突变导致的\nD: 图乙所对应细胞产生的其中一个子细胞的基因型有 4 种可能性\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某雄性哺乳动物的基因型为 Dd, 图甲是该动物体内不同时期的细胞中染色体数与\n\n核 DNA 分子数的关系图。若某细胞在形成细胞(7)的过程中, D 和 $\\mathrm{d}$ 所在染色体出现\n\n了如图乙所示的变化, 即当染色体的端粒断裂后, 姐妹染色单体会在断裂处发生融合,融合的染色体在细胞分裂后期由于纺锤丝的牵引, 形成染色体桥并在两个着丝粒之间随机断裂，形成的两条子染色体移到两极。下列说法错误的是（ ）\n\n[图1]\n\n核DNA分子数/个\n\n甲\n\n[图2]\n\nA: 图甲中细胞(3)(4)(5)肯定含有两个染色体组\nB: 图乙中染色体桥形成发生在有丝分裂\nC: 图乙中姐妹染色单体上出现等位基因只可能是基因突变导致的\nD: 图乙所对应细胞产生的其中一个子细胞的基因型有 4 种可能性\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-17.jpg?height=382&width=662&top_left_y=1328&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-17.jpg?height=377&width=1411&top_left_y=1862&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1322",
        "problem": "The percentage of energy transferred from one trophic level is called the ecological efficiency.\n\n[figure1]\n\nThe figure at left summarises the annual production (in million tonnes) in the North Sea (Moeller Christensen and Nystroem, 1977). The arrows show the direction of the energy flow.\n\nCalculate the ecological efficiency of the true primary consumers that are feeding solely on phytoplankton.\nA. $19.1 \\%$\nA: $19.1 \\%$\nB: $18.1 \\%$\nC: $10.4 \\%$\nD: $7.5 \\%$\nE: $8.7 \\%$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe percentage of energy transferred from one trophic level is called the ecological efficiency.\n\n[figure1]\n\nThe figure at left summarises the annual production (in million tonnes) in the North Sea (Moeller Christensen and Nystroem, 1977). The arrows show the direction of the energy flow.\n\nCalculate the ecological efficiency of the true primary consumers that are feeding solely on phytoplankton.\nA. $19.1 \\%$\n\nA: $19.1 \\%$\nB: $18.1 \\%$\nC: $10.4 \\%$\nD: $7.5 \\%$\nE: $8.7 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-09.jpg?height=1168&width=726&top_left_y=1432&top_left_x=151"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_310",
        "problem": "lacZ、lacY、lacA 是大肠杆菌体内与乳糖代谢有关的三个结构基因（其中 lacY 编码 $\\beta$-半乳糖苷透性酶，可将乳糖运入细胞）。上游的操纵基因（O）对结构基因起着“开关”的作用，直接控制结构基因的表达。图 1 表示环境中无乳糖时，结构基因的表达被关\n闭”的调节机制; 图 2 表示环境中有乳糖时, 结构基因的表达被“打开”的调节机制。下列有关叙述正确的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 乳糖可与阻遏蛋白结合并改变其构象, 导致结构基因无法正常转录\nB: $\\beta$-半乳糖苷透性酶可能位于细胞膜上, 合成后需要经过内质网和高尔基体的加工\nC: 过程(1)碱基配对方式与(2)过程不完全相同, 参与(2)过程的氨基酸可被多种 tRNA 转运\nD: lacZ、lacY、lacA 转录的模板链为 $\\beta$ 链, 转录产生的 mRNA 上有多个启动子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nlacZ、lacY、lacA 是大肠杆菌体内与乳糖代谢有关的三个结构基因（其中 lacY 编码 $\\beta$-半乳糖苷透性酶，可将乳糖运入细胞）。上游的操纵基因（O）对结构基因起着“开关”的作用，直接控制结构基因的表达。图 1 表示环境中无乳糖时，结构基因的表达被关\n闭”的调节机制; 图 2 表示环境中有乳糖时, 结构基因的表达被“打开”的调节机制。下列有关叙述正确的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 乳糖可与阻遏蛋白结合并改变其构象, 导致结构基因无法正常转录\nB: $\\beta$-半乳糖苷透性酶可能位于细胞膜上, 合成后需要经过内质网和高尔基体的加工\nC: 过程(1)碱基配对方式与(2)过程不完全相同, 参与(2)过程的氨基酸可被多种 tRNA 转运\nD: lacZ、lacY、lacA 转录的模板链为 $\\beta$ 链, 转录产生的 mRNA 上有多个启动子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-25.jpg?height=325&width=687&top_left_y=320&top_left_x=339",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-25.jpg?height=300&width=731&top_left_y=341&top_left_x=1048"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_813",
        "problem": "斑马鱼 17 号染色体上有两对等位基因 $\\mathrm{D} / \\mathrm{d} 、 \\mathrm{G} / \\mathrm{g}$, 基因型为 $\\mathrm{dd}$ 的斑马鱼胚胎期会发出红色苂光, 带有 $\\mathrm{G}$ 基因的斑马鱼胚胎期能够发出绿色荧光。用个体 $\\mathrm{M}$ 和 $\\mathrm{N}$ 进行如下杂交实验。下列叙述错误的是（ ）\n\n亲代:\n\n[图1]\n\n(肧胎期无荧光)\n\n## 子代胚胎: 绿色 红色\n\n菼光苂光[图2]\n\n无苂光 红・绿苂光\n\n[图3]\n\n个体 $\\mathrm{N}$ 的17号染色体上相关基因组成\nA: 根据实验结果可以推断出亲代 $\\mathrm{M} 、 \\mathrm{~N}$ 的基因型分别为 Ddgg 和 DdGg\nB: 亲代 $\\mathrm{N}$ 的初级精（卵）母细胞在减数分裂过程中可能发生了互换\nC: 子代中只发绿色荧光的胚胎基因型为 DdGg 或 DDGg\nD: 子代斑马鱼中胚胎期发绿色荧光的数量最多, 发红. 绿荧光的数量最少\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n斑马鱼 17 号染色体上有两对等位基因 $\\mathrm{D} / \\mathrm{d} 、 \\mathrm{G} / \\mathrm{g}$, 基因型为 $\\mathrm{dd}$ 的斑马鱼胚胎期会发出红色苂光, 带有 $\\mathrm{G}$ 基因的斑马鱼胚胎期能够发出绿色荧光。用个体 $\\mathrm{M}$ 和 $\\mathrm{N}$ 进行如下杂交实验。下列叙述错误的是（ ）\n\n亲代:\n\n[图1]\n\n(肧胎期无荧光)\n\n## 子代胚胎: 绿色 红色\n\n菼光苂光[图2]\n\n无苂光 红・绿苂光\n\n[图3]\n\n个体 $\\mathrm{N}$ 的17号染色体上相关基因组成\n\nA: 根据实验结果可以推断出亲代 $\\mathrm{M} 、 \\mathrm{~N}$ 的基因型分别为 Ddgg 和 DdGg\nB: 亲代 $\\mathrm{N}$ 的初级精（卵）母细胞在减数分裂过程中可能发生了互换\nC: 子代中只发绿色荧光的胚胎基因型为 DdGg 或 DDGg\nD: 子代斑马鱼中胚胎期发绿色荧光的数量最多, 发红. 绿荧光的数量最少\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-69.jpg?height=168&width=257&top_left_y=1658&top_left_x=588",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-69.jpg?height=266&width=400&top_left_y=1649&top_left_x=885",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-69.jpg?height=151&width=234&top_left_y=1718&top_left_x=1436"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_332",
        "problem": "已知基因 $\\mathrm{a}$ 和基因 $\\mathrm{b}$ 均可独立导致人体患血友病, $\\mathrm{a}$ 位于性染色体上。下图 1 是某\n家系血友病的遗传图谱(不考虑 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段), 图 2 表示该家系部分成员与血友病有关的基因的电泳结果(A、B、 $a 、 b$ 基因均只电泳出一个条带)。下列相关叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 条带(3)代表基因 b，条带(4)代表基因 a\nB: 3 号个体的基因型为 $b b X^{A} X^{A}$ 或 $b b X^{A} X^{a}$\nC: 4 号个体的 B 基因仅由父亲提供\nD: 5 号与基因型和 3 号相同的女性婚配, 他们生育一患血友病孩子的概率是 $1 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知基因 $\\mathrm{a}$ 和基因 $\\mathrm{b}$ 均可独立导致人体患血友病, $\\mathrm{a}$ 位于性染色体上。下图 1 是某\n家系血友病的遗传图谱(不考虑 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段), 图 2 表示该家系部分成员与血友病有关的基因的电泳结果(A、B、 $a 、 b$ 基因均只电泳出一个条带)。下列相关叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 条带(3)代表基因 b，条带(4)代表基因 a\nB: 3 号个体的基因型为 $b b X^{A} X^{A}$ 或 $b b X^{A} X^{a}$\nC: 4 号个体的 B 基因仅由父亲提供\nD: 5 号与基因型和 3 号相同的女性婚配, 他们生育一患血友病孩子的概率是 $1 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-64.jpg?height=312&width=604&top_left_y=435&top_left_x=358",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-64.jpg?height=303&width=599&top_left_y=431&top_left_x=977"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1213",
        "problem": "THE RENA OIL SPILL - BIOLOGY WHEN DISASTER STRIKES\n\n[figure1]\n\n\nThe grounding of the CV Rena on October 5, 2011 was New Zealand's worst maritime environmental disaster when the container ship spilled approximately 350 tonnes of heavy fuel oil into the ocean. The container ship ran aground on Otaiti (Astrolabe Reef) just off Mōtitīi Island in the Bay of Plenty, releasing oil into a relatively pristine marine environment. Te Mauri Moana, a group of scientists led by the University of Waikato as part of the Government's \\$ 4.2 million Rena Long-term Environmental Recovery Plan, undertook one of the most comprehensive, multi-disciplinary studies ever done following a marine pollution event. Their findings are summarised in the report; \"Rena Environmental Recovery Monitoring Programme 2011-2013.\"\n\nBy international standards the oil spill was relatively minor but it occurred in an otherwise uncontaminated coastline in an area renowned for its beauty and highly valued for tourism, recreation and fisheries. Iwi, government, commercial stakeholders and the public were rightly concerned there would be long-lasting negative impacts on beaches, reefs and fisheries. Te Mauri Moana reports on fifteen monitoring and research programmes focused on the range of ecosystems and habitats within the Bay of Plenty.\n\nChemical contamination of the sediments and fauna of Otaiti Reef and Môtitī Island and the surrounding soft sediments was examined to determine if the discharge of fuel oil (and later release of container debris) from the Rena resulted in significant chemical contamination of the marine environment and marine organisms. Amongst other contaminants the presence of polycyclic aromatic hydrocarbons (PAHs) from fuel oil was examined in sediments and from selected organisms.\n\nThe maps below shows the PAH levels in sediments collected from A. on Otaiti Reef and B. off-reef.\n\n\nA.\n\n[figure2]\n\nB.\n\n[figure3]\n\n[figure4]\n\nThe graph at left shows the PAH levels in sea urchins collected from Otaiti Reef and nearby islands. The numbers in brackets indicate the number of samples collected at each site. All sea urchins on Otaiti reef showed elevated PAH levels in the gonad with an average level of $0.057 \\mathrm{mg} \\mathrm{kg}^{-1}$. Gut levels averaged $1.58 \\mathrm{mg} \\mathrm{kg}^{-1}$ but at Astro 2 and 3 were orders of magnitude higher than at other sites on Otaiti Reef. PAH levels in the gonad and gut from urchins from Mōtîti Island averaged $0.026 \\mathrm{mg} \\mathrm{kg}^{-1}$ and $0.03 \\mathrm{mg} \\mathrm{kg}^{-1}$ respectively and were similar to levels found elsewhere, including East Cape.On-reef sediment levels of PAH are greatest at?\nA: Astro $2 / 3$\nB: Astro 4\nC: Astro 5\nD: Astro 6\nE: Astro 7\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nTHE RENA OIL SPILL - BIOLOGY WHEN DISASTER STRIKES\n\n[figure1]\n\n\nThe grounding of the CV Rena on October 5, 2011 was New Zealand's worst maritime environmental disaster when the container ship spilled approximately 350 tonnes of heavy fuel oil into the ocean. The container ship ran aground on Otaiti (Astrolabe Reef) just off Mōtitīi Island in the Bay of Plenty, releasing oil into a relatively pristine marine environment. Te Mauri Moana, a group of scientists led by the University of Waikato as part of the Government's \\$ 4.2 million Rena Long-term Environmental Recovery Plan, undertook one of the most comprehensive, multi-disciplinary studies ever done following a marine pollution event. Their findings are summarised in the report; \"Rena Environmental Recovery Monitoring Programme 2011-2013.\"\n\nBy international standards the oil spill was relatively minor but it occurred in an otherwise uncontaminated coastline in an area renowned for its beauty and highly valued for tourism, recreation and fisheries. Iwi, government, commercial stakeholders and the public were rightly concerned there would be long-lasting negative impacts on beaches, reefs and fisheries. Te Mauri Moana reports on fifteen monitoring and research programmes focused on the range of ecosystems and habitats within the Bay of Plenty.\n\nChemical contamination of the sediments and fauna of Otaiti Reef and Môtitī Island and the surrounding soft sediments was examined to determine if the discharge of fuel oil (and later release of container debris) from the Rena resulted in significant chemical contamination of the marine environment and marine organisms. Amongst other contaminants the presence of polycyclic aromatic hydrocarbons (PAHs) from fuel oil was examined in sediments and from selected organisms.\n\nThe maps below shows the PAH levels in sediments collected from A. on Otaiti Reef and B. off-reef.\n\n\nA.\n\n[figure2]\n\nB.\n\n[figure3]\n\n[figure4]\n\nThe graph at left shows the PAH levels in sea urchins collected from Otaiti Reef and nearby islands. The numbers in brackets indicate the number of samples collected at each site. All sea urchins on Otaiti reef showed elevated PAH levels in the gonad with an average level of $0.057 \\mathrm{mg} \\mathrm{kg}^{-1}$. Gut levels averaged $1.58 \\mathrm{mg} \\mathrm{kg}^{-1}$ but at Astro 2 and 3 were orders of magnitude higher than at other sites on Otaiti Reef. PAH levels in the gonad and gut from urchins from Mōtîti Island averaged $0.026 \\mathrm{mg} \\mathrm{kg}^{-1}$ and $0.03 \\mathrm{mg} \\mathrm{kg}^{-1}$ respectively and were similar to levels found elsewhere, including East Cape.\n\nproblem:\nOn-reef sediment levels of PAH are greatest at?\n\nA: Astro $2 / 3$\nB: Astro 4\nC: Astro 5\nD: Astro 6\nE: Astro 7\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=556&width=1834&top_left_y=276&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=657&width=894&top_left_y=1876&top_left_x=113",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-24.jpg?height=548&width=854&top_left_y=1942&top_left_x=1092",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-25.jpg?height=591&width=802&top_left_y=201&top_left_x=136"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_561",
        "problem": "如图为某高等动物的一个细胞减数分裂过程图, (1) (6)表示细胞。不考虑基因突变,下列相关说法错误的是（ ）\n\n[图1]\nA: (6)的基因型可能为 $\\mathrm{AB}$ 或 $\\mathrm{Ab}$\nB: (5)的基因型可能为 $a b Y$ 或 $a B Y$\nC: (4)和(5)的基因型不一定相同\nD: (3)中只含有一条 X 染色体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图为某高等动物的一个细胞减数分裂过程图, (1) (6)表示细胞。不考虑基因突变,下列相关说法错误的是（ ）\n\n[图1]\n\nA: (6)的基因型可能为 $\\mathrm{AB}$ 或 $\\mathrm{Ab}$\nB: (5)的基因型可能为 $a b Y$ 或 $a B Y$\nC: (4)和(5)的基因型不一定相同\nD: (3)中只含有一条 X 染色体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-73.jpg?height=660&width=1173&top_left_y=818&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1077",
        "problem": "Arrangement of gene sequences in representative 50kb segments of two organisms is shown. Genes above the line are transcribed left to right \\& below the line to the left.\n\nSegment I\n\n[figure1]\n\nSegment II\n\n[figure2]\n\nOpen boxes represent exons while hatched ones introns.\n\nWhich of the following is correct?\nA: Each segment indicates single stranded DNA in $5^{\\prime}$ to $3^{\\prime}$ orientation (left to right).\nB: Segment I is likely to represent a prokaryote \\& segment II, a eukaryote such as worm.\nC: Segment II is less likely to have stop codons as compared to segment I.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nArrangement of gene sequences in representative 50kb segments of two organisms is shown. Genes above the line are transcribed left to right \\& below the line to the left.\n\nSegment I\n\n[figure1]\n\nSegment II\n\n[figure2]\n\nOpen boxes represent exons while hatched ones introns.\n\nWhich of the following is correct?\n\nA: Each segment indicates single stranded DNA in $5^{\\prime}$ to $3^{\\prime}$ orientation (left to right).\nB: Segment I is likely to represent a prokaryote \\& segment II, a eukaryote such as worm.\nC: Segment II is less likely to have stop codons as compared to segment I.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-15.jpg?height=209&width=1133&top_left_y=421&top_left_x=621",
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-15.jpg?height=201&width=1136&top_left_y=626&top_left_x=603"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_73",
        "problem": "Žliobaite et al. (2017) point out that for a given taxon, the pattern below is usually observed in the fossil record (red line) defined as the number of localities where a taxon is found. They propose that this pattern is a reflection of the evolutionary process, as shown by blue arrows representing the effects of environment and competition.\n\n[figure1]\nA: According to the model the peak of a taxon's history is only identifiable after its extinction.\nB: It has been observed that when different species reach their maximum population size in an island, competition and species richness is also at their maximum. This is consistent with the effect of environment and competition as shown in figure.\nC: The extinction of a taxon should not depend on its age.\nD: When a taxon is rare, its extinction is more likely to be the result of abiotic factors\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nŽliobaite et al. (2017) point out that for a given taxon, the pattern below is usually observed in the fossil record (red line) defined as the number of localities where a taxon is found. They propose that this pattern is a reflection of the evolutionary process, as shown by blue arrows representing the effects of environment and competition.\n\n[figure1]\n\nA: According to the model the peak of a taxon's history is only identifiable after its extinction.\nB: It has been observed that when different species reach their maximum population size in an island, competition and species richness is also at their maximum. This is consistent with the effect of environment and competition as shown in figure.\nC: The extinction of a taxon should not depend on its age.\nD: When a taxon is rare, its extinction is more likely to be the result of abiotic factors\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-57.jpg?height=443&width=970&top_left_y=584&top_left_x=543"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_708",
        "problem": "鸡的性别决定方式为 ZW 型, 其胫色浅、深为一对相对性状, 由 $\\mathrm{I} / \\mathrm{i}$ 基因控制。研究人员将纯种藏鸡与纯种白来航鸡进行了杂交实验, 统计结果如下表所示。下列分析,不正确的是（ ）\n\n| 杂交组合 | 亲本 |  | $F_{1}$ 性状表现和数目 |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 父本 | 母本 | 深色雄 | 深色雌 | 浅色雄 | 浅色雌 |\n| I | 藏鸡 | 白来航鸡 | 0 | 142 | 156 | 0 |\n| II | 白来航鸡 | 藏鸡 | 0 | 0 | 42 | 35 |\nA: 亲本白来航鸡的胫色为隐性性状\nB: I和II为正反交实验，控制胫色基因位于 Z 染色体上\nC: I 中父本和母本的基因型分别为 $Z^{i} Z^{i}$ 和 $Z^{I} W$\nD: II中 $F_{1}$ 雌雄交配所得 $F_{2}$ 理论上为深色雌:浅色雄:浅色雌 $=1: 2: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n鸡的性别决定方式为 ZW 型, 其胫色浅、深为一对相对性状, 由 $\\mathrm{I} / \\mathrm{i}$ 基因控制。研究人员将纯种藏鸡与纯种白来航鸡进行了杂交实验, 统计结果如下表所示。下列分析,不正确的是（ ）\n\n| 杂交组合 | 亲本 |  | $F_{1}$ 性状表现和数目 |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 父本 | 母本 | 深色雄 | 深色雌 | 浅色雄 | 浅色雌 |\n| I | 藏鸡 | 白来航鸡 | 0 | 142 | 156 | 0 |\n| II | 白来航鸡 | 藏鸡 | 0 | 0 | 42 | 35 |\n\nA: 亲本白来航鸡的胫色为隐性性状\nB: I和II为正反交实验，控制胫色基因位于 Z 染色体上\nC: I 中父本和母本的基因型分别为 $Z^{i} Z^{i}$ 和 $Z^{I} W$\nD: II中 $F_{1}$ 雌雄交配所得 $F_{2}$ 理论上为深色雌:浅色雄:浅色雌 $=1: 2: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_185",
        "problem": "In snail, the direction of cleavage, and therefore the snail shell coiling, is controlled by a single gene, in which the right-coiling allele, $\\mathrm{D}$, is dominant to the left-coiling allele, $\\mathrm{d}$. Below table shows the results of a set of mating experiment.\n\n|  | Genotype | Phenotype |\n| :---: | :---: | :---: |\n| $\\mathrm{DD} \\varphi \\mathrm{xdd}$ | Dd | All right-coiling |\n| $\\mathrm{DD} \\precsim \\mathrm{x} d \\mathrm{~d}$ ìš° | Dd | All left-coiling |\n| Dd $\\times$ Dd | 1DD:2Dd:1dd | All right-coiling |\nA: The direction of cleavage is determined by the genotype of the developing snail.\nB: If the egg cytoplasm of a recessive homozygous right-coiling mother is injected into the eggs of the \"dd\" mother, the resulting embryos coil to the left.\nC: Injection of the egg cytoplasm from a heterozygous left-coiling mother into the eggs of the right coiling heterozygous mother, does not affect the right-coiling in the embryos.\nD: \"DD\" or \"Dd\" mothers place a coiling determinant factor inside the egg cytoplasm.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn snail, the direction of cleavage, and therefore the snail shell coiling, is controlled by a single gene, in which the right-coiling allele, $\\mathrm{D}$, is dominant to the left-coiling allele, $\\mathrm{d}$. Below table shows the results of a set of mating experiment.\n\n|  | Genotype | Phenotype |\n| :---: | :---: | :---: |\n| $\\mathrm{DD} \\varphi \\mathrm{xdd}$ | Dd | All right-coiling |\n| $\\mathrm{DD} \\precsim \\mathrm{x} d \\mathrm{~d}$ ìš° | Dd | All left-coiling |\n| Dd $\\times$ Dd | 1DD:2Dd:1dd | All right-coiling |\n\nA: The direction of cleavage is determined by the genotype of the developing snail.\nB: If the egg cytoplasm of a recessive homozygous right-coiling mother is injected into the eggs of the \"dd\" mother, the resulting embryos coil to the left.\nC: Injection of the egg cytoplasm from a heterozygous left-coiling mother into the eggs of the right coiling heterozygous mother, does not affect the right-coiling in the embryos.\nD: \"DD\" or \"Dd\" mothers place a coiling determinant factor inside the egg cytoplasm.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_244",
        "problem": "Long had ten strains of Escherichia coli with different mutations in their lac operon. Analyzing their DNA, he found that each strain is one of five mutation types: lacZ, lacY; lacI; lacIS (the repressor $\\mathrm{LacI}^{S}$ can bind to the operator but cannot bind to the inducer), or $\\operatorname{lac}^{c}$ (the repressor cannot bind to mutated operator $\\operatorname{lac}^{c}$ ). Long also knew that strain number 6 is a lacZ mutant. The lac operon is shown below.\n\n[figure1]\n\nLong isolated DNA containing the lac operon from each strain (donor) and transformed it into other strains, thus making the strain merodiploid (recipient). Thereafter recipient strains were cultured on minimal medium that contained lactose as the only carbon source. Growth of the strains was recorded in the table below (+ means that the strain is growing, - means that the strain is not growing)\n\n[figure2]\n\nDonor colonies\nA: Strain number 7 is a lac $Z$ mutant.\nB: Strain number 3 is a lac $Y$ mutant.\nC: Strains number 2 and 4 are of the same mutation type.\nD: If strain number 5 receives DNA from itself, the transformed bacteria cannot grow.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nLong had ten strains of Escherichia coli with different mutations in their lac operon. Analyzing their DNA, he found that each strain is one of five mutation types: lacZ, lacY; lacI; lacIS (the repressor $\\mathrm{LacI}^{S}$ can bind to the operator but cannot bind to the inducer), or $\\operatorname{lac}^{c}$ (the repressor cannot bind to mutated operator $\\operatorname{lac}^{c}$ ). Long also knew that strain number 6 is a lacZ mutant. The lac operon is shown below.\n\n[figure1]\n\nLong isolated DNA containing the lac operon from each strain (donor) and transformed it into other strains, thus making the strain merodiploid (recipient). Thereafter recipient strains were cultured on minimal medium that contained lactose as the only carbon source. Growth of the strains was recorded in the table below (+ means that the strain is growing, - means that the strain is not growing)\n\n[figure2]\n\nDonor colonies\n\nA: Strain number 7 is a lac $Z$ mutant.\nB: Strain number 3 is a lac $Y$ mutant.\nC: Strains number 2 and 4 are of the same mutation type.\nD: If strain number 5 receives DNA from itself, the transformed bacteria cannot grow.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-074.jpg?height=691&width=1531&top_left_y=842&top_left_x=271",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-075.jpg?height=851&width=1017&top_left_y=180&top_left_x=545"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_386",
        "problem": "棉铃虫是严重危害棉花的一种害虫。科研工作者发现毒蛋白基因 B 和胰蛋白酶抑制剂基因 D, 两种基因均可导致棉䍅虫死亡。现将 $\\mathrm{B}$ 和 $\\mathrm{D}$ 基因同时导入棉花的一条染色体上获得抗虫棉。棉花的短果枝由基因 $\\mathrm{A}$ 控制, 研究者获得了多个基因型为 $\\mathrm{AaBD}$ 的短果枝抗虫棉植株, AaBD 植株与纯合的 aa 长果枝不抗虫植株杂交得到 $F_{1}$ (不考虑减数分裂时的互换）。下列说法不正确的是（）\nA: 若 $F_{1}$ 中短果枝抗虫: 长果枝不抗虫 $=1$ : 1 , 则 $B 、 D$ 基因与 $\\mathrm{A}$ 基因位于同一条染色体上\nB: 若 $F_{1}$ 的表型比例为 $1: 1: 1: 1$, 则 $F_{1}$ 产生的配子的基因型为 $A B 、 A D 、 a B 、 a D$\nC: 若 $F_{1}$ 的表型比例为 1:1:1:1, 则果枝基因和抗虫基因分别位于两对同源染色体上\nD: 若 $F_{1}$ 中短果枝不抗虫: 长果枝抗虫 $=1: 1$, 则 $F_{1}$ 产生的配子的基因型为 $\\mathrm{A}$ 和 $\\mathrm{aBD}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n棉铃虫是严重危害棉花的一种害虫。科研工作者发现毒蛋白基因 B 和胰蛋白酶抑制剂基因 D, 两种基因均可导致棉䍅虫死亡。现将 $\\mathrm{B}$ 和 $\\mathrm{D}$ 基因同时导入棉花的一条染色体上获得抗虫棉。棉花的短果枝由基因 $\\mathrm{A}$ 控制, 研究者获得了多个基因型为 $\\mathrm{AaBD}$ 的短果枝抗虫棉植株, AaBD 植株与纯合的 aa 长果枝不抗虫植株杂交得到 $F_{1}$ (不考虑减数分裂时的互换）。下列说法不正确的是（）\n\nA: 若 $F_{1}$ 中短果枝抗虫: 长果枝不抗虫 $=1$ : 1 , 则 $B 、 D$ 基因与 $\\mathrm{A}$ 基因位于同一条染色体上\nB: 若 $F_{1}$ 的表型比例为 $1: 1: 1: 1$, 则 $F_{1}$ 产生的配子的基因型为 $A B 、 A D 、 a B 、 a D$\nC: 若 $F_{1}$ 的表型比例为 1:1:1:1, 则果枝基因和抗虫基因分别位于两对同源染色体上\nD: 若 $F_{1}$ 中短果枝不抗虫: 长果枝抗虫 $=1: 1$, 则 $F_{1}$ 产生的配子的基因型为 $\\mathrm{A}$ 和 $\\mathrm{aBD}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_548",
        "problem": "甲病、乙病为单基因遗传病。图 1 为某家系的遗传系谱图, 其中 4 号个体不携带致病基因。对家系中部分成员进行甲病的基因检测，将含有相关基因的 DNA 片段酶切后电泳分离，结果如图 2 所示。下列相关说法错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 若 11 号为男孩, 则其同时患甲乙两种病的概率是 $1 / 16$\nB: 若 11 号患乙病且性染色体组成为 XXY, 原因是 8 号产生卵细胞时减数分裂II异常\nC: 甲病是常染色体隐性遗传病, 图 2 中的 X 可能为图 1 中的 2 号、 3 号或 9 号\nD: 甲病致病基因存在 1 个酶切位点, 而正常基因没有该位点\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n甲病、乙病为单基因遗传病。图 1 为某家系的遗传系谱图, 其中 4 号个体不携带致病基因。对家系中部分成员进行甲病的基因检测，将含有相关基因的 DNA 片段酶切后电泳分离，结果如图 2 所示。下列相关说法错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 若 11 号为男孩, 则其同时患甲乙两种病的概率是 $1 / 16$\nB: 若 11 号患乙病且性染色体组成为 XXY, 原因是 8 号产生卵细胞时减数分裂II异常\nC: 甲病是常染色体隐性遗传病, 图 2 中的 X 可能为图 1 中的 2 号、 3 号或 9 号\nD: 甲病致病基因存在 1 个酶切位点, 而正常基因没有该位点\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-088.jpg?height=260&width=466&top_left_y=1800&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-088.jpg?height=271&width=666&top_left_y=1778&top_left_x=1129"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1526",
        "problem": "Peaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nWhich results are expected if fuzz is produced by a single recessive allele?\nA: $\\mathrm{A}$\nB: $\\quad$ B\nC: $\\quad \\mathrm{C}$\nD: B and D\nE: $\\quad$ E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPeaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nTo investigate why trees produce peaches or nectarines, the following experiments were carried out.\n\n|  | Scenario | Result |\n| :---: | :---: | :---: |\n| A | Seeds of peach trees grown in <br> hot or cool conditions | Peaches appear in hot conditions, <br> nectarines in cool conditions |\n| B | Peach tree is crossed with <br> nectarine tree | All offspring produce peaches |\n| C | Peach tree is crossed with <br> nectarine tree | All offspring produce nectarines |\n| D | Peach tree is crossed with <br> nectarine tree | Some offspring produce peaches and some <br> offspring produce nectarines |\n| $E$ | Nectarine trees are left alone for <br> many years | At a certain age, trees produce peaches <br> instead |\n\nWhich results are expected if fuzz is produced by a single recessive allele?\n\nA: $\\mathrm{A}$\nB: $\\quad$ B\nC: $\\quad \\mathrm{C}$\nD: B and D\nE: $\\quad$ E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-45.jpg?height=782&width=1231&top_left_y=474&top_left_x=241"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1311",
        "problem": "A student filled two Petri dishes with a clear cornstarch gel, then marked the letter \" $X$ \" invisibly onto the gel in Petri dish 1 with a damp cotton swab. He then placed saliva from his mouth onto a second cotton swab and used that swab to mark the letter \" $\\mathrm{X}$ \" invisibly onto the gel in Petri dish 2.\n\n[figure1]\n\nPetri dish 1\n\n[figure2]\n\nPetri dish 2\n\nFifteen minutes later, the student rinsed both Petri dishes with a dilute solution of iodine to indicate the presence of starch. The entire surface of Petri dish 1 turned blue-black, indicating starch. Most of the surface of Petri dish 2 was blue-black, except that the letter \"X\" was clear, as shown above.\n\nThe most probable explanation of the clear \" $\\mathrm{X}$ \" is that?\nA: The starch in the gel was absorbed by the damp cotton swab.\nB: The iodine reacted with a chemical in the saliva and broke down.\nC: A chemical in the saliva broke down the starch in the gel.\nD: The saliva prevented the iodine from contacting the starch in the gel.\nE: The cotton swab removed the iodine from the areas it touched.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA student filled two Petri dishes with a clear cornstarch gel, then marked the letter \" $X$ \" invisibly onto the gel in Petri dish 1 with a damp cotton swab. He then placed saliva from his mouth onto a second cotton swab and used that swab to mark the letter \" $\\mathrm{X}$ \" invisibly onto the gel in Petri dish 2.\n\n[figure1]\n\nPetri dish 1\n\n[figure2]\n\nPetri dish 2\n\nFifteen minutes later, the student rinsed both Petri dishes with a dilute solution of iodine to indicate the presence of starch. The entire surface of Petri dish 1 turned blue-black, indicating starch. Most of the surface of Petri dish 2 was blue-black, except that the letter \"X\" was clear, as shown above.\n\nThe most probable explanation of the clear \" $\\mathrm{X}$ \" is that?\n\nA: The starch in the gel was absorbed by the damp cotton swab.\nB: The iodine reacted with a chemical in the saliva and broke down.\nC: A chemical in the saliva broke down the starch in the gel.\nD: The saliva prevented the iodine from contacting the starch in the gel.\nE: The cotton swab removed the iodine from the areas it touched.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-17.jpg?height=491&width=465&top_left_y=1556&top_left_x=864",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-17.jpg?height=477&width=474&top_left_y=1572&top_left_x=1365"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_960",
        "problem": "Which statement best explains why the ascending limb of the loop of Henle is thicker than the descending limb?\nA: Many cells with aquaporin are present in the ascending limb\nB: The ascending limb receives a better blood supply\nC: The triple-pumps $\\left(\\mathrm{Na}^{+} / \\mathrm{K}^{+} / \\mathrm{Cl}\\right)$ of the ascending limb consume a lot of ATP and need many mitochondria.\nD: Unexplained proliferation of cells around the ascending limb\nE: Descending limb has evolved completely while the thickness of the ascending limb is a vestigial feature.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich statement best explains why the ascending limb of the loop of Henle is thicker than the descending limb?\n\nA: Many cells with aquaporin are present in the ascending limb\nB: The ascending limb receives a better blood supply\nC: The triple-pumps $\\left(\\mathrm{Na}^{+} / \\mathrm{K}^{+} / \\mathrm{Cl}\\right)$ of the ascending limb consume a lot of ATP and need many mitochondria.\nD: Unexplained proliferation of cells around the ascending limb\nE: Descending limb has evolved completely while the thickness of the ascending limb is a vestigial feature.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_558",
        "problem": "SSR 是 DNA 分子中的简单重复序列, 非同源染色体上的 SSR 重复单位不同, 不同品种的同源染色体的 SSR 重复次数不同, 常用于染色体特异性标记。研究人员利用 X 射线诱导野生型拟南芥品系 $(2 n=10)$ 培育出单基因隐性突变品系, 让其与野生型杂交获得 $F_{1}, F_{1}$ 自交获得 $F_{2}$ 。提取 $F_{2}$ 突变体中的 15 株个体的 DNA, 利用 2 号、5 号染色体上 SSR 进行 PCR 扩增, 结果如下图。据图分析错误的是 ( )\n\n[图1]\nA: 由图可知, 上述隐性突变基因位于 2 号染色体上\nB: 8 号个体的产生可能与 $\\mathrm{F}_{1}$ 减数分裂时四分体中非姐妹染色单体交换片段有关\nC: 若对 $F_{2}$ 全部突变体的 2 号染色体 SSR 进行扩增, 结果有 3 种且比例为 1:2:1\nD: 图中 1 号个体与 $\\mathrm{F}_{1}$ 杂交, 子代 5 号染色体 SSR 扩增结果有 2 种且比例为 1: 1\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nSSR 是 DNA 分子中的简单重复序列, 非同源染色体上的 SSR 重复单位不同, 不同品种的同源染色体的 SSR 重复次数不同, 常用于染色体特异性标记。研究人员利用 X 射线诱导野生型拟南芥品系 $(2 n=10)$ 培育出单基因隐性突变品系, 让其与野生型杂交获得 $F_{1}, F_{1}$ 自交获得 $F_{2}$ 。提取 $F_{2}$ 突变体中的 15 株个体的 DNA, 利用 2 号、5 号染色体上 SSR 进行 PCR 扩增, 结果如下图。据图分析错误的是 ( )\n\n[图1]\n\nA: 由图可知, 上述隐性突变基因位于 2 号染色体上\nB: 8 号个体的产生可能与 $\\mathrm{F}_{1}$ 减数分裂时四分体中非姐妹染色单体交换片段有关\nC: 若对 $F_{2}$ 全部突变体的 2 号染色体 SSR 进行扩增, 结果有 3 种且比例为 1:2:1\nD: 图中 1 号个体与 $\\mathrm{F}_{1}$ 杂交, 子代 5 号染色体 SSR 扩增结果有 2 种且比例为 1: 1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-40.jpg?height=360&width=1456&top_left_y=1302&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1206",
        "problem": "A population in Hardy-Weinberg equilibrium has a frequency of a particular recessive allele of 0.2 . The percentage of heterozygotes in the population would be\nA: 0.16\nB: 0.8\nC: 16\nD: 32\nE: 64\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA population in Hardy-Weinberg equilibrium has a frequency of a particular recessive allele of 0.2 . The percentage of heterozygotes in the population would be\n\nA: 0.16\nB: 0.8\nC: 16\nD: 32\nE: 64\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1147",
        "problem": "Which of the following is NOT true of a typical co-operative breeding system in birds?\nA: It involves breeding pairs plus helpers\nB: the group defends an all-purpose territory\nC: helpers are genetically related to the breeding pair\nD: helpers are usually only females\nE: all of the above are true\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is NOT true of a typical co-operative breeding system in birds?\n\nA: It involves breeding pairs plus helpers\nB: the group defends an all-purpose territory\nC: helpers are genetically related to the breeding pair\nD: helpers are usually only females\nE: all of the above are true\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1459",
        "problem": "Like many mammals, cat species use the $X Y$ sex-determination system where males have the sex chromosomes XY and females have the sex chromosomes XX. In a specific breed of cats called tortoiseshell cats, the major gene for fur colour is located on the Xchromosome. There are two fur colour alleles for this gene: one for orange fur and one for black fur.\n\nIn a specific litter of kittens, there are three kittens born with the following fur colours: pure orange fur, pure black fur, and one with both orange and black fur.\n\n[figure1]\n\nWhich of the following statements is most likely to be true?\nA: The kitten with both orange and black fur is female\nB: The kitten with both orange and black fur is male\nC: The pure colour kittens (the pure black fur and pure orange fur) are male\nD: The pure colour kittens (the pure black fur and pure orange fur) are female\nE: The kitten's sexes are randomly determined so cannot be determined\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nLike many mammals, cat species use the $X Y$ sex-determination system where males have the sex chromosomes XY and females have the sex chromosomes XX. In a specific breed of cats called tortoiseshell cats, the major gene for fur colour is located on the Xchromosome. There are two fur colour alleles for this gene: one for orange fur and one for black fur.\n\nIn a specific litter of kittens, there are three kittens born with the following fur colours: pure orange fur, pure black fur, and one with both orange and black fur.\n\n[figure1]\n\nWhich of the following statements is most likely to be true?\n\nA: The kitten with both orange and black fur is female\nB: The kitten with both orange and black fur is male\nC: The pure colour kittens (the pure black fur and pure orange fur) are male\nD: The pure colour kittens (the pure black fur and pure orange fur) are female\nE: The kitten's sexes are randomly determined so cannot be determined\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-31.jpg?height=325&width=468&top_left_y=674&top_left_x=794"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_676",
        "problem": "下图为某家族的遗传系谱图，甲病相关基因用 $\\mathrm{A} 、 \\mathrm{a}$ 表示，乙病相关基因用 $\\mathrm{B} 、 \\mathrm{~b}$\n\n[图1]\nA: 乙病为伴 X 染色体隐性遗传病\nB: $\\mathrm{II}_{5}$ 的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{B}}$ 或 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$\nC: $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$, 再生一个孩子, 只患一种病的概率为 $5 / 16$\nD: $\\mathrm{III}_{8}$ 和一个正常男性婚配, 生一个男孩患乙病的概率为 $1 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为某家族的遗传系谱图，甲病相关基因用 $\\mathrm{A} 、 \\mathrm{a}$ 表示，乙病相关基因用 $\\mathrm{B} 、 \\mathrm{~b}$\n\n[图1]\n\nA: 乙病为伴 X 染色体隐性遗传病\nB: $\\mathrm{II}_{5}$ 的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{B}}$ 或 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$\nC: $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$, 再生一个孩子, 只患一种病的概率为 $5 / 16$\nD: $\\mathrm{III}_{8}$ 和一个正常男性婚配, 生一个男孩患乙病的概率为 $1 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-032.jpg?height=488&width=1016&top_left_y=344&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_16",
        "problem": "Frequencies of the human $\\mathrm{ABO}$ blood group alleles in a population are $\\mathrm{p}\\left(I^{A}\\right)=40 \\%$, $\\mathrm{p}\\left(I^{B}\\right)=40 \\%$ and $\\mathrm{p}(i)=20 \\%$. If the population is in Hardy-Weinberg equilibrium\nA: In this population, the number of persons with the blood groups A and B should be equal.\nB: In this population, the number of persons with the blood groups $A$ and $A B$ should be equal.\nC: In this population the frequency of persons with anti-B antibodies is $64 \\%$.\nD: Locus $\\mathrm{ABO}$ is localized on an autosomal chromosome because the blood group frequencies are the same for men and women.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFrequencies of the human $\\mathrm{ABO}$ blood group alleles in a population are $\\mathrm{p}\\left(I^{A}\\right)=40 \\%$, $\\mathrm{p}\\left(I^{B}\\right)=40 \\%$ and $\\mathrm{p}(i)=20 \\%$. If the population is in Hardy-Weinberg equilibrium\n\nA: In this population, the number of persons with the blood groups A and B should be equal.\nB: In this population, the number of persons with the blood groups $A$ and $A B$ should be equal.\nC: In this population the frequency of persons with anti-B antibodies is $64 \\%$.\nD: Locus $\\mathrm{ABO}$ is localized on an autosomal chromosome because the blood group frequencies are the same for men and women.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_550",
        "problem": "为探究高温对拟南芥 $(2 n=10)$ 花药减数分裂过程中染色体的影响, 实验人员观察减数分裂某时期的显微图像, 并用特异性 DNA 探针标记着丝粒（荧光点）, 结果如图所示。下列叙述错误的是（）\n\n[图1]\nA: 正常组在 A 时期发生四分体中非姐妹染色单体间的互换\nB: 高温抑制同源染色体联会, 对同源染色体分离无影响\nC: B 时期高温组细胞可能发生了姐妹染色单体的分离\nD: 高温组的花粉受精形成的植株可能出现染色体变异\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n为探究高温对拟南芥 $(2 n=10)$ 花药减数分裂过程中染色体的影响, 实验人员观察减数分裂某时期的显微图像, 并用特异性 DNA 探针标记着丝粒（荧光点）, 结果如图所示。下列叙述错误的是（）\n\n[图1]\n\nA: 正常组在 A 时期发生四分体中非姐妹染色单体间的互换\nB: 高温抑制同源染色体联会, 对同源染色体分离无影响\nC: B 时期高温组细胞可能发生了姐妹染色单体的分离\nD: 高温组的花粉受精形成的植株可能出现染色体变异\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-63.jpg?height=734&width=1287&top_left_y=815&top_left_x=339"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1419",
        "problem": "Review the following map and answer the 5 subsequent questions.\n\n[figure1]\n\nEcologists have been studying the small archipelago shown in the map above. They have been comparing the ecology of the mainland with those on the nearby islands and have come up with a few discoveries. The area has not been known to have been inhabited by humans, so it remains very natural. The mainland contains a low mountain range approximately $50 \\mathrm{~km}$ inland from the coast (highest peak approximately $400 \\mathrm{~m}$ above sea level) and the prevailing winds are westerlies.\n\nConsidering the smaller islands only, which one would be expected to have the greatest biodiversity?\nA: 2\nB: 4\nC: 5\nD: 6\nE: 0 (they would all be equal)\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nReview the following map and answer the 5 subsequent questions.\n\n[figure1]\n\nEcologists have been studying the small archipelago shown in the map above. They have been comparing the ecology of the mainland with those on the nearby islands and have come up with a few discoveries. The area has not been known to have been inhabited by humans, so it remains very natural. The mainland contains a low mountain range approximately $50 \\mathrm{~km}$ inland from the coast (highest peak approximately $400 \\mathrm{~m}$ above sea level) and the prevailing winds are westerlies.\n\nConsidering the smaller islands only, which one would be expected to have the greatest biodiversity?\n\nA: 2\nB: 4\nC: 5\nD: 6\nE: 0 (they would all be equal)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-39.jpg?height=925&width=1479&top_left_y=451&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_594",
        "problem": "表观遗传是指基因的碱基序列保持不变, 但基因表达和表型发生变化的现象。下列有关叙述不正确的是（ ）\nA: 表观遗传现象虽然基因的碱基序列没有改变，但属于可遗传的变异\nB: DNA 甲基化可能导致 DNA 聚合酶不能结合到 DNA 双链上，抑制基因表达\nC: 吸烟会导致精子中 DNA 的甲基化水平升高, 从而影响基因表达\nD: 除 DNA 的甲基化外, 组蛋白的甲基化和乙酰化也可能导致表观遗传现象\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n表观遗传是指基因的碱基序列保持不变, 但基因表达和表型发生变化的现象。下列有关叙述不正确的是（ ）\n\nA: 表观遗传现象虽然基因的碱基序列没有改变，但属于可遗传的变异\nB: DNA 甲基化可能导致 DNA 聚合酶不能结合到 DNA 双链上，抑制基因表达\nC: 吸烟会导致精子中 DNA 的甲基化水平升高, 从而影响基因表达\nD: 除 DNA 的甲基化外, 组蛋白的甲基化和乙酰化也可能导致表观遗传现象\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1332",
        "problem": "Partial amino acid sequences for a particular protein in three animal species are shown below. Each letter in the sequence stands for an amino acid. For example, $Q$ stands for glutamine, and $L$ stands for leucine.\n\n| Species | Amino Acid <br> Sequence |\n| :---: | :---: |\n| Green junglefowl <br> (bird) | QHEPHERKRM |\n| Nile crocodile <br> (reptile) | SHDPAQQKRL |\n| Domestic chicken <br> (bird) | QHEPHKRKRM |\n\nWhich of the following statements best explains how these sequence data are used as evidence to determine evolutionary relationships?\nA: All species translate the amino acid sequences of their proteins in a similar way.\nB: The species that are most closely related have the most similar amino acid sequences.\nC: Individual organisms acquire changes in their amino acid sequences over their lifetimes.\nD: The organisms that evolved at the same time in geologic history have identical amino acid sequences.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPartial amino acid sequences for a particular protein in three animal species are shown below. Each letter in the sequence stands for an amino acid. For example, $Q$ stands for glutamine, and $L$ stands for leucine.\n\n| Species | Amino Acid <br> Sequence |\n| :---: | :---: |\n| Green junglefowl <br> (bird) | QHEPHERKRM |\n| Nile crocodile <br> (reptile) | SHDPAQQKRL |\n| Domestic chicken <br> (bird) | QHEPHKRKRM |\n\nWhich of the following statements best explains how these sequence data are used as evidence to determine evolutionary relationships?\n\nA: All species translate the amino acid sequences of their proteins in a similar way.\nB: The species that are most closely related have the most similar amino acid sequences.\nC: Individual organisms acquire changes in their amino acid sequences over their lifetimes.\nD: The organisms that evolved at the same time in geologic history have identical amino acid sequences.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1107",
        "problem": "Two statements regarding evolution are made below.\n\nI. Rates of evolution are typically very slow because natural selection is usually $\\quad P$ selection.\n\nII. The plant population growing on high-zinc-soil is able to grow at concentrations which are otherwise lethal to plants of the same species. This is  $\\quad Q$  selection.\n\n$P$ and $Q$ refer to:\nA: directional and disruptive selection respectively.\nB: stabilizing and directional selection respectively.\nC: directional selection.\nD: stabilizing selection.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTwo statements regarding evolution are made below.\n\nI. Rates of evolution are typically very slow because natural selection is usually $\\quad P$ selection.\n\nII. The plant population growing on high-zinc-soil is able to grow at concentrations which are otherwise lethal to plants of the same species. This is  $\\quad Q$  selection.\n\n$P$ and $Q$ refer to:\n\nA: directional and disruptive selection respectively.\nB: stabilizing and directional selection respectively.\nC: directional selection.\nD: stabilizing selection.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_839",
        "problem": "果蝇的长翅 (A) 与残翅 (a)、黑身（B）与黄身（b）为两对相对性状，且位于同一对常染色体上。一只基因型为 $\\mathrm{AaBb}$ 的雌果蝇在减数分裂过程中有 $20 \\%$ 的初级卵母细胞发生图示行为，但雄果蝇均不发生此行为，且基因型为 $\\mathrm{ab}$ 的雄配子中一半不育。基因型为 $\\mathrm{AaBb}$ 的雌雄个体杂交, 后代出现了一定数量的残翅黄身个体, 理论上子代中残翅黄身个体占比为（）\n[图1]\nA: $15 \\%$\nB: $25 \\%$\nC: $22.5 \\%$\nD: $27.5 \\%$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的长翅 (A) 与残翅 (a)、黑身（B）与黄身（b）为两对相对性状，且位于同一对常染色体上。一只基因型为 $\\mathrm{AaBb}$ 的雌果蝇在减数分裂过程中有 $20 \\%$ 的初级卵母细胞发生图示行为，但雄果蝇均不发生此行为，且基因型为 $\\mathrm{ab}$ 的雄配子中一半不育。基因型为 $\\mathrm{AaBb}$ 的雌雄个体杂交, 后代出现了一定数量的残翅黄身个体, 理论上子代中残翅黄身个体占比为（）\n[图1]\n\nA: $15 \\%$\nB: $25 \\%$\nC: $22.5 \\%$\nD: $27.5 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-57.jpg?height=440&width=1450&top_left_y=1228&top_left_x=352"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_565",
        "problem": "苯丙酮尿症是单基因遗传病。某种限制酶对致病基因和正常基因所在 DNA 片段以\n\n图 $\\mathrm{a}$ 和图 $\\mathrm{b}$ 方式进行切割, 形成 $23 \\mathrm{~kb}$ 或 $19 \\mathrm{~kb}$ 的片段, 某一个基因只能形成一种片段。\n\n某夫妻育有患病女儿, 为确定再次怀孕的胎儿是否患病, 进行家系分析和 DNA 检查,\n\n结果如图 $\\mathrm{c}$ 和图 $\\mathrm{d}$ 。不考虑 XY 同源区段及其他变异。下列对该家系的分析错误的是\n[图1]\n\nc\n[图2]\n\nd\nA: 该病的遗传方式为常染色体隐性遗传\nB: (1)号和(4)号的基因型可能是不相同的\nC: 经过诊断分析(4)号个体应该表现为正常\nD: (4)号 $19 \\mathrm{~kb}$ 片段中不一定含有正常基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n苯丙酮尿症是单基因遗传病。某种限制酶对致病基因和正常基因所在 DNA 片段以\n\n图 $\\mathrm{a}$ 和图 $\\mathrm{b}$ 方式进行切割, 形成 $23 \\mathrm{~kb}$ 或 $19 \\mathrm{~kb}$ 的片段, 某一个基因只能形成一种片段。\n\n某夫妻育有患病女儿, 为确定再次怀孕的胎儿是否患病, 进行家系分析和 DNA 检查,\n\n结果如图 $\\mathrm{c}$ 和图 $\\mathrm{d}$ 。不考虑 XY 同源区段及其他变异。下列对该家系的分析错误的是\n[图1]\n\nc\n[图2]\n\nd\n\nA: 该病的遗传方式为常染色体隐性遗传\nB: (1)号和(4)号的基因型可能是不相同的\nC: 经过诊断分析(4)号个体应该表现为正常\nD: (4)号 $19 \\mathrm{~kb}$ 片段中不一定含有正常基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-75.jpg?height=668&width=664&top_left_y=1529&top_left_x=364",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-75.jpg?height=660&width=674&top_left_y=1530&top_left_x=1111"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1128",
        "problem": "Three closely related species of genus Clusia are found in moist forests, dry deciduous forests and dry rocky coast of Trinidad island. In order to study their physiological response, they were raised under identical conditions in growth chambers. For one of these species, measurements were made under well-watered conditions ( 0 day) and after 5 days and 10 days without further irrigation. The results are shown: Graph with hollow circles is response of young leaves while solid line indicates mature leaves of the same plant.\n\n[figure1]\n\nWhich of the following is correct?\nA: Young leaves are unable to fix $\\mathrm{CO}_{2}$ in the night of 0 day while the mature leaves followed CAM pathway.\nB: Peak indicated as \"P\" in the graph is result of activation of \"RUBISCO\" enzyme.\nC: Young leaves are utilizing C3 pathway in the beginning of the experiment and then progressively assimilate $\\mathrm{CO}_{2}$ by CAM pathway.\nD: The response shown in the graph is most likely of a species that grows in moist forest.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThree closely related species of genus Clusia are found in moist forests, dry deciduous forests and dry rocky coast of Trinidad island. In order to study their physiological response, they were raised under identical conditions in growth chambers. For one of these species, measurements were made under well-watered conditions ( 0 day) and after 5 days and 10 days without further irrigation. The results are shown: Graph with hollow circles is response of young leaves while solid line indicates mature leaves of the same plant.\n\n[figure1]\n\nWhich of the following is correct?\n\nA: Young leaves are unable to fix $\\mathrm{CO}_{2}$ in the night of 0 day while the mature leaves followed CAM pathway.\nB: Peak indicated as \"P\" in the graph is result of activation of \"RUBISCO\" enzyme.\nC: Young leaves are utilizing C3 pathway in the beginning of the experiment and then progressively assimilate $\\mathrm{CO}_{2}$ by CAM pathway.\nD: The response shown in the graph is most likely of a species that grows in moist forest.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-18.jpg?height=637&width=1480&top_left_y=213&top_left_x=369"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_164",
        "problem": "SIR model was developed by Kermack and McKendrick (1927) to explain dynamics of an infectious diseases. In this model, there are only three subgroups in the population including susceptible, infected, and recovered (S, I, and R respectively).\n\nThe model assumes that everyone is born susceptible and there is no passive immunity for the infants, everyone who recovers from the disease is immune, the probability of getting the infection is the same for every susceptible person, and people in each group die with their own per capita death rate $\\left(\\mathrm{m}_{\\mathrm{s}}, \\mathrm{m}_{\\mathrm{I}}\\right.$, and $\\mathrm{m}_{\\mathrm{R}}$ ).\n\nIn this model, if a susceptible person contacts with an infected one, infection will be transmitted by probability of $A$ and the rate of recovery is $B$.\nA: For the disease to remain in the population without adding any new source of infection, A should be at least equal to $\\mathrm{B}$.\nB: The smaller the minimal infectious dose (the amount of pathogen required to cause an infection), the higher the $\\mathrm{A}$ is.\nC: If the disease is very lethal, it is more likely for it to be self-limited (the epidemic will finish all by itself).\nD: If the infection period is longer, a higher proportion of population is infected. True False\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nSIR model was developed by Kermack and McKendrick (1927) to explain dynamics of an infectious diseases. In this model, there are only three subgroups in the population including susceptible, infected, and recovered (S, I, and R respectively).\n\nThe model assumes that everyone is born susceptible and there is no passive immunity for the infants, everyone who recovers from the disease is immune, the probability of getting the infection is the same for every susceptible person, and people in each group die with their own per capita death rate $\\left(\\mathrm{m}_{\\mathrm{s}}, \\mathrm{m}_{\\mathrm{I}}\\right.$, and $\\mathrm{m}_{\\mathrm{R}}$ ).\n\nIn this model, if a susceptible person contacts with an infected one, infection will be transmitted by probability of $A$ and the rate of recovery is $B$.\n\nA: For the disease to remain in the population without adding any new source of infection, A should be at least equal to $\\mathrm{B}$.\nB: The smaller the minimal infectious dose (the amount of pathogen required to cause an infection), the higher the $\\mathrm{A}$ is.\nC: If the disease is very lethal, it is more likely for it to be self-limited (the epidemic will finish all by itself).\nD: If the infection period is longer, a higher proportion of population is infected. True False\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1458",
        "problem": "A population of dragonfly larvae (Leucorrhinia intacta) is separated into two groups. In both groups, larvae populations are put inside a cage with no food limitation. The first group is exposed to a fish predator that can swim freely, but cannot enter the cage. The second group is a control with no fish. The proportion of larvae surviving and the proportion of live larvae failing to metamorphose in the two groups are shown below:\n[figure1]\n\nWhich of the following is correct?\nA: One of the causes of high failure rate of metamorphosis of the larvae upon exposure to a non-lethal predator is cannibalism.\nB: The high mortality of larvae in the first group is due to predator-induced stress.\nC: In the predator treatment, the percentage of individuals that survived the larval stage completed emergence to the adult stage is higher than the percentage of those in the fishless treatment.\nD: The survival of dragonfly before metamorphosis is dependent on the predator while those during metamorphosis is not.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA population of dragonfly larvae (Leucorrhinia intacta) is separated into two groups. In both groups, larvae populations are put inside a cage with no food limitation. The first group is exposed to a fish predator that can swim freely, but cannot enter the cage. The second group is a control with no fish. The proportion of larvae surviving and the proportion of live larvae failing to metamorphose in the two groups are shown below:\n[figure1]\n\nWhich of the following is correct?\n\nA: One of the causes of high failure rate of metamorphosis of the larvae upon exposure to a non-lethal predator is cannibalism.\nB: The high mortality of larvae in the first group is due to predator-induced stress.\nC: In the predator treatment, the percentage of individuals that survived the larval stage completed emergence to the adult stage is higher than the percentage of those in the fishless treatment.\nD: The survival of dragonfly before metamorphosis is dependent on the predator while those during metamorphosis is not.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-12.jpg?height=674&width=1398&top_left_y=1302&top_left_x=271"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_83",
        "problem": "The figure 1 illustrated below shows oxygen consumption (respiration) in aqueous suspension of intact animal mitochondria with additions of ADP or chemical compounds (dinitrophenol (DNP) or N,N'dicyclohexylcarbodiimide (DCCD)). The suspension already contains respiratory substrates, oxygen, and inorganic phosphate.\n\n[figure1]\n\nFigure 1. Oxygen consumption of mitochondria in suspension. Identical aliquots of ADP were added in both experiments.\nA: The mitochondria are able to incorporate exogenously added ADP.\nB: Before addition of chemical compounds, the mitochondria respire only when ATP can be produced.\nC: The reason why DNP stimulates oxygen consumption is that ATP synthesis is stimulated by DNP.\nD: DCCD inhibits ATP synthesis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe figure 1 illustrated below shows oxygen consumption (respiration) in aqueous suspension of intact animal mitochondria with additions of ADP or chemical compounds (dinitrophenol (DNP) or N,N'dicyclohexylcarbodiimide (DCCD)). The suspension already contains respiratory substrates, oxygen, and inorganic phosphate.\n\n[figure1]\n\nFigure 1. Oxygen consumption of mitochondria in suspension. Identical aliquots of ADP were added in both experiments.\n\nA: The mitochondria are able to incorporate exogenously added ADP.\nB: Before addition of chemical compounds, the mitochondria respire only when ATP can be produced.\nC: The reason why DNP stimulates oxygen consumption is that ATP synthesis is stimulated by DNP.\nD: DCCD inhibits ATP synthesis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-06.jpg?height=611&width=1099&top_left_y=765&top_left_x=433"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_652",
        "problem": "FrancisGalton 爵士观察了 Juscelinomys 鼠的两个性状: 逃生洞穴的存在概率(图 1)和入口洞穴的长度(图 2)。先取 $\\mathrm{AB}$ 两个亲本杂交, 子一代 $\\left(\\mathrm{F}_{1}\\right)$ ) 再与亲本 $\\mathrm{A}$ 进行回交 (BC), 在 BC 鼠中入口长度于逃生洞穴的有无没有相关性。下列结论中不正确的是\n\n[图1]\nA: 挖掘逃生通道的等位基因对缺一个少逃生通道的基因为显性\nB: 子一代 $\\left(\\mathrm{F}_{1}\\right)$ 可以通过观察洞穴的方法从 $\\mathrm{AB}$ 亲本中分辨出来\nC: 入口通道的长度可能由多对基因决定\nD: 决定入口洞穴长度的基因和决定逃生通道有无的基因位于非同源染色体上\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nFrancisGalton 爵士观察了 Juscelinomys 鼠的两个性状: 逃生洞穴的存在概率(图 1)和入口洞穴的长度(图 2)。先取 $\\mathrm{AB}$ 两个亲本杂交, 子一代 $\\left(\\mathrm{F}_{1}\\right)$ ) 再与亲本 $\\mathrm{A}$ 进行回交 (BC), 在 BC 鼠中入口长度于逃生洞穴的有无没有相关性。下列结论中不正确的是\n\n[图1]\n\nA: 挖掘逃生通道的等位基因对缺一个少逃生通道的基因为显性\nB: 子一代 $\\left(\\mathrm{F}_{1}\\right)$ 可以通过观察洞穴的方法从 $\\mathrm{AB}$ 亲本中分辨出来\nC: 入口通道的长度可能由多对基因决定\nD: 决定入口洞穴长度的基因和决定逃生通道有无的基因位于非同源染色体上\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-24.jpg?height=494&width=560&top_left_y=1284&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1186",
        "problem": "During the $19^{\\text {th }}$ century an accidental hybridisation occurred between Spartina anglica $(2 n=56)$ and $S$.\n\nalterniflora $(2 n=70)$. The result was $S$. townsendii, a fertile hybrid. How many chromosomes would be expected in cells of the hybrid?\nA: 56\nB: 63\nC: 70\nD: 126\nE: 252\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDuring the $19^{\\text {th }}$ century an accidental hybridisation occurred between Spartina anglica $(2 n=56)$ and $S$.\n\nalterniflora $(2 n=70)$. The result was $S$. townsendii, a fertile hybrid. How many chromosomes would be expected in cells of the hybrid?\n\nA: 56\nB: 63\nC: 70\nD: 126\nE: 252\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_284",
        "problem": "As it is shown in the figure, the sensory neurons can modify the function of adjacent sensory pathway. This inhibitory effect is conducted via intermediate neurons at their synapse with second-order neurons. The width of neurons is correlated with their activity.\n\n[figure1]\nA: The firing rate of all second-order neurons is higher than to first-order neurons.\nB: The highest relative changes between second and first-order neurons will happen in Path B.\nC: The difference between the firing rate of second order neurons in $A$ and $B$ pathways will be higher than the difference between firing rate of the first order\nD: This mechanism may help to localize the sensory stimulus more accurately.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAs it is shown in the figure, the sensory neurons can modify the function of adjacent sensory pathway. This inhibitory effect is conducted via intermediate neurons at their synapse with second-order neurons. The width of neurons is correlated with their activity.\n\n[figure1]\n\nA: The firing rate of all second-order neurons is higher than to first-order neurons.\nB: The highest relative changes between second and first-order neurons will happen in Path B.\nC: The difference between the firing rate of second order neurons in $A$ and $B$ pathways will be higher than the difference between firing rate of the first order\nD: This mechanism may help to localize the sensory stimulus more accurately.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-25.jpg?height=870&width=1551&top_left_y=542&top_left_x=261"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_783",
        "problem": "果蝇的眼色受 $\\mathrm{A} 、 \\mathrm{a}$ 和 $\\mathrm{B} 、 \\mathrm{~b}$ 两对等位基因的控制, 色素的产生必须有显性基因 A;第二个显性基因 B 使色素呈紫色, 隐性基因 $\\mathrm{b}$ 使色素呈红色, 不产生色素的个体眼睛呈白色。现有一对红眼雌性与白眼雄性果蝇杂交, $F_{1}$ 为紫眼雌性与红眼雄性。 $F_{1}$ 雌雄果蝇交配后, $\\mathrm{F}_{2}$ 中紫眼：红眼： 白眼=3：3：2. 下列相关分析错误的是（）\nA: 眼色的遗传符合基因的自由组合定律\nB: $\\mathrm{F}_{1}$ 代雌果蝇的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$\nC: $F_{2}$ 中紫眼果蝇的基因型共有 4 种\nD: $F_{2}$ 雌果蝇中杂合子的比例 $2 / 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的眼色受 $\\mathrm{A} 、 \\mathrm{a}$ 和 $\\mathrm{B} 、 \\mathrm{~b}$ 两对等位基因的控制, 色素的产生必须有显性基因 A;第二个显性基因 B 使色素呈紫色, 隐性基因 $\\mathrm{b}$ 使色素呈红色, 不产生色素的个体眼睛呈白色。现有一对红眼雌性与白眼雄性果蝇杂交, $F_{1}$ 为紫眼雌性与红眼雄性。 $F_{1}$ 雌雄果蝇交配后, $\\mathrm{F}_{2}$ 中紫眼：红眼： 白眼=3：3：2. 下列相关分析错误的是（）\n\nA: 眼色的遗传符合基因的自由组合定律\nB: $\\mathrm{F}_{1}$ 代雌果蝇的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$\nC: $F_{2}$ 中紫眼果蝇的基因型共有 4 种\nD: $F_{2}$ 雌果蝇中杂合子的比例 $2 / 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1544",
        "problem": "*Secondary metabolism* of plants is not essential for survival, but can be important. Many secondary metabolites, such as nicotine and caffeine, help them resist damage from herbivorous insects. Glucosinolate is accumulated in the leaves of *Arabidopsis* and repels insects. Segments from the centre or edges of leaves were arranged around caterpillars.\n[figure1]\n\nThe leaves caterpillars chose were recorded. This was repeated for wild type (Wild) plants and mutants incapable of synthesizing glucosinolate.\n\n[figure2]\n\nWhich is true?\nA: Glucosinolate accumulates more in the edge of leaves.\nB: Glucosianate accumulates equally across leaves.\nC: The inside of leaves is more nutritious for caterpillars.\nD: The caterpillars are expected to choose wild-type leaves in preference to mutant leaves.\nE: The caterpillars never eat leaves containing glucosinolate\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n*Secondary metabolism* of plants is not essential for survival, but can be important. Many secondary metabolites, such as nicotine and caffeine, help them resist damage from herbivorous insects. Glucosinolate is accumulated in the leaves of *Arabidopsis* and repels insects. Segments from the centre or edges of leaves were arranged around caterpillars.\n[figure1]\n\nThe leaves caterpillars chose were recorded. This was repeated for wild type (Wild) plants and mutants incapable of synthesizing glucosinolate.\n\n[figure2]\n\nWhich is true?\n\nA: Glucosinolate accumulates more in the edge of leaves.\nB: Glucosianate accumulates equally across leaves.\nC: The inside of leaves is more nutritious for caterpillars.\nD: The caterpillars are expected to choose wild-type leaves in preference to mutant leaves.\nE: The caterpillars never eat leaves containing glucosinolate\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-23.jpg?height=626&width=1054&top_left_y=560&top_left_x=244",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-23.jpg?height=783&width=1016&top_left_y=1379&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1185",
        "problem": "## RATITE EVOLUTION\n\nRatites are named for their flat \"raft-like\" sternum that lacks a keel and thus cannot provide sufficient anchorage for flight muscles to effectively generate the power required to fly. Cassowary (Australia and Papua New Guinea), Emu (Australia), Rhea (South America), Ostrich (Africa), Kiwi (New Zealand), and extinct Moa (New Zealand) are all examples of ground dwelling, flightless ratites. Given their large, flightless nature, biologists have proposed that they have all descended from a common ancestor present in ancient Gondwana before it broke apart. The exact timing and evolutionary origin of the Moa and Kiwi in New Zealand have been greatly debated in the scientific literature.\n\n[figure1]\nfrom Antarctica and India\n\n- 130 mya South America breaks free of Africa\n- 80 mya Zealandia breaks free\n- 60 mya New Zealand separates from Australia\n- 40 mya Australia separates from Antarctica\n- 30 mya South America breaks from West Antarctica (Antarctica freezes over)\n\n[figure2]\n\nFuture Zealandia\n\n## Flying Cousins\n\nRatites are one of only two groups of birds belonging to the \"old jaw\" Paleognaths originating in Gondwana, the other comprises the 47 living Tinamou species of Central and South America. Tinamou species are generally ground dwelling, though they do have wings that allow for limited flight.\n\nIt is generally accepted that loss of flight in birds is due to the development of successful foraging behaviours and diminished predation on eggs and nests.\n\n## Fossil Evidence\n\nExamination of the oldest known Moa fossils, dating from 19 million years ago, determined they had thickened leg bones and no wing structures at all. At least two species of flightless Moa were present in New Zealand at this time.\n\nThe oldest Tinamou fossils in South America are 10 million years old.\n\nThe oldest Kiwi fossil dates back 1 million years. Though diminished in size, kiwis have wings and flight feathers. Kiwis are the only known bird to have nostrils at the end of their beaks and the only ratite with two functioning ovaries.According to the estimated break up of Gondwanaland the BEST hypothesis is that?\nA: Ostrich and Cassowary are the most distantly related ratites.\nB: Tinamou and Cassowary are the most distantly related ratites.\nC: Kiwi evolved from Moa.\nD: Tinamou and Moa are the most closely related ratites.\nE: Moa evolved from Ostriches.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## RATITE EVOLUTION\n\nRatites are named for their flat \"raft-like\" sternum that lacks a keel and thus cannot provide sufficient anchorage for flight muscles to effectively generate the power required to fly. Cassowary (Australia and Papua New Guinea), Emu (Australia), Rhea (South America), Ostrich (Africa), Kiwi (New Zealand), and extinct Moa (New Zealand) are all examples of ground dwelling, flightless ratites. Given their large, flightless nature, biologists have proposed that they have all descended from a common ancestor present in ancient Gondwana before it broke apart. The exact timing and evolutionary origin of the Moa and Kiwi in New Zealand have been greatly debated in the scientific literature.\n\n[figure1]\nfrom Antarctica and India\n\n- 130 mya South America breaks free of Africa\n- 80 mya Zealandia breaks free\n- 60 mya New Zealand separates from Australia\n- 40 mya Australia separates from Antarctica\n- 30 mya South America breaks from West Antarctica (Antarctica freezes over)\n\n[figure2]\n\nFuture Zealandia\n\n## Flying Cousins\n\nRatites are one of only two groups of birds belonging to the \"old jaw\" Paleognaths originating in Gondwana, the other comprises the 47 living Tinamou species of Central and South America. Tinamou species are generally ground dwelling, though they do have wings that allow for limited flight.\n\nIt is generally accepted that loss of flight in birds is due to the development of successful foraging behaviours and diminished predation on eggs and nests.\n\n## Fossil Evidence\n\nExamination of the oldest known Moa fossils, dating from 19 million years ago, determined they had thickened leg bones and no wing structures at all. At least two species of flightless Moa were present in New Zealand at this time.\n\nThe oldest Tinamou fossils in South America are 10 million years old.\n\nThe oldest Kiwi fossil dates back 1 million years. Though diminished in size, kiwis have wings and flight feathers. Kiwis are the only known bird to have nostrils at the end of their beaks and the only ratite with two functioning ovaries.\n\nproblem:\nAccording to the estimated break up of Gondwanaland the BEST hypothesis is that?\n\nA: Ostrich and Cassowary are the most distantly related ratites.\nB: Tinamou and Cassowary are the most distantly related ratites.\nC: Kiwi evolved from Moa.\nD: Tinamou and Moa are the most closely related ratites.\nE: Moa evolved from Ostriches.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-24.jpg?height=711&width=585&top_left_y=689&top_left_x=135",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-24.jpg?height=708&width=1259&top_left_y=688&top_left_x=747"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_81",
        "problem": "Which of the following statements about speciation is correct?\nA: Sympatric speciation occurs more gradually and more slowly than allopatric speciation.\nB: The divergence of two maggot fly races is an example of allopatric speciation due to mating time differences.\nC: The evolution of cultivated wheat is associated with polyploidization. This is an example of sympatric speciation.\nD: Allopatric speciation is usually associated with stronger secondary reproductive barriers than sympatric speciation.\nE: Different species of Drosophila inhabit the different islands of Hawaii. This is an example of sympatric speciation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following statements about speciation is correct?\n\nA: Sympatric speciation occurs more gradually and more slowly than allopatric speciation.\nB: The divergence of two maggot fly races is an example of allopatric speciation due to mating time differences.\nC: The evolution of cultivated wheat is associated with polyploidization. This is an example of sympatric speciation.\nD: Allopatric speciation is usually associated with stronger secondary reproductive barriers than sympatric speciation.\nE: Different species of Drosophila inhabit the different islands of Hawaii. This is an example of sympatric speciation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_716",
        "problem": "CMTI 腓骨肌萎缩症由等位基因 $\\mathrm{A} / \\mathrm{a}$ 控制.鱼鳞病由等位基因 $\\mathrm{B} / \\mathrm{b}$ 控制。图 1 表示某家族遗传系谱图。图 2 表示乙家庭中的部分成员鱼鳞病基因电泳图, 下列说法正确的是\n[图1]\n\n图1\n[图2]\n\n## 图2\nA: 据图判断鱼鳞病为伴 X 染色体显性遗传病\nB: 图 1 中 $\\mathrm{II}_{6}$ 的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{B}}$ 或 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$\nC: 基因 B 发生碱基对的增添形成等位基因 b\nD: $\\mathrm{II}_{6}$ 和 $\\mathrm{II}_{7}$ 再生一个患病孩子的概率为 $7 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nCMTI 腓骨肌萎缩症由等位基因 $\\mathrm{A} / \\mathrm{a}$ 控制.鱼鳞病由等位基因 $\\mathrm{B} / \\mathrm{b}$ 控制。图 1 表示某家族遗传系谱图。图 2 表示乙家庭中的部分成员鱼鳞病基因电泳图, 下列说法正确的是\n[图1]\n\n图1\n[图2]\n\n## 图2\n\nA: 据图判断鱼鳞病为伴 X 染色体显性遗传病\nB: 图 1 中 $\\mathrm{II}_{6}$ 的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{B}}$ 或 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}$\nC: 基因 B 发生碱基对的增添形成等位基因 b\nD: $\\mathrm{II}_{6}$ 和 $\\mathrm{II}_{7}$ 再生一个患病孩子的概率为 $7 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-084.jpg?height=416&width=1332&top_left_y=206&top_left_x=381",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-084.jpg?height=284&width=804&top_left_y=794&top_left_x=404"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_638",
        "problem": "蚕豆病是红细胞 G6PD（葡萄糖-6-磷酸脱氢酶）缺乏者进食蚕豆或接触蚕豆花粉后\n发生的急性溶血性疾病，是一种单基因遗传病，患者中男性约占 90\\%。研究表明，GA、 $\\mathrm{GB} 、 \\mathrm{~g}$ 互为等位基因, 只有 $\\mathrm{GA} 、 \\mathrm{~GB}$ 能控制合成 G6PD。图 1 为某家族蚕豆病遗传系谱图, 图 2 为该家族部分成员相关基因的电泳图谱。以下相关叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 依据信息推测蚕豆病的遗传方式为伴 X染色体隐性遗传\nB: 该病基因通过控制酶的合成来控制代谢过程，进而控制生物体的性状\nC: II-7 患病, 可能是 DNA 碱基发生了甲基化修饰, 这种修饰可遗传给后代\nD: 通过遗传咨询和产前诊断的手段, 可以对遗传病进行检测、预防和治疗\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n蚕豆病是红细胞 G6PD（葡萄糖-6-磷酸脱氢酶）缺乏者进食蚕豆或接触蚕豆花粉后\n发生的急性溶血性疾病，是一种单基因遗传病，患者中男性约占 90\\%。研究表明，GA、 $\\mathrm{GB} 、 \\mathrm{~g}$ 互为等位基因, 只有 $\\mathrm{GA} 、 \\mathrm{~GB}$ 能控制合成 G6PD。图 1 为某家族蚕豆病遗传系谱图, 图 2 为该家族部分成员相关基因的电泳图谱。以下相关叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 依据信息推测蚕豆病的遗传方式为伴 X染色体隐性遗传\nB: 该病基因通过控制酶的合成来控制代谢过程，进而控制生物体的性状\nC: II-7 患病, 可能是 DNA 碱基发生了甲基化修饰, 这种修饰可遗传给后代\nD: 通过遗传咨询和产前诊断的手段, 可以对遗传病进行检测、预防和治疗\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-21.jpg?height=320&width=894&top_left_y=405&top_left_x=344",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-21.jpg?height=223&width=539&top_left_y=391&top_left_x=1244"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_248",
        "problem": "When carbon isotopes $\\left({ }^{13} \\mathrm{C}\\right.$ and ${ }^{12} \\mathrm{C}$ ) are analyzed, plants can be categorized into two groups (Figure 1), based on the isotope fractionation $\\left(\\delta^{13} \\mathrm{C}\\right.$, equation 1$)$. This is because of the slight differences in molecular mass between ${ }^{13} \\mathrm{CO}_{2}$ and ${ }^{12} \\mathrm{CO}_{2}$, although there are no known chemical differences between them. In photosynthesis, two types of carboxylase enzymes fix carbon from $\\mathrm{CO}_{2}$ in the two groups, provided that $\\mathrm{CO}_{2}$ is converted to $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ by an enzyme carbonic anhydrase.\n\nReaction 1: $\\mathrm{C}_{3} \\mathrm{H}_{5} \\mathrm{O}_{6} \\mathrm{P}+\\mathrm{H}_{2} \\mathrm{CO}_{3} \\rightarrow \\mathrm{C}_{4} \\mathrm{H}_{4} \\mathrm{O}_{5}+\\mathrm{H}_{3} \\mathrm{PO}_{4} \\quad$ (C4 plants)\n\nReaction 2: $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{O}_{11} \\mathrm{P}_{2}+\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow 2 \\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{O}_{7} \\mathrm{P} \\quad(\\mathrm{C} 3$ plants)\n\n[figure1]\n\nSample: a plant material Standard: the reference represents the typical carbon on the Earth\n\n[figure2]\n\nFigure 1 Distribution of the carbon isotope fractionation $\\left(\\delta^{13} \\mathrm{C}\\right.$ value) of various plants.\nA: The relative difference in molecular mass due to the carbon isotopes is larger in $\\mathrm{CO}_{2}$ than $\\mathrm{H}_{2} \\mathrm{CO}_{3}$.\nB: Reaction 1 is catalyzed by ribulose 1,5-bisphosphate carboxylase/oxygenase (RubisCO).\nC: Both groups of plants discriminate between the isotopes.\nD: Rice belongs to group 1 and corn (maize) belongs to group 2 . .....\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nWhen carbon isotopes $\\left({ }^{13} \\mathrm{C}\\right.$ and ${ }^{12} \\mathrm{C}$ ) are analyzed, plants can be categorized into two groups (Figure 1), based on the isotope fractionation $\\left(\\delta^{13} \\mathrm{C}\\right.$, equation 1$)$. This is because of the slight differences in molecular mass between ${ }^{13} \\mathrm{CO}_{2}$ and ${ }^{12} \\mathrm{CO}_{2}$, although there are no known chemical differences between them. In photosynthesis, two types of carboxylase enzymes fix carbon from $\\mathrm{CO}_{2}$ in the two groups, provided that $\\mathrm{CO}_{2}$ is converted to $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ by an enzyme carbonic anhydrase.\n\nReaction 1: $\\mathrm{C}_{3} \\mathrm{H}_{5} \\mathrm{O}_{6} \\mathrm{P}+\\mathrm{H}_{2} \\mathrm{CO}_{3} \\rightarrow \\mathrm{C}_{4} \\mathrm{H}_{4} \\mathrm{O}_{5}+\\mathrm{H}_{3} \\mathrm{PO}_{4} \\quad$ (C4 plants)\n\nReaction 2: $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{O}_{11} \\mathrm{P}_{2}+\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow 2 \\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{O}_{7} \\mathrm{P} \\quad(\\mathrm{C} 3$ plants)\n\n[figure1]\n\nSample: a plant material Standard: the reference represents the typical carbon on the Earth\n\n[figure2]\n\nFigure 1 Distribution of the carbon isotope fractionation $\\left(\\delta^{13} \\mathrm{C}\\right.$ value) of various plants.\n\nA: The relative difference in molecular mass due to the carbon isotopes is larger in $\\mathrm{CO}_{2}$ than $\\mathrm{H}_{2} \\mathrm{CO}_{3}$.\nB: Reaction 1 is catalyzed by ribulose 1,5-bisphosphate carboxylase/oxygenase (RubisCO).\nC: Both groups of plants discriminate between the isotopes.\nD: Rice belongs to group 1 and corn (maize) belongs to group 2 . .....\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-07.jpg?height=227&width=1024&top_left_y=992&top_left_x=305",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-07.jpg?height=703&width=1194&top_left_y=1393&top_left_x=428"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_713",
        "problem": "用㱧豆进行遗传试验时, 下列操作错误的是 ( )\nA: 杂交时, 须在开花前除去母本的雄莈\nB: 自交时, 雌芯和雄荵都无需除去\nC: 杂交时, 须在开花前除去母本的雌菭\nD: 人工授粉后，应套袋\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n用㱧豆进行遗传试验时, 下列操作错误的是 ( )\n\nA: 杂交时, 须在开花前除去母本的雄莈\nB: 自交时, 雌芯和雄荵都无需除去\nC: 杂交时, 须在开花前除去母本的雌菭\nD: 人工授粉后，应套袋\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_495",
        "problem": "某植物为二倍体雌雄同株同花植物, 自然状态下可以自花受粉或异花受粉。其花色受 $\\mathrm{A}$ (红色)、 $\\mathrm{A}^{\\mathrm{P}}$ (斑红色)、 $\\mathrm{A}^{\\mathrm{T}}$ (条红色)、a（白色） 4 个复等位基因控制, 4 个复等位基因的显隐性关系为 $\\mathrm{A}>\\mathrm{A}^{\\mathrm{P}}>\\mathrm{A}^{\\mathrm{T}}>\\mathrm{a}$ 。 $\\mathrm{A}^{\\mathrm{T}}$ 是一种“自私基因”，在产生配子时会导致同株一定比例的其他花粉死亡，使其有更多的机会遗传下去。基因型为 $\\mathrm{A}^{\\mathrm{T}} \\mathrm{a}$ 的植株自交， $\\mathrm{F}_{1}$中条红色: 白色 $=5: 1$. 下列叙述正确的是 ( )\nA: 花色基因的遗传遵循孟德尔自由组合定律\nB: 两株花色不同的植株杂交, 子代花色最多有 4 种\nC: 等比例的 $\\mathrm{AA}^{\\mathrm{P}}$ 与 $\\mathrm{A}^{\\mathrm{T}} \\mathrm{a}$ 植株随机交配, $\\mathrm{F}_{1}$ 中含“自私基因”的植株所占比例为 $15 / 28$\nD: 基因型为 $\\mathrm{A}^{\\mathrm{T}} \\mathrm{a}$ 的植株自交, $\\mathrm{F}_{1}$ 条红色植株中能稳定遗传的占 $2 / 5$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某植物为二倍体雌雄同株同花植物, 自然状态下可以自花受粉或异花受粉。其花色受 $\\mathrm{A}$ (红色)、 $\\mathrm{A}^{\\mathrm{P}}$ (斑红色)、 $\\mathrm{A}^{\\mathrm{T}}$ (条红色)、a（白色） 4 个复等位基因控制, 4 个复等位基因的显隐性关系为 $\\mathrm{A}>\\mathrm{A}^{\\mathrm{P}}>\\mathrm{A}^{\\mathrm{T}}>\\mathrm{a}$ 。 $\\mathrm{A}^{\\mathrm{T}}$ 是一种“自私基因”，在产生配子时会导致同株一定比例的其他花粉死亡，使其有更多的机会遗传下去。基因型为 $\\mathrm{A}^{\\mathrm{T}} \\mathrm{a}$ 的植株自交， $\\mathrm{F}_{1}$中条红色: 白色 $=5: 1$. 下列叙述正确的是 ( )\n\nA: 花色基因的遗传遵循孟德尔自由组合定律\nB: 两株花色不同的植株杂交, 子代花色最多有 4 种\nC: 等比例的 $\\mathrm{AA}^{\\mathrm{P}}$ 与 $\\mathrm{A}^{\\mathrm{T}} \\mathrm{a}$ 植株随机交配, $\\mathrm{F}_{1}$ 中含“自私基因”的植株所占比例为 $15 / 28$\nD: 基因型为 $\\mathrm{A}^{\\mathrm{T}} \\mathrm{a}$ 的植株自交, $\\mathrm{F}_{1}$ 条红色植株中能稳定遗传的占 $2 / 5$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1219",
        "problem": "Mary's father has five daughters: 1. Nana, 2. Nemu, 3. Nino, 4. Nomi. What is the name of the fifth daughter?\nA: Nune\nB: Nume\nC: Nunu\nD: Numu\nE: Mary\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMary's father has five daughters: 1. Nana, 2. Nemu, 3. Nino, 4. Nomi. What is the name of the fifth daughter?\n\nA: Nune\nB: Nume\nC: Nunu\nD: Numu\nE: Mary\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1203",
        "problem": "This is a graph of the menstrual cycle showing the hormones FSH and oestradiol. The labels at the bottom refer to the phase of the cycle.\n\n[figure1]\nA: ![]([figure2])\nB: ![]([figure3])\nC: ![]([figure4])\nD: ![]([figure5])\nE: ![]([figure6])\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThis is a graph of the menstrual cycle showing the hormones FSH and oestradiol. The labels at the bottom refer to the phase of the cycle.\n\n[figure1]\n\nA: ![]([figure2])\nB: ![]([figure3])\nC: ![]([figure4])\nD: ![]([figure5])\nE: ![]([figure6])\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-03.jpg?height=731&width=939&top_left_y=1348&top_left_x=133",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-03.jpg?height=60&width=186&top_left_y=1455&top_left_x=1186",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-03.jpg?height=123&width=154&top_left_y=1469&top_left_x=1622",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-03.jpg?height=234&width=180&top_left_y=1542&top_left_x=1189",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-03.jpg?height=257&width=145&top_left_y=1668&top_left_x=1618",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-03.jpg?height=111&width=149&top_left_y=1909&top_left_x=1202"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_112",
        "problem": "Which one of the following graphs shows the relative change in the amount of mitochondrial DNA of a cell undergoing mitosis?\n[figure1]\nA: A\nB: B\nC: C\nD: D\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following graphs shows the relative change in the amount of mitochondrial DNA of a cell undergoing mitosis?\n[figure1]\n\nA: A\nB: B\nC: C\nD: D\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-11.jpg?height=1960&width=1328&top_left_y=480&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1420",
        "problem": "The following graph illustrates the change in membrane potential $(\\mathrm{mV})$ during an action potential.\n\n[figure1]\n\nA drug that prevents the opening of voltage-gated sodium channels is administered. What effect will this have on the action potential?\nA: Repolarisation will occur more slowly.\nB: The threshold voltage required to produce an action potential will increase.\nC: No action potential will be generated, regardless of the size of the stimulus.\nD: Voltage-gated potassium channels will open instead of voltage-gated sodium channels.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph illustrates the change in membrane potential $(\\mathrm{mV})$ during an action potential.\n\n[figure1]\n\nA drug that prevents the opening of voltage-gated sodium channels is administered. What effect will this have on the action potential?\n\nA: Repolarisation will occur more slowly.\nB: The threshold voltage required to produce an action potential will increase.\nC: No action potential will be generated, regardless of the size of the stimulus.\nD: Voltage-gated potassium channels will open instead of voltage-gated sodium channels.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-14.jpg?height=651&width=694&top_left_y=443&top_left_x=270"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_175",
        "problem": "Cooperation is a complicated behaviour that has evolved in animals. Conditional cooperation, such as direct and indirect reciprocity and partner choice, are some of these complicated behaviours. The simple rule which can illustrate how such behaviour evolves is the correlation between the behaviour of interacting individuals (graph below). In fact, such correlation results in cooperative individuals receiving more cooperation, so if there is a cheater, it receives less cooperation, which is a kind of punishment.\n\n[figure1]\n\nThe relationship between deception and cooperation in non-human primates (After McNally, et al., 2013).\nA: Conditional strategies and mechanisms that enforce cooperation make the evolution of deception less likely.\nB: Although overall net benefit of the cooperative behaviour is higher than the cost of being deceived, the existence of defectors doesn't allow cooperation to persist.\nC: We can expect deception to evolve as long as it does not significantly increase the cost of cooperation.\nD: Deceptive behaviour can occur only in novel habitats as there is no time to adapt to the new environmental conditions.\nE: The relationship between cooperativeness and deception rate in primates can be explained by the relatively large social group size in this taxon.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nCooperation is a complicated behaviour that has evolved in animals. Conditional cooperation, such as direct and indirect reciprocity and partner choice, are some of these complicated behaviours. The simple rule which can illustrate how such behaviour evolves is the correlation between the behaviour of interacting individuals (graph below). In fact, such correlation results in cooperative individuals receiving more cooperation, so if there is a cheater, it receives less cooperation, which is a kind of punishment.\n\n[figure1]\n\nThe relationship between deception and cooperation in non-human primates (After McNally, et al., 2013).\n\nA: Conditional strategies and mechanisms that enforce cooperation make the evolution of deception less likely.\nB: Although overall net benefit of the cooperative behaviour is higher than the cost of being deceived, the existence of defectors doesn't allow cooperation to persist.\nC: We can expect deception to evolve as long as it does not significantly increase the cost of cooperation.\nD: Deceptive behaviour can occur only in novel habitats as there is no time to adapt to the new environmental conditions.\nE: The relationship between cooperativeness and deception rate in primates can be explained by the relatively large social group size in this taxon.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-50.jpg?height=603&width=834&top_left_y=675&top_left_x=611"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1331",
        "problem": "The following diagram represents the pathways conveying the sensory information in the human body to the sensory part of the brain.\n\n[figure1]\n\nIn brief, the pathway conveying touch and pressure (shown in grey) passes from the left side of the body to the spinal cord and ascends to the medulla oblongata, where it crosses over to the other side and reaches the sensory regions of the right brain. For the pain and temperature pathway (shown in black), the sensory information passes from the left side of the body to the right side of the spinal cord before ascending up into the medulla and finally onto the right part of the brain. The same process occurs for pain and temperature, and touch and pressure pathways from the right side of the body.Following from the previous question, what would happen if the medulla oblongata was completely damaged?\nA: There would be no loss of sensations.\nB: There would be no sensations coming from the left side of the body only.\nC: There would be no sensations coming from the right side of the body only.\nD: There would be a complete loss of sensation.\nE: None of the options is correct.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nThe following diagram represents the pathways conveying the sensory information in the human body to the sensory part of the brain.\n\n[figure1]\n\nIn brief, the pathway conveying touch and pressure (shown in grey) passes from the left side of the body to the spinal cord and ascends to the medulla oblongata, where it crosses over to the other side and reaches the sensory regions of the right brain. For the pain and temperature pathway (shown in black), the sensory information passes from the left side of the body to the right side of the spinal cord before ascending up into the medulla and finally onto the right part of the brain. The same process occurs for pain and temperature, and touch and pressure pathways from the right side of the body.\n\nproblem:\nFollowing from the previous question, what would happen if the medulla oblongata was completely damaged?\n\nA: There would be no loss of sensations.\nB: There would be no sensations coming from the left side of the body only.\nC: There would be no sensations coming from the right side of the body only.\nD: There would be a complete loss of sensation.\nE: None of the options is correct.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-14.jpg?height=845&width=1516&top_left_y=431&top_left_x=224"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_580",
        "problem": "已知早金莲的花长受三对等位基因控制, 这三对基因分别位于三对同源染色体上,作用相等且具有叠加性。最长花长为 $30 \\mathrm{~mm}$ 的旱金莲与最短花长为 $12 \\mathrm{~mm}$ 的旱金莲相互授粉, 子代花长均为 $21 \\mathrm{mnn}$ 。花长为 $24 \\mathrm{~mm}$ 的植株自交, 后代出现性状分离, 其中花长为 $24 \\mathrm{~mm}$ 的个体所占比例是\nA: $1 / 16$\nB: $1 / 8$\nC: $5 / 16$\nD: $3 / 8$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知早金莲的花长受三对等位基因控制, 这三对基因分别位于三对同源染色体上,作用相等且具有叠加性。最长花长为 $30 \\mathrm{~mm}$ 的旱金莲与最短花长为 $12 \\mathrm{~mm}$ 的旱金莲相互授粉, 子代花长均为 $21 \\mathrm{mnn}$ 。花长为 $24 \\mathrm{~mm}$ 的植株自交, 后代出现性状分离, 其中花长为 $24 \\mathrm{~mm}$ 的个体所占比例是\n\nA: $1 / 16$\nB: $1 / 8$\nC: $5 / 16$\nD: $3 / 8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_781",
        "problem": "现有三个纯合品系 $A 、 B 、 C$ (已将抗病基因 $\\mathrm{H}$ 导入胞内)。为确定不同品系 $\\mathrm{H}$ 基因所在染色体位置, 研究人员进行了下表杂交实验（各类型配子活性正常且相同）。下列相关叙述错误的是（ ）\n\n| 杂交组合 | $F_{1}$ |  | $F_{2}$ |\n| :--- | :--- | :--- | :--- |\n| 组合一; $A \\times B$ | 全部抗病 | $F_{1}$ 自交 | 抗病 449, 非抗病 30 |\n| 组合二: $A \\times C$ | 全部抗病 |  | 抗病 301, 非抗病 19 |\n\n\n| 组合三: $\\mathrm{B} \\times \\mathrm{C}$ | 全部抗病 |  | 非抗病约 $2 \\%$, 其余均抗病 |\n| :--- | :--- | :--- | :--- |\nA: 品系 $\\mathrm{A}$ 与品系 $\\mathrm{B}$ 中 $\\mathrm{H}$ 基因位于非同源染色体上\nB: 组合二 $F_{2}$ 的抗病个体中约 $1 / 5$ 个体自交后代不发生性状分离\nC: 品系 $\\mathrm{B}$ 与品系 $\\mathrm{C}$ 中的 $\\mathrm{H}$ 基因可能位于一对同源染色体上\nD: 组合三 $F_{2}$ 的非抗病个体的产生可能与染色体交叉互换有关\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n现有三个纯合品系 $A 、 B 、 C$ (已将抗病基因 $\\mathrm{H}$ 导入胞内)。为确定不同品系 $\\mathrm{H}$ 基因所在染色体位置, 研究人员进行了下表杂交实验（各类型配子活性正常且相同）。下列相关叙述错误的是（ ）\n\n| 杂交组合 | $F_{1}$ |  | $F_{2}$ |\n| :--- | :--- | :--- | :--- |\n| 组合一; $A \\times B$ | 全部抗病 | $F_{1}$ 自交 | 抗病 449, 非抗病 30 |\n| 组合二: $A \\times C$ | 全部抗病 |  | 抗病 301, 非抗病 19 |\n\n\n| 组合三: $\\mathrm{B} \\times \\mathrm{C}$ | 全部抗病 |  | 非抗病约 $2 \\%$, 其余均抗病 |\n| :--- | :--- | :--- | :--- |\n\nA: 品系 $\\mathrm{A}$ 与品系 $\\mathrm{B}$ 中 $\\mathrm{H}$ 基因位于非同源染色体上\nB: 组合二 $F_{2}$ 的抗病个体中约 $1 / 5$ 个体自交后代不发生性状分离\nC: 品系 $\\mathrm{B}$ 与品系 $\\mathrm{C}$ 中的 $\\mathrm{H}$ 基因可能位于一对同源染色体上\nD: 组合三 $F_{2}$ 的非抗病个体的产生可能与染色体交叉互换有关\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-041.jpg?height=277&width=440&top_left_y=1181&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-041.jpg?height=271&width=277&top_left_y=1184&top_left_x=798",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-041.jpg?height=271&width=279&top_left_y=1184&top_left_x=1114"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_923",
        "problem": "2022 年北京冬奥会吉祥物“冰墩墩”, 大熊猫是其设计原型。大熊猫最初是食肉动物,经过进化, 其 $99 \\%$ 的食物都来源于竹子。现在一个较大的熊猫种群中雌雄数量相等, 且雌雄之间可以自由交配，若该种群中 B 的基因频率为 40\\%, b 的基因频率为 $60 \\%$, 下列有关说法错误的是（ ）\nA: 大熊猫种群中全部 B 和 $\\mathrm{b}$ 的总和不能构成大熊猫种群的基因库\nB: 大熊猫由以肉为食进化为以竹子为食的实质是种群基因频率的定向改变\nC: 若该对等位基因位于常染色体上，则显性个体中出现杂合雌熊猫概率为 37.5\\%\nD: 若该对等位基因只位于 $\\mathrm{X}$ 染色体上，则 $\\mathrm{X}^{\\mathrm{b}} \\mathrm{Y}$ 的基因型频率为 $60 \\%$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2022 年北京冬奥会吉祥物“冰墩墩”, 大熊猫是其设计原型。大熊猫最初是食肉动物,经过进化, 其 $99 \\%$ 的食物都来源于竹子。现在一个较大的熊猫种群中雌雄数量相等, 且雌雄之间可以自由交配，若该种群中 B 的基因频率为 40\\%, b 的基因频率为 $60 \\%$, 下列有关说法错误的是（ ）\n\nA: 大熊猫种群中全部 B 和 $\\mathrm{b}$ 的总和不能构成大熊猫种群的基因库\nB: 大熊猫由以肉为食进化为以竹子为食的实质是种群基因频率的定向改变\nC: 若该对等位基因位于常染色体上，则显性个体中出现杂合雌熊猫概率为 37.5\\%\nD: 若该对等位基因只位于 $\\mathrm{X}$ 染色体上，则 $\\mathrm{X}^{\\mathrm{b}} \\mathrm{Y}$ 的基因型频率为 $60 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_94",
        "problem": "The graphs below show sucrose (Suc)- and/or indole 3-acetic acid (IAA, an auxin)-induced cell growth (Figure $a$ ) and the kinetics of IAA-induced cell elongation and cell wall acidification in coleoptiles (Figure $b$ ). Based on these results, together with the fact that these processes are delayed by cold treatments or inhibitors of protein synthesis, the \"acid-growth hypothesis\" was proposed as the best model to explain auxin-induced cell growth.\n\n$a$\n\n[figure1]\n\n$b$\n\n[figure2]\n\nWhich of the following statements is most accurate?\nA: IAA-driven protons, pumped into the cell wall, are utilized to synthesize the ATP required for cell elongation.\nB: IAA-induced acidification of the cell wall is an ATP-dependent process, and can be delayed by a treatment of a metabolic inhibitor.\nC: IAA-induced loosening of the cell wall is mainly caused by an acidification-induced weakening of the covalent bonds in cell wall proteins.\nD: IAA- or sucrose-induced cell elongation shares a common action mechanism, such as an increase in the cell wall acidity and the following change in turgor pressure.\nE: Cell wall acidification and stimulation in the elongation is an IAA-specific process, thus it is not induced by treatment with Fusicoccoin, an activator of the proton pump, in the absence of the IAA.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graphs below show sucrose (Suc)- and/or indole 3-acetic acid (IAA, an auxin)-induced cell growth (Figure $a$ ) and the kinetics of IAA-induced cell elongation and cell wall acidification in coleoptiles (Figure $b$ ). Based on these results, together with the fact that these processes are delayed by cold treatments or inhibitors of protein synthesis, the \"acid-growth hypothesis\" was proposed as the best model to explain auxin-induced cell growth.\n\n$a$\n\n[figure1]\n\n$b$\n\n[figure2]\n\nWhich of the following statements is most accurate?\n\nA: IAA-driven protons, pumped into the cell wall, are utilized to synthesize the ATP required for cell elongation.\nB: IAA-induced acidification of the cell wall is an ATP-dependent process, and can be delayed by a treatment of a metabolic inhibitor.\nC: IAA-induced loosening of the cell wall is mainly caused by an acidification-induced weakening of the covalent bonds in cell wall proteins.\nD: IAA- or sucrose-induced cell elongation shares a common action mechanism, such as an increase in the cell wall acidity and the following change in turgor pressure.\nE: Cell wall acidification and stimulation in the elongation is an IAA-specific process, thus it is not induced by treatment with Fusicoccoin, an activator of the proton pump, in the absence of the IAA.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-17.jpg?height=757&width=605&top_left_y=778&top_left_x=360",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-17.jpg?height=674&width=899&top_left_y=871&top_left_x=1001"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_608",
        "problem": "下列为某一遗传病的家系图, 已知 I-1 为携带者。可以准确判断的是 ( )[图1]\nA: 该病为常染色体隐性遗传\nB: II-4 是携带者\nC: II-6 是携带者的概率为 $1 / 2$\nD: III -8 是正常纯合子的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列为某一遗传病的家系图, 已知 I-1 为携带者。可以准确判断的是 ( )[图1]\n\nA: 该病为常染色体隐性遗传\nB: II-4 是携带者\nC: II-6 是携带者的概率为 $1 / 2$\nD: III -8 是正常纯合子的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-15.jpg?height=235&width=377&top_left_y=268&top_left_x=180"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_693",
        "problem": "不同品种的水稻杂交种常有育性下降的现象。研究发现, 在花粉发育过程中, $\\mathrm{T}$ 或 $\\mathrm{G}$ 基因能表达对花粉发育重要的蛋白质, $\\mathrm{t}$ 和 $\\mathrm{g}$ 基因无法表达有功能的蛋白质。研究人员将基因型为 TTgg 的栽培稻和基因型 $\\mathrm{ttGG}$ 的野生稻杂交到 $\\mathrm{F}_{1}$, 将 $\\mathrm{F}_{1}$ 自交时发现某种花粉(占总配子数 $1 / 4$ )的花粉发育不正常导致不能受精。选取 $F_{2}$ 部分植株, 通过 PCR 扩增相关基因后, 电泳检测结果图如下。已知图中(1)(2)个体花粉发育完全正常, 下列说法错误的是 ( )\n\n[图1]\nA: $\\mathrm{T} / \\mathrm{t}$ 和 $\\mathrm{G} / \\mathrm{g}$ 基因的遗传遵循基因的自由组合定律\nB: 基因型为 $\\operatorname{tg}$ 的花粉发育不正常\nC: $F_{1}$ 自交后代花粉发育全部正常的个体占比为 $7 / 12$\nD: (3) (4) (6)中花粉正常发育数最少的是 (4)\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n不同品种的水稻杂交种常有育性下降的现象。研究发现, 在花粉发育过程中, $\\mathrm{T}$ 或 $\\mathrm{G}$ 基因能表达对花粉发育重要的蛋白质, $\\mathrm{t}$ 和 $\\mathrm{g}$ 基因无法表达有功能的蛋白质。研究人员将基因型为 TTgg 的栽培稻和基因型 $\\mathrm{ttGG}$ 的野生稻杂交到 $\\mathrm{F}_{1}$, 将 $\\mathrm{F}_{1}$ 自交时发现某种花粉(占总配子数 $1 / 4$ )的花粉发育不正常导致不能受精。选取 $F_{2}$ 部分植株, 通过 PCR 扩增相关基因后, 电泳检测结果图如下。已知图中(1)(2)个体花粉发育完全正常, 下列说法错误的是 ( )\n\n[图1]\n\nA: $\\mathrm{T} / \\mathrm{t}$ 和 $\\mathrm{G} / \\mathrm{g}$ 基因的遗传遵循基因的自由组合定律\nB: 基因型为 $\\operatorname{tg}$ 的花粉发育不正常\nC: $F_{1}$ 自交后代花粉发育全部正常的个体占比为 $7 / 12$\nD: (3) (4) (6)中花粉正常发育数最少的是 (4)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-89.jpg?height=312&width=708&top_left_y=158&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-89.jpg?height=54&width=1393&top_left_y=1389&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_178",
        "problem": "Perception of the emotional stimuli is a reciprocal process between the brain and the heart. In an experimental task, pictures depicting facial expressions were quickly and sequentially presented to the subjects at different phases of the cardiac cycle. Participants then rated the emotional intensity of the faces in a scale of 0 to 100. Given the role of the amygdala in the process of fear, the amygdala response was measured in normal individuals and subjects with pure autonomic failure (PAF) at the time of presenting facial expression images (using the fMRI method). The results of the study are as follows. (Consider $\\mathrm{p}<0.05$ indicates significant difference between groups).\n\n[figure1]\n\nSystole (S)\n\n[figure2]\n\n[figure3]\n(Late) diastole (D)\n[figure4]\n[figure5]\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPerception of the emotional stimuli is a reciprocal process between the brain and the heart. In an experimental task, pictures depicting facial expressions were quickly and sequentially presented to the subjects at different phases of the cardiac cycle. Participants then rated the emotional intensity of the faces in a scale of 0 to 100. Given the role of the amygdala in the process of fear, the amygdala response was measured in normal individuals and subjects with pure autonomic failure (PAF) at the time of presenting facial expression images (using the fMRI method). The results of the study are as follows. (Consider $\\mathrm{p}<0.05$ indicates significant difference between groups).\n\n[figure1]\n\nSystole (S)\n\n[figure2]\n\n[figure3]\n(Late) diastole (D)\n[figure4]\n[figure5]\n\nA: A\nB: B\nC: C\nD: D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-20.jpg?height=677&width=759&top_left_y=718&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-20.jpg?height=417&width=714&top_left_y=1613&top_left_x=380",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-20.jpg?height=314&width=580&top_left_y=774&top_left_x=1138",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-20.jpg?height=460&width=510&top_left_y=1589&top_left_x=1178",
            "https://i.postimg.cc/50DLscQ0/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1248",
        "problem": "Interestingly, a recent paper by Evans et al. 2012 in Antarctic Science suggests that both Antarctic icefish and Arctic cods have evolved essentially identical AFGPs, which are synthesized and recycled in similar ways. This is an example of convergent evolution (where distantly related organisms independently evolve similar traits) in response to the identical problem of how to deal with internal ice in freezing environments.\n\nAFGPs are synthesized in the exocrine pancreas in both groups of fish. They are then discharged into the gastrointestinal tract (gut) to inhibit the growth of ingested ice. AFGPs bound to ice are lost with the faeces or if unbound, absorbed from the gut in the rectum. AFGPs circulate in the blood and interstitial fluids where they are available to bind to ice crystals that may form. It is thought that AFGPs are phagocytosed by macrophages (engulfed by the cell) and build up in the macrophages of the spleen where they remain bound to ice crystals until a warming event occurs. AFGPs in the blood are ultimately secreted into the bile and re-enter the gut when bile is secreted for digestion. Arctic cods, unlike the Antarctic icefish, also synthesize AFGP in the liver.\n\n[figure1]Which of the following statements is NOT correct?\nA: Ice ingested with food is attached to an AFGP molecule and excreted with the faeces.\nB: In winter, ice bound to AFGPs accumulated in the macrophages of the spleen.\nC: AFGPs are recycled, conserving energy.\nD: AFGPs are found in interstitial fluid, blood, pancreas and gut fluids.\nE: Antarctic icefish synthesize AFGPs in both the exocrine pancreas and the liver.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nInterestingly, a recent paper by Evans et al. 2012 in Antarctic Science suggests that both Antarctic icefish and Arctic cods have evolved essentially identical AFGPs, which are synthesized and recycled in similar ways. This is an example of convergent evolution (where distantly related organisms independently evolve similar traits) in response to the identical problem of how to deal with internal ice in freezing environments.\n\nAFGPs are synthesized in the exocrine pancreas in both groups of fish. They are then discharged into the gastrointestinal tract (gut) to inhibit the growth of ingested ice. AFGPs bound to ice are lost with the faeces or if unbound, absorbed from the gut in the rectum. AFGPs circulate in the blood and interstitial fluids where they are available to bind to ice crystals that may form. It is thought that AFGPs are phagocytosed by macrophages (engulfed by the cell) and build up in the macrophages of the spleen where they remain bound to ice crystals until a warming event occurs. AFGPs in the blood are ultimately secreted into the bile and re-enter the gut when bile is secreted for digestion. Arctic cods, unlike the Antarctic icefish, also synthesize AFGP in the liver.\n\n[figure1]\n\nproblem:\nWhich of the following statements is NOT correct?\n\nA: Ice ingested with food is attached to an AFGP molecule and excreted with the faeces.\nB: In winter, ice bound to AFGPs accumulated in the macrophages of the spleen.\nC: AFGPs are recycled, conserving energy.\nD: AFGPs are found in interstitial fluid, blood, pancreas and gut fluids.\nE: Antarctic icefish synthesize AFGPs in both the exocrine pancreas and the liver.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-31.jpg?height=1053&width=780&top_left_y=290&top_left_x=1072"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1488",
        "problem": "The adult cells of baobab trees have 168 chromosomes, compared to 46 in humans. These contain four copies of the genome, compared to two copies in humans.\n\n[figure1]\n\nDuring meiosis, approximately how many different combinations of chromosomes could be produced in human gametes.\nA: 900\nB: 9000\nC: 90000\nD: 900000\nE: 9000000\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe adult cells of baobab trees have 168 chromosomes, compared to 46 in humans. These contain four copies of the genome, compared to two copies in humans.\n\n[figure1]\n\nDuring meiosis, approximately how many different combinations of chromosomes could be produced in human gametes.\n\nA: 900\nB: 9000\nC: 90000\nD: 900000\nE: 9000000\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-49.jpg?height=1228&width=925&top_left_y=474&top_left_x=240"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_512",
        "problem": "如图是患甲、乙两种遗传病的系谱图，且已知II-4 无致病基因。下列有关分析正确的是 ( )\n\n[图1]\nA: 甲病的遗传方式为常染色体隐性遗传\nB: 乙病的遗传方式为常染色体隐性遗传\nC: I-1 与III- 1 的基因型相同的概率为 $1 / 2$\nD: 如果III-2 与III-3 婚配, 生出正常孩子的概率为 $7 / 32$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图是患甲、乙两种遗传病的系谱图，且已知II-4 无致病基因。下列有关分析正确的是 ( )\n\n[图1]\n\nA: 甲病的遗传方式为常染色体隐性遗传\nB: 乙病的遗传方式为常染色体隐性遗传\nC: I-1 与III- 1 的基因型相同的概率为 $1 / 2$\nD: 如果III-2 与III-3 婚配, 生出正常孩子的概率为 $7 / 32$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-087.jpg?height=294&width=472&top_left_y=1715&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_306",
        "problem": "用苂光物质标记小鼠 $(2 n=40)$ 性腺细胞中的染色体端粒和染色体上的基因, 每个端粒都标记为黄色、 $\\mathrm{A} / \\mathrm{a}$ 基因标记为红色、 $\\mathrm{B} / \\mathrm{b}$ 基因标记为绿色（两对基因独立遗传）,不考虑基因突变和染色体变异, 有关分析错误的是（）\nA: 端粒是染色体两端的特殊序列的 DNA-蛋白质复合体\nB: 若标记有丝分裂中期的细胞, 则黄色荧光点为 160 个\nC: 减数分裂中一个四分体的苂光颜色可能有 1 种或 2 种\nD: 减数分裂II中期的细胞中, 一条染色体上可能有三种荧光颜色\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n用苂光物质标记小鼠 $(2 n=40)$ 性腺细胞中的染色体端粒和染色体上的基因, 每个端粒都标记为黄色、 $\\mathrm{A} / \\mathrm{a}$ 基因标记为红色、 $\\mathrm{B} / \\mathrm{b}$ 基因标记为绿色（两对基因独立遗传）,不考虑基因突变和染色体变异, 有关分析错误的是（）\n\nA: 端粒是染色体两端的特殊序列的 DNA-蛋白质复合体\nB: 若标记有丝分裂中期的细胞, 则黄色荧光点为 160 个\nC: 减数分裂中一个四分体的苂光颜色可能有 1 种或 2 种\nD: 减数分裂II中期的细胞中, 一条染色体上可能有三种荧光颜色\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1256",
        "problem": "The data below show how the incubation period of trout eggs in well-aerated water varies with temperature.\n\n| Temperature, ${ }^{\\circ} \\mathrm{C}$ | 2 | 5 | 10 | 15 |\n| :--- | :--- | :--- | :--- | :--- |\n| Incubation period <br> (days) | 205 | 118 | 4 | 27 |\n\nThe incubation period at $8{ }^{\\circ} \\mathrm{C}$ would be approximately\nA: 66 days\nB: 61 days\nC: 56 days\nD: 51 days\nE: 42 days\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe data below show how the incubation period of trout eggs in well-aerated water varies with temperature.\n\n| Temperature, ${ }^{\\circ} \\mathrm{C}$ | 2 | 5 | 10 | 15 |\n| :--- | :--- | :--- | :--- | :--- |\n| Incubation period <br> (days) | 205 | 118 | 4 | 27 |\n\nThe incubation period at $8{ }^{\\circ} \\mathrm{C}$ would be approximately\n\nA: 66 days\nB: 61 days\nC: 56 days\nD: 51 days\nE: 42 days\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1541",
        "problem": "In bacteria, one transcription factor often controls multiple genes. The expression of genes consumes energy so groups of genes controlled by the same factor are finely adapted to help the survival strategy of the bacterium. Floating individuals move vigorously in search of nutrients, but bacteria in biofilms rarely move.\n\nWhich genes turn on in lone individuals?\nA: Genes for secreted defensive compounds, Genes for plasmid sharing\nB: Genes for secreted defensive compounds, Flagella forming genes, Nutrient sensors\nC: Flagella forming genes, Genes for surviving starvation, Nutrient sensors\nD: Flagella forming genes, Genes for surviving starvation, Genes for plasmid sharing\nE: Genes for secreted defensive compounds, Genes for plasmid sharing, Flagella forming genes\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn bacteria, one transcription factor often controls multiple genes. The expression of genes consumes energy so groups of genes controlled by the same factor are finely adapted to help the survival strategy of the bacterium. Floating individuals move vigorously in search of nutrients, but bacteria in biofilms rarely move.\n\nWhich genes turn on in lone individuals?\n\nA: Genes for secreted defensive compounds, Genes for plasmid sharing\nB: Genes for secreted defensive compounds, Flagella forming genes, Nutrient sensors\nC: Flagella forming genes, Genes for surviving starvation, Nutrient sensors\nD: Flagella forming genes, Genes for surviving starvation, Genes for plasmid sharing\nE: Genes for secreted defensive compounds, Genes for plasmid sharing, Flagella forming genes\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1236",
        "problem": "The graph at right shows the conflict rate of primates, measured throughout the day and night over 10 weeks of a study period.\n\n[figure1]\n\nWeek\n\nHow many fights occurred in total?\nA: 9\nB:  59\nC: 12\nD: 63\nE: 14\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph at right shows the conflict rate of primates, measured throughout the day and night over 10 weeks of a study period.\n\n[figure1]\n\nWeek\n\nHow many fights occurred in total?\n\nA: 9\nB:  59\nC: 12\nD: 63\nE: 14\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-18.jpg?height=660&width=1188&top_left_y=1286&top_left_x=751"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1345",
        "problem": "Phloem tissue, taken from the storage root of carrot plants, and cultured on a nutrient medium, will produce completely new carrot plants. This is possible because the phloem cells\nA: are not fully differentiated\nB: synthesise the necessary growth hormone\nC: are genetically totipotent\nD: are normally meristematic\nE: retain the capacity to divide after differentiation\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPhloem tissue, taken from the storage root of carrot plants, and cultured on a nutrient medium, will produce completely new carrot plants. This is possible because the phloem cells\n\nA: are not fully differentiated\nB: synthesise the necessary growth hormone\nC: are genetically totipotent\nD: are normally meristematic\nE: retain the capacity to divide after differentiation\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_30",
        "problem": "The following experiments are designed to investigate the differentiation mechanism of skeletal muscle.\n\n<Experiment 1> Cultured mouse muscle cells were chemically induced to fuse with undifferentiated human cells.\n\nResult 1: Many of the fused cells had human muscle-specific proteins.\n\nResult 2: Unfused cells had no human muscle-specific proteins.\n\n<Experiment 2> Cytoplasmic portions of human muscle cells were injected into undifferentiated mouse stem cell.\n\nResult: The cells injected with the cytoplasm of human muscle cells transiently expressed mouse muscle-specific genes. However, the expression of muscle specific gene was disappeared after 24 hours.\n\nWhat do these experiments suggest?\nA: The nucleus of muscle cell should be fused with human cell nucleus to induce human muscle-specific proteins.\nB: The expression of muscle specific gene in human undifferentiated cell is suppressed by cytoplasmic factor.\nC: The continuous production of cytoplasmic factor(s) is indispensible for maintaining the differentiation state of the muscle cell.\nD: The cytoplasm of muscle cell induced a mutation of DNA to differentiate into muscle cell.\nE: The induction of muscle differentiation is a species-specific phenomenon.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following experiments are designed to investigate the differentiation mechanism of skeletal muscle.\n\n<Experiment 1> Cultured mouse muscle cells were chemically induced to fuse with undifferentiated human cells.\n\nResult 1: Many of the fused cells had human muscle-specific proteins.\n\nResult 2: Unfused cells had no human muscle-specific proteins.\n\n<Experiment 2> Cytoplasmic portions of human muscle cells were injected into undifferentiated mouse stem cell.\n\nResult: The cells injected with the cytoplasm of human muscle cells transiently expressed mouse muscle-specific genes. However, the expression of muscle specific gene was disappeared after 24 hours.\n\nWhat do these experiments suggest?\n\nA: The nucleus of muscle cell should be fused with human cell nucleus to induce human muscle-specific proteins.\nB: The expression of muscle specific gene in human undifferentiated cell is suppressed by cytoplasmic factor.\nC: The continuous production of cytoplasmic factor(s) is indispensible for maintaining the differentiation state of the muscle cell.\nD: The cytoplasm of muscle cell induced a mutation of DNA to differentiate into muscle cell.\nE: The induction of muscle differentiation is a species-specific phenomenon.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1326",
        "problem": "\"Square trees\" are being investigated to make forestry more sustainable. The underlying concept involves modifying the genome of trees so that the trunks grow in a 3D 'cuboid' shape instead of a cylinder. If the arrows on the cross sections of the trees shown are the same size, and the trunks are the same height, what is the percentage wood gained in tree on the right when compared with tree on the left? (Cuboid volume $=$ base $x$ height $x$ width), (Cylinder volume $=\\pi r^{2} x$ height).\n\n[figure1]\nA: $21 \\%$\nB: $10 \\%$\nC: $34 \\%$\nD: $5 \\%$\nE: $31.4 \\%$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n\"Square trees\" are being investigated to make forestry more sustainable. The underlying concept involves modifying the genome of trees so that the trunks grow in a 3D 'cuboid' shape instead of a cylinder. If the arrows on the cross sections of the trees shown are the same size, and the trunks are the same height, what is the percentage wood gained in tree on the right when compared with tree on the left? (Cuboid volume $=$ base $x$ height $x$ width), (Cylinder volume $=\\pi r^{2} x$ height).\n\n[figure1]\n\nA: $21 \\%$\nB: $10 \\%$\nC: $34 \\%$\nD: $5 \\%$\nE: $31.4 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-12.jpg?height=474&width=468&top_left_y=2236&top_left_x=794"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_149",
        "problem": "Thalassemia, the most common inherited disorder of hemoglobin, is caused by loss or substantial reduction of one of the globin chains. This results in lowered levels of functional hemoglobin and decreased function of red blood cells, which lead to anemia. In $\\alpha$-thalassemia, the $\\alpha$ chain of hemoglobin is not produced in sufficient quantity and consequently, hemoglobin tetramers form that contain only the $\\beta$ chain. In $\\beta$-thalassemia, the $\\beta$ chain of hemoglobin is not produced in sufficient quantity and the $\\alpha$ chains form insoluble aggregates that precipitate inside immature red blood cells and prevent differentiation into mature cells.\n\nThe normal haploid human genome has one $\\beta$ chain and two $\\alpha$ chain coding genes. Presence of four alleles for $\\alpha$ chain compared to two alleles for $\\beta$ chain in the cells of normal individuals is expected to results in excess amounts of $\\alpha$ chain and production of $\\alpha$ aggregates. However, $\\alpha$ aggregates do not exist in the cells of normal individuals. One mechanism for maintaining $\\alpha$ chains in soluble form was revealed by the discovery of an $11-\\mathrm{kD}$ a protein in red blood cells called $\\alpha$-hemoglobin stabilizing protein (AHSP). This protein forms a soluble complex specifically with $\\alpha$ chain monomers as they are synthesized. The crystal structure of a complex between AHSP and $\\alpha$-hemoglobin reveals that AHSP binds to the same face of $\\alpha$-globin as does $\\beta$-globin and ensures the proper folding of $\\alpha$-globin as it is produced. $\\beta$-globin displaces AHSP when it is expressed.\nA: Higher incidence of severe $\\beta$-thalassemia compared to severe $\\alpha$-thalassemia could be explained by difference in copy numbers of $\\alpha$ and $\\beta$ genes.\nB: $\\alpha$-globin/ $\\beta$-globin ratio is an appropriate marker for screening of $\\beta$-thalassemia.\nC: $\\alpha$-Hemoglobin has a higher affinity for $\\beta$-hemoglobin than for AHSP.\nD: AHSP deleterious mutations are expected to mimic $\\beta$-thalassemia phenotype with respect to $\\alpha$ - chain aggregation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThalassemia, the most common inherited disorder of hemoglobin, is caused by loss or substantial reduction of one of the globin chains. This results in lowered levels of functional hemoglobin and decreased function of red blood cells, which lead to anemia. In $\\alpha$-thalassemia, the $\\alpha$ chain of hemoglobin is not produced in sufficient quantity and consequently, hemoglobin tetramers form that contain only the $\\beta$ chain. In $\\beta$-thalassemia, the $\\beta$ chain of hemoglobin is not produced in sufficient quantity and the $\\alpha$ chains form insoluble aggregates that precipitate inside immature red blood cells and prevent differentiation into mature cells.\n\nThe normal haploid human genome has one $\\beta$ chain and two $\\alpha$ chain coding genes. Presence of four alleles for $\\alpha$ chain compared to two alleles for $\\beta$ chain in the cells of normal individuals is expected to results in excess amounts of $\\alpha$ chain and production of $\\alpha$ aggregates. However, $\\alpha$ aggregates do not exist in the cells of normal individuals. One mechanism for maintaining $\\alpha$ chains in soluble form was revealed by the discovery of an $11-\\mathrm{kD}$ a protein in red blood cells called $\\alpha$-hemoglobin stabilizing protein (AHSP). This protein forms a soluble complex specifically with $\\alpha$ chain monomers as they are synthesized. The crystal structure of a complex between AHSP and $\\alpha$-hemoglobin reveals that AHSP binds to the same face of $\\alpha$-globin as does $\\beta$-globin and ensures the proper folding of $\\alpha$-globin as it is produced. $\\beta$-globin displaces AHSP when it is expressed.\n\nA: Higher incidence of severe $\\beta$-thalassemia compared to severe $\\alpha$-thalassemia could be explained by difference in copy numbers of $\\alpha$ and $\\beta$ genes.\nB: $\\alpha$-globin/ $\\beta$-globin ratio is an appropriate marker for screening of $\\beta$-thalassemia.\nC: $\\alpha$-Hemoglobin has a higher affinity for $\\beta$-hemoglobin than for AHSP.\nD: AHSP deleterious mutations are expected to mimic $\\beta$-thalassemia phenotype with respect to $\\alpha$ - chain aggregation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_41",
        "problem": "The bacterium Bradyrhizobium japonicum can infect soybean (Glycine max) roots and form nodules. The nitrogen fixation catalyzed by nitrogenase occurs in the nodules and the nitrogenase activity can be measured easily by acetylene reduction instead of nitrogen reduction. Scientists generated a defective mutation of $\\mathrm{NAD}^{+}$-dependent malic enzyme, the enzyme that generates pyruvate and $\\mathrm{NADH}$, and infected soybean seedling roots with wildtype and mutant bacteria. The seedlings were grown in nitrogen-free media. After 14 and 28 days of inoculation, the number and weight of nodules in the seedlings and acetylene reduction activity were recorded (Fig.Q13).\n[figure1]\n\nFig.Q13. Nodule number and dry weight and acetylene reduction acitivity of soybean. Soybean nodules infected with wild-type B. japonicum (open bars) and the dme mutant (solid bars) are presented.\nA: Nitrogen fixation activity in nodules of the same treament at 28 days after inoculation is higher than that at 14 days after inoculation.\nB: Both number and size of nodules increase with time from 14 to 28 days after inoculation with $B$. japonicum.\nC: The reduction in nitrogen-fixing activity of nodules infected by the mutant at 28 days after inoculation compared to those at 14 days after inoculation is due to the reduction of nitrogenase activity and nodule formation.\nD: Nitrogen fixation in B. japonicum -induced nodule is down-regulated by $\\mathrm{NAD}^{+}$dependent malic enzyme.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe bacterium Bradyrhizobium japonicum can infect soybean (Glycine max) roots and form nodules. The nitrogen fixation catalyzed by nitrogenase occurs in the nodules and the nitrogenase activity can be measured easily by acetylene reduction instead of nitrogen reduction. Scientists generated a defective mutation of $\\mathrm{NAD}^{+}$-dependent malic enzyme, the enzyme that generates pyruvate and $\\mathrm{NADH}$, and infected soybean seedling roots with wildtype and mutant bacteria. The seedlings were grown in nitrogen-free media. After 14 and 28 days of inoculation, the number and weight of nodules in the seedlings and acetylene reduction activity were recorded (Fig.Q13).\n[figure1]\n\nFig.Q13. Nodule number and dry weight and acetylene reduction acitivity of soybean. Soybean nodules infected with wild-type B. japonicum (open bars) and the dme mutant (solid bars) are presented.\n\nA: Nitrogen fixation activity in nodules of the same treament at 28 days after inoculation is higher than that at 14 days after inoculation.\nB: Both number and size of nodules increase with time from 14 to 28 days after inoculation with $B$. japonicum.\nC: The reduction in nitrogen-fixing activity of nodules infected by the mutant at 28 days after inoculation compared to those at 14 days after inoculation is due to the reduction of nitrogenase activity and nodule formation.\nD: Nitrogen fixation in B. japonicum -induced nodule is down-regulated by $\\mathrm{NAD}^{+}$dependent malic enzyme.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-030.jpg?height=1132&width=1162&top_left_y=1184&top_left_x=480"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_764",
        "problem": "二倍体花椒皮刺的大小受一对等位基因 A、a 控制, 基因型为 AA 的植株表现为大皮刺, $\\mathrm{Aa}$ 为小皮刺, $\\mathrm{a}$ 为无皮刺。皮刺颜色（紫色和绿色）受另一对等位基因 $\\mathrm{R} 、 \\mathrm{r}$ 控制, $\\mathrm{R}$ 对 $\\mathrm{r}$ 为完全显性, 两对基因独立遗传。下列有关叙述错误的是（）\nA: 若基因型为 $\\mathrm{AaRr}$ 的亲本自交, 则子代共有 9 种基因型\nB: 若基因型为 $\\mathrm{AaRr}$ 的亲本自交, 则子代共有 6 种表现型\nC: 若基因型为 AaRr 与 Aarr 的亲本杂交, 则子代是紫色皮刺的植株占 $3 / 8$\nD: 若基因型为 $\\mathrm{AaRr}$ 的个体测交，则子代表现型有 3 种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n二倍体花椒皮刺的大小受一对等位基因 A、a 控制, 基因型为 AA 的植株表现为大皮刺, $\\mathrm{Aa}$ 为小皮刺, $\\mathrm{a}$ 为无皮刺。皮刺颜色（紫色和绿色）受另一对等位基因 $\\mathrm{R} 、 \\mathrm{r}$ 控制, $\\mathrm{R}$ 对 $\\mathrm{r}$ 为完全显性, 两对基因独立遗传。下列有关叙述错误的是（）\n\nA: 若基因型为 $\\mathrm{AaRr}$ 的亲本自交, 则子代共有 9 种基因型\nB: 若基因型为 $\\mathrm{AaRr}$ 的亲本自交, 则子代共有 6 种表现型\nC: 若基因型为 AaRr 与 Aarr 的亲本杂交, 则子代是紫色皮刺的植株占 $3 / 8$\nD: 若基因型为 $\\mathrm{AaRr}$ 的个体测交，则子代表现型有 3 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_977",
        "problem": "Sexual selection is a case of natural selection that describes evolution due not to variable survival rates due to fitness, but rather due to variable reproductive rates stemming from characteristics that allow an individual to successfully attract a mate. These traits are called secondary sexual characters. Which answer is NOT a correct statement about secondary sexual characteristics?\nA: The peacock's tail is a secondary sexual character.\nB: Reproductive organs are secondary sexual characters.\nC: Bird song can be used for mate attraction.\nD: The roaring of deer can be used for mate attraction.\nE: Sexual selection can select for traits that decrease individuals' likelihood of survival.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSexual selection is a case of natural selection that describes evolution due not to variable survival rates due to fitness, but rather due to variable reproductive rates stemming from characteristics that allow an individual to successfully attract a mate. These traits are called secondary sexual characters. Which answer is NOT a correct statement about secondary sexual characteristics?\n\nA: The peacock's tail is a secondary sexual character.\nB: Reproductive organs are secondary sexual characters.\nC: Bird song can be used for mate attraction.\nD: The roaring of deer can be used for mate attraction.\nE: Sexual selection can select for traits that decrease individuals' likelihood of survival.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1404",
        "problem": "Anna is studying a population of seabirds which roost on one particular island off the coast of Tasmania. She hypothesises that seabirds with longer beaks lay heavier eggs.\n\nAnna recorded the beak length and egg weight of a sample of 40 birds. Her results and the population of seabirds are plotted below\n\n[figure1]\n\nWhich of the following is supported by Anna's data?\nA: The population has average egg weight of roughly 43 grams\nB: Birds with longer beaks tend to lay heavier eggs\nC: Birds with longer beaks tend to lay lighter eggs\nD: Birds with longer beaks have a competitive advantage\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAnna is studying a population of seabirds which roost on one particular island off the coast of Tasmania. She hypothesises that seabirds with longer beaks lay heavier eggs.\n\nAnna recorded the beak length and egg weight of a sample of 40 birds. Her results and the population of seabirds are plotted below\n\n[figure1]\n\nWhich of the following is supported by Anna's data?\n\nA: The population has average egg weight of roughly 43 grams\nB: Birds with longer beaks tend to lay heavier eggs\nC: Birds with longer beaks tend to lay lighter eggs\nD: Birds with longer beaks have a competitive advantage\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-05.jpg?height=791&width=1314&top_left_y=1752&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_273",
        "problem": "Influenza A genome consists of 8 separate single stranded RNA molecules, which encode a total of 11 viral proteins. Influenza A viruses are categorized by their two surface antigens, the hemagglutinin $(\\mathrm{H})$, of which there are 18 different subtypes (H118); and neuraminidase $(\\mathrm{N})$, of which there are 11 different subtypes (N1-11) (Fig.Q.58A). The influenza A virus life cycle is presented in Fig. Q.58B.\n[figure1]\n\nFig.Q.58. Influenza A virus: (A) virus structure and (B) virus life cycle.\nA: Influenza A viruses exhibit rapid evolutionary dynamics because the genome is segmented.\nB: In theory, there are 144 types of influenza A viruses.\nC: Influenza A viruses exhibit high mutation rates because the genome is single strand RNA.\nD: Influenza A viruses will be active only if RNA-dependent RNA polymerase is present in the virion.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nInfluenza A genome consists of 8 separate single stranded RNA molecules, which encode a total of 11 viral proteins. Influenza A viruses are categorized by their two surface antigens, the hemagglutinin $(\\mathrm{H})$, of which there are 18 different subtypes (H118); and neuraminidase $(\\mathrm{N})$, of which there are 11 different subtypes (N1-11) (Fig.Q.58A). The influenza A virus life cycle is presented in Fig. Q.58B.\n[figure1]\n\nFig.Q.58. Influenza A virus: (A) virus structure and (B) virus life cycle.\n\nA: Influenza A viruses exhibit rapid evolutionary dynamics because the genome is segmented.\nB: In theory, there are 144 types of influenza A viruses.\nC: Influenza A viruses exhibit high mutation rates because the genome is single strand RNA.\nD: Influenza A viruses will be active only if RNA-dependent RNA polymerase is present in the virion.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-018.jpg?height=1836&width=1608&top_left_y=647&top_left_x=244"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1057",
        "problem": "Pedigree analysis helps determine the pattern of inheritance of a trait among related individuals. One such pedigree is given.\n\n[figure1]\n\nSquares indicate males and circles females. Filled symbols indicate presence of trait while circles with a dot indicate carriers.\n\nThe inheritance pattern shown in the given pedigree is:\nA: X-linked recessive\nB: Autosomal recessive\nC: Sex limited trait\nD: Sex influenced trait\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPedigree analysis helps determine the pattern of inheritance of a trait among related individuals. One such pedigree is given.\n\n[figure1]\n\nSquares indicate males and circles females. Filled symbols indicate presence of trait while circles with a dot indicate carriers.\n\nThe inheritance pattern shown in the given pedigree is:\n\nA: X-linked recessive\nB: Autosomal recessive\nC: Sex limited trait\nD: Sex influenced trait\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-08.jpg?height=463&width=913&top_left_y=972&top_left_x=671"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_127",
        "problem": "Bighorn sheep (Ovis canadensis), the males of which are famous for their magnificent curl of horns, live in North America. Their hunting was restricted in 1970. This restriction made trophy rams (males with large and fully-curved horns) extremely valuable, sometime costing over $\\$ 100,000$ to be hunted. Funds raised in this way were used for preserving bighorn habitats.\n\nColtman and his colleagues (2003) showed that there is a relationship between year and decrease of mean mass and mean horn length of bighorn sheep in Alberta, Canada where trophy hunting was conducted over 30 years.\n[figure1]\nA: The observed changes in the mean mass and horn length imply a reduction in the bighorn sheep population.\nB: Phenotype-based selective harvest can change population characters if it targets heritable traits.\nC: If variation in horn size is mainly due to additive genetic interactions, the heritability (the portion of the phenotypic variance explain by genotype variance) for this trait would decrease over time.\nD: By hunting males with longest horns, the variance in the reproductive success of the males increases.\nE: The trends establish genetic correlation between horn length and mass.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBighorn sheep (Ovis canadensis), the males of which are famous for their magnificent curl of horns, live in North America. Their hunting was restricted in 1970. This restriction made trophy rams (males with large and fully-curved horns) extremely valuable, sometime costing over $\\$ 100,000$ to be hunted. Funds raised in this way were used for preserving bighorn habitats.\n\nColtman and his colleagues (2003) showed that there is a relationship between year and decrease of mean mass and mean horn length of bighorn sheep in Alberta, Canada where trophy hunting was conducted over 30 years.\n[figure1]\n\nA: The observed changes in the mean mass and horn length imply a reduction in the bighorn sheep population.\nB: Phenotype-based selective harvest can change population characters if it targets heritable traits.\nC: If variation in horn size is mainly due to additive genetic interactions, the heritability (the portion of the phenotypic variance explain by genotype variance) for this trait would decrease over time.\nD: By hunting males with longest horns, the variance in the reproductive success of the males increases.\nE: The trends establish genetic correlation between horn length and mass.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-41.jpg?height=1230&width=850&top_left_y=730&top_left_x=608"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_715",
        "problem": "如图表示基因型为 YyRr 的二倍体生物某细胞的核 DNA 含量的变化曲线。下列有关叙述不正确的是 ( )\n\n[图1]\nA: D 时期细胞中的染色体组数目是 $E \\sim I$ 时期细胞中的两倍\nB: L 时期基因组成可能为 YYRR、YYrr、yyRR、yyrr\nC: E、 $M$ 时期，由于遗传物质的解旋，染色体逐渐恢复为染色质\nD: 在 G 和 I 时期, 细胞可产生可遗传的变异\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图表示基因型为 YyRr 的二倍体生物某细胞的核 DNA 含量的变化曲线。下列有关叙述不正确的是 ( )\n\n[图1]\n\nA: D 时期细胞中的染色体组数目是 $E \\sim I$ 时期细胞中的两倍\nB: L 时期基因组成可能为 YYRR、YYrr、yyRR、yyrr\nC: E、 $M$ 时期，由于遗传物质的解旋，染色体逐渐恢复为染色质\nD: 在 G 和 I 时期, 细胞可产生可遗传的变异\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-64.jpg?height=248&width=891&top_left_y=978&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1531",
        "problem": "At a temple in Bali, macaques steal things from tourists and barter them for food. The macaques steal high-end items like smartphones more often than easily-accessible items like hats. The macaques then wait until the tourist offers them fruit. The macaques only return the most valuable items when a large amount of fruit has been offered, whereas cheaper items are returned for less fruit.\n\n[figure1]\n\nHow do the macaques know which items are valuable?\nA: Reasoning\nB: Instinct\nC: Association/conditioning\nD: Habituation\nE: Imprinting\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAt a temple in Bali, macaques steal things from tourists and barter them for food. The macaques steal high-end items like smartphones more often than easily-accessible items like hats. The macaques then wait until the tourist offers them fruit. The macaques only return the most valuable items when a large amount of fruit has been offered, whereas cheaper items are returned for less fruit.\n\n[figure1]\n\nHow do the macaques know which items are valuable?\n\nA: Reasoning\nB: Instinct\nC: Association/conditioning\nD: Habituation\nE: Imprinting\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-15.jpg?height=1397&width=945&top_left_y=555&top_left_x=227"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1530",
        "problem": "Blue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nWhich method is better for estimating the change in average whale size?\nA: Mark-recapture.\nB: Counting from images.\nC: They are equal.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBlue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nWhich method is better for estimating the change in average whale size?\n\nA: Mark-recapture.\nB: Counting from images.\nC: They are equal.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-36.jpg?height=948&width=1673&top_left_y=657&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_980",
        "problem": "Correctly identify the tissue types A, B, C in the image below:\n[figure1]\nA: Sclerenchyma, parenchyma, collenchyma\nB: Collenchyma, parenchyma, sclerenchyma\nC: Collenchyma, sclerenchyma, parenchyma\nD: Parenchyma, collenchyma, sclerenchyma\nE: Parenchyma, sclerenchyma, collenchyma\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCorrectly identify the tissue types A, B, C in the image below:\n[figure1]\n\nA: Sclerenchyma, parenchyma, collenchyma\nB: Collenchyma, parenchyma, sclerenchyma\nC: Collenchyma, sclerenchyma, parenchyma\nD: Parenchyma, collenchyma, sclerenchyma\nE: Parenchyma, sclerenchyma, collenchyma\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_c11579df664240d6158ag-04.jpg?height=596&width=1046&top_left_y=899&top_left_x=560"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1272",
        "problem": "Dandelions are triploid and their pollen is sterile. This is most likely to be because\nA: the germinating pollen contains three gametic nuclei\nB: the stigmas lack a specific chemical necessary for pollen tube formation\nC: polyploids are infertile\nD: the meiotic mechanism breaks down\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDandelions are triploid and their pollen is sterile. This is most likely to be because\n\nA: the germinating pollen contains three gametic nuclei\nB: the stigmas lack a specific chemical necessary for pollen tube formation\nC: polyploids are infertile\nD: the meiotic mechanism breaks down\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1525",
        "problem": "The complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nWhich factor is the rate limiting factor, so tweaking its levels alters the activity of the complement system?\nA: $\\quad \\mathrm{C} 3$\nB: $\\quad \\mathrm{B}$\nC: $\\mathrm{D}$\nD: $\\quad \\mathrm{C} 5-9$\nE: $\\quad \\mathrm{C} 1-4$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nWhich factor is the rate limiting factor, so tweaking its levels alters the activity of the complement system?\n\nA: $\\quad \\mathrm{C} 3$\nB: $\\quad \\mathrm{B}$\nC: $\\mathrm{D}$\nD: $\\quad \\mathrm{C} 5-9$\nE: $\\quad \\mathrm{C} 1-4$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=790&width=1714&top_left_y=486&top_left_x=228",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=930&width=1806&top_left_y=1335&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_806",
        "problem": "“卵子死亡”是我国科学家新发现的一种人类单基因遗传病。该病致病基因由人体正\n\n常 PANX1 基因中一个碱基对的替换形成，其表达产物会引起 PANX1 通道异常激活,导致不孕(“卵子死亡”)。图 a 是一个“卵子死亡”患者的家系图 图 b 显示的是正常 PANX1 基因和突变 PANX1 基因对应片段中某限制酶的酶切位点; 分别提取家系中部分成员的 DNA, 经过酶切、电泳等步骤, 再用特异性探针做分子杂交, 结果见图 c。下列有关叙述错误的是（ )\n\n[图1]\n\n图a\n\n[图2]\n\n“个”或 “ $\\downarrow$ ”表示酶切位点; kb：千碱基对\n\n图b\n\n[图3]\n\n图c\nA: 理论上 PANX1 基因可存在多个突变位点，是因为基因突变具有随机性\nB: 该病属于常染色体显性遗传病\nC: $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 个体婚配生出“卵子死亡”遗传病患者的概率为 0\nD: $\\mathrm{II}_{6}$ 与正常女性婚配生出“卵子死亡”遗传病患者的概率为 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n“卵子死亡”是我国科学家新发现的一种人类单基因遗传病。该病致病基因由人体正\n\n常 PANX1 基因中一个碱基对的替换形成，其表达产物会引起 PANX1 通道异常激活,导致不孕(“卵子死亡”)。图 a 是一个“卵子死亡”患者的家系图 图 b 显示的是正常 PANX1 基因和突变 PANX1 基因对应片段中某限制酶的酶切位点; 分别提取家系中部分成员的 DNA, 经过酶切、电泳等步骤, 再用特异性探针做分子杂交, 结果见图 c。下列有关叙述错误的是（ )\n\n[图1]\n\n图a\n\n[图2]\n\n“个”或 “ $\\downarrow$ ”表示酶切位点; kb：千碱基对\n\n图b\n\n[图3]\n\n图c\n\nA: 理论上 PANX1 基因可存在多个突变位点，是因为基因突变具有随机性\nB: 该病属于常染色体显性遗传病\nC: $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 个体婚配生出“卵子死亡”遗传病患者的概率为 0\nD: $\\mathrm{II}_{6}$ 与正常女性婚配生出“卵子死亡”遗传病患者的概率为 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-06.jpg?height=254&width=554&top_left_y=1552&top_left_x=360",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-06.jpg?height=305&width=691&top_left_y=1475&top_left_x=1002",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-06.jpg?height=297&width=1211&top_left_y=1962&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_404",
        "problem": "哺乳动物细胞中的每对同源染色体上都有来源标记, 以标明该染色体源自父母中的哪一方。DNA 甲基化是标记的主要方式, 这些标记区域称为印记控制区。在 Igf2 基因和 H19 基因之间有一印记控制区（ICR），ICR 区域甲基化后不能结合增强子阻遏蛋白 CTCF, 进而影响基因的表达。该印记控制区对 Igf2 基因和 H19 基因的控制如图所示。下列有关叙述正确的是（）\n\n[图1]\n\n## CTCF:增强子阻遏蛋白 ICR:印记控制区\n\n[图2]\nA: 被甲基化的印记控制区 ICR 不能遗传给后代\nB: 父方和母方的 ICR 区域的碱基排列顺序不同\nC: Igf2 基因只能在雄性中表达, H19 基因只能在雌性中表达\nD: 相同的基因，来自父方或母方产生的遗传效应可能不同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n哺乳动物细胞中的每对同源染色体上都有来源标记, 以标明该染色体源自父母中的哪一方。DNA 甲基化是标记的主要方式, 这些标记区域称为印记控制区。在 Igf2 基因和 H19 基因之间有一印记控制区（ICR），ICR 区域甲基化后不能结合增强子阻遏蛋白 CTCF, 进而影响基因的表达。该印记控制区对 Igf2 基因和 H19 基因的控制如图所示。下列有关叙述正确的是（）\n\n[图1]\n\n## CTCF:增强子阻遏蛋白 ICR:印记控制区\n\n[图2]\n\nA: 被甲基化的印记控制区 ICR 不能遗传给后代\nB: 父方和母方的 ICR 区域的碱基排列顺序不同\nC: Igf2 基因只能在雄性中表达, H19 基因只能在雌性中表达\nD: 相同的基因，来自父方或母方产生的遗传效应可能不同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-48.jpg?height=511&width=979&top_left_y=904&top_left_x=356",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-48.jpg?height=83&width=802&top_left_y=1489&top_left_x=541"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_327",
        "problem": "在胁迫条件下, 植物体内活性氧（ROS）增加, 导致线粒体膜的通透性增加, 诱导细胞发生程序性死亡 $(\\mathrm{PCD})$ 。研究人员篮选出 1 个叶绿体功能缺陷突变体 $(\\bmod 1), \\bmod 1$中的 MOD1 基因缺陷, 导致叶绿体中的脂肪酸合酶合成缺陷, 存在明显的 ROS 积累,诱导植物发生 PCD (如图所示); 若直接使用苹果酸处理人类 HeLa 细胞, 也能够诱导 ROS 产生和细胞死亡, 下列有关说法错误的是（ ）\n\n[图1]\nA: 电子传递链复合体 (mETC) 的功能缺失能使 mod 1 的 PCD 表型恢复正常\nB: 编码酶 1、转运蛋白 D 和酶 2 的任一基因发生突变, 均可抑制 mod 1 发生 PCD\nC: 叶绿体中 MOD1 功能缺失, 导致苹果酸从叶绿体向线粒体的转运减少, 产生过量 ROS，引发 PCD\nD: 从细胞质到线粒体的细胞死亡调控机制在植物与动物间存在一定的保守性, 为生物进化提供了证据\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在胁迫条件下, 植物体内活性氧（ROS）增加, 导致线粒体膜的通透性增加, 诱导细胞发生程序性死亡 $(\\mathrm{PCD})$ 。研究人员篮选出 1 个叶绿体功能缺陷突变体 $(\\bmod 1), \\bmod 1$中的 MOD1 基因缺陷, 导致叶绿体中的脂肪酸合酶合成缺陷, 存在明显的 ROS 积累,诱导植物发生 PCD (如图所示); 若直接使用苹果酸处理人类 HeLa 细胞, 也能够诱导 ROS 产生和细胞死亡, 下列有关说法错误的是（ ）\n\n[图1]\n\nA: 电子传递链复合体 (mETC) 的功能缺失能使 mod 1 的 PCD 表型恢复正常\nB: 编码酶 1、转运蛋白 D 和酶 2 的任一基因发生突变, 均可抑制 mod 1 发生 PCD\nC: 叶绿体中 MOD1 功能缺失, 导致苹果酸从叶绿体向线粒体的转运减少, 产生过量 ROS，引发 PCD\nD: 从细胞质到线粒体的细胞死亡调控机制在植物与动物间存在一定的保守性, 为生物进化提供了证据\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-69.jpg?height=794&width=831&top_left_y=1299&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1244",
        "problem": "Radioactive phosphate $\\left({ }^{32} \\mathrm{PO}_{4}\\right)$ was applied to a plot of natural grassland. During the next 35 days, the radioactive content of samples from four species of arthropod was measured. The graph shows the relative amounts of radioactivity found in the four species, $\\mathrm{W}, \\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$.\n\n[figure1]\n\nWhich one of the rows $A-E$ represents the organisms in the graph?\nA: $\\mathbf{W}$: decomposer, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: herbivore, $\\mathbf{Z}$: carnivore\nB: $\\mathbf{W}$: herbivore, $\\mathbf{X}$: carnivore, $\\mathbf{Y}$: decomposer, $\\mathbf{Z}$: herbivore\nC: $\\mathbf{W}$: carnivore, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: decomposer, $\\mathbf{Z}$: herbivore\nD: $\\mathbf{W}$: herbivore, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: decomposer, $\\mathbf{Z}$: decomposer \nE: $\\mathbf{W}$: herbivore, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: carnivore, $\\mathbf{Z}$: decomposer \n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRadioactive phosphate $\\left({ }^{32} \\mathrm{PO}_{4}\\right)$ was applied to a plot of natural grassland. During the next 35 days, the radioactive content of samples from four species of arthropod was measured. The graph shows the relative amounts of radioactivity found in the four species, $\\mathrm{W}, \\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$.\n\n[figure1]\n\nWhich one of the rows $A-E$ represents the organisms in the graph?\n\nA: $\\mathbf{W}$: decomposer, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: herbivore, $\\mathbf{Z}$: carnivore\nB: $\\mathbf{W}$: herbivore, $\\mathbf{X}$: carnivore, $\\mathbf{Y}$: decomposer, $\\mathbf{Z}$: herbivore\nC: $\\mathbf{W}$: carnivore, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: decomposer, $\\mathbf{Z}$: herbivore\nD: $\\mathbf{W}$: herbivore, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: decomposer, $\\mathbf{Z}$: decomposer \nE: $\\mathbf{W}$: herbivore, $\\mathbf{X}$: herbivore, $\\mathbf{Y}$: carnivore, $\\mathbf{Z}$: decomposer \n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-07.jpg?height=414&width=1056&top_left_y=1047&top_left_x=500"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_873",
        "problem": "某家系的遗传系谱图及部分个体基因型如下图所示, $\\mathrm{A}_{1} 、 \\mathrm{~A}_{2} 、 \\mathrm{~A}_{3}$ 是位于 $\\mathrm{X}$ 染色体上的等位基因。下列推断正确的是（ ）\n\n[图1]\nA: $\\mathrm{II}_{2}$ 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 2}$ 的概率是 $1 / 4$\nB: $\\mathrm{III}_{1}$ 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{Y}$ 的概率是 $1 / 4$\nC: $\\mathrm{III}_{2}$ 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 2}$ 的概率是 $1 / 8$\nD: $\\mathrm{IV}_{1}$ 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 2}$ 的概率是 $1 / 8$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某家系的遗传系谱图及部分个体基因型如下图所示, $\\mathrm{A}_{1} 、 \\mathrm{~A}_{2} 、 \\mathrm{~A}_{3}$ 是位于 $\\mathrm{X}$ 染色体上的等位基因。下列推断正确的是（ ）\n\n[图1]\n\nA: $\\mathrm{II}_{2}$ 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 2}$ 的概率是 $1 / 4$\nB: $\\mathrm{III}_{1}$ 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{Y}$ 的概率是 $1 / 4$\nC: $\\mathrm{III}_{2}$ 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 2}$ 的概率是 $1 / 8$\nD: $\\mathrm{IV}_{1}$ 基因型为 $\\mathrm{X}^{\\mathrm{A} 1} \\mathrm{X}^{\\mathrm{A} 2}$ 的概率是 $1 / 8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-07.jpg?height=454&width=489&top_left_y=1304&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1112",
        "problem": "A bacterial cell is lac gal $^{+} \\&$ contains $\\mathrm{F}$ plasmid that is $l a c^{+}$. The plasmid has temperature sensitive mutation in its replication system such that above $40^{\\circ} \\mathrm{C}$ the replication is blocked. A few statements are made about the growth of these organisms in different conditions. \n\nWhich of the following is correct?\nA: The cells will grow normally in media containing galactose as the sole source of carbon \\& incubation temperature as $42^{\\circ} \\mathrm{C}$.\nB: The cells will grow normally only if media contains both galactose \\& lactose irrespective of temperature.\nC: The cells will grow normally in media containing lactose as the sole source of carbon \\& incubation temperature of $42^{\\circ} \\mathrm{C}$.\nD: If the cells are grown at $37^{\\circ} \\mathrm{C}$, lactose should be added to the medium and if the cells are grown at $42^{\\circ} \\mathrm{C}$, galactose should be added to the medium.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA bacterial cell is lac gal $^{+} \\&$ contains $\\mathrm{F}$ plasmid that is $l a c^{+}$. The plasmid has temperature sensitive mutation in its replication system such that above $40^{\\circ} \\mathrm{C}$ the replication is blocked. A few statements are made about the growth of these organisms in different conditions. \n\nWhich of the following is correct?\n\nA: The cells will grow normally in media containing galactose as the sole source of carbon \\& incubation temperature as $42^{\\circ} \\mathrm{C}$.\nB: The cells will grow normally only if media contains both galactose \\& lactose irrespective of temperature.\nC: The cells will grow normally in media containing lactose as the sole source of carbon \\& incubation temperature of $42^{\\circ} \\mathrm{C}$.\nD: If the cells are grown at $37^{\\circ} \\mathrm{C}$, lactose should be added to the medium and if the cells are grown at $42^{\\circ} \\mathrm{C}$, galactose should be added to the medium.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_472",
        "problem": "帕金森综合征是一种神经退行性疾病，研究发现患者普遍存在溶酶体膜蛋白 TMEM175 变异, TMEM175 蛋白的第 41 位氨基酸由天冬氨酸（GCG、GCA、GCC、 GCU）突变为丙氨酸（GAC、GAU），蛋白异常从而影响溶酶体的功能，如图所示。下列说法正确的是（ ）\n[图1]\nA: 正常 TMEM175 蛋白可将 $\\mathrm{H}^{+}$以主动转运的方式运出, 维持溶酶体内 $\\mathrm{pH}$ 约为 4.6\nB: 帕金森综合征患者 TMEM175 基因结构改变而致病, 体现基因突变的有害性\nC: TMEM175 基因内部核苷酸发生替换, 模板链上对应的位点由 C 突变为 A\nD: 可通过基因敲除技术，敲除突变的 TMEM175 基因, 缓减帕金森症状\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n帕金森综合征是一种神经退行性疾病，研究发现患者普遍存在溶酶体膜蛋白 TMEM175 变异, TMEM175 蛋白的第 41 位氨基酸由天冬氨酸（GCG、GCA、GCC、 GCU）突变为丙氨酸（GAC、GAU），蛋白异常从而影响溶酶体的功能，如图所示。下列说法正确的是（ ）\n[图1]\n\nA: 正常 TMEM175 蛋白可将 $\\mathrm{H}^{+}$以主动转运的方式运出, 维持溶酶体内 $\\mathrm{pH}$ 约为 4.6\nB: 帕金森综合征患者 TMEM175 基因结构改变而致病, 体现基因突变的有害性\nC: TMEM175 基因内部核苷酸发生替换, 模板链上对应的位点由 C 突变为 A\nD: 可通过基因敲除技术，敲除突变的 TMEM175 基因, 缓减帕金森症状\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-030.jpg?height=450&width=1400&top_left_y=140&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_366",
        "problem": "决定小鼠毛色为黑（B）褐（b）色、有（s）/无（S）白斑的两对等位基因分别位于两对同源染色体上。基因型为 BbSs 的小鼠间相互交配，后代中出现黑色有白斑小鼠的比例是（ ）\nA: $1 / 16$\nB: $3 / 16$\nC: $7 / 16$\nD: $9 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n决定小鼠毛色为黑（B）褐（b）色、有（s）/无（S）白斑的两对等位基因分别位于两对同源染色体上。基因型为 BbSs 的小鼠间相互交配，后代中出现黑色有白斑小鼠的比例是（ ）\n\nA: $1 / 16$\nB: $3 / 16$\nC: $7 / 16$\nD: $9 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1144",
        "problem": "The graph show egg production at different population densities of the barnacle (Semibalanus balanoides) at three rates of water flow. Which of the following inferences is supported by these data?\n\n[figure1]\nA: The greater the population density, the more fertile the organisms become.\nB: Beyond a certain population density, overcrowding reduces the rate of egg production by individuals.\nC: An inverse relationship exists between the rate of water flow and egg production.\nD: A linear relationship exists between population density and egg production at all population densities.\nE: The effect of water flow on egg production is greater at a density of 15 individuals per $\\mathrm{cm}^{2}$ than at a density of 5 per $\\mathrm{cm}^{2}$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph show egg production at different population densities of the barnacle (Semibalanus balanoides) at three rates of water flow. Which of the following inferences is supported by these data?\n\n[figure1]\n\nA: The greater the population density, the more fertile the organisms become.\nB: Beyond a certain population density, overcrowding reduces the rate of egg production by individuals.\nC: An inverse relationship exists between the rate of water flow and egg production.\nD: A linear relationship exists between population density and egg production at all population densities.\nE: The effect of water flow on egg production is greater at a density of 15 individuals per $\\mathrm{cm}^{2}$ than at a density of 5 per $\\mathrm{cm}^{2}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-08.jpg?height=502&width=788&top_left_y=2076&top_left_x=154"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1441",
        "problem": "Sucrose is produced in leaves and translocated short and long distances through veins to non-photosynthetic organs such as roots, stems, flowers and fruits. Sucrose transport is essential for the distribution of carbohydrates in plants. Two principal pathways include symplast and apoplast by which sucrose molecules are transported in the phloem of leaves as shown in the Figure. In the symplastic pathway, sucrose moves from cell to cell via plasmodesmata, which are continuous junctions between neighbouring plant cells. The apoplastic pathway involves a step where sucrose is momentarily transported outside of a plant cell plasma membrane.\n\nA\n\n[figure1]\n\nB\n[figure2]\n\nFigure. Diagram of the whole plant phloem network. M - Mesophyll, BS - Bundle sheath, MS - Mestome sheath, PP - Phloem parenchyma, VP - Vascular parenchyma, CC - Companion cell, TST - Thick walled sieve element, ST - Sieve element.\n\nWhich statement about phloem is true? It carries:\nA: nitrate ions from the leaves to the roots\nB: sugars from the roots to the leaves\nC: nutrients to growing regions of the plant\nD: water from the roots to the leaves\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSucrose is produced in leaves and translocated short and long distances through veins to non-photosynthetic organs such as roots, stems, flowers and fruits. Sucrose transport is essential for the distribution of carbohydrates in plants. Two principal pathways include symplast and apoplast by which sucrose molecules are transported in the phloem of leaves as shown in the Figure. In the symplastic pathway, sucrose moves from cell to cell via plasmodesmata, which are continuous junctions between neighbouring plant cells. The apoplastic pathway involves a step where sucrose is momentarily transported outside of a plant cell plasma membrane.\n\nA\n\n[figure1]\n\nB\n[figure2]\n\nFigure. Diagram of the whole plant phloem network. M - Mesophyll, BS - Bundle sheath, MS - Mestome sheath, PP - Phloem parenchyma, VP - Vascular parenchyma, CC - Companion cell, TST - Thick walled sieve element, ST - Sieve element.\n\nWhich statement about phloem is true? It carries:\n\nA: nitrate ions from the leaves to the roots\nB: sugars from the roots to the leaves\nC: nutrients to growing regions of the plant\nD: water from the roots to the leaves\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-11.jpg?height=582&width=597&top_left_y=760&top_left_x=250",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-11.jpg?height=598&width=946&top_left_y=751&top_left_x=840"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_413",
        "problem": "豌豆种群中偶尔会出现一种三体植株 (多一条 2 号染色体)。在减数分裂时, 该植株（基因型为 Aaa）2 号染色体的任意两条移向细胞一极，剩下一条移向另一极。下列有关叙述正确的是（ ）\nA: 三体踠豆植株的形成属于染色体结构变异\nB: 三体踠豆植株能产生四种类型的配子，其中 a 配子的比例为 $1 / 4$\nC: 该植株自交，产生 AAa 基因型子代的概率为 $1 / 36$\nD: 该植株减数分裂过程中，有 4 个 $\\mathrm{a}$ 基因的细胞可能是减数第二次分裂时期\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n豌豆种群中偶尔会出现一种三体植株 (多一条 2 号染色体)。在减数分裂时, 该植株（基因型为 Aaa）2 号染色体的任意两条移向细胞一极，剩下一条移向另一极。下列有关叙述正确的是（ ）\n\nA: 三体踠豆植株的形成属于染色体结构变异\nB: 三体踠豆植株能产生四种类型的配子，其中 a 配子的比例为 $1 / 4$\nC: 该植株自交，产生 AAa 基因型子代的概率为 $1 / 36$\nD: 该植株减数分裂过程中，有 4 个 $\\mathrm{a}$ 基因的细胞可能是减数第二次分裂时期\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1182",
        "problem": "There is consensus that factors such as geographic range influence species survivorship (species duration) in the fossil record with species with large geographical ranges surviving for longer periods. There is less agreement about the influence of factors such as body size and larval type. Crampton et al. 2010 looked at a variety of biotic influences on species duration. The relationships between duration, geographic range and these other traits are illustrated below for both bivalves (clams) and gastropods (snails).\n\n[figure1]\n\nFrom these relationships it can be inferred that in gastropods:\nA: There is a direct relationship between a planktotrophic larval phase and a large geographical range.\nB: Larval type influences species survivorship but not geographical range.\nC: A planktotrophic larval phase has no influence on geographic range.\nD: A planktotrophic larval phase is associated with large range size and long duration in a three-way interaction.\nE: Larval type has no influence on species survivorship.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThere is consensus that factors such as geographic range influence species survivorship (species duration) in the fossil record with species with large geographical ranges surviving for longer periods. There is less agreement about the influence of factors such as body size and larval type. Crampton et al. 2010 looked at a variety of biotic influences on species duration. The relationships between duration, geographic range and these other traits are illustrated below for both bivalves (clams) and gastropods (snails).\n\n[figure1]\n\nFrom these relationships it can be inferred that in gastropods:\n\nA: There is a direct relationship between a planktotrophic larval phase and a large geographical range.\nB: Larval type influences species survivorship but not geographical range.\nC: A planktotrophic larval phase has no influence on geographic range.\nD: A planktotrophic larval phase is associated with large range size and long duration in a three-way interaction.\nE: Larval type has no influence on species survivorship.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-23.jpg?height=651&width=1171&top_left_y=503&top_left_x=157"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_124",
        "problem": "One of the questions that long ago arose with respect to splicing was whether the temporal order of removal of introns in transcripts of genes with multiple introns is the same as the physical order of the introns.\n\nTo address this question with respect to the ovomucoid encoding gene $(5.6 \\mathrm{Kbp}$ ) that has seven introns (A-G), Northern blotting was performed on RNA isolated from nuclei of ovomucoid expressing cells. The probe used in the Northern blotting was labelled DNA of the ovomucoid encoding gene. An image of the Northern blot is presented below. RNA was extracted from each major band seen in gel, and the presence or absence of specific introns in RNA molecules of various band was assessed using intronspecific probes. Finally, the RNA molecules were grouped on the basis of number of introns removed by splicing and calculations that revealed identity of introns lost in various processed RNA molecules were made. The results of these calculations are also presented below.\n\n## Nuclear RNA\n\n[figure1]\n\n## Northern blot\nA: ovomucoid primary transcripts is not random.\nB: The data in the table show that intron $\\mathrm{E}$ is usually but not always removed from primary transcripts before intron $\\mathrm{G}$ is removed.\nC: The data in the table indicate that introns A, B, and C are usually removed from primary transcripts before introns $\\mathrm{D}$ and $\\mathrm{G}$.\nD: The data presented suggest that at the time of analysis, there is generally a progressive increase in concentrations of more highly processed transcripts.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nOne of the questions that long ago arose with respect to splicing was whether the temporal order of removal of introns in transcripts of genes with multiple introns is the same as the physical order of the introns.\n\nTo address this question with respect to the ovomucoid encoding gene $(5.6 \\mathrm{Kbp}$ ) that has seven introns (A-G), Northern blotting was performed on RNA isolated from nuclei of ovomucoid expressing cells. The probe used in the Northern blotting was labelled DNA of the ovomucoid encoding gene. An image of the Northern blot is presented below. RNA was extracted from each major band seen in gel, and the presence or absence of specific introns in RNA molecules of various band was assessed using intronspecific probes. Finally, the RNA molecules were grouped on the basis of number of introns removed by splicing and calculations that revealed identity of introns lost in various processed RNA molecules were made. The results of these calculations are also presented below.\n\n## Nuclear RNA\n\n[figure1]\n\n## Northern blot\n\nA: ovomucoid primary transcripts is not random.\nB: The data in the table show that intron $\\mathrm{E}$ is usually but not always removed from primary transcripts before intron $\\mathrm{G}$ is removed.\nC: The data in the table indicate that introns A, B, and C are usually removed from primary transcripts before introns $\\mathrm{D}$ and $\\mathrm{G}$.\nD: The data presented suggest that at the time of analysis, there is generally a progressive increase in concentrations of more highly processed transcripts.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-06.jpg?height=683&width=1566&top_left_y=949&top_left_x=245"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_381",
        "problem": "控制色觉的基因位于 $\\mathrm{X}$ 染色体上，正常色觉基因 $\\mathrm{B}$ 对色弱基因 $\\mathrm{B}$ ，色盲基因 $\\mathrm{b}$ 为显性, 色弱基因 B-对色盲基因 $\\mathrm{b}$ 为显性。下图左为某家族系谱图, 右为同种限制酶处理第二代成员色觉基因的结果，序号(1)～(5)表示电泳条带。下列叙述不正确的是（）\n\n[图1]\nA: 条带(2)(3)代表色弱基因, 条带(2)(4)(5)代表色盲基因\nB: 正常色觉基因上无所用限制酶的酶切位点\nC: I-1 对应的电泳条带应是(2)(3) (4)\nD: II- 3 与正常人结婚，子代表现为色弱的概率是 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n控制色觉的基因位于 $\\mathrm{X}$ 染色体上，正常色觉基因 $\\mathrm{B}$ 对色弱基因 $\\mathrm{B}$ ，色盲基因 $\\mathrm{b}$ 为显性, 色弱基因 B-对色盲基因 $\\mathrm{b}$ 为显性。下图左为某家族系谱图, 右为同种限制酶处理第二代成员色觉基因的结果，序号(1)～(5)表示电泳条带。下列叙述不正确的是（）\n\n[图1]\n\nA: 条带(2)(3)代表色弱基因, 条带(2)(4)(5)代表色盲基因\nB: 正常色觉基因上无所用限制酶的酶切位点\nC: I-1 对应的电泳条带应是(2)(3) (4)\nD: II- 3 与正常人结婚，子代表现为色弱的概率是 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-58.jpg?height=348&width=1291&top_left_y=908&top_left_x=357",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-59.jpg?height=52&width=1300&top_left_y=248&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_84",
        "problem": "This picture illustrates monthly changes in the human ovary during the reproductive cycle.\n\n[figure1]\n\nWhich of the following statements most accurately describes each structure?\nA: Before puberty, the oocyte (a) has not started the process of meiosis.\nB: The hormone produced by structure (b) causes thinning of the uterine cervical mucus to allow passage of sperm.\nC: During ovulation, structure (c) stays at the interphase between meiosis I and meiosis II.\nD: The hormone produced by structure (d) stimulates the pituitary gland to secrete luteinizing hormone.\nE: The hormone produced by structure ( $e$ ) causes the proliferation of the uterine endometrium.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThis picture illustrates monthly changes in the human ovary during the reproductive cycle.\n\n[figure1]\n\nWhich of the following statements most accurately describes each structure?\n\nA: Before puberty, the oocyte (a) has not started the process of meiosis.\nB: The hormone produced by structure (b) causes thinning of the uterine cervical mucus to allow passage of sperm.\nC: During ovulation, structure (c) stays at the interphase between meiosis I and meiosis II.\nD: The hormone produced by structure (d) stimulates the pituitary gland to secrete luteinizing hormone.\nE: The hormone produced by structure ( $e$ ) causes the proliferation of the uterine endometrium.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-32.jpg?height=679&width=1193&top_left_y=580&top_left_x=543"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1056",
        "problem": "In order to study the properties of membrane proteins, they are required to be isolated and purified. In one such experiment, two methods were followed to isolate membrane proteins from erythrocytes.\n\nI. Varying the $\\mathrm{pH}$ of the buffer in which the membrane fractions were isolated. The supernatant collected and analysed. Protein ' $X$ ' was found.\n\nII. Addition of detergents to the membrane preparation. Soluble fractions analysed. Protein ' $Y$ ' was found.\n\nWhen proteins $X$ and $Y$ were studied for their structural properties, it was found that polar amino acids predominated (property 1 ) on the surface of one of them while the non-polar on the other (property 2).\n\nWhich of the following is correct?\nA: $\\mathrm{X}$ is likely to be bound to membrane by electrostatic attraction.\nB: $\\mathrm{Y}$ is likely to show property 2 .\nC: $\\mathrm{X}$ is likely to be the integral membrane protein.\nD: $Y$ is most likely to be a cytoskeletal protein.\nE: Protein $\\mathrm{X}$ is likely to be present in cell exterior (exoplasmic face) and $\\mathrm{Y}$ in the cyotsolic face.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn order to study the properties of membrane proteins, they are required to be isolated and purified. In one such experiment, two methods were followed to isolate membrane proteins from erythrocytes.\n\nI. Varying the $\\mathrm{pH}$ of the buffer in which the membrane fractions were isolated. The supernatant collected and analysed. Protein ' $X$ ' was found.\n\nII. Addition of detergents to the membrane preparation. Soluble fractions analysed. Protein ' $Y$ ' was found.\n\nWhen proteins $X$ and $Y$ were studied for their structural properties, it was found that polar amino acids predominated (property 1 ) on the surface of one of them while the non-polar on the other (property 2).\n\nWhich of the following is correct?\n\nA: $\\mathrm{X}$ is likely to be bound to membrane by electrostatic attraction.\nB: $\\mathrm{Y}$ is likely to show property 2 .\nC: $\\mathrm{X}$ is likely to be the integral membrane protein.\nD: $Y$ is most likely to be a cytoskeletal protein.\nE: Protein $\\mathrm{X}$ is likely to be present in cell exterior (exoplasmic face) and $\\mathrm{Y}$ in the cyotsolic face.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1024",
        "problem": "Although many fish are exothermic, some fish are endothermic. Which of the following is/are most likely endothermic fish\n\nI. Bluefin tuna\n\nII. Salmon\n\nIII. Great white shark\nA: II\nB: I, II\nC: I, III\nD: II, III\nE: I, II, III\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAlthough many fish are exothermic, some fish are endothermic. Which of the following is/are most likely endothermic fish\n\nI. Bluefin tuna\n\nII. Salmon\n\nIII. Great white shark\n\nA: II\nB: I, II\nC: I, III\nD: II, III\nE: I, II, III\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_361",
        "problem": "生物等位基因的显隐性关系会根据依据的标准不同而有所改变, 下表是镰刀形细胞贫血症的相关情况, 下图是一个血友病(用 $\\mathrm{B} 、 \\mathrm{~b}$ 表示) 和镰刀形细胞贫血症遗传系谱图,\n\n其中 $\\mathrm{II}_{4}$ 血液中含有镰刀形细胞。镰刀形细胞贫血症的相关情况表, 下列叙述错误的是\n\n| 基因型 | $\\mathrm{Hb}^{\\mathrm{A}} \\mathrm{Hb}^{\\mathrm{A}}$ | $\\mathrm{Hb}^{\\mathrm{A}} \\mathrm{Hb}^{\\mathrm{S}}$ | $\\mathrm{Hb}^{\\mathrm{S}} \\mathrm{Hb}^{\\mathrm{S}}$ |\n| :--- | :--- | :--- | :--- |\n| 有无临床症状 | 无症状 | 无症状 | 贫血患者 |\n| 镰刀形细胞数多少 | 无 | 少 | 多 |\n\n[图1]\nA: $\\mathrm{II}_{1}$ 血液中有镰刀形细胞的概率是 $2 / 3$\nB: $\\mathrm{II}_{5}$ 的基因型为 $\\mathrm{Hb}^{\\mathrm{A}} \\mathrm{Hb}^{\\mathrm{A}} \\mathrm{X}^{\\mathrm{B}} Y$ 或者 $\\mathrm{Hb}^{\\mathrm{A}} \\mathrm{Hb}^{\\mathrm{S}} \\mathrm{X}^{\\mathrm{B}} Y$\nC: 若 $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生一个小孩, 该小孩患血友病且血液中含镰刀形细胞的概率为 $1 / 24$\nD: $\\mathrm{III}_{1}$ 细胞中的染色体最多有 22 条来自 $\\mathrm{I}_{1}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n生物等位基因的显隐性关系会根据依据的标准不同而有所改变, 下表是镰刀形细胞贫血症的相关情况, 下图是一个血友病(用 $\\mathrm{B} 、 \\mathrm{~b}$ 表示) 和镰刀形细胞贫血症遗传系谱图,\n\n其中 $\\mathrm{II}_{4}$ 血液中含有镰刀形细胞。镰刀形细胞贫血症的相关情况表, 下列叙述错误的是\n\n| 基因型 | $\\mathrm{Hb}^{\\mathrm{A}} \\mathrm{Hb}^{\\mathrm{A}}$ | $\\mathrm{Hb}^{\\mathrm{A}} \\mathrm{Hb}^{\\mathrm{S}}$ | $\\mathrm{Hb}^{\\mathrm{S}} \\mathrm{Hb}^{\\mathrm{S}}$ |\n| :--- | :--- | :--- | :--- |\n| 有无临床症状 | 无症状 | 无症状 | 贫血患者 |\n| 镰刀形细胞数多少 | 无 | 少 | 多 |\n\n[图1]\n\nA: $\\mathrm{II}_{1}$ 血液中有镰刀形细胞的概率是 $2 / 3$\nB: $\\mathrm{II}_{5}$ 的基因型为 $\\mathrm{Hb}^{\\mathrm{A}} \\mathrm{Hb}^{\\mathrm{A}} \\mathrm{X}^{\\mathrm{B}} Y$ 或者 $\\mathrm{Hb}^{\\mathrm{A}} \\mathrm{Hb}^{\\mathrm{S}} \\mathrm{X}^{\\mathrm{B}} Y$\nC: 若 $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生一个小孩, 该小孩患血友病且血液中含镰刀形细胞的概率为 $1 / 24$\nD: $\\mathrm{III}_{1}$ 细胞中的染色体最多有 22 条来自 $\\mathrm{I}_{1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-061.jpg?height=265&width=817&top_left_y=1815&top_left_x=354",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-062.jpg?height=765&width=1083&top_left_y=1065&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_21",
        "problem": "To understand the effects of several factors on plants (Agrimonia rostellata and Trillium erectum) in forest ecosystems, students transplanted seedlings into experimental sites and observed the proportion of surviving seedlings growing with native or non-native vegetation, with or without slugs, and with low or high earthworm density. The results are shown in the figure below.\n\n[figure1]\n\n- Slug present\n\n(b)\n[figure2]\n\nEarthworm density\n\nFigure Q.95. Proportion of surviving seedlings. Agrimonia rostellata $(a, b)$ and Trillium erectum (c, d)\nA: There are either positive effects or interactive effects with slug exclusion.\nB: Slug effects are dependent on other stressors, especially on interactions with nonnative plants and earthworms.\nC: Earthworms have positive effects on Agrimonia rostellata and Trillium erectum.\nD: Non-native plants and slugs synergistically decrease seedling survival through increased competition and consumption.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nTo understand the effects of several factors on plants (Agrimonia rostellata and Trillium erectum) in forest ecosystems, students transplanted seedlings into experimental sites and observed the proportion of surviving seedlings growing with native or non-native vegetation, with or without slugs, and with low or high earthworm density. The results are shown in the figure below.\n\n[figure1]\n\n- Slug present\n\n(b)\n[figure2]\n\nEarthworm density\n\nFigure Q.95. Proportion of surviving seedlings. Agrimonia rostellata $(a, b)$ and Trillium erectum (c, d)\n\nA: There are either positive effects or interactive effects with slug exclusion.\nB: Slug effects are dependent on other stressors, especially on interactions with nonnative plants and earthworms.\nC: Earthworms have positive effects on Agrimonia rostellata and Trillium erectum.\nD: Non-native plants and slugs synergistically decrease seedling survival through increased competition and consumption.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-099.jpg?height=1081&width=739&top_left_y=692&top_left_x=381",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-099.jpg?height=944&width=590&top_left_y=752&top_left_x=1141"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_859",
        "problem": "某同学用卡片模拟杂交实验, 表示亲本的卡片在信封中的放置情况如图, 其中信封 4 中 $\\mathrm{Y}$ 的数量待定, 每次抓取卡片并记录后, 将卡片放回原信封, 重复 100 次。已知 $\\mathrm{A} 、 \\mathrm{a}$ 位于常染色体上，群体中 $\\mathrm{A}$ 的基因频率为 $0.8: \\mathrm{B} 、 \\mathrm{~b}$ 位于 $\\mathrm{X}$ 染色体的非同源区段，群体中 $\\mathrm{X}^{\\mathrm{B}}$ 的基因频率为 0.6 。群体中无配子致死和基因型致死现象。\n\n| 信封1 | 信封2 | 信封3 | 信封4 |\n| :---: | :---: | :---: | :---: |\n| A 40 张 | A 40 张 | $\\mathrm{X}^{\\mathrm{B}} 30$ 张 | $X^{B} 30$ 张 |\n| a 10 张 | a 10 张 | $\\mathrm{X}^{\\mathrm{b}} 20$ 张 | Y ? 张 |\n\n下列叙述正确的是（ ）\nA: 随机从信封 $1 、 2$ 中各取 1 张卡片组合在一起，模拟 $\\mathrm{Aa}$ 个体随机交配产生子代\nB: 信封 4 可模拟雄性生殖器官, 图中 $Y$ 的数量与信封 3 中 $X$ 的总数不同\nC: 随机从信封 $1 、 4$ 中各取 1 张卡片组合在一起, 模拟的是雌雄配子的自由组合\nD: 随机从信封 $3 、 4$ 中各取 1 张卡片组合在一起, 所得结果中女性纯合子的概率为 0.26\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某同学用卡片模拟杂交实验, 表示亲本的卡片在信封中的放置情况如图, 其中信封 4 中 $\\mathrm{Y}$ 的数量待定, 每次抓取卡片并记录后, 将卡片放回原信封, 重复 100 次。已知 $\\mathrm{A} 、 \\mathrm{a}$ 位于常染色体上，群体中 $\\mathrm{A}$ 的基因频率为 $0.8: \\mathrm{B} 、 \\mathrm{~b}$ 位于 $\\mathrm{X}$ 染色体的非同源区段，群体中 $\\mathrm{X}^{\\mathrm{B}}$ 的基因频率为 0.6 。群体中无配子致死和基因型致死现象。\n\n| 信封1 | 信封2 | 信封3 | 信封4 |\n| :---: | :---: | :---: | :---: |\n| A 40 张 | A 40 张 | $\\mathrm{X}^{\\mathrm{B}} 30$ 张 | $X^{B} 30$ 张 |\n| a 10 张 | a 10 张 | $\\mathrm{X}^{\\mathrm{b}} 20$ 张 | Y ? 张 |\n\n下列叙述正确的是（ ）\n\nA: 随机从信封 $1 、 2$ 中各取 1 张卡片组合在一起，模拟 $\\mathrm{Aa}$ 个体随机交配产生子代\nB: 信封 4 可模拟雄性生殖器官, 图中 $Y$ 的数量与信封 3 中 $X$ 的总数不同\nC: 随机从信封 $1 、 4$ 中各取 1 张卡片组合在一起, 模拟的是雌雄配子的自由组合\nD: 随机从信封 $3 、 4$ 中各取 1 张卡片组合在一起, 所得结果中女性纯合子的概率为 0.26\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_79",
        "problem": "Three morphs of a polymorphic species of Catacola moths differ only in the patterns on the forewings.\n\nSix experienced blue jays (Cyanocitta cristata) searched for prey on a computer screen in a series of trials.\n\n[figure1]\n\nEach trial involved a screen showing the presence or absence of a moth from three distinct morphs of a moth population. If the bird found the moth, it was rewarded with food. Each bird had 36 prey and 84 no-prey trials and lasted for 50 days. Three replicates were carried out, with the second having a larger population of morph 2 and third replicate starting off with a larger relative abundance of morph 3.\n[figure2]\n\n## Generation number\n\na. Population numbers of the three morphs.\n\nb. Prey detectability\n\n(Bond \\& Kamil, 1998 - doi:10.1038/26961)\nA: Morph 1 was the most cryptic morph.\nB: Relative numbers that escaped detection determine the abundance of each prey type.\nC: Preferential feeding behavior of the blue-jays of the most prevalent morph maximizes their foraging success.\nD: Polymorphisms are maintained in the population through frequency-dependent selection by predators.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThree morphs of a polymorphic species of Catacola moths differ only in the patterns on the forewings.\n\nSix experienced blue jays (Cyanocitta cristata) searched for prey on a computer screen in a series of trials.\n\n[figure1]\n\nEach trial involved a screen showing the presence or absence of a moth from three distinct morphs of a moth population. If the bird found the moth, it was rewarded with food. Each bird had 36 prey and 84 no-prey trials and lasted for 50 days. Three replicates were carried out, with the second having a larger population of morph 2 and third replicate starting off with a larger relative abundance of morph 3.\n[figure2]\n\n## Generation number\n\na. Population numbers of the three morphs.\n\nb. Prey detectability\n\n(Bond \\& Kamil, 1998 - doi:10.1038/26961)\n\nA: Morph 1 was the most cryptic morph.\nB: Relative numbers that escaped detection determine the abundance of each prey type.\nC: Preferential feeding behavior of the blue-jays of the most prevalent morph maximizes their foraging success.\nD: Polymorphisms are maintained in the population through frequency-dependent selection by predators.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-076.jpg?height=832&width=917&top_left_y=649&top_left_x=567",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-077.jpg?height=1034&width=1616&top_left_y=234&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_918",
        "problem": "某雌果蝇 $(2 n=8)$ 甲的性染色体由一条等臂染色体 (两条 $X$ 染色体相连形成) 和一条 $\\mathrm{Y}$ 染色体组成, 该果蝇可产生两种可育配子。经诱变处理后的乙果蝇与甲果蝇进行杂交，过程如下图所示。下列叙述正确的是（）\n\n[图1]\nA: 甲果蝇体细胞中的染色体最多为 18 条\nB: 若乙果蝇的 X 染色体上发生显性致死突变, 则 $\\mathrm{F}_{1}$ 均为雌性\nC: 丙果蝇的性状与甲果蝇完全相同\nD: 若丁表现出突变性状，则突变基因位于 X、Y 的同源区段\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某雌果蝇 $(2 n=8)$ 甲的性染色体由一条等臂染色体 (两条 $X$ 染色体相连形成) 和一条 $\\mathrm{Y}$ 染色体组成, 该果蝇可产生两种可育配子。经诱变处理后的乙果蝇与甲果蝇进行杂交，过程如下图所示。下列叙述正确的是（）\n\n[图1]\n\nA: 甲果蝇体细胞中的染色体最多为 18 条\nB: 若乙果蝇的 X 染色体上发生显性致死突变, 则 $\\mathrm{F}_{1}$ 均为雌性\nC: 丙果蝇的性状与甲果蝇完全相同\nD: 若丁表现出突变性状，则突变基因位于 X、Y 的同源区段\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-047.jpg?height=580&width=1197&top_left_y=1800&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_20",
        "problem": "Intraguild predation is the killing and sometimes eating of potential competitors. Reciprocal intraguild predation with size structure, is one type of intraguild predation where both consumers feed on each other's juveniles. In this case, P1 has piercing-sucking type mouthparts, and P2 has chewing type mouthparts. In the figure below \"R\" represents resource, \"P\" represents (reciprocal) intraguild predator, \"j\" represents juveniles and \"a\" represents adults.\nA: If $\\mathrm{P} 1 \\mathrm{j}$ population increases, then $\\mathrm{P} 2 \\mathrm{a}$ will produce more eggs.\nB: If $\\mathrm{P} 2 \\mathrm{j}$ population decreases, then $\\mathrm{P} 2 \\mathrm{a}$ population will increase in next generation.\nC: If $\\mathrm{R}$ increases, then $\\mathrm{P} 2 \\mathrm{a}$ doesn't feed on $\\mathrm{P} 1 \\mathrm{j}$.\nD: If $\\mathrm{R}$ decreases, then $\\mathrm{P} 2 \\mathrm{a}$ population will increase in next generation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIntraguild predation is the killing and sometimes eating of potential competitors. Reciprocal intraguild predation with size structure, is one type of intraguild predation where both consumers feed on each other's juveniles. In this case, P1 has piercing-sucking type mouthparts, and P2 has chewing type mouthparts. In the figure below \"R\" represents resource, \"P\" represents (reciprocal) intraguild predator, \"j\" represents juveniles and \"a\" represents adults.\n\nA: If $\\mathrm{P} 1 \\mathrm{j}$ population increases, then $\\mathrm{P} 2 \\mathrm{a}$ will produce more eggs.\nB: If $\\mathrm{P} 2 \\mathrm{j}$ population decreases, then $\\mathrm{P} 2 \\mathrm{a}$ population will increase in next generation.\nC: If $\\mathrm{R}$ increases, then $\\mathrm{P} 2 \\mathrm{a}$ doesn't feed on $\\mathrm{P} 1 \\mathrm{j}$.\nD: If $\\mathrm{R}$ decreases, then $\\mathrm{P} 2 \\mathrm{a}$ population will increase in next generation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_919",
        "problem": "图 1 表示某二倍体动物细胞减数分裂过程中某项指标的数量变化曲线; 图 2 表示对该动物精巢切片显微观察后绘制的两幅细胞分裂示意图, 下列分析错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n甲\n\n[图3]\n\n乙\n\n图2\nA: 图 1 曲线可表示细胞中同源染色体对数的数量变化\nB: 细胞甲为次级精母细胞, 处于图 1 中的 $\\mathrm{BC}$ 段（不含 $\\mathrm{B}$ 点）, 可能存在等位基因\nC: 图 1 曲线中 $\\mathrm{BC}$ 段细胞中染色体数量可能与 $\\mathrm{AB}$ 段细胞中染色体数量相同\nD: 若细胞乙的 DNA 两条链都被 ${ }^{32} \\mathrm{P}$ 标记, 使其在不含放射性的培养液中继续培养至第二次有丝分裂中期，部分染色体可能没有放射性\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 1 表示某二倍体动物细胞减数分裂过程中某项指标的数量变化曲线; 图 2 表示对该动物精巢切片显微观察后绘制的两幅细胞分裂示意图, 下列分析错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n甲\n\n[图3]\n\n乙\n\n图2\n\nA: 图 1 曲线可表示细胞中同源染色体对数的数量变化\nB: 细胞甲为次级精母细胞, 处于图 1 中的 $\\mathrm{BC}$ 段（不含 $\\mathrm{B}$ 点）, 可能存在等位基因\nC: 图 1 曲线中 $\\mathrm{BC}$ 段细胞中染色体数量可能与 $\\mathrm{AB}$ 段细胞中染色体数量相同\nD: 若细胞乙的 DNA 两条链都被 ${ }^{32} \\mathrm{P}$ 标记, 使其在不含放射性的培养液中继续培养至第二次有丝分裂中期，部分染色体可能没有放射性\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-65.jpg?height=294&width=622&top_left_y=1875&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-65.jpg?height=231&width=217&top_left_y=1889&top_left_x=951",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-65.jpg?height=225&width=234&top_left_y=1892&top_left_x=1159"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_58",
        "problem": "Which of the following statements is correct concerning gas exchange organs in animals?\nA: In starfish, the gill plays a role in gas exchange, but the tube feet do not play a role in that process.\nB: In grasshoppers, well-developed muscles surrounding the tracheae control movement of air inward and outward through an external opening.\nC: In fish, blood flows through the gill-filament capillaries in the same direction as that of water exiting from the mouth and pharynx to the outside.\nD: In birds, during exhalation both air sacs deflate, forcing air to the outside, whereas the lung is filled with air.\nE: In humans, surfactants are required to increase the surface tension in the trace amount of fluid coating the inner alveolar surface; in the absence of surfactants, the alveoli collapse during exhalation, blocking the entry of air during inhalation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following statements is correct concerning gas exchange organs in animals?\n\nA: In starfish, the gill plays a role in gas exchange, but the tube feet do not play a role in that process.\nB: In grasshoppers, well-developed muscles surrounding the tracheae control movement of air inward and outward through an external opening.\nC: In fish, blood flows through the gill-filament capillaries in the same direction as that of water exiting from the mouth and pharynx to the outside.\nD: In birds, during exhalation both air sacs deflate, forcing air to the outside, whereas the lung is filled with air.\nE: In humans, surfactants are required to increase the surface tension in the trace amount of fluid coating the inner alveolar surface; in the absence of surfactants, the alveoli collapse during exhalation, blocking the entry of air during inhalation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1037",
        "problem": "You perform a Western blot on two proteins of similar molecular weights $(\\sim 50 \\mathrm{kDa})$ and find only one band developed on your SDS-PAGE gel. How would you modify your assay to distinguish the two proteins?\nA: Use a lower concentration of acrylamide to raise the resolution of your gel\nB: Use a non-ionic detergent to denature your protein\nC: Focus your sample isoelectrically on a $\\mathrm{pH}$ gradient\nD: Remove reducing agents like mecaptoethanol or dithiothreitol\nE: Switch the anode and cathode on your gel\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nYou perform a Western blot on two proteins of similar molecular weights $(\\sim 50 \\mathrm{kDa})$ and find only one band developed on your SDS-PAGE gel. How would you modify your assay to distinguish the two proteins?\n\nA: Use a lower concentration of acrylamide to raise the resolution of your gel\nB: Use a non-ionic detergent to denature your protein\nC: Focus your sample isoelectrically on a $\\mathrm{pH}$ gradient\nD: Remove reducing agents like mecaptoethanol or dithiothreitol\nE: Switch the anode and cathode on your gel\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_734",
        "problem": "某种小鼠的体型有野生型和侏儒型, 由一对基因控制, 为研究其遗传机制, 用相应的纯种小鼠进行了甲与乙两组杂交实验, 过程及结果如下表, 下列选项能够解释该实验现象的是 ( )\n\n| 实验组合 | 亲本表型 |  | $\\mathrm{F}_{1}$ 表型 | $F_{1}$ 雌雄个体相互交配后 <br> 获得的 $F_{1}$ 表型及比例 |\n| :---: | :---: | :---: | :---: | :---: |\n|  | ð | q |  |  |\n| 甲 | 侏儒型 | 野生型 | 侏儒型 | 野生型: 侏儒型 $=1: 1$ |\n| 乙 | 野生型 | 侏儒型 | 野生型 | 野生型: 侏儒型 $=1: 1$ |\nA: 控制该性状的基因在线粒体 DNA 上\nB: 控制该性状的基因只位于 X 染色体上，且侏儒型为隐性性状\nC: 控制该性状的基因位于常染色体上, 且来源于母本的野生型、侏儒型基因均不表达\nD: 控制该性状的基因位于 $\\mathrm{X}$ 和 $\\mathrm{Y}$ 染色体的同源区段上，且野生型为隐性性状\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种小鼠的体型有野生型和侏儒型, 由一对基因控制, 为研究其遗传机制, 用相应的纯种小鼠进行了甲与乙两组杂交实验, 过程及结果如下表, 下列选项能够解释该实验现象的是 ( )\n\n| 实验组合 | 亲本表型 |  | $\\mathrm{F}_{1}$ 表型 | $F_{1}$ 雌雄个体相互交配后 <br> 获得的 $F_{1}$ 表型及比例 |\n| :---: | :---: | :---: | :---: | :---: |\n|  | ð | q |  |  |\n| 甲 | 侏儒型 | 野生型 | 侏儒型 | 野生型: 侏儒型 $=1: 1$ |\n| 乙 | 野生型 | 侏儒型 | 野生型 | 野生型: 侏儒型 $=1: 1$ |\n\nA: 控制该性状的基因在线粒体 DNA 上\nB: 控制该性状的基因只位于 X 染色体上，且侏儒型为隐性性状\nC: 控制该性状的基因位于常染色体上, 且来源于母本的野生型、侏儒型基因均不表达\nD: 控制该性状的基因位于 $\\mathrm{X}$ 和 $\\mathrm{Y}$ 染色体的同源区段上，且野生型为隐性性状\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_393",
        "problem": "在雄果蝇减数分裂中, 位于同一对染色体上的基因不会发生互换, 雌果蝇则可能发生互换。已知果蝇的灰身与黑身由等位基因（B、b）控制, 长翅与残翅由等位基因\n\n（V、v）控制，进行如下杂交实验。下列相关叙述错误的是（）\n\n| 实 | $P$ | $F_{1}$ | $F_{2}$ ( $F_{1}$ 中雄、雌个体相互交配获得） |\n| :--- | :--- | :--- | :--- |\n| 一 | 纯种灰身长翅 $\\times$ 黑身 <br> 残翅 | 灰身长 <br> 翅 | 灰身长翅: 黑身残翅: 灰身残翅:黑身长翅 $=14: 4: 1: 1$ |\nA: 上述两对基因位于一对同源染色体上\nB: $\\mathrm{F}_{1}$ 雄果蝇产生两种配子, 比例为 $\\mathrm{BV}: \\mathrm{bv}=1: 1$\nC: $\\mathrm{F}_{1}$ 雌果蝇产生四种配子, 比例为 $\\mathrm{BV}: \\mathrm{bv}: \\mathrm{Bv}: \\mathrm{bV}=4: 4: 1: 1$\nD: 若 $F_{1}$ 某卵母细胞发生了互换, 产生的卵细胞是 Bv, 则与之来自同一个次级卵母细胞的极体必然是 bV\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在雄果蝇减数分裂中, 位于同一对染色体上的基因不会发生互换, 雌果蝇则可能发生互换。已知果蝇的灰身与黑身由等位基因（B、b）控制, 长翅与残翅由等位基因\n\n（V、v）控制，进行如下杂交实验。下列相关叙述错误的是（）\n\n| 实 | $P$ | $F_{1}$ | $F_{2}$ ( $F_{1}$ 中雄、雌个体相互交配获得） |\n| :--- | :--- | :--- | :--- |\n| 一 | 纯种灰身长翅 $\\times$ 黑身 <br> 残翅 | 灰身长 <br> 翅 | 灰身长翅: 黑身残翅: 灰身残翅:黑身长翅 $=14: 4: 1: 1$ |\n\nA: 上述两对基因位于一对同源染色体上\nB: $\\mathrm{F}_{1}$ 雄果蝇产生两种配子, 比例为 $\\mathrm{BV}: \\mathrm{bv}=1: 1$\nC: $\\mathrm{F}_{1}$ 雌果蝇产生四种配子, 比例为 $\\mathrm{BV}: \\mathrm{bv}: \\mathrm{Bv}: \\mathrm{bV}=4: 4: 1: 1$\nD: 若 $F_{1}$ 某卵母细胞发生了互换, 产生的卵细胞是 Bv, 则与之来自同一个次级卵母细胞的极体必然是 bV\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1159",
        "problem": "During an investigation of mitochondrial metabolism, isolated mitochondria were maintained in a solution of 0.4 molar sucrose. The sucrose solution was used\nA: as a respiratory substrate\nB: to buffer the $\\mathrm{pH}$ of the medium\nC: as a carbon source of Krebs (tricarboxylic acid) cycle intermediates\nD: to prevent the mitochondrial membranes rupturing\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDuring an investigation of mitochondrial metabolism, isolated mitochondria were maintained in a solution of 0.4 molar sucrose. The sucrose solution was used\n\nA: as a respiratory substrate\nB: to buffer the $\\mathrm{pH}$ of the medium\nC: as a carbon source of Krebs (tricarboxylic acid) cycle intermediates\nD: to prevent the mitochondrial membranes rupturing\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_741",
        "problem": "利用苂光标记法对基因型为 $E X^{F} Y$ 的果蝇 $(2 n=8)$ 精原细胞的基因进行苂光标记,基因 E、e、 F 分别被红色、绿色、蓝色荧光标记。一个正在发生分裂的细胞中呈现大小、形态不同的 4 种染色体, 若不考虑基因突变和染色体畸变, 其苂光点的数量和颜色不可能是（ ）\nA: 1 个红色、 1 个绿色\nB: 2 个绿色、 2 个蓝色\nC: 2 个红色、 2 个绿色\nD: 1 个红色、 1 个绿色、 2 个蓝色\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n利用苂光标记法对基因型为 $E X^{F} Y$ 的果蝇 $(2 n=8)$ 精原细胞的基因进行苂光标记,基因 E、e、 F 分别被红色、绿色、蓝色荧光标记。一个正在发生分裂的细胞中呈现大小、形态不同的 4 种染色体, 若不考虑基因突变和染色体畸变, 其苂光点的数量和颜色不可能是（ ）\n\nA: 1 个红色、 1 个绿色\nB: 2 个绿色、 2 个蓝色\nC: 2 个红色、 2 个绿色\nD: 1 个红色、 1 个绿色、 2 个蓝色\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_217",
        "problem": "Which statement correctly describes the differentiation and development of cells and organs in flowering plants?\nA: Organomorphogenesis involves cell movement as one of the important mechanisms.\nB: Post-embryogenesis is a growth process, as all of the plant organs are pre-formed during embryogenesis.\nC: Totipotency of plant tissues provides the original source of power to develop a complete plant by re-differentiation, without going through the de-differentiation process.\nD: The direction of cell division determines cell type and function.\nE: Lineage information obtained by genetic inheritance overides environmental factors in determining the time for organ development.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich statement correctly describes the differentiation and development of cells and organs in flowering plants?\n\nA: Organomorphogenesis involves cell movement as one of the important mechanisms.\nB: Post-embryogenesis is a growth process, as all of the plant organs are pre-formed during embryogenesis.\nC: Totipotency of plant tissues provides the original source of power to develop a complete plant by re-differentiation, without going through the de-differentiation process.\nD: The direction of cell division determines cell type and function.\nE: Lineage information obtained by genetic inheritance overides environmental factors in determining the time for organ development.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_45",
        "problem": "Pedigrees 1-4 show the inheritance of four different rare disorders. It is known that the disease in pedigree 4 is $\\mathrm{X}$-linked recessive.\n\n[figure1]\nA: The disorder in pedigree 1 is most likely caused by a recessive allele.\nB: Person $\\mathrm{III}_{2}$ and $\\mathrm{III}_{7}$ in pedigree 2 have the same genotype.\nC: Pedigree 3 shows the inheritance of a rare disorder can be caused by a recessive allele on X-chromosome.\nD: If the affected man and his unaffected wife in pedigree 4 have a son then the probability of this son be affected is 0.125 .\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPedigrees 1-4 show the inheritance of four different rare disorders. It is known that the disease in pedigree 4 is $\\mathrm{X}$-linked recessive.\n\n[figure1]\n\nA: The disorder in pedigree 1 is most likely caused by a recessive allele.\nB: Person $\\mathrm{III}_{2}$ and $\\mathrm{III}_{7}$ in pedigree 2 have the same genotype.\nC: Pedigree 3 shows the inheritance of a rare disorder can be caused by a recessive allele on X-chromosome.\nD: If the affected man and his unaffected wife in pedigree 4 have a son then the probability of this son be affected is 0.125 .\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-089.jpg?height=1778&width=1661&top_left_y=390&top_left_x=252"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_569",
        "problem": "某种二倍体昆虫的性别决定方式为 XY 型, 存在斑翅与正常翅（由一对等位基因 $A 、 a$ 控制 )、桃色眼与黑色眼（由一对等位基因 $B 、 b$ 控制）两对相对性状。为了研究这两对相对性状的遗传机制, 实验小组选择斑翅桃色眼雌性昆虫与正常翅黑色眼雄性昆虫杂交, 得到 $F_{1}$ 代雌性昆虫为正常翅桃色眼: 正常翅黑色眼 $=1: 1$, 雄性昆虫为斑翅桃色眼: 玟翅黑色眼 $=1: 1, F_{1}$ 相互交配产生 $F_{2}$, 得到 $F_{2}$ 中正常翅黑色眼: 正常翅桃色眼: 斑翅黑色眼:玟翅桃色眼 $=2: 3: 2: 3$ 。下列说法错误的是（ ）\nA: 根据杂交实验结果, 可以确定翅型中的斑翅属于隐性性状\nB: BB 纯合时可能存在致死效应，且黑色眼昆虫的基因型是 Bb\nC: 昆虫种群中, 正常翅黑色眼昆虫的基因型共有 3 种\nD: $F_{2}$ 中斑翅桃色眼昆虫的基因型是 $b b X^{a} X^{a}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种二倍体昆虫的性别决定方式为 XY 型, 存在斑翅与正常翅（由一对等位基因 $A 、 a$ 控制 )、桃色眼与黑色眼（由一对等位基因 $B 、 b$ 控制）两对相对性状。为了研究这两对相对性状的遗传机制, 实验小组选择斑翅桃色眼雌性昆虫与正常翅黑色眼雄性昆虫杂交, 得到 $F_{1}$ 代雌性昆虫为正常翅桃色眼: 正常翅黑色眼 $=1: 1$, 雄性昆虫为斑翅桃色眼: 玟翅黑色眼 $=1: 1, F_{1}$ 相互交配产生 $F_{2}$, 得到 $F_{2}$ 中正常翅黑色眼: 正常翅桃色眼: 斑翅黑色眼:玟翅桃色眼 $=2: 3: 2: 3$ 。下列说法错误的是（ ）\n\nA: 根据杂交实验结果, 可以确定翅型中的斑翅属于隐性性状\nB: BB 纯合时可能存在致死效应，且黑色眼昆虫的基因型是 Bb\nC: 昆虫种群中, 正常翅黑色眼昆虫的基因型共有 3 种\nD: $F_{2}$ 中斑翅桃色眼昆虫的基因型是 $b b X^{a} X^{a}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_776",
        "problem": "控制果蝇红眼（A）和白眼（a）的基因位于 X 染色体。果蝇的性染色体组成与性别和育性的关系如表所示。杂交实验结果如图所示。下列关于 $F_{1}$ 和 $F_{2}$ 中例外个体的分析, 错误的是 ( )\n\n| 性染色体组成 | 性别及育性 |\n| :---: | :---: |\n| $\\mathrm{XX} 、 \\mathrm{XXY}$ | 甲可育 |\n| $\\mathrm{XY} 、 \\mathrm{XYY}$ | ð可育 |\n| $\\mathrm{XO}$ | ð不育 |\n| $\\mathrm{XXX} 、 \\mathrm{YO} 、 \\mathrm{YY}$ | 胚胎期死亡 |\n\n[图1]\nA: 不育雄果蝇的基因型为 $\\mathrm{X}^{\\mathrm{A}} \\mathrm{O}$\nB: $F_{1}$ 白眼雌果蝇产生的卵细胞基因型是 $X^{a}$\nC: $\\mathrm{F}_{1}$ 及 $\\mathrm{F}_{2}$ 中例外个体的出现源于母本减数分裂异常\nD: 可通过显微镜观察例外个体的染色体组成分析例外现象的成因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n控制果蝇红眼（A）和白眼（a）的基因位于 X 染色体。果蝇的性染色体组成与性别和育性的关系如表所示。杂交实验结果如图所示。下列关于 $F_{1}$ 和 $F_{2}$ 中例外个体的分析, 错误的是 ( )\n\n| 性染色体组成 | 性别及育性 |\n| :---: | :---: |\n| $\\mathrm{XX} 、 \\mathrm{XXY}$ | 甲可育 |\n| $\\mathrm{XY} 、 \\mathrm{XYY}$ | ð可育 |\n| $\\mathrm{XO}$ | ð不育 |\n| $\\mathrm{XXX} 、 \\mathrm{YO} 、 \\mathrm{YY}$ | 胚胎期死亡 |\n\n[图1]\n\nA: 不育雄果蝇的基因型为 $\\mathrm{X}^{\\mathrm{A}} \\mathrm{O}$\nB: $F_{1}$ 白眼雌果蝇产生的卵细胞基因型是 $X^{a}$\nC: $\\mathrm{F}_{1}$ 及 $\\mathrm{F}_{2}$ 中例外个体的出现源于母本减数分裂异常\nD: 可通过显微镜观察例外个体的染色体组成分析例外现象的成因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-76.jpg?height=577&width=1126&top_left_y=168&top_left_x=385"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_50",
        "problem": "Parathyroid hormone (PTH) plays an important role in the regulation of plasma calcium and phosphate levels. Figure Q. 72 shows the changes in levels of PTH, $\\mathrm{Ca}^{2+}$, and phosphate $(\\mathrm{Pi})$ in plasma of mice injected with a specific inhibitor of PTH secretion.\n[figure1]\nA: If Line I shows the level of PTH, then Line II and Line III would likely be showing the levels of $\\mathrm{Pi}$ and $\\mathrm{Ca}^{2+}$, respectively.\nB: PTH knock-out mice would have higher Pi levels in their urine compared with the wild type mice on the same diet.\nC: Eating a calcium-rich diet decreases the plasma level of vitamin D (active form) in healthy people.\nD: People with calcium-sensing receptor suppression have higher levels of plasma $\\mathrm{Ca}^{2+}$ compared with healthy people on the same diet.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nParathyroid hormone (PTH) plays an important role in the regulation of plasma calcium and phosphate levels. Figure Q. 72 shows the changes in levels of PTH, $\\mathrm{Ca}^{2+}$, and phosphate $(\\mathrm{Pi})$ in plasma of mice injected with a specific inhibitor of PTH secretion.\n[figure1]\n\nA: If Line I shows the level of PTH, then Line II and Line III would likely be showing the levels of $\\mathrm{Pi}$ and $\\mathrm{Ca}^{2+}$, respectively.\nB: PTH knock-out mice would have higher Pi levels in their urine compared with the wild type mice on the same diet.\nC: Eating a calcium-rich diet decreases the plasma level of vitamin D (active form) in healthy people.\nD: People with calcium-sensing receptor suppression have higher levels of plasma $\\mathrm{Ca}^{2+}$ compared with healthy people on the same diet.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-054.jpg?height=946&width=1212&top_left_y=660&top_left_x=408"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_320",
        "problem": "科研人员利用不耐高温的野生型水稻培育出两种耐高温水稻甲（aaBB）和乙\n\n（AAbb）, 并且用这两种耐高温水稻和野生型水稻进行了三组杂交实验, 实验结果如表。不考虑互换，下列分析错误的是（ ）\n\n| 杂交组合 |  | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :---: | :---: | :---: | :---: |\n| (1) | 甲×野生型 | $\\mathrm{F}_{1}$ 自交 | 不耐高温: 耐高温=3: 1 |\n| (2) | 乙×野生型 |  | 不耐高温: 耐高温=3: 1 |\n| (3) | 甲 $\\times$ 乙 |  | 不耐高温: 耐高温=1: 1 |\nA: (1)一(3)组杂交实验的 $\\mathrm{F}_{1}$ 基因型不同，表型相同\nB: 控制水稻耐高温的两对等位基因的遗传不遵循自由组合定律\nC: 若(1)组 $F_{2}$ 中不耐高温的水稻自交, 则子代耐高温水稻占 $1 / 6$\nD: (3)组 $F_{2}$ 中不耐高温的水稻和耐高温的水稻基因型均只有一种\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n科研人员利用不耐高温的野生型水稻培育出两种耐高温水稻甲（aaBB）和乙\n\n（AAbb）, 并且用这两种耐高温水稻和野生型水稻进行了三组杂交实验, 实验结果如表。不考虑互换，下列分析错误的是（ ）\n\n| 杂交组合 |  | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :---: | :---: | :---: | :---: |\n| (1) | 甲×野生型 | $\\mathrm{F}_{1}$ 自交 | 不耐高温: 耐高温=3: 1 |\n| (2) | 乙×野生型 |  | 不耐高温: 耐高温=3: 1 |\n| (3) | 甲 $\\times$ 乙 |  | 不耐高温: 耐高温=1: 1 |\n\nA: (1)一(3)组杂交实验的 $\\mathrm{F}_{1}$ 基因型不同，表型相同\nB: 控制水稻耐高温的两对等位基因的遗传不遵循自由组合定律\nC: 若(1)组 $F_{2}$ 中不耐高温的水稻自交, 则子代耐高温水稻占 $1 / 6$\nD: (3)组 $F_{2}$ 中不耐高温的水稻和耐高温的水稻基因型均只有一种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_344",
        "problem": "在二倍体生物中, 单体 $(2 n-1)$ 指体细胞中某对同源染色体缺失一条的个体, 三体 $(2 n+1)$ 指体细胞中某对同源染色体多一条的个体，单体和三体都可以用于基因定位。某大豆突变株表现为黄叶 ( $\\mathrm{yy}$ ), 为探究 $\\mathrm{Y} / \\mathrm{y}$ 是否位于 7 号染色体上, 用该突变株分别与单体 (7 号染色体缺失一条)、三体 (7 号染色体多一条) 绿叶纯合植株杂交得 $\\mathrm{F}_{1}$,让 $F_{1}$ 自交得 $F_{2}$ 。已知单体和三体产生的配子均可育, 而一对同源染色体均缺失的个体致死。以下叙述错误的是（ ）\nA: 通过突变株与单体杂交的 $F_{1}$ 即可进行基因定位, 而与三体杂交的 $F_{1}$ 则不能\nB: 若突变株与单体杂交的 $\\mathrm{F}_{2}$ 黄叶:绿叶 $=3$ : 4 , 则 $\\mathrm{Y} / \\mathrm{y}$ 基因位于 7 号染色体上\nC: 若突变株与三体杂交的 $\\mathrm{F}_{2}$ 黄叶: 绿叶 $=5: 31$, 则 $\\mathrm{Y} / \\mathrm{y}$ 基因位于 7 号染色体上\nD: 若 $Y / y$ 基因不位于 7 号染色体, 则突变株与单体或三体杂交的 $F_{2}$ 全为黄叶: 绿叶 $=1: 3$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在二倍体生物中, 单体 $(2 n-1)$ 指体细胞中某对同源染色体缺失一条的个体, 三体 $(2 n+1)$ 指体细胞中某对同源染色体多一条的个体，单体和三体都可以用于基因定位。某大豆突变株表现为黄叶 ( $\\mathrm{yy}$ ), 为探究 $\\mathrm{Y} / \\mathrm{y}$ 是否位于 7 号染色体上, 用该突变株分别与单体 (7 号染色体缺失一条)、三体 (7 号染色体多一条) 绿叶纯合植株杂交得 $\\mathrm{F}_{1}$,让 $F_{1}$ 自交得 $F_{2}$ 。已知单体和三体产生的配子均可育, 而一对同源染色体均缺失的个体致死。以下叙述错误的是（ ）\n\nA: 通过突变株与单体杂交的 $F_{1}$ 即可进行基因定位, 而与三体杂交的 $F_{1}$ 则不能\nB: 若突变株与单体杂交的 $\\mathrm{F}_{2}$ 黄叶:绿叶 $=3$ : 4 , 则 $\\mathrm{Y} / \\mathrm{y}$ 基因位于 7 号染色体上\nC: 若突变株与三体杂交的 $\\mathrm{F}_{2}$ 黄叶: 绿叶 $=5: 31$, 则 $\\mathrm{Y} / \\mathrm{y}$ 基因位于 7 号染色体上\nD: 若 $Y / y$ 基因不位于 7 号染色体, 则突变株与单体或三体杂交的 $F_{2}$ 全为黄叶: 绿叶 $=1: 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_742",
        "problem": "同位素标记法是生物学常用的一种研究方法, 下列相关叙述正确的是 ( )\nA: 给脉鼠供应 ${ }^{18} \\mathrm{O}_{2}$, 其呼出气体中会含有 $\\mathrm{C}^{18} \\mathrm{O}_{2}$\nB: 给小麦供应 ${ }^{14} \\mathrm{CO}_{2},{ }^{14} \\mathrm{C}$ 不会出现在线粒体中\nC: 用含有 ${ }^{35} \\mathrm{~S}$ 标记的普通培养基来培养噬菌体，可获得 ${ }^{35} \\mathrm{~S}$ 标记的噬菌体\nD: 设法让洋葱根尖吸收含 ${ }^{3} \\mathrm{H}$ 标记的尿嘧啶核糖核苷酸, 只能在分生区细胞中检测到放射性\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n同位素标记法是生物学常用的一种研究方法, 下列相关叙述正确的是 ( )\n\nA: 给脉鼠供应 ${ }^{18} \\mathrm{O}_{2}$, 其呼出气体中会含有 $\\mathrm{C}^{18} \\mathrm{O}_{2}$\nB: 给小麦供应 ${ }^{14} \\mathrm{CO}_{2},{ }^{14} \\mathrm{C}$ 不会出现在线粒体中\nC: 用含有 ${ }^{35} \\mathrm{~S}$ 标记的普通培养基来培养噬菌体，可获得 ${ }^{35} \\mathrm{~S}$ 标记的噬菌体\nD: 设法让洋葱根尖吸收含 ${ }^{3} \\mathrm{H}$ 标记的尿嘧啶核糖核苷酸, 只能在分生区细胞中检测到放射性\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1015",
        "problem": "Many metabolic pathways involve multi-step reactions. Consider the following pathway, where E represents different enzymes, and A, B, C, D and F represent substrates and products of the pathway.\n\n$$\n\\begin{aligned}\n& \\begin{array}{llll}\n\\mathrm{E}_{1} & \\mathrm{E}_{2} & \\mathrm{E}_{3} & \\mathrm{E}_{4}\n\\end{array} \\\\\n& \\mathrm{~A} \\rightarrow \\mathrm{B} \\rightarrow \\mathrm{C} \\rightarrow \\mathrm{D} \\rightarrow \\mathrm{F}\n\\end{aligned}\n$$\n\nFeedback inhibition of this pathway may involve\nA: The product of the final reaction, $\\mathrm{F}$, interacting with and inhibiting $\\mathrm{E}_{1}$\nB: F interacting with and inhibiting product $B$\nC: Product $\\mathrm{B}$ interacting with and inhibiting $\\mathrm{E}_{4}$\nD: Product $\\mathrm{C}$ interacting with and inhibiting $\\mathrm{E}_{4}$\nE: $\\mathrm{E}_{3}$ interacting with and inhibiting $\\mathrm{E}_{2}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMany metabolic pathways involve multi-step reactions. Consider the following pathway, where E represents different enzymes, and A, B, C, D and F represent substrates and products of the pathway.\n\n$$\n\\begin{aligned}\n& \\begin{array}{llll}\n\\mathrm{E}_{1} & \\mathrm{E}_{2} & \\mathrm{E}_{3} & \\mathrm{E}_{4}\n\\end{array} \\\\\n& \\mathrm{~A} \\rightarrow \\mathrm{B} \\rightarrow \\mathrm{C} \\rightarrow \\mathrm{D} \\rightarrow \\mathrm{F}\n\\end{aligned}\n$$\n\nFeedback inhibition of this pathway may involve\n\nA: The product of the final reaction, $\\mathrm{F}$, interacting with and inhibiting $\\mathrm{E}_{1}$\nB: F interacting with and inhibiting product $B$\nC: Product $\\mathrm{B}$ interacting with and inhibiting $\\mathrm{E}_{4}$\nD: Product $\\mathrm{C}$ interacting with and inhibiting $\\mathrm{E}_{4}$\nE: $\\mathrm{E}_{3}$ interacting with and inhibiting $\\mathrm{E}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1298",
        "problem": "The tamarin, a New World monkey, lives in groups containing a dominant female who suppresses ovulation in subordinates, causing her to be the only one capable of reproduction. The dominant female births multiple offspring fertilized by more than one male. The males have a high investment in the care of the offspring with males often carrying infants on their backs, even when they are not the father. Both related and unrelated males cooperate to care for the young. Which of the following terms best describes the behavior of the female tamarins?\nA: Monogamy\nB: Polygamy\nC: Polygyny\nD: Polyandry\nE: Promiscuity\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe tamarin, a New World monkey, lives in groups containing a dominant female who suppresses ovulation in subordinates, causing her to be the only one capable of reproduction. The dominant female births multiple offspring fertilized by more than one male. The males have a high investment in the care of the offspring with males often carrying infants on their backs, even when they are not the father. Both related and unrelated males cooperate to care for the young. Which of the following terms best describes the behavior of the female tamarins?\n\nA: Monogamy\nB: Polygamy\nC: Polygyny\nD: Polyandry\nE: Promiscuity\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1337",
        "problem": "Conservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.The major identified risks to Maui's dolphins are:\n\n1. Trawl fishing\n2. Set netting\n3. Seismic surveying\n4. Mining and oil activities\n5. Noise leading to trauma\n6. Non-trauma noise effects\n\nWhich of these risks are estimated to be above the PBR and could result in gradual extinction of the Maui's dolphins if the current rates are allowed to continue.\nA: 1,2,3,4,5 and 6\nB: 2,3,4,5 and 6 only.\nC: 1,2 and 4 only.\nD: 1, 2, 4 and 6 only.\nE: 2, 4 and 6 only.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nConservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.\n\nproblem:\nThe major identified risks to Maui's dolphins are:\n\n1. Trawl fishing\n2. Set netting\n3. Seismic surveying\n4. Mining and oil activities\n5. Noise leading to trauma\n6. Non-trauma noise effects\n\nWhich of these risks are estimated to be above the PBR and could result in gradual extinction of the Maui's dolphins if the current rates are allowed to continue.\n\nA: 1,2,3,4,5 and 6\nB: 2,3,4,5 and 6 only.\nC: 1,2 and 4 only.\nD: 1, 2, 4 and 6 only.\nE: 2, 4 and 6 only.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_668",
        "problem": "基因转录出的初始 RNA 要经过加工才能发挥作用。初始 RNA 经不同方式的剪接形成不同的 mRNA。研究人员从同一个体的造血干细胞和浆细胞中分别提取它们的全部 mRNA (分别标记为 L-mRNA 和 P-mRNA), 并以此为模板合成相应的单链 DNA (分别标记为 L-cDNA 和 P-cDNA)。下列有关叙述错误的是（）\nA: 转录产物的不同剪接使一个基因编码多种不同结构的多肽成为可能\nB: 将 P-cDNA 与 L-cDNA 混合后, 会出现双链 DNA 现象\nC: L-cDNA 和 P-cDNA 就是能转录出初始 RNA 的基因的模板链\nD: 能与 L-cDNA 互补的 P-mRNA 中含有编码 ATP 酶的 mRNA\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n基因转录出的初始 RNA 要经过加工才能发挥作用。初始 RNA 经不同方式的剪接形成不同的 mRNA。研究人员从同一个体的造血干细胞和浆细胞中分别提取它们的全部 mRNA (分别标记为 L-mRNA 和 P-mRNA), 并以此为模板合成相应的单链 DNA (分别标记为 L-cDNA 和 P-cDNA)。下列有关叙述错误的是（）\n\nA: 转录产物的不同剪接使一个基因编码多种不同结构的多肽成为可能\nB: 将 P-cDNA 与 L-cDNA 混合后, 会出现双链 DNA 现象\nC: L-cDNA 和 P-cDNA 就是能转录出初始 RNA 的基因的模板链\nD: 能与 L-cDNA 互补的 P-mRNA 中含有编码 ATP 酶的 mRNA\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1375",
        "problem": "[figure1]\n\nThe organelle labelled $B$ plays a role in the structural integrity of the cell. When filled with water, ions, minerals, and nutrients this organelle exerts a pressure to maintain the structural integrity. The term that relates to this pressure is:\nA: Osmotic pressure\nB: Turgor pressure\nC: Cell pressure\nD: Hydrostatic pressure\nE: Structural pressure\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nThe organelle labelled $B$ plays a role in the structural integrity of the cell. When filled with water, ions, minerals, and nutrients this organelle exerts a pressure to maintain the structural integrity. The term that relates to this pressure is:\n\nA: Osmotic pressure\nB: Turgor pressure\nC: Cell pressure\nD: Hydrostatic pressure\nE: Structural pressure\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-03.jpg?height=434&width=485&top_left_y=314&top_left_x=777"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_811",
        "problem": "为达到实验目的, 需要选用合适的实验材料进行实验。下列实验目的与实验材料的对应, 不合理的是\n|  | 实验材料 | 实验目的 |\n| :---: | :---: | :---: |\n| A | 大蒜根尖分生区细胞 | 观察细胞的质壁分离与复原 |\n| B | 蝗虫的精巢细胞 | 观察细胞的减数分裂 |\n| C | 哺乳动物的红细胞 | 观察细胞的吸水和失水 |\n| D | 人口腔上皮细胞 | 观察 DNA、RNA 在细胞中的分布 |\nA: A\nB: $B$\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n为达到实验目的, 需要选用合适的实验材料进行实验。下列实验目的与实验材料的对应, 不合理的是\n|  | 实验材料 | 实验目的 |\n| :---: | :---: | :---: |\n| A | 大蒜根尖分生区细胞 | 观察细胞的质壁分离与复原 |\n| B | 蝗虫的精巢细胞 | 观察细胞的减数分裂 |\n| C | 哺乳动物的红细胞 | 观察细胞的吸水和失水 |\n| D | 人口腔上皮细胞 | 观察 DNA、RNA 在细胞中的分布 |\n\nA: A\nB: $B$\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1396",
        "problem": "Differential staining is an important technique used in bacterial microbiology. One such stain is the Gram stain, which uses a crystal violet dye that binds to a cell wall layer called peptidoglycan. Thick layers of peptidoglycan stain dark purple in Gram-positive bacteria because the stain is retained. Thinner layers of peptidoglycan do not retain the stain effectively in Gram-negative bacteria and are therefore not stained the same deep purple. This is important because the staining result and subsequent classification inform medical staff on how best to treat an infection with this bacterium.\n\nBased on the information about differential staining, choose the CORRECT answer:\nA: A Gram-negative and a Gram-positive bacterium will stain the same\nB: Gram-positive bacteria are harder to kill\nC: Gram-positive bacteria can be identified by a thick purple stained cell wall\nD: Gram-negative bacteria have no peptidoglycan layer in their cell wall\nE: The gram stain can be used to classify bacteria by their genus as well as their cell wall properties\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDifferential staining is an important technique used in bacterial microbiology. One such stain is the Gram stain, which uses a crystal violet dye that binds to a cell wall layer called peptidoglycan. Thick layers of peptidoglycan stain dark purple in Gram-positive bacteria because the stain is retained. Thinner layers of peptidoglycan do not retain the stain effectively in Gram-negative bacteria and are therefore not stained the same deep purple. This is important because the staining result and subsequent classification inform medical staff on how best to treat an infection with this bacterium.\n\nBased on the information about differential staining, choose the CORRECT answer:\n\nA: A Gram-negative and a Gram-positive bacterium will stain the same\nB: Gram-positive bacteria are harder to kill\nC: Gram-positive bacteria can be identified by a thick purple stained cell wall\nD: Gram-negative bacteria have no peptidoglycan layer in their cell wall\nE: The gram stain can be used to classify bacteria by their genus as well as their cell wall properties\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_744",
        "problem": "两种柳穿鱼植株杂交, $F_{1}$ 均开两侧对称花, $F_{1}$ 自交产生的 $F_{2}$ 中开两侧对称花 34 株,开辐射对称花的 5 株。进一步研究发现，两种柳穿鱼植株的 Leyc 基因碱基序列相同,只是在开两侧对称花植株中表达，在开辐射对称花植株中不表达，二者 Lcyc 基因的甲基化情况如下图所示。下列叙述正确的是（）\n\n[图1]\nA: 控制两侧对称和辐射对称花的基因所含遗传信息不同\nB: $F_{2}$ 表型比说明柳穿鱼花型的遗传遵循基因的分离定律\nC: 控制辐射对称花的 Lcyc 基因的甲基化程度相对较高\nD: 推测甲基化的程度与 Lcyc 基因的表达程度成正相关\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n两种柳穿鱼植株杂交, $F_{1}$ 均开两侧对称花, $F_{1}$ 自交产生的 $F_{2}$ 中开两侧对称花 34 株,开辐射对称花的 5 株。进一步研究发现，两种柳穿鱼植株的 Leyc 基因碱基序列相同,只是在开两侧对称花植株中表达，在开辐射对称花植株中不表达，二者 Lcyc 基因的甲基化情况如下图所示。下列叙述正确的是（）\n\n[图1]\n\nA: 控制两侧对称和辐射对称花的基因所含遗传信息不同\nB: $F_{2}$ 表型比说明柳穿鱼花型的遗传遵循基因的分离定律\nC: 控制辐射对称花的 Lcyc 基因的甲基化程度相对较高\nD: 推测甲基化的程度与 Lcyc 基因的表达程度成正相关\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-56.jpg?height=277&width=1325&top_left_y=1940&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_705",
        "problem": "下列为甲型血友病、糖原沉积病I型及虽豆病的遗传系谱图。已知甲家族不具有糖原沉积病I型致病基因, 乙家族不具有蚕豆病致病基因，I-2 不含蚕豆病致病基因。人群中患糖原沉积病I型的概率为 $1 / 2500$ 。下列叙述错误的是（ ）\n[图1]\nA: II-1 与II-6 基因型相同的概率为 49/51\nB: II-5 与II-6 再生一个两病兼患的孩子概率为 $1 / 612$\nC: IV-2 的性染色体组成为 XXY, 则产生异常生殖细胞的最可能是母亲\nD: 为确保IV-3 的优生, III-3 可通过羊膜腔穿刺获取少许羊水, 离心后进行羊水细胞和生物化学方面的检查\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列为甲型血友病、糖原沉积病I型及虽豆病的遗传系谱图。已知甲家族不具有糖原沉积病I型致病基因, 乙家族不具有蚕豆病致病基因，I-2 不含蚕豆病致病基因。人群中患糖原沉积病I型的概率为 $1 / 2500$ 。下列叙述错误的是（ ）\n[图1]\n\nA: II-1 与II-6 基因型相同的概率为 49/51\nB: II-5 与II-6 再生一个两病兼患的孩子概率为 $1 / 612$\nC: IV-2 的性染色体组成为 XXY, 则产生异常生殖细胞的最可能是母亲\nD: 为确保IV-3 的优生, III-3 可通过羊膜腔穿刺获取少许羊水, 离心后进行羊水细胞和生物化学方面的检查\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-057.jpg?height=402&width=1194&top_left_y=1141&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1418",
        "problem": "In Australia, the $25^{\\text {th }}$ of April is ANZAC day. ANZAC Day is a national day of remembrance and honours all who served and died in all wars, conflicts, and peacekeeping operations.\n\nIn the United States however, the $25^{\\text {th }}$ of April is National DNA Day. The day commemorates the completion of the Human Genome Project in 2003 and the discovery of DNA's double helix structure in 1953.\n\nThe human genome is?\nA: All of our genes\nB: All of our DNA\nC: All of the DNA and RNA found in the nucleus of a cell\nD: All of our RNA\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn Australia, the $25^{\\text {th }}$ of April is ANZAC day. ANZAC Day is a national day of remembrance and honours all who served and died in all wars, conflicts, and peacekeeping operations.\n\nIn the United States however, the $25^{\\text {th }}$ of April is National DNA Day. The day commemorates the completion of the Human Genome Project in 2003 and the discovery of DNA's double helix structure in 1953.\n\nThe human genome is?\n\nA: All of our genes\nB: All of our DNA\nC: All of the DNA and RNA found in the nucleus of a cell\nD: All of our RNA\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_481",
        "problem": "人的某条染色体上 $A 、 B 、 C$ 三个基因紧密排列，且不发生互换。这三个基因各有上百个等位基因（用 $\\mathrm{A}_{1} \\sim \\mathrm{A}_{n}, \\mathrm{~B}_{1}-\\mathrm{B}_{\\mathrm{n}}, \\mathrm{C}_{1} \\sim \\mathrm{C}_{\\mathrm{n}}$ 表示)。家庭成员基因组成如表所示，下列分析正确的是（ ）\n\n| 家庭成员 | 父亲 | 母亲 | 儿子 | 女儿 |\n| :---: | :---: | :---: | :---: | :---: |\n| 基因组成 | $\\mathrm{A}_{23} \\mathrm{~A}_{25} \\mathrm{~B}_{7} \\mathrm{~B}_{35} \\mathrm{C}_{2} \\mathrm{C}_{4}$ | $\\mathrm{~A}_{3} \\mathrm{~A}_{24} \\mathrm{~B}_{8} \\mathrm{~B}_{44} \\mathrm{C}_{5} \\mathrm{C}_{9}$ | $\\mathrm{~A}_{24} \\mathrm{~A}_{25} \\mathrm{~B}_{7} \\mathrm{~B}_{8} \\mathrm{C}_{4} \\mathrm{C}_{5}$ | $\\mathrm{~A}_{3} \\mathrm{~A}_{23} \\mathrm{~B}_{35} \\mathrm{~B}_{44} \\mathrm{C}_{2} \\mathrm{C}_{9}$ |\nA: 基因 $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C}$ 的遗传方式是伴 $\\mathrm{Y}$ 染色体遗传\nB: 父亲的其中一条染色体上基因组成是 $\\mathrm{A}_{25} \\mathrm{~B}_{7} \\mathrm{C}_{2}$\nC: 基因 $\\mathrm{A}_{1} \\sim \\mathrm{A}_{n}$ 的遗传符合分离定律，基因 $\\mathrm{B}$ 与基因 $\\mathrm{C}$ 的遗传符合自由组合定律\nD: 若此夫妻第 3 个孩子的 $\\mathrm{A}$ 基因组成为 $\\mathrm{A}_{23} \\mathrm{~A}_{24}$, 则其 $\\mathrm{C}$ 基因组成为 $\\mathrm{C}_{2} \\mathrm{C}_{5}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人的某条染色体上 $A 、 B 、 C$ 三个基因紧密排列，且不发生互换。这三个基因各有上百个等位基因（用 $\\mathrm{A}_{1} \\sim \\mathrm{A}_{n}, \\mathrm{~B}_{1}-\\mathrm{B}_{\\mathrm{n}}, \\mathrm{C}_{1} \\sim \\mathrm{C}_{\\mathrm{n}}$ 表示)。家庭成员基因组成如表所示，下列分析正确的是（ ）\n\n| 家庭成员 | 父亲 | 母亲 | 儿子 | 女儿 |\n| :---: | :---: | :---: | :---: | :---: |\n| 基因组成 | $\\mathrm{A}_{23} \\mathrm{~A}_{25} \\mathrm{~B}_{7} \\mathrm{~B}_{35} \\mathrm{C}_{2} \\mathrm{C}_{4}$ | $\\mathrm{~A}_{3} \\mathrm{~A}_{24} \\mathrm{~B}_{8} \\mathrm{~B}_{44} \\mathrm{C}_{5} \\mathrm{C}_{9}$ | $\\mathrm{~A}_{24} \\mathrm{~A}_{25} \\mathrm{~B}_{7} \\mathrm{~B}_{8} \\mathrm{C}_{4} \\mathrm{C}_{5}$ | $\\mathrm{~A}_{3} \\mathrm{~A}_{23} \\mathrm{~B}_{35} \\mathrm{~B}_{44} \\mathrm{C}_{2} \\mathrm{C}_{9}$ |\n\nA: 基因 $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C}$ 的遗传方式是伴 $\\mathrm{Y}$ 染色体遗传\nB: 父亲的其中一条染色体上基因组成是 $\\mathrm{A}_{25} \\mathrm{~B}_{7} \\mathrm{C}_{2}$\nC: 基因 $\\mathrm{A}_{1} \\sim \\mathrm{A}_{n}$ 的遗传符合分离定律，基因 $\\mathrm{B}$ 与基因 $\\mathrm{C}$ 的遗传符合自由组合定律\nD: 若此夫妻第 3 个孩子的 $\\mathrm{A}$ 基因组成为 $\\mathrm{A}_{23} \\mathrm{~A}_{24}$, 则其 $\\mathrm{C}$ 基因组成为 $\\mathrm{C}_{2} \\mathrm{C}_{5}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_618",
        "problem": "果蝇 $(2 \\mathrm{~N}=8)$ 的某一个精原细胞进行减数分裂时发生了一次异常, 现分别检测了其分裂进行至 $\\mathrm{T}_{1} 、 \\mathrm{~T}_{2} 、 \\mathrm{~T}_{3}$ 时期的三个细胞中染色体、核 DNA、染色单体的数量，结果如图, 下列叙述正确的是（ ）\n\n[图1]\nA: (1)(3)分别表示核 DNA、染色单体的数量\nB: $\\mathrm{T}_{2}$ 时期可发生同源非姐妹染色体单体互换\nC: $\\mathrm{T}_{3}$ 时期的细胞中 $\\mathrm{Y}$ 染色体的数量可能为 $0 、 2$\nD: 最终形成的 4 个精细胞染色体数均异常\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇 $(2 \\mathrm{~N}=8)$ 的某一个精原细胞进行减数分裂时发生了一次异常, 现分别检测了其分裂进行至 $\\mathrm{T}_{1} 、 \\mathrm{~T}_{2} 、 \\mathrm{~T}_{3}$ 时期的三个细胞中染色体、核 DNA、染色单体的数量，结果如图, 下列叙述正确的是（ ）\n\n[图1]\n\nA: (1)(3)分别表示核 DNA、染色单体的数量\nB: $\\mathrm{T}_{2}$ 时期可发生同源非姐妹染色体单体互换\nC: $\\mathrm{T}_{3}$ 时期的细胞中 $\\mathrm{Y}$ 染色体的数量可能为 $0 、 2$\nD: 最终形成的 4 个精细胞染色体数均异常\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-28.jpg?height=585&width=988&top_left_y=2106&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_331",
        "problem": "葫芦科植物喷瓜的性别由染色体上一组复等位基因决定: 雄性基因 $\\mathrm{a}^{\\mathrm{D}}$ 、雌雄同体基因 $\\mathrm{a}^{+}$和雌性基因 $\\mathrm{a}^{\\mathrm{d}}$, 其中 $\\mathrm{a}^{\\mathrm{D}}$ 对 $\\mathrm{a}^{+}$和 $\\mathrm{a}^{\\mathrm{d}}$ 为显性, $\\mathrm{a}^{+}$对 $\\mathrm{a}^{\\mathrm{d}}$ 为显性。喷瓜的叶色由另外一对染色体上的一对等位基因 ( $\\mathrm{B} / \\mathrm{b})$ 控制, $\\mathrm{BB}$ 表现深绿; $\\mathrm{Bb}$ 表现浅绿; $\\mathrm{bb}$ 呈黄色, 幼苗阶段死亡。现有两株基因型 $\\mathrm{a}^{+} \\mathrm{a}^{+} \\mathrm{Bb}$ 和 $\\mathrm{a}^{+} \\mathrm{a}^{\\mathrm{d}} \\mathrm{Bb}$ 的喷瓜随机授粉产生 $\\mathrm{F}_{1}$, 取 $\\mathrm{F}_{1}$ 两性植株自交产生 $F_{2}$ ，下列关于子代成熟植株的叙述，正确的是（ ）\nA: $F_{1}$ 深绿色与浅绿色之比为 $2: 1$\nB: $F_{1}$ 深绿色基因型有 4 种\nC: $F_{2}$ 深绿色与浅绿色之比为 3:2\nD: $\\mathrm{F}_{2}$ 中 $\\mathrm{a}^{+}$基因频率为 $50 \\%$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n葫芦科植物喷瓜的性别由染色体上一组复等位基因决定: 雄性基因 $\\mathrm{a}^{\\mathrm{D}}$ 、雌雄同体基因 $\\mathrm{a}^{+}$和雌性基因 $\\mathrm{a}^{\\mathrm{d}}$, 其中 $\\mathrm{a}^{\\mathrm{D}}$ 对 $\\mathrm{a}^{+}$和 $\\mathrm{a}^{\\mathrm{d}}$ 为显性, $\\mathrm{a}^{+}$对 $\\mathrm{a}^{\\mathrm{d}}$ 为显性。喷瓜的叶色由另外一对染色体上的一对等位基因 ( $\\mathrm{B} / \\mathrm{b})$ 控制, $\\mathrm{BB}$ 表现深绿; $\\mathrm{Bb}$ 表现浅绿; $\\mathrm{bb}$ 呈黄色, 幼苗阶段死亡。现有两株基因型 $\\mathrm{a}^{+} \\mathrm{a}^{+} \\mathrm{Bb}$ 和 $\\mathrm{a}^{+} \\mathrm{a}^{\\mathrm{d}} \\mathrm{Bb}$ 的喷瓜随机授粉产生 $\\mathrm{F}_{1}$, 取 $\\mathrm{F}_{1}$ 两性植株自交产生 $F_{2}$ ，下列关于子代成熟植株的叙述，正确的是（ ）\n\nA: $F_{1}$ 深绿色与浅绿色之比为 $2: 1$\nB: $F_{1}$ 深绿色基因型有 4 种\nC: $F_{2}$ 深绿色与浅绿色之比为 3:2\nD: $\\mathrm{F}_{2}$ 中 $\\mathrm{a}^{+}$基因频率为 $50 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_761",
        "problem": "下图为甲、乙两种单基因遗传病的遗传家系图，其中 I-2 只携带一种致病基因。人群中甲病基因频率为 $1 / 15$, 乙病基因频率为 $1 / 10$ 。下列叙述正确的是（ ）\n\n[图1]\nA: 甲病为伴性遗传, 乙病为常染色体遗传\nB: II-3 体细胞内最多含有 6 个致病基因\nC: III-1 与一正常女性结婚, 生下患病小孩的概率为 6/11\nD: III- 7 与一甲病男性结婚，生下正常小孩的概率为 49/116\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为甲、乙两种单基因遗传病的遗传家系图，其中 I-2 只携带一种致病基因。人群中甲病基因频率为 $1 / 15$, 乙病基因频率为 $1 / 10$ 。下列叙述正确的是（ ）\n\n[图1]\n\nA: 甲病为伴性遗传, 乙病为常染色体遗传\nB: II-3 体细胞内最多含有 6 个致病基因\nC: III-1 与一正常女性结婚, 生下患病小孩的概率为 6/11\nD: III- 7 与一甲病男性结婚，生下正常小孩的概率为 49/116\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-009.jpg?height=305&width=828&top_left_y=1869&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_109",
        "problem": "J. B. S. Haldane (1892-1964), one of founders of the modern evolutionary biology, stated that if among hybrid offspring one sex is absent, rare, or sterile, that sex is the heterogametic one, i.e., in an X-Y sex-determination system, XY hybrids are preferentially sterile or inviable. Presgraves and Orr (1998) decided to test Haldane's rule on taxa lacking a hemizygous X.\n\nThey focused on the mosquitoes from the genera Aedes and Anopheles. In Aedes, males are XY and females are XX, but both X and Y chromosomes carry complete sets of homologous genes, while in Anopheles Y chromosomes are degenerate. They compiled the available data on the sterility of the hybrids resulted from crossing pairs of species in these taxa (tables below).\n\nHint: Assume each character is controlled by many loci.\n\nData on Aedes\n\n| $\\mathbf{A}$ | B | A crossed with B |\n| :---: | :---: | :---: |\n| Ae. zoosophus | Ae. triseriatus | Only males are sterile |\n| Ae. triseriatus | Ae. brelandi | Only males are sterile. |\n| Ae. sollicitans | {Ae. taeniorhynchus Both sexes are sterile. <br> s Ae. nigromaculatus Both sexes are sterile.} |  |\n| Ae. taeniorhync |  |  |\n| Data on Anopheles |  |  |\n| $\\bar{A}$ | B | A crossed with B |\n| An. albitarsus | An. daeneorum | Only males are sterile. |\n| An. crucians | An. bradleyi | Only males are sterile. |\n| An. freeborni | An. occidentalis | Only males are sterile. |\n| An. freeborni | An. atroparvus | Only males are sterile. |\nA: If the males in genus Anopheles were under strong selection, we would expect the same pattern of hybrid sterility as presented here.\nB: The hybrid sterility patterns observed in both genera are consistent with the idea that alleles involved in hybrid sterility are recessive.\nC: If the pattern of sterility in Aedes were identical to that of Anopheles, it would disprove Haldane's rule.\nD: The pattern of hybrid sterility in Aedes suggest variation in $\\mathrm{Y}$ chromosome functionality in this genus.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nJ. B. S. Haldane (1892-1964), one of founders of the modern evolutionary biology, stated that if among hybrid offspring one sex is absent, rare, or sterile, that sex is the heterogametic one, i.e., in an X-Y sex-determination system, XY hybrids are preferentially sterile or inviable. Presgraves and Orr (1998) decided to test Haldane's rule on taxa lacking a hemizygous X.\n\nThey focused on the mosquitoes from the genera Aedes and Anopheles. In Aedes, males are XY and females are XX, but both X and Y chromosomes carry complete sets of homologous genes, while in Anopheles Y chromosomes are degenerate. They compiled the available data on the sterility of the hybrids resulted from crossing pairs of species in these taxa (tables below).\n\nHint: Assume each character is controlled by many loci.\n\nData on Aedes\n\n| $\\mathbf{A}$ | B | A crossed with B |\n| :---: | :---: | :---: |\n| Ae. zoosophus | Ae. triseriatus | Only males are sterile |\n| Ae. triseriatus | Ae. brelandi | Only males are sterile. |\n| Ae. sollicitans | {Ae. taeniorhynchus Both sexes are sterile. <br> s Ae. nigromaculatus Both sexes are sterile.} |  |\n| Ae. taeniorhync |  |  |\n| Data on Anopheles |  |  |\n| $\\bar{A}$ | B | A crossed with B |\n| An. albitarsus | An. daeneorum | Only males are sterile. |\n| An. crucians | An. bradleyi | Only males are sterile. |\n| An. freeborni | An. occidentalis | Only males are sterile. |\n| An. freeborni | An. atroparvus | Only males are sterile. |\n\nA: If the males in genus Anopheles were under strong selection, we would expect the same pattern of hybrid sterility as presented here.\nB: The hybrid sterility patterns observed in both genera are consistent with the idea that alleles involved in hybrid sterility are recessive.\nC: If the pattern of sterility in Aedes were identical to that of Anopheles, it would disprove Haldane's rule.\nD: The pattern of hybrid sterility in Aedes suggest variation in $\\mathrm{Y}$ chromosome functionality in this genus.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_937",
        "problem": "一踠豆杂合子（Aa）植株自交时, 下列叙述错误的是（ ）\nA: 若自交后代基因型比例是 2: 3: 1, 可能是含有隐性基因的花粉 50\\%的死亡造成\nB: 若自交后代的基因型比例是 2: 2: 1, 可能是隐性个体有 50\\%的死亡造成\nC: 若自交后代的基因型比例是 4: 4: 1, 可能是含有隐性基因的配子有 50\\%的死亡造成\nD: 若自交后代的基因型比例是 1：2：1, 可能是花粉有 50\\%的死亡造成\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一踠豆杂合子（Aa）植株自交时, 下列叙述错误的是（ ）\n\nA: 若自交后代基因型比例是 2: 3: 1, 可能是含有隐性基因的花粉 50\\%的死亡造成\nB: 若自交后代的基因型比例是 2: 2: 1, 可能是隐性个体有 50\\%的死亡造成\nC: 若自交后代的基因型比例是 4: 4: 1, 可能是含有隐性基因的配子有 50\\%的死亡造成\nD: 若自交后代的基因型比例是 1：2：1, 可能是花粉有 50\\%的死亡造成\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_87",
        "problem": "Nitrogen assimilation plays an important role in plant metabolism as well as in plant cell development. Plant cells can acquire inorganic nitrogen in the form of ammonium $\\left(\\mathrm{NH}_{4}{ }^{+}\\right)$and nitrate $\\left(\\mathrm{NO}_{3}{ }^{\\circ}\\right)$. When entering the plant cells through membrane-bound nitrate transporter (NRT), $\\mathrm{NO}_{3}{ }^{-}$can be reduced to $\\mathrm{NO}_{2}^{-}$by nitrate reductase (NR) and subsequently to $\\mathrm{NH}_{4}{ }^{+}$and amino acids (AA). In addition, $\\mathrm{NO}_{2}{ }^{-}$can be converted into nitric oxide (NO), then forming S-nitrosoglutathione (GSNO) by reaction with glutathione (GSH), and finally into oxidized glutathione (GSSG) and $\\mathrm{NH}_{4}{ }^{+}$under catalysis of S-nitrosoglutathione reductase 1 (GSNOR1).\n\n[figure1]\n\nFig.Q65. A schematic model for the control of nitrogen assimilation in plants through NO signalling.\nA: In the nitrogen metabolism process of the plant cells, NO is a one of the end products but plays a key role signaling regulation of $\\mathrm{NH}_{4}^{+}$formation and $\\mathrm{NO}_{3}{ }^{-}$ assimilation.\nB: $\\mathrm{NH}^{4+}$ level in chloroplasts of plant cells is controlled by activity of GSNO and GSNOR1.\nC: Reduction of $\\mathrm{NO}^{2-}$ ions mainly occur in cytosol\nD: NO feedback regulates flux through nitrate assimilation pathway and controls its bioavailability by modulating its own metabolism.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nNitrogen assimilation plays an important role in plant metabolism as well as in plant cell development. Plant cells can acquire inorganic nitrogen in the form of ammonium $\\left(\\mathrm{NH}_{4}{ }^{+}\\right)$and nitrate $\\left(\\mathrm{NO}_{3}{ }^{\\circ}\\right)$. When entering the plant cells through membrane-bound nitrate transporter (NRT), $\\mathrm{NO}_{3}{ }^{-}$can be reduced to $\\mathrm{NO}_{2}^{-}$by nitrate reductase (NR) and subsequently to $\\mathrm{NH}_{4}{ }^{+}$and amino acids (AA). In addition, $\\mathrm{NO}_{2}{ }^{-}$can be converted into nitric oxide (NO), then forming S-nitrosoglutathione (GSNO) by reaction with glutathione (GSH), and finally into oxidized glutathione (GSSG) and $\\mathrm{NH}_{4}{ }^{+}$under catalysis of S-nitrosoglutathione reductase 1 (GSNOR1).\n\n[figure1]\n\nFig.Q65. A schematic model for the control of nitrogen assimilation in plants through NO signalling.\n\nA: In the nitrogen metabolism process of the plant cells, NO is a one of the end products but plays a key role signaling regulation of $\\mathrm{NH}_{4}^{+}$formation and $\\mathrm{NO}_{3}{ }^{-}$ assimilation.\nB: $\\mathrm{NH}^{4+}$ level in chloroplasts of plant cells is controlled by activity of GSNO and GSNOR1.\nC: Reduction of $\\mathrm{NO}^{2-}$ ions mainly occur in cytosol\nD: NO feedback regulates flux through nitrate assimilation pathway and controls its bioavailability by modulating its own metabolism.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-037.jpg?height=868&width=1360&top_left_y=989&top_left_x=421"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_121",
        "problem": "Mice that were trisomic for each of the 20 different chromosomes were monitored during embryonic development. Their survival time was plotted against the size of trisomic chromosome in Fig.Q.88.\n\n[figure1]\n\nFig.Q. 88\nA: Chromosome 19 is likely the smallest chromosomes with respect to the numbers of transcripts that they encode.\nB: The total amount of additional genetic material determines the severity of the defects associated with the chromosome imbalance.\nC: Gene density on chromosome 1 is probably lower than that on chromosome 10 .\nD: Genes on chromosomes 12 are probably more important for embryo development than those on chromosome 13.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nMice that were trisomic for each of the 20 different chromosomes were monitored during embryonic development. Their survival time was plotted against the size of trisomic chromosome in Fig.Q.88.\n\n[figure1]\n\nFig.Q. 88\n\nA: Chromosome 19 is likely the smallest chromosomes with respect to the numbers of transcripts that they encode.\nB: The total amount of additional genetic material determines the severity of the defects associated with the chromosome imbalance.\nC: Gene density on chromosome 1 is probably lower than that on chromosome 10 .\nD: Genes on chromosomes 12 are probably more important for embryo development than those on chromosome 13.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-086.jpg?height=918&width=1350&top_left_y=467&top_left_x=405"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_698",
        "problem": "下列四幅图表示了在“肺炎双球菌转化实验”和“筮菌体侵染细菌的实验”（搅拌强度、时长等都合理）中相关含量的变化，相关叙述正确的是（）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n丙\n\n[图3]\n\n丁\nA: 图甲表示在“32P 标记的噬菌体侵染细菌实验”中，上清液放射性含量的变化\nB: 图乙表示在“35S 标记的噬菌体侵染细菌实验”中, 上清液放射性含量的变化\nC: 图丙表示“肺炎双球菌体外转化实验”中加入 $\\mathrm{S}$ 型细菌 DNA 后, $\\mathrm{R}$ 型与 $\\mathrm{S}$ 型细菌的数量变化\nD: 图丁表示“肺炎双球菌体外转化实验”中加入 $\\mathrm{S}$ 型细菌 DNA 后, $\\mathrm{R}$ 型与 $\\mathrm{S}$ 型细菌的数量变化\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列四幅图表示了在“肺炎双球菌转化实验”和“筮菌体侵染细菌的实验”（搅拌强度、时长等都合理）中相关含量的变化，相关叙述正确的是（）\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n丙\n\n[图3]\n\n丁\n\nA: 图甲表示在“32P 标记的噬菌体侵染细菌实验”中，上清液放射性含量的变化\nB: 图乙表示在“35S 标记的噬菌体侵染细菌实验”中, 上清液放射性含量的变化\nC: 图丙表示“肺炎双球菌体外转化实验”中加入 $\\mathrm{S}$ 型细菌 DNA 后, $\\mathrm{R}$ 型与 $\\mathrm{S}$ 型细菌的数量变化\nD: 图丁表示“肺炎双球菌体外转化实验”中加入 $\\mathrm{S}$ 型细菌 DNA 后, $\\mathrm{R}$ 型与 $\\mathrm{S}$ 型细菌的数量变化\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-095.jpg?height=314&width=323&top_left_y=1693&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-095.jpg?height=334&width=609&top_left_y=1695&top_left_x=752",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-095.jpg?height=245&width=260&top_left_y=1782&top_left_x=1498"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1196",
        "problem": "SAVING THE MAUI'S DOLPHINS - A STORY OF BIOLOGY, POLICY \\& CONSERVATION\n\n[figure1]\n\nWWF-New Zealand\n\nhttp://wwf.panda.org/wwf news/?206249/NZ-govt-fails-Mauisdolphins-on-global-stage\n\n[figure2]\n\nhttp://uww.doc.govt.nz/conservation/native-animals/marinemammals/dolphins/hectors-dolphin/docs-work/hectors-and-mauisdolphin-incident-databasel\n\nMaui's dolphin (Cephalorhynchus hectori maui) is one of the world's rarest dolphins and is found only on the west coast of the North Island of New Zealand (Resource Pack, Figure 1). It is a sub-species of Hector's dolphin (Cephalorhynchus hectori). The Maui's dolphin is protected by the West Coast North Island Marine Mammal Sanctuary (WCNIMMS) which restricts seabed mining activities, acoustic seismic survey work and commercial and recreational set netting. The boundaries of this sanctuary extend alongshore from Maunganui Bluff in Northland to Oakura Beach, Taranaki, in the south and from mean high water springs to the 12 nautical mile (nm) territorial sea limit. The total area of the sanctuary is approximately 1,200,086 hectares covering 2,164 km of coastline. (Resource Pack, Figure 1).\n\nMaui's dolphin is 'critically endangered' (IUCN Red List), with the population dropping from around 1000 individuals in 1970 to 111 in 2004 according to research by Assoc. Prof. Dr Liz Slooten and others from Otago University. Recent research by the Department of Conservation (DOC) suggests there are now even fewer Maui's dolphins remaining. Maui's dolphins are relatively short-lived (approximately 25 years), and are slow breeders. Females do not have their first calf until they are about seven or eight years old, and have a new calf only every two to four years. This means the species may be threatened by even occasional deaths caused by human activity. Fishing, particularly set netting, is the greatest known human threat to Maui's dolphins and thought to be responsible for about 75 per cent of reported deaths with a known cause. Other human threats include marine tourism, vessel traffic, mining, construction, coastal development pollution, sedimentation, oil spills, plastic bags, marine farming and climate change.\n\nIn recent weeks the need to develop effective management strategies for this species to prevent its extinction has hit the news. Liz Slooten presented the latest research to the International Whaling Commission in May this year, showing that the current protection measures are not sufficient to avoid the extinction of Maui's dolphin. WWF-New Zealand, Greenpeace, and Forest \\& Bird and international conservation groups including NABU and WDC in 2012 and 2013 are all campaigning to protect the Maui's dolphin and hundreds of angry protesters marched to the office of the Energy and Resources Minister Simon Bridges' office in Tauranga to protest the government allowing oil exploration within the West Coast North Island Marine Mammal Sanctuary in their annual tender process 'Block Offer 2014' (Resource Pack, Figure 3). This government decision came just two weeks after the International Whaling Committee (IWC) criticised New Zealand for not taking the necessary steps to save the Maui's dolphin. The IWC noted that a $350 \\mathrm{~km}^{2}$ set net restriction had been added to the WCNIMMS but commented that these measures fell significantly short of those required to reverse the Maui's dolphin decline as recommended by the IWC in 2012 and 2013. The IWC reiterated its extreme concern about the continued decline of such a small population \"as the humaninduced death of even one dolphin would increase the extinction risk for this subspecies\". In 2013 it strongly recommended that the NZ government should:\n\n\"take immediate management actions that will eliminate bycatch of Maui's dolphins. This includes full closures of any fisheries within the range of Maui's dolphins that are known to pose a risk of bycatch of small cetaceans (i.e. set net and trawl fisheries)\". Ensuring full protection of Maui's dolphins in all areas throughout their habitat, together with an ample buffer zone, would minimise the risk of bycatch and maximise the chances of population increase\". and \"commit to specific population increase targets and timelines\".\n\nBecause this recommendation was ignored by the NZ government, in 2014, the IWC recommended that the protected area should be extended south to Whanganui, offshore to 20 nautical miles and should include the harbours.\n\nThe resource pack contains maps of Maui's and Hector's dolphin sightings from 1970 - July 2013, Protection measures for Maui's dolphins on the West Coast North Island and the 'Block Offer 2014' for the Offshore Release Area: Taranaki Basin 14TAR-R1. It also has a table of all Maui's dolphin sightings from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti. Much of this area is included in the Block Offer 2014' Taranaki Basin 14TAR-R1 area. This table is modified from the Maui's dolphin sightings database:\n\nhttp://www.doc.govt.nz/conservation/native-animals/marine-mammals/dolphins/mauis-dolphin/docs-work/mauisdolphin-sightings/\n\nConservation Minister Nick Smith was questioned in parliament about the decision to open part of the sanctuary for exploration. \"The block offer is nowhere near where the Maui's live,\" Dr Smith said. \"There hasn't been a single observation of a Maui's dolphin, and the oil and gas industry hasn't been involved in a single Maui's dolphin incident in Taranaki over the past 40 years despite 23 wells being drilled\" (Source: http://www.3news.co.nz/Greenpeace-get23000-signatures-in-petition-to-sack-Bridges/tabid/423/articleID/349373/Default.aspx).What evidence suggests that Maui's dolphins do interact with the current oil and gas industry and may do so at increased levels if exploration occurs within the West Coast North Island Marine Mammal Sanctuary\nA: There have been four sightings of Maui's dolphins from oil platforms in Taranaki.\nB: Maui's dolphins have been sighted in Port Taranaki.\nC: Commercial fisherman have reported sightings of Maui's dolphins.\nD: Both $A$ and $B$ provide evidence.\nE: A, B and C provide evidence.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nSAVING THE MAUI'S DOLPHINS - A STORY OF BIOLOGY, POLICY \\& CONSERVATION\n\n[figure1]\n\nWWF-New Zealand\n\nhttp://wwf.panda.org/wwf news/?206249/NZ-govt-fails-Mauisdolphins-on-global-stage\n\n[figure2]\n\nhttp://uww.doc.govt.nz/conservation/native-animals/marinemammals/dolphins/hectors-dolphin/docs-work/hectors-and-mauisdolphin-incident-databasel\n\nMaui's dolphin (Cephalorhynchus hectori maui) is one of the world's rarest dolphins and is found only on the west coast of the North Island of New Zealand (Resource Pack, Figure 1). It is a sub-species of Hector's dolphin (Cephalorhynchus hectori). The Maui's dolphin is protected by the West Coast North Island Marine Mammal Sanctuary (WCNIMMS) which restricts seabed mining activities, acoustic seismic survey work and commercial and recreational set netting. The boundaries of this sanctuary extend alongshore from Maunganui Bluff in Northland to Oakura Beach, Taranaki, in the south and from mean high water springs to the 12 nautical mile (nm) territorial sea limit. The total area of the sanctuary is approximately 1,200,086 hectares covering 2,164 km of coastline. (Resource Pack, Figure 1).\n\nMaui's dolphin is 'critically endangered' (IUCN Red List), with the population dropping from around 1000 individuals in 1970 to 111 in 2004 according to research by Assoc. Prof. Dr Liz Slooten and others from Otago University. Recent research by the Department of Conservation (DOC) suggests there are now even fewer Maui's dolphins remaining. Maui's dolphins are relatively short-lived (approximately 25 years), and are slow breeders. Females do not have their first calf until they are about seven or eight years old, and have a new calf only every two to four years. This means the species may be threatened by even occasional deaths caused by human activity. Fishing, particularly set netting, is the greatest known human threat to Maui's dolphins and thought to be responsible for about 75 per cent of reported deaths with a known cause. Other human threats include marine tourism, vessel traffic, mining, construction, coastal development pollution, sedimentation, oil spills, plastic bags, marine farming and climate change.\n\nIn recent weeks the need to develop effective management strategies for this species to prevent its extinction has hit the news. Liz Slooten presented the latest research to the International Whaling Commission in May this year, showing that the current protection measures are not sufficient to avoid the extinction of Maui's dolphin. WWF-New Zealand, Greenpeace, and Forest \\& Bird and international conservation groups including NABU and WDC in 2012 and 2013 are all campaigning to protect the Maui's dolphin and hundreds of angry protesters marched to the office of the Energy and Resources Minister Simon Bridges' office in Tauranga to protest the government allowing oil exploration within the West Coast North Island Marine Mammal Sanctuary in their annual tender process 'Block Offer 2014' (Resource Pack, Figure 3). This government decision came just two weeks after the International Whaling Committee (IWC) criticised New Zealand for not taking the necessary steps to save the Maui's dolphin. The IWC noted that a $350 \\mathrm{~km}^{2}$ set net restriction had been added to the WCNIMMS but commented that these measures fell significantly short of those required to reverse the Maui's dolphin decline as recommended by the IWC in 2012 and 2013. The IWC reiterated its extreme concern about the continued decline of such a small population \"as the humaninduced death of even one dolphin would increase the extinction risk for this subspecies\". In 2013 it strongly recommended that the NZ government should:\n\n\"take immediate management actions that will eliminate bycatch of Maui's dolphins. This includes full closures of any fisheries within the range of Maui's dolphins that are known to pose a risk of bycatch of small cetaceans (i.e. set net and trawl fisheries)\". Ensuring full protection of Maui's dolphins in all areas throughout their habitat, together with an ample buffer zone, would minimise the risk of bycatch and maximise the chances of population increase\". and \"commit to specific population increase targets and timelines\".\n\nBecause this recommendation was ignored by the NZ government, in 2014, the IWC recommended that the protected area should be extended south to Whanganui, offshore to 20 nautical miles and should include the harbours.\n\nThe resource pack contains maps of Maui's and Hector's dolphin sightings from 1970 - July 2013, Protection measures for Maui's dolphins on the West Coast North Island and the 'Block Offer 2014' for the Offshore Release Area: Taranaki Basin 14TAR-R1. It also has a table of all Maui's dolphin sightings from Pariokariwa Point to New Plymouth and New Plymouth South to Cape Terawhiti. Much of this area is included in the Block Offer 2014' Taranaki Basin 14TAR-R1 area. This table is modified from the Maui's dolphin sightings database:\n\nhttp://www.doc.govt.nz/conservation/native-animals/marine-mammals/dolphins/mauis-dolphin/docs-work/mauisdolphin-sightings/\n\nConservation Minister Nick Smith was questioned in parliament about the decision to open part of the sanctuary for exploration. \"The block offer is nowhere near where the Maui's live,\" Dr Smith said. \"There hasn't been a single observation of a Maui's dolphin, and the oil and gas industry hasn't been involved in a single Maui's dolphin incident in Taranaki over the past 40 years despite 23 wells being drilled\" (Source: http://www.3news.co.nz/Greenpeace-get23000-signatures-in-petition-to-sack-Bridges/tabid/423/articleID/349373/Default.aspx).\n\nproblem:\nWhat evidence suggests that Maui's dolphins do interact with the current oil and gas industry and may do so at increased levels if exploration occurs within the West Coast North Island Marine Mammal Sanctuary\n\nA: There have been four sightings of Maui's dolphins from oil platforms in Taranaki.\nB: Maui's dolphins have been sighted in Port Taranaki.\nC: Commercial fisherman have reported sightings of Maui's dolphins.\nD: Both $A$ and $B$ provide evidence.\nE: A, B and C provide evidence.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-30.jpg?height=503&width=682&top_left_y=277&top_left_x=113",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-30.jpg?height=497&width=848&top_left_y=277&top_left_x=1049"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_230",
        "problem": "Mosquitos are vectors for transmitting human diseases, and the application of insecticides to water bodies is occasionally conducted to control mosquito populations. In a mosquito population, there are two alleles of a locus that affect susceptibility to pesticides, (s): susceptible, and (r): resistant. The resistance is completely recessive. The table below shows the change in the number of individuals with different genotypes before (Pre1990), during (1990-2000; shown by an arrow), and after (2005-2015) pesticide application.\n\n|  | $\\mathrm{s} / \\mathrm{s}$ | $\\mathrm{s} / \\mathrm{r}$ | $\\mathrm{r} / \\mathrm{r}$ |\n| ---: | ---: | ---: | ---: |\n| Pre-1990 | 222 | 3 | 0 |\n| 1990 | 31 | 12 | 4 |\n| 1995 | 26 | 35 | 41 |\n| 2000 | 2 | 12 | 126 |\n| 2005 | 74 | 64 | 44 |\n| 2010 | 165 | 45 | 20 |\n| 2015 | 210 | 12 | 1 |\nA: No resistance allele was present before insecticide application.\nB: During pesticide application, natural selection favored the resistance allele.\nC: Resistant individuals ( $r / r$ ) are likely to have lower fitness than others ( $s / r, s / s)$ in the absence of pesticide application.\nD: From 1990 to 1995, the frequency of the resistant allele increased more than 10 times.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nMosquitos are vectors for transmitting human diseases, and the application of insecticides to water bodies is occasionally conducted to control mosquito populations. In a mosquito population, there are two alleles of a locus that affect susceptibility to pesticides, (s): susceptible, and (r): resistant. The resistance is completely recessive. The table below shows the change in the number of individuals with different genotypes before (Pre1990), during (1990-2000; shown by an arrow), and after (2005-2015) pesticide application.\n\n|  | $\\mathrm{s} / \\mathrm{s}$ | $\\mathrm{s} / \\mathrm{r}$ | $\\mathrm{r} / \\mathrm{r}$ |\n| ---: | ---: | ---: | ---: |\n| Pre-1990 | 222 | 3 | 0 |\n| 1990 | 31 | 12 | 4 |\n| 1995 | 26 | 35 | 41 |\n| 2000 | 2 | 12 | 126 |\n| 2005 | 74 | 64 | 44 |\n| 2010 | 165 | 45 | 20 |\n| 2015 | 210 | 12 | 1 |\n\nA: No resistance allele was present before insecticide application.\nB: During pesticide application, natural selection favored the resistance allele.\nC: Resistant individuals ( $r / r$ ) are likely to have lower fitness than others ( $s / r, s / s)$ in the absence of pesticide application.\nD: From 1990 to 1995, the frequency of the resistant allele increased more than 10 times.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_585",
        "problem": "水稻开两性花且花小、去雄困难, 我国科研人员发现了两个光周期依赖型的水稻雄性不育突变株甲（aaBB）、乙（AAbb），各相关基因型的育性与光周期关系如下表所示，已知基因 $\\mathrm{A} / \\mathrm{a}$ 和 $\\mathrm{B} / \\mathrm{b}$ 位于非同源染色体上, 现利用甲和乙在不同日照条件下杂交产生\n\n$F_{1}, F_{1}$ 自交产生 $F_{2}$ 。下列相关叙述错误的是 ( )\n\n| 基因型 | 日照条件 |  |\n| :---: | :---: | :---: |\n|  | 长日照 | 短日照 |\n| $\\mathrm{AA} 、 \\mathrm{Aa}$ | 育性正常 | 育性正常 |\n| $\\mathrm{aa}$ | 育性正常 | 完全雄性不育 |\n| $\\mathrm{BB} 、 \\mathrm{Bb}$ | 育性正常 | 育性正常 |\n| $\\mathrm{bb}$ | 雄性半不育 (花粉 $50 \\%$ 可育) | 育性正常 |\nA: 光周期依赖型的水稻雄性不育说明基因不是影响生物性状的唯一因素\nB: 长日照条件下杂交, $\\mathrm{F}_{2}$ 的性状及分离比为育性正常：雄性半不育 $=3: 1$\nC: 短日照条件下杂交, $\\mathrm{F}_{2}$ 的性状及分离比为育性正常：完全雄性不育 $=1: 3$\nD: 无论长或短日照, $\\mathrm{AaBb}$ 的个体形成的 $\\mathrm{AB} 、 \\mathrm{Ab}$ 花粉与卵细胞结合的概率是均等的\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n水稻开两性花且花小、去雄困难, 我国科研人员发现了两个光周期依赖型的水稻雄性不育突变株甲（aaBB）、乙（AAbb），各相关基因型的育性与光周期关系如下表所示，已知基因 $\\mathrm{A} / \\mathrm{a}$ 和 $\\mathrm{B} / \\mathrm{b}$ 位于非同源染色体上, 现利用甲和乙在不同日照条件下杂交产生\n\n$F_{1}, F_{1}$ 自交产生 $F_{2}$ 。下列相关叙述错误的是 ( )\n\n| 基因型 | 日照条件 |  |\n| :---: | :---: | :---: |\n|  | 长日照 | 短日照 |\n| $\\mathrm{AA} 、 \\mathrm{Aa}$ | 育性正常 | 育性正常 |\n| $\\mathrm{aa}$ | 育性正常 | 完全雄性不育 |\n| $\\mathrm{BB} 、 \\mathrm{Bb}$ | 育性正常 | 育性正常 |\n| $\\mathrm{bb}$ | 雄性半不育 (花粉 $50 \\%$ 可育) | 育性正常 |\n\nA: 光周期依赖型的水稻雄性不育说明基因不是影响生物性状的唯一因素\nB: 长日照条件下杂交, $\\mathrm{F}_{2}$ 的性状及分离比为育性正常：雄性半不育 $=3: 1$\nC: 短日照条件下杂交, $\\mathrm{F}_{2}$ 的性状及分离比为育性正常：完全雄性不育 $=1: 3$\nD: 无论长或短日照, $\\mathrm{AaBb}$ 的个体形成的 $\\mathrm{AB} 、 \\mathrm{Ab}$ 花粉与卵细胞结合的概率是均等的\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1485",
        "problem": "Peaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nIn which of the following scenarios are some plants known to evolve fuzzy/furry coatings?\nA: Reduce water loss in dry environments.\nB: Reduce UV damage in mountains.\nC: Increase $\\mathrm{CO}_{2}$ uptake.\nD: Increase surface area for photosynthesis.\nE: $\\quad \\mathrm{a}$ and $b$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPeaches and nectarines are produced by the same species of tree. Peaches have a fuzzy coating but nectarines do not.\n\n[figure1]\n\nIn which of the following scenarios are some plants known to evolve fuzzy/furry coatings?\n\nA: Reduce water loss in dry environments.\nB: Reduce UV damage in mountains.\nC: Increase $\\mathrm{CO}_{2}$ uptake.\nD: Increase surface area for photosynthesis.\nE: $\\quad \\mathrm{a}$ and $b$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-45.jpg?height=782&width=1231&top_left_y=474&top_left_x=241"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_846",
        "problem": "某基因型为 $A^{2} X^{b} Y$ 的精原细胞（2n=16）所有 DNA 分子双链均用 ${ }^{15} \\mathrm{~N}$ 标记, 置于含 ${ }^{14} \\mathrm{~N}$ 的培养基中培养。经过 1 次有丝分裂后, 再分别完成减数分裂, 发现了一个 $\\mathrm{AX}^{\\mathrm{b}} \\mathrm{Y}$试卷第 16 页，共 88 页\n的异常精细胞。若无其他突变和互换发生, 下列说法正确的是（ ）\nA: 与该异常精细胞同时产生的另外 3 个精细胞的基因型为 $\\mathrm{aX}{ }^{\\mathrm{b}} \\mathrm{Y} 、 \\mathrm{~A} 、 \\mathrm{~A}$\nB: 产生该异常精细胞的初级精母细胞中被标记的染色体有 16 条, 且所有的核 DNA 分子都含 ${ }^{15} \\mathrm{~N}$\nC: 产生该异常精细胞的次级精母细胞在减数第二次分裂后期, 有 9 个核 DNA 含 ${ }^{15} \\mathrm{~N}$\nD: 分裂产生的某个精细胞中的核 DNA 不可能都含 ${ }^{15} \\mathrm{~N}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某基因型为 $A^{2} X^{b} Y$ 的精原细胞（2n=16）所有 DNA 分子双链均用 ${ }^{15} \\mathrm{~N}$ 标记, 置于含 ${ }^{14} \\mathrm{~N}$ 的培养基中培养。经过 1 次有丝分裂后, 再分别完成减数分裂, 发现了一个 $\\mathrm{AX}^{\\mathrm{b}} \\mathrm{Y}$试卷第 16 页，共 88 页\n的异常精细胞。若无其他突变和互换发生, 下列说法正确的是（ ）\n\nA: 与该异常精细胞同时产生的另外 3 个精细胞的基因型为 $\\mathrm{aX}{ }^{\\mathrm{b}} \\mathrm{Y} 、 \\mathrm{~A} 、 \\mathrm{~A}$\nB: 产生该异常精细胞的初级精母细胞中被标记的染色体有 16 条, 且所有的核 DNA 分子都含 ${ }^{15} \\mathrm{~N}$\nC: 产生该异常精细胞的次级精母细胞在减数第二次分裂后期, 有 9 个核 DNA 含 ${ }^{15} \\mathrm{~N}$\nD: 分裂产生的某个精细胞中的核 DNA 不可能都含 ${ }^{15} \\mathrm{~N}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1115",
        "problem": "Complete digestion of a plasmid with a restriction enzyme means that all the sites for this enzyme in all the plasmid molecules are cut. Partial digestion means that only some of the sites for this enzyme are cut.\n\nFor expression of a recombinant protein, you wish to clone its gene under the promoter $\\mathrm{P}$ in the plasmid shown below. The direction of the arrow shows the direction of transcription. (ori represents origin of replication and $a m p^{r}$ represents the gene for ampicillin resistance)\n\n[figure1]\n\nWhich of the following treatments can be used to successfully clone and express the gene?\nA: Complete digestion with Xhol.\nB: Partial digestion with EcoRI.\nC: Complete digestion with BamHI.\nD: Partial digestion with BamHI.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nComplete digestion of a plasmid with a restriction enzyme means that all the sites for this enzyme in all the plasmid molecules are cut. Partial digestion means that only some of the sites for this enzyme are cut.\n\nFor expression of a recombinant protein, you wish to clone its gene under the promoter $\\mathrm{P}$ in the plasmid shown below. The direction of the arrow shows the direction of transcription. (ori represents origin of replication and $a m p^{r}$ represents the gene for ampicillin resistance)\n\n[figure1]\n\nWhich of the following treatments can be used to successfully clone and express the gene?\n\nA: Complete digestion with Xhol.\nB: Partial digestion with EcoRI.\nC: Complete digestion with BamHI.\nD: Partial digestion with BamHI.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-18.jpg?height=420&width=770&top_left_y=104&top_left_x=683"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_438",
        "problem": "减数分裂的四分体时期, 同源染色体的非姐妹染色单体之间经常发生缠绕, 并交换相应的片段。某动物在减数分裂产生精子时, 有 $4 \\%$ 的精原细胞发生如图所示的交换, $\\mathrm{A} / \\mathrm{a}$和 $\\mathrm{B} / \\mathrm{b}$ 表示相应基因。下列有关说法正确的是（ ）\n\n[图1]\nA: 如图所示的变异类型属于染色体结构变异\nB: 基因 $\\mathrm{A}$ 与 $\\mathrm{a}$ 都是在减数分裂第一次分裂后期分离的\nC: 由图可知, 该种变异类型没有增加配子的种类\nD: 产生的精子基因型及比例为 $\\mathrm{AB}: \\mathrm{Ab}: \\mathrm{aB}: \\mathrm{ab}=49: 1: 1: 49$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n减数分裂的四分体时期, 同源染色体的非姐妹染色单体之间经常发生缠绕, 并交换相应的片段。某动物在减数分裂产生精子时, 有 $4 \\%$ 的精原细胞发生如图所示的交换, $\\mathrm{A} / \\mathrm{a}$和 $\\mathrm{B} / \\mathrm{b}$ 表示相应基因。下列有关说法正确的是（ ）\n\n[图1]\n\nA: 如图所示的变异类型属于染色体结构变异\nB: 基因 $\\mathrm{A}$ 与 $\\mathrm{a}$ 都是在减数分裂第一次分裂后期分离的\nC: 由图可知, 该种变异类型没有增加配子的种类\nD: 产生的精子基因型及比例为 $\\mathrm{AB}: \\mathrm{Ab}: \\mathrm{aB}: \\mathrm{ab}=49: 1: 1: 49$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-46.jpg?height=245&width=1151&top_left_y=914&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_388",
        "problem": "某雌雄异株植物, 叶片形状有细长、圆宽和锯齿等类型。为了研究其遗传机制, 进行了杂交实验, 结果见下表, 下列叙述正确的是( )\n\n| 杂交 <br> 编号 | 母本植株数目 <br> 表现型 | 父本植株 <br> 数目 <br> 表现型 | $\\mathbf{F}_{1}$ 植株数目 <br> 表现型 |  | $\\mathbf{F}_{2}$ 植株数目 <br> 表现型 |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| I | 80 <br> 锯齿 | 82 <br> 圆宽 | 82 <br> 锯齿 | $\\mathbf{8 1}$ <br> 细长우 | 81 <br> 圆宽 | 242 <br> 锯齿 | 243 <br> 细长 |\n| II | 92 <br> 圆宽 | 90 <br> 锯齿 | 93 <br> 细长 | 92 <br> 细长웅 | 93 <br> 圆宽 | 91 <br> 铌齿 | 275 <br> 细长 |\nA: 选取杂交 I 的 $F_{2}$ 中所有的圆宽叶植株随机杂交, 杂交 1 代中所有植株均为圆宽叶, 雌雄株比例为 4:3, 其中雌株有 2 种基因型, 比例为 $1: 1$\nB: 选取杂交 II 的 $F_{2}$ 中所有的圆宽叶植株随机杂交, 杂交 1 代中所有植株均为圆宽叶, 雌雄株比例为 $4: 3$, 其中雌株有 2 种基因型, 比例为 $3: 1$\nC: 选取杂交 I 的 $F_{1}$ 中锯齿叶植株, 与杂交 11 的圆宽叶亲本杂交, 杂交 1 代中有锯齿叶和细长叶两种, 比例为 $1: 1$,其中雌株有 2 种基因型, 比例为 $3: 1$\nD: 选取杂交 11 的 $F_{2}$ 中所有的锯齿叶植株随机杂交, 杂交 1 代中所有植株均为锯齿叶, 雌雄株比例为 4:3, 其中雌株有 2 种基因型, 比例为 $1: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某雌雄异株植物, 叶片形状有细长、圆宽和锯齿等类型。为了研究其遗传机制, 进行了杂交实验, 结果见下表, 下列叙述正确的是( )\n\n| 杂交 <br> 编号 | 母本植株数目 <br> 表现型 | 父本植株 <br> 数目 <br> 表现型 | $\\mathbf{F}_{1}$ 植株数目 <br> 表现型 |  | $\\mathbf{F}_{2}$ 植株数目 <br> 表现型 |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| I | 80 <br> 锯齿 | 82 <br> 圆宽 | 82 <br> 锯齿 | $\\mathbf{8 1}$ <br> 细长우 | 81 <br> 圆宽 | 242 <br> 锯齿 | 243 <br> 细长 |\n| II | 92 <br> 圆宽 | 90 <br> 锯齿 | 93 <br> 细长 | 92 <br> 细长웅 | 93 <br> 圆宽 | 91 <br> 铌齿 | 275 <br> 细长 |\n\nA: 选取杂交 I 的 $F_{2}$ 中所有的圆宽叶植株随机杂交, 杂交 1 代中所有植株均为圆宽叶, 雌雄株比例为 4:3, 其中雌株有 2 种基因型, 比例为 $1: 1$\nB: 选取杂交 II 的 $F_{2}$ 中所有的圆宽叶植株随机杂交, 杂交 1 代中所有植株均为圆宽叶, 雌雄株比例为 $4: 3$, 其中雌株有 2 种基因型, 比例为 $3: 1$\nC: 选取杂交 I 的 $F_{1}$ 中锯齿叶植株, 与杂交 11 的圆宽叶亲本杂交, 杂交 1 代中有锯齿叶和细长叶两种, 比例为 $1: 1$,其中雌株有 2 种基因型, 比例为 $3: 1$\nD: 选取杂交 11 的 $F_{2}$ 中所有的锯齿叶植株随机杂交, 杂交 1 代中所有植株均为锯齿叶, 雌雄株比例为 4:3, 其中雌株有 2 种基因型, 比例为 $1: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1389",
        "problem": "Cellular respiration is a complex process involving many steps. Generally, these steps are grouped into 3 main phases: glycolysis, the citric acid cycle, and the electron transport chain. The process of glycolysis can be summarised as follows:\n[figure1]\n\nGiven the necessity of cellular respiration to life, which of the following features would you NOT expect?\nA: Negative feedback loops\nB: Positive feedback loops\nC: High levels of regulation\nD: Susceptibility to mutation\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCellular respiration is a complex process involving many steps. Generally, these steps are grouped into 3 main phases: glycolysis, the citric acid cycle, and the electron transport chain. The process of glycolysis can be summarised as follows:\n[figure1]\n\nGiven the necessity of cellular respiration to life, which of the following features would you NOT expect?\n\nA: Negative feedback loops\nB: Positive feedback loops\nC: High levels of regulation\nD: Susceptibility to mutation\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-34.jpg?height=1204&width=832&top_left_y=532&top_left_x=286"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1464",
        "problem": "The graph below displays the results of an ecological experiment conducted at the University of Sydney. Researchers removed limpets and/or urchins from study areas and recorded the percentage seaweed cover over the next 18 months. Limpets and urchins both feed on seaweed. The results of this experiment showed that urchins have a much greater effect than limpets in restricting seaweed distribution.\n\n[figure1]\n\nUnfortunately, the labels of the plot lines have disappeared. In no particular order, the labels are:\n\n1. No limpets or urchins removed (Control)\n2. Both limpets and urchins removed\n3. Only urchins removed\n4. Only limpets removed\n\nWhich of the following options correctly labels each line?\nA: Purple - 1, Green - 3, Blue - 4, Red - 2\nB: Purple - 2, Green - 3, Blue - 4, Red - 1\nC: Purple - 2, Green - 4, Blue - 3, Red - 1\nD: Purple - 2, Green - 3, Blue - 1, Red - 4\nE: Purple-1, Green-2, Blue - 3, Red - 4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph below displays the results of an ecological experiment conducted at the University of Sydney. Researchers removed limpets and/or urchins from study areas and recorded the percentage seaweed cover over the next 18 months. Limpets and urchins both feed on seaweed. The results of this experiment showed that urchins have a much greater effect than limpets in restricting seaweed distribution.\n\n[figure1]\n\nUnfortunately, the labels of the plot lines have disappeared. In no particular order, the labels are:\n\n1. No limpets or urchins removed (Control)\n2. Both limpets and urchins removed\n3. Only urchins removed\n4. Only limpets removed\n\nWhich of the following options correctly labels each line?\n\nA: Purple - 1, Green - 3, Blue - 4, Red - 2\nB: Purple - 2, Green - 3, Blue - 4, Red - 1\nC: Purple - 2, Green - 4, Blue - 3, Red - 1\nD: Purple - 2, Green - 3, Blue - 1, Red - 4\nE: Purple-1, Green-2, Blue - 3, Red - 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-26.jpg?height=879&width=1259&top_left_y=1665&top_left_x=404"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1110",
        "problem": "Norwegian scientists performed a cross-fostering experiment. They removed just born great-tit chicks and placed them in the nests of blue-tits and vice versa. These chicks were observed till their adulthood and their pair-making behavior was noted. The graph shows pairing success of these cross-fostered birds along with non-fostered birds.\n\n[figure1]\n\nWhich of the following is correct?\nA: The experimental results show imprinting behavior in the two bird species which is an example of a learned behavior.\nB: Since one species could successfully find mates in spite of being reared by different species shows that courtship behavior is controlled only by genes and not affected by the environment.\nC: Since both the species show a very small percent of mates of species that was their foster parents, shows that this behavior is genetically acquired.\nD: The ability of the two species to consider the foster species as their own varied. This indicates that this behavior is genetically influenced.\nE: The percent pairing success for blue-tits shows that foster parenting has no effect on the courtship behavior.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nNorwegian scientists performed a cross-fostering experiment. They removed just born great-tit chicks and placed them in the nests of blue-tits and vice versa. These chicks were observed till their adulthood and their pair-making behavior was noted. The graph shows pairing success of these cross-fostered birds along with non-fostered birds.\n\n[figure1]\n\nWhich of the following is correct?\n\nA: The experimental results show imprinting behavior in the two bird species which is an example of a learned behavior.\nB: Since one species could successfully find mates in spite of being reared by different species shows that courtship behavior is controlled only by genes and not affected by the environment.\nC: Since both the species show a very small percent of mates of species that was their foster parents, shows that this behavior is genetically acquired.\nD: The ability of the two species to consider the foster species as their own varied. This indicates that this behavior is genetically influenced.\nE: The percent pairing success for blue-tits shows that foster parenting has no effect on the courtship behavior.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-29.jpg?height=751&width=1583&top_left_y=1451&top_left_x=301"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1548",
        "problem": "The animals below are found in the same food web, but fit into different food chains. One chain takes place close to the ocean surface, and one begins with dead animals resting on the seabed.\n\nWhich food chain has the most energy flowing through it?\nA: Ocean surface\nB: Seabed\nC: They are equal\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe animals below are found in the same food web, but fit into different food chains. One chain takes place close to the ocean surface, and one begins with dead animals resting on the seabed.\n\nWhich food chain has the most energy flowing through it?\n\nA: Ocean surface\nB: Seabed\nC: They are equal\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_277",
        "problem": "\"Secondary metabolism\" in microorganisms and plants is not essential for their survival, but is a metabolic process that plays an important role depending on species or in environmental adaptation. Many secondary metabolites accumulated by plants, such as nicotine and caffeine, play a role in resistance to damage from herbivorous insects.\n\nGlucosinolate, which is accumulated in the leaves of Arabidopsis thaliana, is a repellent for herbivorous insects (Helicoverpa armigera). The leaves of the wild type (Wild) and the leaves of the mutant (Mutant) incapable of synthesizing glucosinolate are arranged as shown in Figure 1.\n\n## a\n\n[figure1]\n\nb\n\n[figure2]\n\nFigure 1 (a) Experimental design, (b) Result of choice by insect larvae.\n\nThe following conclusions can be assumed from this experiment.\nA: In the Arabidopsis wild strain, glucosinolate is accumulated more in the outer lamina of the leaves.\nB: In this mutant, glucosinolate is evenly accumulated at any region of the leaf.\nC: Arabidopsis accumulates only glucosinolate as a repellent in its leaves.\nD: For Arabidopsis, inner lamina is likely to be more physiologically important than outer lamina.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n\"Secondary metabolism\" in microorganisms and plants is not essential for their survival, but is a metabolic process that plays an important role depending on species or in environmental adaptation. Many secondary metabolites accumulated by plants, such as nicotine and caffeine, play a role in resistance to damage from herbivorous insects.\n\nGlucosinolate, which is accumulated in the leaves of Arabidopsis thaliana, is a repellent for herbivorous insects (Helicoverpa armigera). The leaves of the wild type (Wild) and the leaves of the mutant (Mutant) incapable of synthesizing glucosinolate are arranged as shown in Figure 1.\n\n## a\n\n[figure1]\n\nb\n\n[figure2]\n\nFigure 1 (a) Experimental design, (b) Result of choice by insect larvae.\n\nThe following conclusions can be assumed from this experiment.\n\nA: In the Arabidopsis wild strain, glucosinolate is accumulated more in the outer lamina of the leaves.\nB: In this mutant, glucosinolate is evenly accumulated at any region of the leaf.\nC: Arabidopsis accumulates only glucosinolate as a repellent in its leaves.\nD: For Arabidopsis, inner lamina is likely to be more physiologically important than outer lamina.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-11.jpg?height=500&width=874&top_left_y=1029&top_left_x=220",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-11.jpg?height=531&width=683&top_left_y=1074&top_left_x=1212"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_662",
        "problem": "左图是某家系甲、乙两种常染色体单基因遗传病的系谱图。右图是该家系中 1 6 号成员两种遗传病相关基因的电泳图（每种条带代表一种基因， ”表示具有相关的基因）。下列有关叙述正确的是（ ）\n\n[图1]\nA: 甲病为隐性遗传病, 乙病为显性遗传病\nB: 条带 2、3 分别代表甲、乙病的致病基因\nC: 控制甲、乙病的基因位于非同源染色体上\nD: III- 8 是正常男孩的概率为 $3 / 8$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n左图是某家系甲、乙两种常染色体单基因遗传病的系谱图。右图是该家系中 1 6 号成员两种遗传病相关基因的电泳图（每种条带代表一种基因， ”表示具有相关的基因）。下列有关叙述正确的是（ ）\n\n[图1]\n\nA: 甲病为隐性遗传病, 乙病为显性遗传病\nB: 条带 2、3 分别代表甲、乙病的致病基因\nC: 控制甲、乙病的基因位于非同源染色体上\nD: III- 8 是正常男孩的概率为 $3 / 8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-18.jpg?height=351&width=1396&top_left_y=1058&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_942",
        "problem": "研究表明, 正常女性细胞核内两条 X 染色体中的一条会随机失活, 浓缩形成染色较深的巴氏小体。肾上腺脑白质营养不良(ALD)是伴 X 染色体隐性遗传病(致病基因用 a 表示), 女性杂合子中有 $5 \\%$ 的个体会患病, 图 1 为某患者家族遗传系谱图。利用图中四位女性细胞中与此病有关的基因片段经能识别特定碱基序列的酶进行切割(如图 2), 产物的电泳结果如图 3 所示。下列叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\nA: 女性杂合子患 ALD 的原因很可能是巴氏小体上的基因无法正常表达\nB: II-2 个体的基因型是 $X^{A} X^{a}$, 患 ALD 的原因是来自母方的 X 染色体形成巴氏小体\nC: a 基因的产生可能是由于 $\\mathrm{A}$ 基因发生碱基替换，酶切后会出现 3 个大小不同的 DNA 片段\nD: 若II-1 和一个基因型与II-4 相同的女性婚配, 后代患 ALD 的概率为 $2.5 \\%$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究表明, 正常女性细胞核内两条 X 染色体中的一条会随机失活, 浓缩形成染色较深的巴氏小体。肾上腺脑白质营养不良(ALD)是伴 X 染色体隐性遗传病(致病基因用 a 表示), 女性杂合子中有 $5 \\%$ 的个体会患病, 图 1 为某患者家族遗传系谱图。利用图中四位女性细胞中与此病有关的基因片段经能识别特定碱基序列的酶进行切割(如图 2), 产物的电泳结果如图 3 所示。下列叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\n\nA: 女性杂合子患 ALD 的原因很可能是巴氏小体上的基因无法正常表达\nB: II-2 个体的基因型是 $X^{A} X^{a}$, 患 ALD 的原因是来自母方的 X 染色体形成巴氏小体\nC: a 基因的产生可能是由于 $\\mathrm{A}$ 基因发生碱基替换，酶切后会出现 3 个大小不同的 DNA 片段\nD: 若II-1 和一个基因型与II-4 相同的女性婚配, 后代患 ALD 的概率为 $2.5 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-10.jpg?height=300&width=725&top_left_y=978&top_left_x=354",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-10.jpg?height=265&width=721&top_left_y=1378&top_left_x=359",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-10.jpg?height=303&width=708&top_left_y=1739&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_157",
        "problem": "Evolution of sexual traits such as horn size, tail feathers and extreme coloration can be determined by sexual selection theory. According to the sexual selection theory male with such attributes gain an advantage over other males to acquire mate. Two processes have been proposed to account for exaggerated traits as the result of sexual selection (figure below).\n\n## Runaway selection\n\n[figure1]\n\nChase-away selection\n\n[figure2]\n\nRunaway selection: Imagine some males with traits (such as tail length) with the higher survival rates than other males in same population. In this ancestral population, some females start to prefer these males. Their offspring will inherit both of these traits (higher tail length in males and females choice for such traits in males). This pattern generates a runaway process until the male trait becomes so exaggerated that it becomes selected against by natural selection.\n\nChase-away selection: A mutation in males provides them with a novel trait that becomes sexually attractive to the females, and changes female mate choice in favour of such novel trait. Males with this mutation can easily mate with the females, while they don't offer any material or genetic benefit to the females. When such choice becomes disadvantageous for the females, the females' threshold increases against such a trait. Males with more extreme forms trait will again attract the females and this process will continue until natural selection selects against this exaggerated trait.\nA: In chase away selection, female choice correlates with male attributes that increase female fitness.\nB: Extreme traits will evolve in males according to the runaway selection and will decrease male life expectancy.\nC: Male attributes that become extreme because of chase-away selection will reduce male life expectancy.\nD: Sensory bias is the primary requirement for chase-away selection to be evolved, but is not necessary for runaway selection.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nEvolution of sexual traits such as horn size, tail feathers and extreme coloration can be determined by sexual selection theory. According to the sexual selection theory male with such attributes gain an advantage over other males to acquire mate. Two processes have been proposed to account for exaggerated traits as the result of sexual selection (figure below).\n\n## Runaway selection\n\n[figure1]\n\nChase-away selection\n\n[figure2]\n\nRunaway selection: Imagine some males with traits (such as tail length) with the higher survival rates than other males in same population. In this ancestral population, some females start to prefer these males. Their offspring will inherit both of these traits (higher tail length in males and females choice for such traits in males). This pattern generates a runaway process until the male trait becomes so exaggerated that it becomes selected against by natural selection.\n\nChase-away selection: A mutation in males provides them with a novel trait that becomes sexually attractive to the females, and changes female mate choice in favour of such novel trait. Males with this mutation can easily mate with the females, while they don't offer any material or genetic benefit to the females. When such choice becomes disadvantageous for the females, the females' threshold increases against such a trait. Males with more extreme forms trait will again attract the females and this process will continue until natural selection selects against this exaggerated trait.\n\nA: In chase away selection, female choice correlates with male attributes that increase female fitness.\nB: Extreme traits will evolve in males according to the runaway selection and will decrease male life expectancy.\nC: Male attributes that become extreme because of chase-away selection will reduce male life expectancy.\nD: Sensory bias is the primary requirement for chase-away selection to be evolved, but is not necessary for runaway selection.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-51.jpg?height=446&width=1313&top_left_y=625&top_left_x=377",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-51.jpg?height=440&width=1311&top_left_y=1119&top_left_x=378"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1304",
        "problem": "Understanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.Considering all the data above, the best conclusion about the diet of the orange-fronted parakeet is?\nA: They have very specific dietary preferences.\nB: They rely primarily on fruit for energy.\nC: Invertebrates are an important food source in spring.\nD: They exhibit dietary flexibility.\nE: They feed on the most abundant plant species in an area.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nUnderstanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.\n\nproblem:\nConsidering all the data above, the best conclusion about the diet of the orange-fronted parakeet is?\n\nA: They have very specific dietary preferences.\nB: They rely primarily on fruit for energy.\nC: Invertebrates are an important food source in spring.\nD: They exhibit dietary flexibility.\nE: They feed on the most abundant plant species in an area.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1210",
        "problem": "## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAAAbout $80 \\%$ of marine debris comes from sources on land and much of this debris is plastic. Charles Moore from Algalita Marine Research Foundation first published an article about marine debris in the November 2003 issue of the Journal Natural History. He showed that the dominant feature in the North Pacific Ocean is the North Pacific Gyre, a large water mass that is rotating in a clockwise direction and can trap debris originating from across the Pacific. Floating debris accumulates in the \"eastern garbage patch\", an area the size of Texas. There is approximately $250 \\mathrm{~g}$ of plastic for every $100 \\mathrm{~m}^{2}$ of sea surface in the \"eastern garbage patch\".\n\nScientists use satellite telemetry to track the movement of albatross. IN 20049 albatross where fitted with satellite tags and their position recorded from July to October.\n\nWith reference to Figure 3 in your Resource Pack, which Albatross is likely to have ingested the MOST plastic debris?\nA: bfal_36634 (red)\nB: bfal_36635 (peach)\nC: bfal_36636 (blue)\nD: bfal_36639 (lime green)\nE: bfal_36641 (pink)\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAA\n\nproblem:\nAbout $80 \\%$ of marine debris comes from sources on land and much of this debris is plastic. Charles Moore from Algalita Marine Research Foundation first published an article about marine debris in the November 2003 issue of the Journal Natural History. He showed that the dominant feature in the North Pacific Ocean is the North Pacific Gyre, a large water mass that is rotating in a clockwise direction and can trap debris originating from across the Pacific. Floating debris accumulates in the \"eastern garbage patch\", an area the size of Texas. There is approximately $250 \\mathrm{~g}$ of plastic for every $100 \\mathrm{~m}^{2}$ of sea surface in the \"eastern garbage patch\".\n\nScientists use satellite telemetry to track the movement of albatross. IN 20049 albatross where fitted with satellite tags and their position recorded from July to October.\n\nWith reference to Figure 3 in your Resource Pack, which Albatross is likely to have ingested the MOST plastic debris?\n\nA: bfal_36634 (red)\nB: bfal_36635 (peach)\nC: bfal_36636 (blue)\nD: bfal_36639 (lime green)\nE: bfal_36641 (pink)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=800&width=617&top_left_y=1299&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=817&width=1097&top_left_y=1299&top_left_x=845"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1352",
        "problem": "Biology is often focused on survival and/or reproduction of individuals. Normal 'wild-type' male nematode worms were compared with two different mutant male strains - one group which did not produce sperm (top graph) and one group which did not mate (bottom graph). Their survivorship curves are below.\n\n[figure1]\n\nAdapted from Krohne , 1998 General Ecology\n\n\nWhat is the best conclusion that can be made from these data?\nA: There is a cost to mating and sperm production that shortens a nematode's life.\nB: There is a cost to mating or sperm production that shortens a nematode's life.\nC: The cost of mating is offset by having many offspring\nD: Nematode females probably eat the males after mating.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBiology is often focused on survival and/or reproduction of individuals. Normal 'wild-type' male nematode worms were compared with two different mutant male strains - one group which did not produce sperm (top graph) and one group which did not mate (bottom graph). Their survivorship curves are below.\n\n[figure1]\n\nAdapted from Krohne , 1998 General Ecology\n\n\nWhat is the best conclusion that can be made from these data?\n\nA: There is a cost to mating and sperm production that shortens a nematode's life.\nB: There is a cost to mating or sperm production that shortens a nematode's life.\nC: The cost of mating is offset by having many offspring\nD: Nematode females probably eat the males after mating.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-17.jpg?height=816&width=772&top_left_y=390&top_left_x=1017"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_992",
        "problem": "Among flowering plants the details of the morphology of the flower were considered the most important characteristics to determine relatedness because:\nA: Flowers are the most conspicuous part of the plant\nB: Flower structure could easily be observed with a magnifying glass\nC: Flower parts were the easiest to physically measure and accurately describe\nD: Small changes in flower structure could dramatically affect the fertility of the plant\nE: Flower size was not altered by soil properties\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAmong flowering plants the details of the morphology of the flower were considered the most important characteristics to determine relatedness because:\n\nA: Flowers are the most conspicuous part of the plant\nB: Flower structure could easily be observed with a magnifying glass\nC: Flower parts were the easiest to physically measure and accurately describe\nD: Small changes in flower structure could dramatically affect the fertility of the plant\nE: Flower size was not altered by soil properties\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1412",
        "problem": "The following table provides information about three different solutions, with respect to the plasma membrane.\n\n| Solution | Diffusible across plasma <br> membrane |\n| :--- | :---: |\n| 0.9\\% sodium chloride <br> solution | No |\n| $5.0 \\%$ dextrose (glucose) <br> solution | No |\n| $1.8 \\%$ urea solution | Yes |\n\nThese solutions can all be treated as iso-osmolar to human cells.\n\nWhich of the following statements about the three solutions is true?\nA: Only $1.8 \\%$ urea solution is isotonic to human cells.\nB: Only $0.9 \\%$ saline and $5.0 \\%$ dextrose are isotonic to human cells.\nC: Only $1.8 \\%$ urea and $0.9 \\%$ saline are isotonic to human cells.\nD: All three solutions are isotonic to human cells.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following table provides information about three different solutions, with respect to the plasma membrane.\n\n| Solution | Diffusible across plasma <br> membrane |\n| :--- | :---: |\n| 0.9\\% sodium chloride <br> solution | No |\n| $5.0 \\%$ dextrose (glucose) <br> solution | No |\n| $1.8 \\%$ urea solution | Yes |\n\nThese solutions can all be treated as iso-osmolar to human cells.\n\nWhich of the following statements about the three solutions is true?\n\nA: Only $1.8 \\%$ urea solution is isotonic to human cells.\nB: Only $0.9 \\%$ saline and $5.0 \\%$ dextrose are isotonic to human cells.\nC: Only $1.8 \\%$ urea and $0.9 \\%$ saline are isotonic to human cells.\nD: All three solutions are isotonic to human cells.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1455",
        "problem": "A phylogenetic tree is a diagram that shows the evolutionary relationships of living things that have descended from a common ancestor. Each branching point in this phylogenetic tree represents the divergence of two groups from a common ancestor.\n\n[figure1]\n\nConsidering the vertebrate tree above, which of the following options correctly orders vertebrates from the least to the most number of branching points experienced in total since the evolutionary divergence that resulted in vertebrates?\nA: Jawless fish < cartilaginous fish $<$ crocodiles $<$ birds\nB: Bony fish $<$ amphibians $<$ snakes $<$ lizards\nC: Amphibians $<$ mammals $<$ turtles $<$ birds\nD: Lizards $<$ turtles $<$ amphibians $<$ birds\nE: Lizards $<$ snakes $<$ crocodiles $<$ birds\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA phylogenetic tree is a diagram that shows the evolutionary relationships of living things that have descended from a common ancestor. Each branching point in this phylogenetic tree represents the divergence of two groups from a common ancestor.\n\n[figure1]\n\nConsidering the vertebrate tree above, which of the following options correctly orders vertebrates from the least to the most number of branching points experienced in total since the evolutionary divergence that resulted in vertebrates?\n\nA: Jawless fish < cartilaginous fish $<$ crocodiles $<$ birds\nB: Bony fish $<$ amphibians $<$ snakes $<$ lizards\nC: Amphibians $<$ mammals $<$ turtles $<$ birds\nD: Lizards $<$ turtles $<$ amphibians $<$ birds\nE: Lizards $<$ snakes $<$ crocodiles $<$ birds\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-37.jpg?height=871&width=1311&top_left_y=370&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_335",
        "problem": "在欧洲人群中, 每 2500 人中就有一人患囊性纤维变性, 这是一种常染色体遗传病,一对健康夫妇有一个患有此病的孩子, 以后该妇女又与一健康男子结婚, 问这对夫妇生一个孩子，孩子患此病的概率是（）\nA: $1 / 25$\nB: $1 / 51$\nC: $1 / 102$\nD: $1 / 250$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在欧洲人群中, 每 2500 人中就有一人患囊性纤维变性, 这是一种常染色体遗传病,一对健康夫妇有一个患有此病的孩子, 以后该妇女又与一健康男子结婚, 问这对夫妇生一个孩子，孩子患此病的概率是（）\n\nA: $1 / 25$\nB: $1 / 51$\nC: $1 / 102$\nD: $1 / 250$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1230",
        "problem": "## DEAR ENEMIES \\& NASTY NEIGHBORS\n\n\"Dear Enemy, I curse you, and hope that something slightly unpleasant happens to you, like an onion falling on your head.\"\n\n## Blackadder I, The Archbishop\n\nMany species show territorial behaviour (actively defending an area containing resources) allowing the species to monopolise resources such as food or mates. Territoriality is costly in terms of energy and time so some species reduce these costs by being less aggressive towards their neighbours than towards unfamiliar strangers, the so called \"dear enemy\" hypothesis. The converse are \"nasty neighbours\" in which species are more, not less aggressive towards their neighbours.\n\nDr Brunton's research group, at the Institute of Natural Sciences at Massey University, has been studying the behaviour of the New Zealand bellbird,\n\n[figure1]\n(Anthornis melanura) on Tiritiri Matangi Island. Both sexes sing prolifically and are known to use vocalisations to recognise individuals. Male and female bellbirds counter-sing in response to the vocalisations of their neighbours in a territorial behaviour that may lead to chasing of an individual bird intruding on a territory.\n\nThe researchers used speakers to play the song of neighbouring females or stranger females at different breeding stages. They recorded the responses of male and female bellbirds, including counter-singing and whether the birds approached the speaker. An asterisk indicates significant differences between the response to neighbours and strangers. Due to small sample sizes, the responses to the neighbours versus strangers were not tested during the incubation stage. Control 1 was silence, control 2 played back the\n\n[figure2]\nsong of a different species.\n[figure3]\n\nb) Counter-singing\n[figure4]\n\nUsing the information and the graphs above answer Questions 54 - 58 on the following page.Considering the differences in the behaviour of male and female bellbirds towards playback of female song, which of the following can be concluded from this research?\nA: Territorial male bellbirds are more likely to move towards a female intruder than their female partner.\nB: Female bellbirds were most territorial during courtship and chick rearing.\nC: Male bellbird territorial behaviours of all types are strongest during courtship.\nD: During the incubation period female bellbirds are more likely to counter-sing in response to an intruding female than are males.\nE: There were no significant differences between the territorial behavior of male and female bellbirds.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## DEAR ENEMIES \\& NASTY NEIGHBORS\n\n\"Dear Enemy, I curse you, and hope that something slightly unpleasant happens to you, like an onion falling on your head.\"\n\n## Blackadder I, The Archbishop\n\nMany species show territorial behaviour (actively defending an area containing resources) allowing the species to monopolise resources such as food or mates. Territoriality is costly in terms of energy and time so some species reduce these costs by being less aggressive towards their neighbours than towards unfamiliar strangers, the so called \"dear enemy\" hypothesis. The converse are \"nasty neighbours\" in which species are more, not less aggressive towards their neighbours.\n\nDr Brunton's research group, at the Institute of Natural Sciences at Massey University, has been studying the behaviour of the New Zealand bellbird,\n\n[figure1]\n(Anthornis melanura) on Tiritiri Matangi Island. Both sexes sing prolifically and are known to use vocalisations to recognise individuals. Male and female bellbirds counter-sing in response to the vocalisations of their neighbours in a territorial behaviour that may lead to chasing of an individual bird intruding on a territory.\n\nThe researchers used speakers to play the song of neighbouring females or stranger females at different breeding stages. They recorded the responses of male and female bellbirds, including counter-singing and whether the birds approached the speaker. An asterisk indicates significant differences between the response to neighbours and strangers. Due to small sample sizes, the responses to the neighbours versus strangers were not tested during the incubation stage. Control 1 was silence, control 2 played back the\n\n[figure2]\nsong of a different species.\n[figure3]\n\nb) Counter-singing\n[figure4]\n\nUsing the information and the graphs above answer Questions 54 - 58 on the following page.\n\nproblem:\nConsidering the differences in the behaviour of male and female bellbirds towards playback of female song, which of the following can be concluded from this research?\n\nA: Territorial male bellbirds are more likely to move towards a female intruder than their female partner.\nB: Female bellbirds were most territorial during courtship and chick rearing.\nC: Male bellbird territorial behaviours of all types are strongest during courtship.\nD: During the incubation period female bellbirds are more likely to counter-sing in response to an intruding female than are males.\nE: There were no significant differences between the territorial behavior of male and female bellbirds.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=357&width=571&top_left_y=524&top_left_x=1368",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=485&width=311&top_left_y=931&top_left_x=1495",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=486&width=1734&top_left_y=1493&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=534&width=1744&top_left_y=2042&top_left_x=108"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1070",
        "problem": "The manner in which species are connected within a food web has implications for community structure and dynamics in addition to direct interactions between prey and predators. One of the ways in which the food webs of different communities differ is their degree of connectance, also known as 'connectance index'.\n\nConnectance is a way of describing how many possible links in a food web are present. They are simply the lines that link consumers and the consumed. A simple formula for calculating connectance index $(\\mathrm{C})$ is:\n\n$$\nC=L /\\left[\\frac{S(S-1)}{2}\\right]\n$$\n\nwhere $L$ is the actual number of links in the food chain and $S$ is the number of species in the food chain.\n\nConsider the food web represented below.\n\n[figure1]\n\nWhat is the connectance index for this food web? (Express the answer in decimal form.)",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe manner in which species are connected within a food web has implications for community structure and dynamics in addition to direct interactions between prey and predators. One of the ways in which the food webs of different communities differ is their degree of connectance, also known as 'connectance index'.\n\nConnectance is a way of describing how many possible links in a food web are present. They are simply the lines that link consumers and the consumed. A simple formula for calculating connectance index $(\\mathrm{C})$ is:\n\n$$\nC=L /\\left[\\frac{S(S-1)}{2}\\right]\n$$\n\nwhere $L$ is the actual number of links in the food chain and $S$ is the number of species in the food chain.\n\nConsider the food web represented below.\n\n[figure1]\n\nWhat is the connectance index for this food web? (Express the answer in decimal form.)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-33.jpg?height=447&width=406&top_left_y=1788&top_left_x=835"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1199",
        "problem": "The pea weevil is a type of insect. The table below shows the average time it takes for pea weevil eggs to hatch at different temperatures.\n\n| Temperature ( $\\left.{ }^{\\circ} \\mathbf{C}\\right)$ | Average Hatching <br> Time (days) |\n| :---: | :---: |\n| 11 | 38 |\n| 14 | 20 |\n| 16 | 16 |\n| 18 | 10 |\n| 22 | 10 |\n| 24 | 7 |\n| 25 | 5 |\n| 27 | 5 |\n| 28 | 7 |\n\nBased on the data, which of the following climatic conditions would promote the highest population growth rate in pea weevils?\nA: Cold springs with temperatures from $11^{\\circ} \\mathrm{C}$ to $16^{\\circ} \\mathrm{C}$\nB: Moderate summers with temperatures from $25^{\\circ} \\mathrm{C}$ to $27^{\\circ} \\mathrm{C}$\nC: Heat waves in which the temperature is sustained well above $28^{\\circ} \\mathrm{C}$\nD: Overnight frosts after which the temperature warms from $0^{\\circ} \\mathrm{C}$ to $11^{\\circ} \\mathrm{C}$\nE: Cold winters with the temperatures below $11^{\\circ} \\mathrm{C}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe pea weevil is a type of insect. The table below shows the average time it takes for pea weevil eggs to hatch at different temperatures.\n\n| Temperature ( $\\left.{ }^{\\circ} \\mathbf{C}\\right)$ | Average Hatching <br> Time (days) |\n| :---: | :---: |\n| 11 | 38 |\n| 14 | 20 |\n| 16 | 16 |\n| 18 | 10 |\n| 22 | 10 |\n| 24 | 7 |\n| 25 | 5 |\n| 27 | 5 |\n| 28 | 7 |\n\nBased on the data, which of the following climatic conditions would promote the highest population growth rate in pea weevils?\n\nA: Cold springs with temperatures from $11^{\\circ} \\mathrm{C}$ to $16^{\\circ} \\mathrm{C}$\nB: Moderate summers with temperatures from $25^{\\circ} \\mathrm{C}$ to $27^{\\circ} \\mathrm{C}$\nC: Heat waves in which the temperature is sustained well above $28^{\\circ} \\mathrm{C}$\nD: Overnight frosts after which the temperature warms from $0^{\\circ} \\mathrm{C}$ to $11^{\\circ} \\mathrm{C}$\nE: Cold winters with the temperatures below $11^{\\circ} \\mathrm{C}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_898",
        "problem": "某二倍体植物有高茎与矮茎、红花与白花两对相对性状，且均各只受一对等位基因控制。现有一高茎红花亲本，其自交后代表现型及比例为高茎红花:高茎白花:矮茎红花:矮茎白花 $=5: 3: 3: 1$, 下列分析错误的是\nA: 控制上述两对相对性状的基因遗传时遵循自由组合定律\nB: 出现 5:3:3:1 的原因是可能存在某种基因型植株致死现象\nC: 出现 5:3:3:1 的原因是可能存在某种基因型配子致死现象\nD: 自交后代中高茎红花均为杂合子\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二倍体植物有高茎与矮茎、红花与白花两对相对性状，且均各只受一对等位基因控制。现有一高茎红花亲本，其自交后代表现型及比例为高茎红花:高茎白花:矮茎红花:矮茎白花 $=5: 3: 3: 1$, 下列分析错误的是\n\nA: 控制上述两对相对性状的基因遗传时遵循自由组合定律\nB: 出现 5:3:3:1 的原因是可能存在某种基因型植株致死现象\nC: 出现 5:3:3:1 的原因是可能存在某种基因型配子致死现象\nD: 自交后代中高茎红花均为杂合子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1132",
        "problem": "One of the most important mechanisms for reduced arterial pressure at high altitude involves shifting the shapes and position of oxygen dissociation curve (ODC). $\\mathrm{Q}$ in the figure indicates normal sea level ODC of hemoglobin.\n\n\n[figure1]\n\nWhich of the following is correct?\nA: (iv) on the curve represents arterial blood while (iii) represents mixed venous blood.\nB: Higher slope of line (i) - (ii) as compared to that of (iii) - (iv) indicates a compensatory mechanism to relieve hypoxic condition.\nC: Graph $\\mathrm{P}$ indicates increased affinity of hemoglobin for oxygen while $\\mathrm{R}$ indicates the decreased affinity.\nD: Greater slope of line (v) - (vi) as compared to (i) - (ii) indicates that greater amount of oxygen is released to tissues to compensate for hypoxic conditions.\nE: Shift of ODC to $R$ is of greater advantage to large mammals while shifting to $P$ would be of greater adaptive values in small sized mammals.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nOne of the most important mechanisms for reduced arterial pressure at high altitude involves shifting the shapes and position of oxygen dissociation curve (ODC). $\\mathrm{Q}$ in the figure indicates normal sea level ODC of hemoglobin.\n\n\n[figure1]\n\nWhich of the following is correct?\n\nA: (iv) on the curve represents arterial blood while (iii) represents mixed venous blood.\nB: Higher slope of line (i) - (ii) as compared to that of (iii) - (iv) indicates a compensatory mechanism to relieve hypoxic condition.\nC: Graph $\\mathrm{P}$ indicates increased affinity of hemoglobin for oxygen while $\\mathrm{R}$ indicates the decreased affinity.\nD: Greater slope of line (v) - (vi) as compared to (i) - (ii) indicates that greater amount of oxygen is released to tissues to compensate for hypoxic conditions.\nE: Shift of ODC to $R$ is of greater advantage to large mammals while shifting to $P$ would be of greater adaptive values in small sized mammals.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-22.jpg?height=734&width=916&top_left_y=755&top_left_x=602"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_556",
        "problem": "脆性 X 染色体综合征是一种伴 X 染色体显性遗传病, 其致病机理是 FMR 1 基因中序列 CGG 的重复次数过多。已知 CGG 重复序列中的 CG 易发生甲基化修饰而导致基因不能表达。图 1 为某脆性 X 染色体综合征患者的家系图, 图 2 为各家庭成员 FMR1 基因的 cDNA（在体外由 mRNA 经逆转录获得）片段的电泳结果图（泳道空表示任何位置均无产物）。下列相关叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: $\\mathrm{I}_{2}$ 为杂合子, $\\mathrm{II}_{4}$ 与 $\\mathrm{I}_{1}$ 的基因型不同\nB: FMR1 致病基因的 cDNA 片段长度约为 $550 \\mathrm{bp}$\nC: $\\mathrm{II}_{4}$ 的 FMR1 基因发生了甲基化修饰导致遗传信息改变\nD: $\\mathrm{II}_{4}$ 与人群中某一正常女性婚配后所生的男孩均不患病\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n脆性 X 染色体综合征是一种伴 X 染色体显性遗传病, 其致病机理是 FMR 1 基因中序列 CGG 的重复次数过多。已知 CGG 重复序列中的 CG 易发生甲基化修饰而导致基因不能表达。图 1 为某脆性 X 染色体综合征患者的家系图, 图 2 为各家庭成员 FMR1 基因的 cDNA（在体外由 mRNA 经逆转录获得）片段的电泳结果图（泳道空表示任何位置均无产物）。下列相关叙述错误的是（）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: $\\mathrm{I}_{2}$ 为杂合子, $\\mathrm{II}_{4}$ 与 $\\mathrm{I}_{1}$ 的基因型不同\nB: FMR1 致病基因的 cDNA 片段长度约为 $550 \\mathrm{bp}$\nC: $\\mathrm{II}_{4}$ 的 FMR1 基因发生了甲基化修饰导致遗传信息改变\nD: $\\mathrm{II}_{4}$ 与人群中某一正常女性婚配后所生的男孩均不患病\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-15.jpg?height=180&width=757&top_left_y=1343&top_left_x=364",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-15.jpg?height=186&width=577&top_left_y=1369&top_left_x=1208"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_604",
        "problem": "水稻的花是两性花, 靠风媒传粉, 且花很小, 导致杂交育种工作量巨大。雄性育性基因 $\\mathrm{M}$ （可育）与 $\\mathrm{m}$ （不育）为一对等位基因， G3 育种技术的原理是通过基因工程技术将外源育性恢复基因 $\\mathrm{M}$ 与外源花粉致死基因 $\\mathrm{F}$ 紧密连锁的 $\\mathrm{M}-\\mathrm{F}$ 导入雄性不育系,获得保持系 (M-Fmm) 水稻, 保持系水稻产生的可育雄配子基因型为 $m$ 。下列相关叙述正确的是（ ）\nA: 基因型为 $\\mathrm{mm}$ 的水稻花粉不育, 可作为父本避免人工去雄的麻烦\nB: 紧密连锁的基因 $M-F$ 导入雄性不育系后, 所产生的雌配子仅有 1 种\nC: 让保持系水稻自交得到 $F_{1}, F_{1}$ 中雄性不育系植株所占比例为 $1 / 2$\nD: 让保持系水稻自由交配至 $F_{2}, F_{2}$ 中保持系植株所占比例为 $3 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n水稻的花是两性花, 靠风媒传粉, 且花很小, 导致杂交育种工作量巨大。雄性育性基因 $\\mathrm{M}$ （可育）与 $\\mathrm{m}$ （不育）为一对等位基因， G3 育种技术的原理是通过基因工程技术将外源育性恢复基因 $\\mathrm{M}$ 与外源花粉致死基因 $\\mathrm{F}$ 紧密连锁的 $\\mathrm{M}-\\mathrm{F}$ 导入雄性不育系,获得保持系 (M-Fmm) 水稻, 保持系水稻产生的可育雄配子基因型为 $m$ 。下列相关叙述正确的是（ ）\n\nA: 基因型为 $\\mathrm{mm}$ 的水稻花粉不育, 可作为父本避免人工去雄的麻烦\nB: 紧密连锁的基因 $M-F$ 导入雄性不育系后, 所产生的雌配子仅有 1 种\nC: 让保持系水稻自交得到 $F_{1}, F_{1}$ 中雄性不育系植株所占比例为 $1 / 2$\nD: 让保持系水稻自由交配至 $F_{2}, F_{2}$ 中保持系植株所占比例为 $3 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1515",
        "problem": "The graph shows pressure of pulses of blood flowing through different vessels.\n\n## Blood pressure\n\n[figure1]\n\nWhy does the pulsatility of the blood reduce and become smooth flow as blood flows away from the heart?\nA: The heart pumps blood continuously, with no moments when blood is not pumped.\nB: Muscles in the blood vessels contract and relax in time with pulses.\nC: Elastic fibres in arteries stretch and recoil in time with pulses.\nD: Fluid is forced out of vessels during pulses so they are reduced over time.\nE: The total resistance of vessels increases away from the heart so pulses cannot flow.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows pressure of pulses of blood flowing through different vessels.\n\n## Blood pressure\n\n[figure1]\n\nWhy does the pulsatility of the blood reduce and become smooth flow as blood flows away from the heart?\n\nA: The heart pumps blood continuously, with no moments when blood is not pumped.\nB: Muscles in the blood vessels contract and relax in time with pulses.\nC: Elastic fibres in arteries stretch and recoil in time with pulses.\nD: Fluid is forced out of vessels during pulses so they are reduced over time.\nE: The total resistance of vessels increases away from the heart so pulses cannot flow.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-13.jpg?height=948&width=1213&top_left_y=491&top_left_x=590"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_730",
        "problem": "果蝇的性别是由早期胚胎的性指数所决定的, 即 X 染色体的数目与常染色体组数的比例（X：A）。X：A=1 时，会激活性别相关基因 $\\mathrm{M}$ 进而发育成为雌性，若 $\\mathrm{M}$ 基因发生突变, 则发育为雄性; $\\mathrm{X}: \\mathrm{A}=0.5$ 时, 即使存在 $\\mathrm{M}$ 基因也会发育为雄性。已知 $\\mathrm{Y}$染色体只决定雄蝇的可育性, $\\mathrm{M}$ 基因仅位于 $\\mathrm{X}$ 染色体上, XXX 和 $\\mathrm{YY}$ 的个体致死。下列说法错误的是 ( )\nA: $X^{M} X^{M} A A$ 和 $X^{M} O A A$ 的果蝇杂交子代雌雄之比为 1: 1\nB: $\\mathrm{X}^{\\mathrm{M}} \\mathrm{X}^{\\mathrm{m}} \\mathrm{AA}$ 和 $\\mathrm{X}^{\\mathrm{m}} \\mathrm{YAA}$ 的果蝇杂交子代雌雄之比为 1:3\nC: $\\mathrm{X}^{\\mathrm{M}} \\mathrm{X}^{\\mathrm{m}} \\mathrm{YAA}$ 和 $\\mathrm{X}^{\\mathrm{m}} \\mathrm{YAA}$ 的果蝇杂交子代雌雄之比为 2: 7\nD: XM $X^{M} A A A$ 的果蝇可能与母本减数分裂 I 异常有关\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n果蝇的性别是由早期胚胎的性指数所决定的, 即 X 染色体的数目与常染色体组数的比例（X：A）。X：A=1 时，会激活性别相关基因 $\\mathrm{M}$ 进而发育成为雌性，若 $\\mathrm{M}$ 基因发生突变, 则发育为雄性; $\\mathrm{X}: \\mathrm{A}=0.5$ 时, 即使存在 $\\mathrm{M}$ 基因也会发育为雄性。已知 $\\mathrm{Y}$染色体只决定雄蝇的可育性, $\\mathrm{M}$ 基因仅位于 $\\mathrm{X}$ 染色体上, XXX 和 $\\mathrm{YY}$ 的个体致死。下列说法错误的是 ( )\n\nA: $X^{M} X^{M} A A$ 和 $X^{M} O A A$ 的果蝇杂交子代雌雄之比为 1: 1\nB: $\\mathrm{X}^{\\mathrm{M}} \\mathrm{X}^{\\mathrm{m}} \\mathrm{AA}$ 和 $\\mathrm{X}^{\\mathrm{m}} \\mathrm{YAA}$ 的果蝇杂交子代雌雄之比为 1:3\nC: $\\mathrm{X}^{\\mathrm{M}} \\mathrm{X}^{\\mathrm{m}} \\mathrm{YAA}$ 和 $\\mathrm{X}^{\\mathrm{m}} \\mathrm{YAA}$ 的果蝇杂交子代雌雄之比为 2: 7\nD: XM $X^{M} A A A$ 的果蝇可能与母本减数分裂 I 异常有关\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-094.jpg?height=49&width=1391&top_left_y=1226&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_892",
        "problem": "下图 1 为甲 $(A 、 a) 、 乙(D 、 d)$ 两种单基因遗传病的家系图, 已知正常人群中甲病基因携带者占 $1 / 40$. 通过电泳图谱方法能够使基因 $\\mathrm{D}$ 显示一个条带, 基因 $\\mathrm{d}$ 则显示为位置不同的另一个条带, 用该方法对上述家系中的部分个体进行分析, 条带的有无及其位置表示为图 2. 下列相关叙述错误的是（）\n[图1]\nA: 图 2 中 DNA 片段为 $150 \\mathrm{bp}$ 代表 D 基因\nB: II-6 的基因型为 $A a X^{D} X^{d}$ 或 $A a X^{D} X^{D}$\nC: II-8 与 III-12 基因型相同的概率为 $1 / 3$\nD: III-10 和人群中正常男性结婚, 生育一个患甲病孩子的概率为 $1 / 80$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图 1 为甲 $(A 、 a) 、 乙(D 、 d)$ 两种单基因遗传病的家系图, 已知正常人群中甲病基因携带者占 $1 / 40$. 通过电泳图谱方法能够使基因 $\\mathrm{D}$ 显示一个条带, 基因 $\\mathrm{d}$ 则显示为位置不同的另一个条带, 用该方法对上述家系中的部分个体进行分析, 条带的有无及其位置表示为图 2. 下列相关叙述错误的是（）\n[图1]\n\nA: 图 2 中 DNA 片段为 $150 \\mathrm{bp}$ 代表 D 基因\nB: II-6 的基因型为 $A a X^{D} X^{d}$ 或 $A a X^{D} X^{D}$\nC: II-8 与 III-12 基因型相同的概率为 $1 / 3$\nD: III-10 和人群中正常男性结婚, 生育一个患甲病孩子的概率为 $1 / 80$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-33.jpg?height=1094&width=1296&top_left_y=172&top_left_x=346"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_559",
        "problem": "2017 年中国科学家利用“聪明的化学方法和操作技巧 (10s 内去核, $15 \\mathrm{~s}$ 内核移\n\n植）”成功培育了体细胞克隆猴，流程如下，下列叙述错误的是（ ）\n\n[图1]\nA: 科学家多年的反复实验带来的娴熟技术使卵细胞受损减少\nB: 核移植前用“灭活病毒短暂处理”主要目的是诱导细胞融合\nC: “聪明的化学方法”主要指图中(1)(2), 通过调节组蛋白修饰而抑制基因表达\nD: 上述技术有助于加速针对阿尔茨海默病、免疫缺陷、肿瘤等的新药研发进程\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2017 年中国科学家利用“聪明的化学方法和操作技巧 (10s 内去核, $15 \\mathrm{~s}$ 内核移\n\n植）”成功培育了体细胞克隆猴，流程如下，下列叙述错误的是（ ）\n\n[图1]\n\nA: 科学家多年的反复实验带来的娴熟技术使卵细胞受损减少\nB: 核移植前用“灭活病毒短暂处理”主要目的是诱导细胞融合\nC: “聪明的化学方法”主要指图中(1)(2), 通过调节组蛋白修饰而抑制基因表达\nD: 上述技术有助于加速针对阿尔茨海默病、免疫缺陷、肿瘤等的新药研发进程\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-29.jpg?height=317&width=1445&top_left_y=2354&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_843",
        "problem": "果蝇的红眼基因 $(\\mathrm{R})$ 对白眼基因 $(\\mathrm{r})$ 为显性，位于 $\\mathrm{X}$ 染色体上; 长翅基因(B)对残翅基因 (b)为显性，位于常染色体上。现有一只红眼长翅果蝇与一只白眼长翅果蝇交配, $F_{1}$ 雄蝇中有 $1 / 8$ 为白眼残翅, 下列叙述错误的是 ( )\nA: 亲本雌蝇的基因型是 $B b X^{R} X^{r}$\nB: $F_{1}$ 中出现长翅雄蝇的概率为 $1 / 16$\nC: 雌、雄亲本产生含 $\\mathrm{X}^{\\mathrm{r}}$ 配子的比例相同\nD: 白眼残翅雌蝇可形成基因型为 $b X^{\\mathrm{r}}$ 的极体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的红眼基因 $(\\mathrm{R})$ 对白眼基因 $(\\mathrm{r})$ 为显性，位于 $\\mathrm{X}$ 染色体上; 长翅基因(B)对残翅基因 (b)为显性，位于常染色体上。现有一只红眼长翅果蝇与一只白眼长翅果蝇交配, $F_{1}$ 雄蝇中有 $1 / 8$ 为白眼残翅, 下列叙述错误的是 ( )\n\nA: 亲本雌蝇的基因型是 $B b X^{R} X^{r}$\nB: $F_{1}$ 中出现长翅雄蝇的概率为 $1 / 16$\nC: 雌、雄亲本产生含 $\\mathrm{X}^{\\mathrm{r}}$ 配子的比例相同\nD: 白眼残翅雌蝇可形成基因型为 $b X^{\\mathrm{r}}$ 的极体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_47",
        "problem": "Four different bacteria were isolated from the gut of a shrimp to be studied about their probiotic potency through the activity to decrease pathogenicity of Vibrio harveyi, a common bacteria infecting shrimp culture. In the first experiment, the four isolated bacteria were inoculated in cross-streak plates to observe inhibition zones against 4 bacterial strains (Fig.9A). In the second experiment, the shrimp survival rate in presence of Vibrio harveyi and each bacterial isolate after 5 days incubation was measured (Fig.9B).\n\n[figure1]\n\nFig.Q.9A. $\\mathrm{K}=$ Control (no bacteria streaked on the dash box), P1-P4 = Probiotic candidates 1-4, $\\mathrm{a}=$ Streptococcus sp. (Gram-positive), $\\mathrm{b}=$ Vibrio sp. (Gram-negative), $\\mathrm{c}=$ Bacillus sp. (Gram-positive), $\\mathrm{d}=$ Salmonella $\\mathrm{sp}$. (Gram-negative)\n\n[figure2]\n\nFig.Q.9B. $U=$ shrimp culture alone, $U+V=$ shrimp culture with addition of Vibrio sp.,\n\n$\\mathrm{U}+\\mathrm{V}+\\mathrm{P} 1-4=$ shrimp culture with addition of Vibrio sp. and a specific probiotic candidate P1-4, respectively.\n\nIndicate in the Answer Sheet if each of the following statements is True or False.\nA: Candidate No. 1 (P1) produced an antimicrobial compound that inhibited Gramnegative and Gram-positive bacteria.\nB: Candidate No. 2 (P2) was able to decrease Vibrio sp. pathogenicity without killing them.\nC: Candidate No. 3 (P3) produced an antimicrobial compound targeting the outer membrane.\nD: Candidate No. 4 (P4) had good effect on the shrimp survival by inbibiting Gramnegative bacteria.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFour different bacteria were isolated from the gut of a shrimp to be studied about their probiotic potency through the activity to decrease pathogenicity of Vibrio harveyi, a common bacteria infecting shrimp culture. In the first experiment, the four isolated bacteria were inoculated in cross-streak plates to observe inhibition zones against 4 bacterial strains (Fig.9A). In the second experiment, the shrimp survival rate in presence of Vibrio harveyi and each bacterial isolate after 5 days incubation was measured (Fig.9B).\n\n[figure1]\n\nFig.Q.9A. $\\mathrm{K}=$ Control (no bacteria streaked on the dash box), P1-P4 = Probiotic candidates 1-4, $\\mathrm{a}=$ Streptococcus sp. (Gram-positive), $\\mathrm{b}=$ Vibrio sp. (Gram-negative), $\\mathrm{c}=$ Bacillus sp. (Gram-positive), $\\mathrm{d}=$ Salmonella $\\mathrm{sp}$. (Gram-negative)\n\n[figure2]\n\nFig.Q.9B. $U=$ shrimp culture alone, $U+V=$ shrimp culture with addition of Vibrio sp.,\n\n$\\mathrm{U}+\\mathrm{V}+\\mathrm{P} 1-4=$ shrimp culture with addition of Vibrio sp. and a specific probiotic candidate P1-4, respectively.\n\nIndicate in the Answer Sheet if each of the following statements is True or False.\n\nA: Candidate No. 1 (P1) produced an antimicrobial compound that inhibited Gramnegative and Gram-positive bacteria.\nB: Candidate No. 2 (P2) was able to decrease Vibrio sp. pathogenicity without killing them.\nC: Candidate No. 3 (P3) produced an antimicrobial compound targeting the outer membrane.\nD: Candidate No. 4 (P4) had good effect on the shrimp survival by inbibiting Gramnegative bacteria.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-022.jpg?height=702&width=1092&top_left_y=841&top_left_x=501",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-022.jpg?height=664&width=1069&top_left_y=1756&top_left_x=424"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1101",
        "problem": "Consider a predator and a prey of comparable mass. During the pursuit of a prey by a predator, a sudden change in direction is a strategy used by the prey to shake off the pursuer. Three possible patterns of pursuit A, B and C by the predator (dotted line) in response to the prey movement (full line) are depicted below; the prey makes two turns in each of the three instances.\n[figure1]\n\nIn pattern A, predator cuts the corner to reach the prey faster.\n\nIn pattern $B$, predator overshoots the corner.\n\nIn pattern $\\mathrm{C}$, predator takes the same path as the prey.\n\nA single turn shown in figure A or B may not necessarily result in capture or escape but several such may. \n\nWhich of the following is correct?\nA: Greater the distance between the two, greater is the likelihood of predator following pattern A.\nB: Lesser the distance between the two, lesser is the likelihood of predator following pattern B.\nC: Greater the response time of the predator, more is the likelihood of predator following pattern $\\mathrm{C}$.\nD: Pattern C is likely to evolve as a stable evolutionary strategy.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nConsider a predator and a prey of comparable mass. During the pursuit of a prey by a predator, a sudden change in direction is a strategy used by the prey to shake off the pursuer. Three possible patterns of pursuit A, B and C by the predator (dotted line) in response to the prey movement (full line) are depicted below; the prey makes two turns in each of the three instances.\n[figure1]\n\nIn pattern A, predator cuts the corner to reach the prey faster.\n\nIn pattern $B$, predator overshoots the corner.\n\nIn pattern $\\mathrm{C}$, predator takes the same path as the prey.\n\nA single turn shown in figure A or B may not necessarily result in capture or escape but several such may. \n\nWhich of the following is correct?\n\nA: Greater the distance between the two, greater is the likelihood of predator following pattern A.\nB: Lesser the distance between the two, lesser is the likelihood of predator following pattern B.\nC: Greater the response time of the predator, more is the likelihood of predator following pattern $\\mathrm{C}$.\nD: Pattern C is likely to evolve as a stable evolutionary strategy.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-28.jpg?height=1594&width=1072&top_left_y=1122&top_left_x=407"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_80",
        "problem": "A study on the effects of lead $(\\mathrm{Pb})$, a toxic heavy metal, on growth and photosynthesis of two microalgae, Chlorella and Scenedesmus was conducted. The figure on the left below shows the growth of these species responded differently to lead concentration after 4 day treatment. From growth rate $(K e)$, generation time $(G)$ of each species at each concentration of lead can be calculated as equation: $\\mathrm{G}=(\\ln 2) / K e$. The right hand Fig.Q. 66 is result of the effect of lead on photosynthesis of these species, indicated by Fv/Fm, a sensitive parameter that decreases when photosynthesis is impaired. The concentration of lead that gives half-maximal response, the $\\mathrm{IC}_{50}$, can be estimated based on response versus lead concentration plots.\n[figure1]\n\nFig.Q66\nA: Estimated $\\mathrm{IC}_{50}$ of growth for the Scenedesmus was higher than that for the Chlorella.\nB: Photosynthetic impairment by $\\mathrm{Pb}$ was likely responsible for the growth decrease in the Chlorella but this was not the scenario in the Scenedesmus\nC: The estimated $\\mathrm{IC}_{50}$ for effects on $\\mathrm{F}_{\\mathrm{v}} / \\mathrm{F}_{\\mathrm{m}}$ was higher than that for growth in Scenedesmus\nD: At lead concentration that $\\log _{10}([\\mathrm{~Pb}]+1)$ is 0.5 , the Scenedesmus reproduced faster than the Chlorella did.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA study on the effects of lead $(\\mathrm{Pb})$, a toxic heavy metal, on growth and photosynthesis of two microalgae, Chlorella and Scenedesmus was conducted. The figure on the left below shows the growth of these species responded differently to lead concentration after 4 day treatment. From growth rate $(K e)$, generation time $(G)$ of each species at each concentration of lead can be calculated as equation: $\\mathrm{G}=(\\ln 2) / K e$. The right hand Fig.Q. 66 is result of the effect of lead on photosynthesis of these species, indicated by Fv/Fm, a sensitive parameter that decreases when photosynthesis is impaired. The concentration of lead that gives half-maximal response, the $\\mathrm{IC}_{50}$, can be estimated based on response versus lead concentration plots.\n[figure1]\n\nFig.Q66\n\nA: Estimated $\\mathrm{IC}_{50}$ of growth for the Scenedesmus was higher than that for the Chlorella.\nB: Photosynthetic impairment by $\\mathrm{Pb}$ was likely responsible for the growth decrease in the Chlorella but this was not the scenario in the Scenedesmus\nC: The estimated $\\mathrm{IC}_{50}$ for effects on $\\mathrm{F}_{\\mathrm{v}} / \\mathrm{F}_{\\mathrm{m}}$ was higher than that for growth in Scenedesmus\nD: At lead concentration that $\\log _{10}([\\mathrm{~Pb}]+1)$ is 0.5 , the Scenedesmus reproduced faster than the Chlorella did.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-039.jpg?height=712&width=1440&top_left_y=1047&top_left_x=202"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_722",
        "problem": "某雌雄同株植物的紫花与红花是一对相对性状, 由一对等位基因控制。现有一批紫花植株和红花植株, 研究人员用它们分别进行如下 4 组实验: 实验 (1): 紫花 $\\times$ 紫花 $\\rightarrow$ 紫花: 红花 $=3: 1$; 实验(2)：紫花 $\\times$ 红花 $\\rightarrow$ 紫花：红花 $=1 ： 0$; 实验(3)：紫花 $\\times$ 红花 $\\rightarrow$ 紫花:红花=1：1; 实验(4)：紫花×紫花 $\\rightarrow$ 紫花：红花=5：1。下列相关叙述正确的是 ( )\nA: 除了杂交实验(3)外, 依据其他 3 组杂交实验都可以判断出紫花对红花为显性\nB: 除了杂交实验(2)外, 其他 3 组杂交实验的亲本中紫花植株都是杂合子\nC: 除了第(1)组实验外, 其他 3 组杂交实验均为测交实验\nD: 第(1)组杂交实验的子代中紫花植株自交, 结果也是紫花：红花=3：1\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某雌雄同株植物的紫花与红花是一对相对性状, 由一对等位基因控制。现有一批紫花植株和红花植株, 研究人员用它们分别进行如下 4 组实验: 实验 (1): 紫花 $\\times$ 紫花 $\\rightarrow$ 紫花: 红花 $=3: 1$; 实验(2)：紫花 $\\times$ 红花 $\\rightarrow$ 紫花：红花 $=1 ： 0$; 实验(3)：紫花 $\\times$ 红花 $\\rightarrow$ 紫花:红花=1：1; 实验(4)：紫花×紫花 $\\rightarrow$ 紫花：红花=5：1。下列相关叙述正确的是 ( )\n\nA: 除了杂交实验(3)外, 依据其他 3 组杂交实验都可以判断出紫花对红花为显性\nB: 除了杂交实验(2)外, 其他 3 组杂交实验的亲本中紫花植株都是杂合子\nC: 除了第(1)组实验外, 其他 3 组杂交实验均为测交实验\nD: 第(1)组杂交实验的子代中紫花植株自交, 结果也是紫花：红花=3：1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_237",
        "problem": "Electrocardiography is a method to study the electrical activity of the heart (ECG). In addition to P, QRS and T waves, there are segments and intervals which are defined to show a specific period of time in a cardiac cycle. A variety of different factors like concentration of electrolytes, drugs and temperature could have an effect on waves and intervals of ECG. A group of researchers have studied the effect of body temperature on QT and RR interval in electrocardiogram of beagle dogs. The result of their study is demonstrated in the figures.\n\nNote that RR interval is the time between two consecutive $R$ waves and $Q T$ interval is the exact time before start of $Q$ wave till the end of the next $T$ wave.\n[figure1]\n\nFigure (left) Relation between RR Interval and body temperature. (right) Relation between QT Interval and body temperature.\n\nGiven that QT interval gets shorten by the shortening of the RR interval, a formula (given below) has been approved to calculate a corrected QT interval (QTc) to allow the comparison of QT interval over time at different heart rates.\n\nHodges formula: $\\mathrm{QTC}=\\mathrm{QT}+1.75$ (heart rate -60 )\nA: Body temperature does not have any effect on heart rate.\nB: Independent of changes of heart rate, QT interval is shortened by rise of body temperature.\nC: Hypocalcaemia will shorten the QTc interval.\nD: In higher body temperature, end diastolic volume of ventricles is increased.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nElectrocardiography is a method to study the electrical activity of the heart (ECG). In addition to P, QRS and T waves, there are segments and intervals which are defined to show a specific period of time in a cardiac cycle. A variety of different factors like concentration of electrolytes, drugs and temperature could have an effect on waves and intervals of ECG. A group of researchers have studied the effect of body temperature on QT and RR interval in electrocardiogram of beagle dogs. The result of their study is demonstrated in the figures.\n\nNote that RR interval is the time between two consecutive $R$ waves and $Q T$ interval is the exact time before start of $Q$ wave till the end of the next $T$ wave.\n[figure1]\n\nFigure (left) Relation between RR Interval and body temperature. (right) Relation between QT Interval and body temperature.\n\nGiven that QT interval gets shorten by the shortening of the RR interval, a formula (given below) has been approved to calculate a corrected QT interval (QTc) to allow the comparison of QT interval over time at different heart rates.\n\nHodges formula: $\\mathrm{QTC}=\\mathrm{QT}+1.75$ (heart rate -60 )\n\nA: Body temperature does not have any effect on heart rate.\nB: Independent of changes of heart rate, QT interval is shortened by rise of body temperature.\nC: Hypocalcaemia will shorten the QTc interval.\nD: In higher body temperature, end diastolic volume of ventricles is increased.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-39.jpg?height=740&width=1586&top_left_y=838&top_left_x=240"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_353",
        "problem": "家蚕 $(2 n=28)$ 的性别决定方式为 $\\mathrm{ZW}$ 型, 一般进行有性生殖, 也能进行没有雄性参与的孤雌生殖。通过构建家蚕某品种的纯系, 可发现并淘汰隐性致死基因和其他不良基因, 育种途径如图所示, $\\mathrm{b}$ 育种途径中的热处理能抑制同源染色体分离。下列分析正确的是（ ）\n\n[图1]\nA: a 过程为减数分裂，基因重组主要发生在 $\\mathrm{d}$ 过程中\nB: $b$ 育种途径得到的子代的基因型与父本的相同\nC: $\\mathrm{c}$ 育种途径得到 $\\mathrm{f}$ 依据的遗传学原理是基因重组\nD: $\\mathrm{e}$ 与 $\\mathrm{f}$ 交配，得到的子代的隐性性状都能表现\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n家蚕 $(2 n=28)$ 的性别决定方式为 $\\mathrm{ZW}$ 型, 一般进行有性生殖, 也能进行没有雄性参与的孤雌生殖。通过构建家蚕某品种的纯系, 可发现并淘汰隐性致死基因和其他不良基因, 育种途径如图所示, $\\mathrm{b}$ 育种途径中的热处理能抑制同源染色体分离。下列分析正确的是（ ）\n\n[图1]\n\nA: a 过程为减数分裂，基因重组主要发生在 $\\mathrm{d}$ 过程中\nB: $b$ 育种途径得到的子代的基因型与父本的相同\nC: $\\mathrm{c}$ 育种途径得到 $\\mathrm{f}$ 依据的遗传学原理是基因重组\nD: $\\mathrm{e}$ 与 $\\mathrm{f}$ 交配，得到的子代的隐性性状都能表现\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-68.jpg?height=522&width=811&top_left_y=1955&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1081",
        "problem": "Which of the following proteins are likely to be active during mitosis?\n\ni. Telomerase\n\nii. Primase\n\niii. Topoisomerase\n\niv. Kinesin\nA: ii, iii \\& iv only\nB: iii \\& iv only\nC: i, ii \\& iii only\nD: only iv\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following proteins are likely to be active during mitosis?\n\ni. Telomerase\n\nii. Primase\n\niii. Topoisomerase\n\niv. Kinesin\n\nA: ii, iii \\& iv only\nB: iii \\& iv only\nC: i, ii \\& iii only\nD: only iv\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_732",
        "problem": "在肿瘤细胞中, 许多抑癌基因通过表观遗传机制被关闭, CDK9 的特异性小分子抑制剂 MC18 可以重新激活这些基因的表达。研究人员在人结肠癌细胞系 YB- 5 中引入绿色荧光蛋白 (GFP) 报告系统, 如图 1 所示。利用小分子药物 MC18 处理 YB-5 细胞系后, 得到的结果如图 2 (DMSO 为小分子药物的溶剂)。MC18 对小鼠（已诱导形成肿瘤）体内肿瘤生长的影响如图 3。下列说法错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\nA: 启动子甲基化可导致抑癌基因不能转录\nB: $\\mathrm{MC} 18$ 可能干扰了肿瘤细胞的细胞周期\nC: MC18 可能能去除启动子上的表观遗传标记\nD: CDK9 是打开抑癌基因的关键物质\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在肿瘤细胞中, 许多抑癌基因通过表观遗传机制被关闭, CDK9 的特异性小分子抑制剂 MC18 可以重新激活这些基因的表达。研究人员在人结肠癌细胞系 YB- 5 中引入绿色荧光蛋白 (GFP) 报告系统, 如图 1 所示。利用小分子药物 MC18 处理 YB-5 细胞系后, 得到的结果如图 2 (DMSO 为小分子药物的溶剂)。MC18 对小鼠（已诱导形成肿瘤）体内肿瘤生长的影响如图 3。下列说法错误的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n图3\n\nA: 启动子甲基化可导致抑癌基因不能转录\nB: $\\mathrm{MC} 18$ 可能干扰了肿瘤细胞的细胞周期\nC: MC18 可能能去除启动子上的表观遗传标记\nD: CDK9 是打开抑癌基因的关键物质\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-054.jpg?height=223&width=508&top_left_y=1008&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-054.jpg?height=343&width=326&top_left_y=911&top_left_x=908",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-054.jpg?height=391&width=548&top_left_y=907&top_left_x=1231"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_879",
        "problem": "某种二倍体玉米植株有雌株、雄株和两性植株三种类型; 受一组复等位基因 $\\mathrm{G}^{+}$、\n\nG、 $g$ 控制, 其中 $\\mathrm{G}^{+}$基因存在时表现为雄株, 不含 $\\mathrm{G}^{+}$基因但含有 $\\mathrm{G}$ 基因时表现为两性植株，只含 $\\mathrm{g}$ 基因时表现为雌株。下列相关描述正确的是（ ）\nA: 基因型为 $\\mathrm{G}^{+} \\mathrm{G}^{+}$的个体自交后代不发生性状分离\nB: 若雌雄植株杂交, 子代可能全为雄株\nC: 若三个基因的频率相等, 则雌株所占比例最小\nD: 让一株两性植株与雌株杂交, 子代最多可出现三种类型植株\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种二倍体玉米植株有雌株、雄株和两性植株三种类型; 受一组复等位基因 $\\mathrm{G}^{+}$、\n\nG、 $g$ 控制, 其中 $\\mathrm{G}^{+}$基因存在时表现为雄株, 不含 $\\mathrm{G}^{+}$基因但含有 $\\mathrm{G}$ 基因时表现为两性植株，只含 $\\mathrm{g}$ 基因时表现为雌株。下列相关描述正确的是（ ）\n\nA: 基因型为 $\\mathrm{G}^{+} \\mathrm{G}^{+}$的个体自交后代不发生性状分离\nB: 若雌雄植株杂交, 子代可能全为雄株\nC: 若三个基因的频率相等, 则雌株所占比例最小\nD: 让一株两性植株与雌株杂交, 子代最多可出现三种类型植株\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1403",
        "problem": "Rhizophores are leafless branches that arise from the stem and grow downward, producing roots at its tip when it reaches the soil. Rhizophores are commonly known as aerial roots. Scientists measured length and height of rhizophores of mangrove plant (Rhizophora mangle, Fig A). They also made cross sections of rhizophores and observed their anatomical characteristics. The rhizophore cross sections were classified in orders according to the number of arches away from the main stem. The results are shown in Fig B and Fig $\\mathrm{C}$.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\n[figure3]\n\nFig: Rhizophores of Rhizophora mangle plants.\n\nA: Rhizophore height and length measurement.\n\nB: Change in height (empty square) and in length/height proportion (full circle) in five sequential orders of rhizophores.\n\nC: Relative proportions of bark (including aerenchyma), xylem and pith along the length of individual first-order rhizophores (left), and at the base of rhizophores of sequential orders (right).\n\nWhich of the following statements is correct?\nA: Rhizophore height and the length/height proportion in the rhizophores decreases with rhizophore order.\nB: Within first-order rhizophores, the xylem proportion in the cross-section is larger when closer to the main stem, and decreases progressively as the rhizophore approaches the ground.\nC: When rhizophore order changes from 1 to 5 , bark and pith proportion decreases, while xylem proportion increases.\nD: The supportive function is likely enhanced in the first-order rhizophores, with lower length/height proportion, and the lowest proportion of xylem.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRhizophores are leafless branches that arise from the stem and grow downward, producing roots at its tip when it reaches the soil. Rhizophores are commonly known as aerial roots. Scientists measured length and height of rhizophores of mangrove plant (Rhizophora mangle, Fig A). They also made cross sections of rhizophores and observed their anatomical characteristics. The rhizophore cross sections were classified in orders according to the number of arches away from the main stem. The results are shown in Fig B and Fig $\\mathrm{C}$.\n\nA\n\n[figure1]\n\nB\n\n[figure2]\n\n[figure3]\n\nFig: Rhizophores of Rhizophora mangle plants.\n\nA: Rhizophore height and length measurement.\n\nB: Change in height (empty square) and in length/height proportion (full circle) in five sequential orders of rhizophores.\n\nC: Relative proportions of bark (including aerenchyma), xylem and pith along the length of individual first-order rhizophores (left), and at the base of rhizophores of sequential orders (right).\n\nWhich of the following statements is correct?\n\nA: Rhizophore height and the length/height proportion in the rhizophores decreases with rhizophore order.\nB: Within first-order rhizophores, the xylem proportion in the cross-section is larger when closer to the main stem, and decreases progressively as the rhizophore approaches the ground.\nC: When rhizophore order changes from 1 to 5 , bark and pith proportion decreases, while xylem proportion increases.\nD: The supportive function is likely enhanced in the first-order rhizophores, with lower length/height proportion, and the lowest proportion of xylem.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-09.jpg?height=454&width=508&top_left_y=681&top_left_x=266",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-09.jpg?height=448&width=377&top_left_y=690&top_left_x=814",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-09.jpg?height=496&width=552&top_left_y=640&top_left_x=1226"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_584",
        "problem": "某些蛋白质合成基因的突变会造成蛋白质功能异常, 细胞可以通过校正基因突变并转录获得特殊的校正 tRNA, 这种校正 tRNA 能识别某些蛋白合成基因突变后转录的密码子。例如, 小鼠细胞中 $\\mathrm{H}$ 蛋白基因发生突变会导致翻译提前终止, 而校正基因突变产生了一种携带亮氨酸但能识别终止密码子的 tRNA, 在 $\\mathrm{H}$ 蛋白合成过程中该校正 tRNA 发挥了重要作用。下列叙述正确的是（ ）\nA: 基因突变后碱基序列改变, 会导致蛋白质结构发生不可逆改变\nB: 基因转录形成的 RNA 都会和核糖体结合成复合体，进行翻译\nC: 突变基因表达时, 校正 tRNA 会导致蛋白质中亮氨酸比例增加\nD: 校正 tRNA 有利于削弱基因突变带来的影响, 维持性状的稳定\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某些蛋白质合成基因的突变会造成蛋白质功能异常, 细胞可以通过校正基因突变并转录获得特殊的校正 tRNA, 这种校正 tRNA 能识别某些蛋白合成基因突变后转录的密码子。例如, 小鼠细胞中 $\\mathrm{H}$ 蛋白基因发生突变会导致翻译提前终止, 而校正基因突变产生了一种携带亮氨酸但能识别终止密码子的 tRNA, 在 $\\mathrm{H}$ 蛋白合成过程中该校正 tRNA 发挥了重要作用。下列叙述正确的是（ ）\n\nA: 基因突变后碱基序列改变, 会导致蛋白质结构发生不可逆改变\nB: 基因转录形成的 RNA 都会和核糖体结合成复合体，进行翻译\nC: 突变基因表达时, 校正 tRNA 会导致蛋白质中亮氨酸比例增加\nD: 校正 tRNA 有利于削弱基因突变带来的影响, 维持性状的稳定\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_515",
        "problem": "1 个精原细胞在含有 ${ }^{32} \\mathrm{p}$ 的培养基中进行一次有丝分裂得到 2 个子细胞, 将一个子细胞转入不含有 ${ }^{32} \\mathrm{p}$ 的普通培养基继续培养, 在进行核 DNA 复制时, 一个 DNA 分子的三个腺嘌呤碱基发生变异后转变为次黄嘌呤（I）。已知次黄嘌呤可与胞嘧啶（C）配对,在不考虑其他变异的情况下本次分裂不可能出现的现象是（ ）\nA: 若为有丝分裂, 子细胞含有 ${ }^{32} \\mathrm{P}$ 的染色体上无次黄嘌呤\nB: 若为有丝分裂, 子细胞中含有次黄嘌呤但不含 ${ }^{32} \\mathrm{P}$ 的细胞占 $1 / 2$\nC: 若为减数分裂, 两个次级精母细胞中都含有次黄嘌呤\nD: 若为减数分裂, 四个精子中含有次黄嘌呤又含 ${ }^{32} \\mathrm{P}$ 的细胞占 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n1 个精原细胞在含有 ${ }^{32} \\mathrm{p}$ 的培养基中进行一次有丝分裂得到 2 个子细胞, 将一个子细胞转入不含有 ${ }^{32} \\mathrm{p}$ 的普通培养基继续培养, 在进行核 DNA 复制时, 一个 DNA 分子的三个腺嘌呤碱基发生变异后转变为次黄嘌呤（I）。已知次黄嘌呤可与胞嘧啶（C）配对,在不考虑其他变异的情况下本次分裂不可能出现的现象是（ ）\n\nA: 若为有丝分裂, 子细胞含有 ${ }^{32} \\mathrm{P}$ 的染色体上无次黄嘌呤\nB: 若为有丝分裂, 子细胞中含有次黄嘌呤但不含 ${ }^{32} \\mathrm{P}$ 的细胞占 $1 / 2$\nC: 若为减数分裂, 两个次级精母细胞中都含有次黄嘌呤\nD: 若为减数分裂, 四个精子中含有次黄嘌呤又含 ${ }^{32} \\mathrm{P}$ 的细胞占 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_236",
        "problem": "The early embryonic cells pass through two stages of differentiation: specification and determination.\n\n[figure1]\nA: These data supports that the fate of the determined cell is not reversible.\nB: The cell that is specified loses its other differentiation potentials\nC: If the graft is removed at the late stage and cultured in isolation, it will give rise to the red spinous sphere (shown in the picture)\nD: If the graft is removed at the late stage and cultured in the presence of the eye inducing factors, the eye-like structure will be developed.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe early embryonic cells pass through two stages of differentiation: specification and determination.\n\n[figure1]\n\nA: These data supports that the fate of the determined cell is not reversible.\nB: The cell that is specified loses its other differentiation potentials\nC: If the graft is removed at the late stage and cultured in isolation, it will give rise to the red spinous sphere (shown in the picture)\nD: If the graft is removed at the late stage and cultured in the presence of the eye inducing factors, the eye-like structure will be developed.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-26.jpg?height=908&width=1268&top_left_y=457&top_left_x=428"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_640",
        "problem": "实验室中某只果蝇的一个精原细胞的一条染色体上某基因的一条脱氧核苷酸链中的\n\n一个碱基 G 替换成了 $\\mathrm{T}$ ，另一条脱氧核苷酸链的碱基未发生替换。下列叙述正确的是\nA: 该精原细胞发生的变异是染色体变异, 可作为果蝇进化的原材料\nB: 该 DNA 分子复制产生的子代 DNA 的嘌呤数小于嘧啶数\nC: 若该精原细胞进行减数分裂, 则产生的精细胞中含突变基因的所占比例为 $1 / 2$\nD: 若该精原细胞进行有丝分裂, 则产生的精原细胞中含突变基因的所占比例是 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n实验室中某只果蝇的一个精原细胞的一条染色体上某基因的一条脱氧核苷酸链中的\n\n一个碱基 G 替换成了 $\\mathrm{T}$ ，另一条脱氧核苷酸链的碱基未发生替换。下列叙述正确的是\n\nA: 该精原细胞发生的变异是染色体变异, 可作为果蝇进化的原材料\nB: 该 DNA 分子复制产生的子代 DNA 的嘌呤数小于嘧啶数\nC: 若该精原细胞进行减数分裂, 则产生的精细胞中含突变基因的所占比例为 $1 / 2$\nD: 若该精原细胞进行有丝分裂, 则产生的精原细胞中含突变基因的所占比例是 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1021",
        "problem": "One fifth of the cardiac output flows through the kidneys. After being filtered by the glomerulus, in what order does the filtrate pass through the following nephric structures?\n1. Ascending limb of the Loop of Henle\n2. Distal convoluted tubule\n3. Descending limb of the Loop of Henle\n4. Proximal convoluted tubule\n5. Collecting duct\nA: $1,3,4,2,5$\nB: $2,1,3,4$\nC: $2,3,1,4,5$\nD: $4,1,3,5$\nE: $4,3,1,2,5$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOne fifth of the cardiac output flows through the kidneys. After being filtered by the glomerulus, in what order does the filtrate pass through the following nephric structures?\n1. Ascending limb of the Loop of Henle\n2. Distal convoluted tubule\n3. Descending limb of the Loop of Henle\n4. Proximal convoluted tubule\n5. Collecting duct\n\nA: $1,3,4,2,5$\nB: $2,1,3,4$\nC: $2,3,1,4,5$\nD: $4,1,3,5$\nE: $4,3,1,2,5$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_757",
        "problem": "研究表明, 人的 $\\mathrm{ABO}$ 血型不仅由位于 9 号染色体上的 $\\mathrm{I}^{\\mathrm{A}} 、 \\mathrm{I}^{\\mathrm{B}} 、 \\mathrm{i}$ 基因决定, 还与位于第 19 号染色体上的 $\\mathrm{H} 、 \\mathrm{~h}$ 基因有关。在人体内, 其原理如图 1 所示。图 2 为一个家族的血型遗传系谱图。下列叙述错误的是\n\n[图1]\n\n图 1\n\n[图2]\n\n图 2\nA: 图 1 相关基因决定 $\\mathrm{ABO}$ 血型的基因型有 18 种，表现型有 4 种\nB: 若 $\\mathrm{III}_{2}$ 与 $\\mathrm{AB}$ 型男子结婚, 生了 $\\mathrm{O}$ 型的男孩, 再生一个女孩型为 $\\mathrm{AB}$ 型的概率为 $3 / 8$\nC: 若 $\\mathrm{II}_{2}$ 的基因型为 $\\mathrm{hhI}^{\\mathrm{B}} \\mathrm{i}$, 那么 $\\mathrm{II}_{1}$ 和 $\\mathrm{II}_{2}$ 再生一个小孩血型为 $\\mathrm{AB}$ 型的概率为 $1 / 8$\nD: 9 号染色体上与 ABO 血型系统相关的等位基因之间既有共显性, 又有完全显性\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究表明, 人的 $\\mathrm{ABO}$ 血型不仅由位于 9 号染色体上的 $\\mathrm{I}^{\\mathrm{A}} 、 \\mathrm{I}^{\\mathrm{B}} 、 \\mathrm{i}$ 基因决定, 还与位于第 19 号染色体上的 $\\mathrm{H} 、 \\mathrm{~h}$ 基因有关。在人体内, 其原理如图 1 所示。图 2 为一个家族的血型遗传系谱图。下列叙述错误的是\n\n[图1]\n\n图 1\n\n[图2]\n\n图 2\n\nA: 图 1 相关基因决定 $\\mathrm{ABO}$ 血型的基因型有 18 种，表现型有 4 种\nB: 若 $\\mathrm{III}_{2}$ 与 $\\mathrm{AB}$ 型男子结婚, 生了 $\\mathrm{O}$ 型的男孩, 再生一个女孩型为 $\\mathrm{AB}$ 型的概率为 $3 / 8$\nC: 若 $\\mathrm{II}_{2}$ 的基因型为 $\\mathrm{hhI}^{\\mathrm{B}} \\mathrm{i}$, 那么 $\\mathrm{II}_{1}$ 和 $\\mathrm{II}_{2}$ 再生一个小孩血型为 $\\mathrm{AB}$ 型的概率为 $1 / 8$\nD: 9 号染色体上与 ABO 血型系统相关的等位基因之间既有共显性, 又有完全显性\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-038.jpg?height=214&width=626&top_left_y=2012&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-038.jpg?height=246&width=371&top_left_y=2010&top_left_x=997"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1033",
        "problem": "In the phylogenetic tree of green plants below, indicate the branch (A to E) where traits I and IV are acquired?\n\n I. Pollen\n\nII. Tracheid\n\nIII. Cuticle\n\nIV. Seed\n\nV. Carpel\n\nVI. Multicellular embryo\n\n[figure1]\nA: A\nB: B\nC: C\nD: D\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the phylogenetic tree of green plants below, indicate the branch (A to E) where traits I and IV are acquired?\n\n I. Pollen\n\nII. Tracheid\n\nIII. Cuticle\n\nIV. Seed\n\nV. Carpel\n\nVI. Multicellular embryo\n\n[figure1]\n\nA: A\nB: B\nC: C\nD: D\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_c11579df664240d6158ag-15.jpg?height=794&width=1111&top_left_y=1229&top_left_x=607"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_978",
        "problem": "Which of the following statements about seeds is FALSE?\nA: Coleorhizae are found in monocotyledonous seeds\nB: Both monocotlyedonous and dicotyledonous seeds have radicles\nC: Both monocotlyedonous and dicotyledonous seeds have a pericarp\nD: The coleoptile is found only in monocotyledonous seeds\nE: The aleurone layer is found only in dicotyledonous seeds\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following statements about seeds is FALSE?\n\nA: Coleorhizae are found in monocotyledonous seeds\nB: Both monocotlyedonous and dicotyledonous seeds have radicles\nC: Both monocotlyedonous and dicotyledonous seeds have a pericarp\nD: The coleoptile is found only in monocotyledonous seeds\nE: The aleurone layer is found only in dicotyledonous seeds\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_855",
        "problem": "下图甲表示某鸟体内细胞分裂过程中, 每条染色体上 DNA 数量的变化曲线, 图乙表示该鸟某个细胞分裂的示意图。下列叙述错误的是（）\n[图1]\nA: 图甲表明细胞分裂过程中，染色体上的 DNA 复制后又彼此分离\nB: 图甲 $\\mathrm{bc}$ 段时期, 该动物细胞中可能进行同源染色体的交叉互换\nC: 图乙细胞为初级精母细胞, 细胞内含两条形态相同的性染色体\nD: 图乙细胞中的同源染色体彼此分离, 其子细胞中不含有等位基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图甲表示某鸟体内细胞分裂过程中, 每条染色体上 DNA 数量的变化曲线, 图乙表示该鸟某个细胞分裂的示意图。下列叙述错误的是（）\n[图1]\n\nA: 图甲表明细胞分裂过程中，染色体上的 DNA 复制后又彼此分离\nB: 图甲 $\\mathrm{bc}$ 段时期, 该动物细胞中可能进行同源染色体的交叉互换\nC: 图乙细胞为初级精母细胞, 细胞内含两条形态相同的性染色体\nD: 图乙细胞中的同源染色体彼此分离, 其子细胞中不含有等位基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-53.jpg?height=454&width=1290&top_left_y=2286&top_left_x=383"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_906",
        "problem": "小鼠 gper 基因的功能与减数分裂有关。研究人员向小鼠卵母细胞中注射与 gper 基因转录出的 mRNA 部分序列互补的 siRNA，观察其对减数分裂的影响，结果如图所示。下列说法错误的是（ ）\n\n[图1]\n\n未注射siRNA\n\n[图2]\n\n注射siRNA\nA: gper 基因中可能存在与 siRNA 相同的碱基序列\nB: gper 基因的表达会受到 siRNA 的抑制\nC: gper 基因的功能是促进卵母细胞均等分裂\nD: gper 基因缺乏的小鼠生育能力下降\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n小鼠 gper 基因的功能与减数分裂有关。研究人员向小鼠卵母细胞中注射与 gper 基因转录出的 mRNA 部分序列互补的 siRNA，观察其对减数分裂的影响，结果如图所示。下列说法错误的是（ ）\n\n[图1]\n\n未注射siRNA\n\n[图2]\n\n注射siRNA\n\nA: gper 基因中可能存在与 siRNA 相同的碱基序列\nB: gper 基因的表达会受到 siRNA 的抑制\nC: gper 基因的功能是促进卵母细胞均等分裂\nD: gper 基因缺乏的小鼠生育能力下降\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-84.jpg?height=431&width=460&top_left_y=2032&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-84.jpg?height=428&width=437&top_left_y=2033&top_left_x=815"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_8",
        "problem": "In order to prevent an excess water loss, stomata respond rapidly to changes in humidity. Transpiration rate per unit leaf area represents the speed of water loss from the plant body. It is proportional to the diffusion rate of water vapor $\\left(\\mathrm{d}_{\\text {water }}\\right)$, the water vapor concentration difference across the leaf epidermis $(\\Delta \\mathrm{w})$, and the relative stomatal aperture. Figure 1 shows relative stomatal apertures in normal air and in Helox air (79:21 mixture of $\\mathrm{He}$ and $\\mathrm{O}_{2}$ with the appropriate concentrations of water vapor and $\\mathrm{CO}_{2}$ added). Relative stomatal apertures were measured in normal air (Air) and in Helox air under three different $\\Delta \\mathrm{w}$ conditions: the same $\\Delta \\mathrm{w}$ as that in normal air (Helox), $2 / 3$ of the normal air $\\Delta \\mathrm{w}\\left(\\operatorname{Helox}^{2 / 3}\\right)$, and $1 / 2.33$ of the normal air $\\Delta \\mathrm{w}\\left(\\operatorname{Helox}^{1 / 2.33}\\right)$. $\\mathrm{d}_{\\text {water }}$ in Helox air is 2.33 times higher than that in normal air, while Helox air does not affect any other factors of transpiration. Note that water vapor diffuses only though the stomata and that the water vapor concentration inside the leaf is always saturated.\n\n[figure1]\n\nFigure 1 Relative stomatal apertures in various conditions\nA: Stomata respond to the absolute humidity of the air.\nB: Transpiration is higher in Helox air than that in normal air at the same humidity.\nC: Stomatal response to low humidity decreases the photosynthetic assimilation rate.\nD: Stomatal response to low humidity keeps the water loss constant.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn order to prevent an excess water loss, stomata respond rapidly to changes in humidity. Transpiration rate per unit leaf area represents the speed of water loss from the plant body. It is proportional to the diffusion rate of water vapor $\\left(\\mathrm{d}_{\\text {water }}\\right)$, the water vapor concentration difference across the leaf epidermis $(\\Delta \\mathrm{w})$, and the relative stomatal aperture. Figure 1 shows relative stomatal apertures in normal air and in Helox air (79:21 mixture of $\\mathrm{He}$ and $\\mathrm{O}_{2}$ with the appropriate concentrations of water vapor and $\\mathrm{CO}_{2}$ added). Relative stomatal apertures were measured in normal air (Air) and in Helox air under three different $\\Delta \\mathrm{w}$ conditions: the same $\\Delta \\mathrm{w}$ as that in normal air (Helox), $2 / 3$ of the normal air $\\Delta \\mathrm{w}\\left(\\operatorname{Helox}^{2 / 3}\\right)$, and $1 / 2.33$ of the normal air $\\Delta \\mathrm{w}\\left(\\operatorname{Helox}^{1 / 2.33}\\right)$. $\\mathrm{d}_{\\text {water }}$ in Helox air is 2.33 times higher than that in normal air, while Helox air does not affect any other factors of transpiration. Note that water vapor diffuses only though the stomata and that the water vapor concentration inside the leaf is always saturated.\n\n[figure1]\n\nFigure 1 Relative stomatal apertures in various conditions\n\nA: Stomata respond to the absolute humidity of the air.\nB: Transpiration is higher in Helox air than that in normal air at the same humidity.\nC: Stomatal response to low humidity decreases the photosynthetic assimilation rate.\nD: Stomatal response to low humidity keeps the water loss constant.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-40.jpg?height=768&width=717&top_left_y=1255&top_left_x=675"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1503",
        "problem": "The NDMAR protein is a membrane channel found on neurones. It binds glutamate then opens to allow ions into the cell. Several drugs (NO, ethanol, ketamine) inhibit NDMAR and can cause powerful psychological dissociation. The effect of several inhibitors is shown.\n\n$$\n\\text { Ion entry }\n$$\n\n$$\n\\text { into neurone }\n$$\n[figure1]\n\nWhich inhibitor has the most powerful psychological effects? *Assume all were tested at relevant concentrations.*\nA: A\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: All of them\nE: None of them\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe NDMAR protein is a membrane channel found on neurones. It binds glutamate then opens to allow ions into the cell. Several drugs (NO, ethanol, ketamine) inhibit NDMAR and can cause powerful psychological dissociation. The effect of several inhibitors is shown.\n\n$$\n\\text { Ion entry }\n$$\n\n$$\n\\text { into neurone }\n$$\n[figure1]\n\nWhich inhibitor has the most powerful psychological effects? *Assume all were tested at relevant concentrations.*\n\nA: A\nB: $\\quad \\mathrm{B}$\nC: $\\quad \\mathrm{C}$\nD: All of them\nE: None of them\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-10.jpg?height=960&width=1476&top_left_y=591&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_598",
        "problem": "自然杀伤 (NK) 细胞是一类具有广泛杀伤能力的天然免疫细胞, 研究人员对粒细胞集落刺激因子 (CSF) 抑制 $\\mathrm{NK}$ 细胞活性的分子机制进行了探索。 $\\mathrm{NK}$ 细胞分泌的干扰素 $\\gamma(I F N \\gamma)$ 可增强免疫反应, 但过量则会引起炎症因子风暴, 危害健康。实验证明, $\\mathrm{S}$ 基因的表达会使 IFN $\\gamma$ 的产生量减少。已知糖皮质激素 (G) 能激活细胞质中的糖皮质激素受体 (GR), 使 $\\mathrm{GR}$ 移动并调控 $\\mathrm{S}$ 基因的表达。下图为检测施加 $\\mathrm{CSF}$ 后 $\\mathrm{NK}$ 细胞中 GR 的变化结果。下列相关叙述正确的是（）\n\n[图1]\nA: CSF 可以抑制 $\\mathrm{S}$ 基因的转录，从而缓解炎症因子风暴\nB: 根据荧光的亮度及位置推测, GR 进入细胞核穿过 2 层膜\nC: CSF 在临床上可以应用于器官移植的抗排异药物的研究\nD: NK 细胞的作用属于特异性免疫, 体现了免疫监视功能\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n自然杀伤 (NK) 细胞是一类具有广泛杀伤能力的天然免疫细胞, 研究人员对粒细胞集落刺激因子 (CSF) 抑制 $\\mathrm{NK}$ 细胞活性的分子机制进行了探索。 $\\mathrm{NK}$ 细胞分泌的干扰素 $\\gamma(I F N \\gamma)$ 可增强免疫反应, 但过量则会引起炎症因子风暴, 危害健康。实验证明, $\\mathrm{S}$ 基因的表达会使 IFN $\\gamma$ 的产生量减少。已知糖皮质激素 (G) 能激活细胞质中的糖皮质激素受体 (GR), 使 $\\mathrm{GR}$ 移动并调控 $\\mathrm{S}$ 基因的表达。下图为检测施加 $\\mathrm{CSF}$ 后 $\\mathrm{NK}$ 细胞中 GR 的变化结果。下列相关叙述正确的是（）\n\n[图1]\n\nA: CSF 可以抑制 $\\mathrm{S}$ 基因的转录，从而缓解炎症因子风暴\nB: 根据荧光的亮度及位置推测, GR 进入细胞核穿过 2 层膜\nC: CSF 在临床上可以应用于器官移植的抗排异药物的研究\nD: NK 细胞的作用属于特异性免疫, 体现了免疫监视功能\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-88.jpg?height=672&width=1145&top_left_y=178&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1309",
        "problem": "Imagine there is giant jellyfish. The jellyfish weighs $10 \\mathrm{~kg}$, and is $99 \\%$ water by weight. The jellyfish washes up on the beach and the water evaporates from the jellyfish until it reaches $98 \\%$ water by weight. How much does the jellyfish now weigh?\nA: $9.9 \\mathrm{~kg}$\nB: $9.8 \\mathrm{~kg}$\nC: $9 \\mathrm{~kg}$\nD: $8 \\mathrm{~kg}$\nE: $5 \\mathrm{~kg}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nImagine there is giant jellyfish. The jellyfish weighs $10 \\mathrm{~kg}$, and is $99 \\%$ water by weight. The jellyfish washes up on the beach and the water evaporates from the jellyfish until it reaches $98 \\%$ water by weight. How much does the jellyfish now weigh?\n\nA: $9.9 \\mathrm{~kg}$\nB: $9.8 \\mathrm{~kg}$\nC: $9 \\mathrm{~kg}$\nD: $8 \\mathrm{~kg}$\nE: $5 \\mathrm{~kg}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_96",
        "problem": "As seen in the left-hand graph, a population of moth species A exhibits individual variation in body color. The environment in which this population lives includes predators, such as birds, which find species A palatable. The environment also includes other moth species unpalatable to birds: one individual from each of three species (1 3) is shown in the right-hand illustration. Species 1, 2, and 3 are similar to different phenotypes found within species A: species 1 to lighter individuals, species 2 to individuals with intermediate phenotype, and species 3 to darker individuals. After capturing and tasting species 1, 2 and 3, birds learn to avoid eating them. Species A is considered a Batesian mimic of the other species. If species 3 becomes most abundant in this habitat, which graph accurately predicts what you would observe in species A? (The dotted line represents the mean value of the original population of species A.)\n[figure1]\n\nSpecies A\nA: [figure2]  Phenotypic variation of body color\nB: [figure3]  Phenotypic variation of body color\nC: [figure4]\nD:  [figure5]\nE: [figure6]  Phenotypic variation of body color\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAs seen in the left-hand graph, a population of moth species A exhibits individual variation in body color. The environment in which this population lives includes predators, such as birds, which find species A palatable. The environment also includes other moth species unpalatable to birds: one individual from each of three species (1 3) is shown in the right-hand illustration. Species 1, 2, and 3 are similar to different phenotypes found within species A: species 1 to lighter individuals, species 2 to individuals with intermediate phenotype, and species 3 to darker individuals. After capturing and tasting species 1, 2 and 3, birds learn to avoid eating them. Species A is considered a Batesian mimic of the other species. If species 3 becomes most abundant in this habitat, which graph accurately predicts what you would observe in species A? (The dotted line represents the mean value of the original population of species A.)\n[figure1]\n\nSpecies A\n\nA: [figure2]  Phenotypic variation of body color\nB: [figure3]  Phenotypic variation of body color\nC: [figure4]\nD:  [figure5]\nE: [figure6]  Phenotypic variation of body color\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-40.jpg?height=814&width=1506&top_left_y=1235&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-41.jpg?height=403&width=756&top_left_y=387&top_left_x=313",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-41.jpg?height=399&width=756&top_left_y=383&top_left_x=1181",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-41.jpg?height=431&width=637&top_left_y=1138&top_left_x=321",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-41.jpg?height=432&width=626&top_left_y=1126&top_left_x=1189",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-41.jpg?height=397&width=668&top_left_y=1823&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_893",
        "problem": "如图表示某二倍体雄性哺乳动物的一个正在分裂的细胞, 图中所示为染色体 (用数字表示）及基因（用字母表示），下列分析中错误的是（ ）\n\n[图1]\nA: 若该图表示减数第二次分裂前期的细胞, 形成该细胞发生了两种变异\nB: 若该图表示有丝分裂的细胞, 则该细胞内有 2 号和 4 号为两条性染色体\nC: 若该图表示初级精母细胞, 该细胞能产生 3 种精子\nD: 若该图表示初级精母细胞, 经过减数分裂产生了一个 $\\mathrm{AaX}^{\\mathrm{B}}$ 的精细胞, 则同时产生的精细胞一 定为 $\\mathrm{X}^{\\mathrm{B}} 、 \\mathrm{AY} 、 \\mathrm{AY}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图表示某二倍体雄性哺乳动物的一个正在分裂的细胞, 图中所示为染色体 (用数字表示）及基因（用字母表示），下列分析中错误的是（ ）\n\n[图1]\n\nA: 若该图表示减数第二次分裂前期的细胞, 形成该细胞发生了两种变异\nB: 若该图表示有丝分裂的细胞, 则该细胞内有 2 号和 4 号为两条性染色体\nC: 若该图表示初级精母细胞, 该细胞能产生 3 种精子\nD: 若该图表示初级精母细胞, 经过减数分裂产生了一个 $\\mathrm{AaX}^{\\mathrm{B}}$ 的精细胞, 则同时产生的精细胞一 定为 $\\mathrm{X}^{\\mathrm{B}} 、 \\mathrm{AY} 、 \\mathrm{AY}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-053.jpg?height=340&width=346&top_left_y=1389&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1328",
        "problem": "The following diagram shows the relative scale of different things. According to this diagram what is the approximate size range of typical viruses.\nA: $0.1 \\mu \\mathrm{m}-0.8 \\mathrm{~mm}$\nB: $0.1 \\mu \\mathrm{m}-0.8 \\mu \\mathrm{m}$\nC: $0.00001 \\mathrm{~m}-0.0008 \\mathrm{~m}$\nD: $10 \\mathrm{~nm}-80 \\mathrm{~nm}$\nE: $10 \\mathrm{~nm}-800 \\mathrm{~nm}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following diagram shows the relative scale of different things. According to this diagram what is the approximate size range of typical viruses.\n\nA: $0.1 \\mu \\mathrm{m}-0.8 \\mathrm{~mm}$\nB: $0.1 \\mu \\mathrm{m}-0.8 \\mu \\mathrm{m}$\nC: $0.00001 \\mathrm{~m}-0.0008 \\mathrm{~m}$\nD: $10 \\mathrm{~nm}-80 \\mathrm{~nm}$\nE: $10 \\mathrm{~nm}-800 \\mathrm{~nm}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_999",
        "problem": "Although the armadillo has a leather coat like a reptile, it is classified as a mammal due to the presence of mammary glands. Which of the following additional features support its inclusion in the class Mammalia?\n\n[figure1]\n\nArmadillo\n\nI. Hair over parts of the body.\n\nII. Presence of pituitary and thyroid gland.\n\nIII. Complete separation of pulmonary and systemic circulation in a 4 chambered heart.\n\nIV. A diaphragm separating thoracic and abdominal cavities.\n\nV. Regulation of body temperature irrespective of ambient temperature.\n\nVI. Enucleated red blood cells.\nA: III and VI\nB: I, IV and V\nC: I and IV\nD: I and II, and III\nE: I, III, IV, and VI\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAlthough the armadillo has a leather coat like a reptile, it is classified as a mammal due to the presence of mammary glands. Which of the following additional features support its inclusion in the class Mammalia?\n\n[figure1]\n\nArmadillo\n\nI. Hair over parts of the body.\n\nII. Presence of pituitary and thyroid gland.\n\nIII. Complete separation of pulmonary and systemic circulation in a 4 chambered heart.\n\nIV. A diaphragm separating thoracic and abdominal cavities.\n\nV. Regulation of body temperature irrespective of ambient temperature.\n\nVI. Enucleated red blood cells.\n\nA: III and VI\nB: I, IV and V\nC: I and IV\nD: I and II, and III\nE: I, III, IV, and VI\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_c11579df664240d6158ag-16.jpg?height=434&width=637&top_left_y=515&top_left_x=554"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1044",
        "problem": "What is the probability of obtaining the given genotype in the offspring AAbbCCdd from the parents AaBbCcDd x AABbCcDd (Assume independent assortment of all gene pairs)?\nA: $1 / 64$\nB: $1 / 128$\nC: $3 / 128$\nD: $9 / 256$\nE: $3 / 256$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the probability of obtaining the given genotype in the offspring AAbbCCdd from the parents AaBbCcDd x AABbCcDd (Assume independent assortment of all gene pairs)?\n\nA: $1 / 64$\nB: $1 / 128$\nC: $3 / 128$\nD: $9 / 256$\nE: $3 / 256$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_65",
        "problem": "Mating experiments with a yeast species that has 16 chromosomes of equal length were performed. Results of mating between two mutant yeast strains (shown with an unhappy phenotype), each with a mutation in a single gene, are shown below. The mutated allele of each gene is recessive with respect to the wild type allele. Unhappy is polygenic and many genes affect the phenotype. Recombination does not occur in these yeast strains.\n\n[figure1]\nA: If mutations of $\\mathrm{m} 1$ and $\\mathrm{m} 2$ were in the same gene, it is possible that all products of meiosis of cross 3 will be unhappy.\nB: If 1 happy to 3 unhappy yeasts were produced from meiosis of cross 3 , then the mutation of $\\mathrm{m} 1$ and the mutation of $\\mathrm{m} 2$ were in different genes.\nC: If 1 happy to lunhappy yeasts were produced from meiosis of cross 3 , then the mutation of $\\mathrm{m} 1$ and the mutation of $\\mathrm{m} 2$ can suppress each other.\nD: If the experiments were performed with various pairs of unhappy mutants, the most frequently observed ratio of happy to unhappy yeasts among meiosis\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nMating experiments with a yeast species that has 16 chromosomes of equal length were performed. Results of mating between two mutant yeast strains (shown with an unhappy phenotype), each with a mutation in a single gene, are shown below. The mutated allele of each gene is recessive with respect to the wild type allele. Unhappy is polygenic and many genes affect the phenotype. Recombination does not occur in these yeast strains.\n\n[figure1]\n\nA: If mutations of $\\mathrm{m} 1$ and $\\mathrm{m} 2$ were in the same gene, it is possible that all products of meiosis of cross 3 will be unhappy.\nB: If 1 happy to 3 unhappy yeasts were produced from meiosis of cross 3 , then the mutation of $\\mathrm{m} 1$ and the mutation of $\\mathrm{m} 2$ were in different genes.\nC: If 1 happy to lunhappy yeasts were produced from meiosis of cross 3 , then the mutation of $\\mathrm{m} 1$ and the mutation of $\\mathrm{m} 2$ can suppress each other.\nD: If the experiments were performed with various pairs of unhappy mutants, the most frequently observed ratio of happy to unhappy yeasts among meiosis\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-13.jpg?height=705&width=1245&top_left_y=630&top_left_x=405"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1059",
        "problem": "In corn, three enzymes catalyze the same reaction. Corresponding genes $\\left(\\mathrm{a}^{+}\\right.$, $\\mathrm{b}^{+} \\& \\mathrm{c}^{+}$) are located on three different chromosomes. The reaction is as follows:\n\nColourless compound $\\xrightarrow{\\mathrm{a}^{+} \\text {or } \\mathrm{b}^{+} \\text {or } \\mathrm{c}^{+}}$red compound\n\nThe normal functioning of any one of these genes is sufficient to convert colourless compound to the red compound. The mutant alleles of $\\mathrm{a}^{+}, \\mathrm{b}^{+} \\& \\mathrm{c}^{+}$are $\\mathrm{a}, \\mathrm{b}$ and $\\mathrm{c}$ respectively.\n\nAnother step involved in the pathway is as follows:\n\n[figure1]\n\nA red $(a+/ a, b+/ b, c+/ c, d+/ d)$ plants are selfed, what proportion of $F 2$ corn is colourless? (Express your answer as a fraction.)",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIn corn, three enzymes catalyze the same reaction. Corresponding genes $\\left(\\mathrm{a}^{+}\\right.$, $\\mathrm{b}^{+} \\& \\mathrm{c}^{+}$) are located on three different chromosomes. The reaction is as follows:\n\nColourless compound $\\xrightarrow{\\mathrm{a}^{+} \\text {or } \\mathrm{b}^{+} \\text {or } \\mathrm{c}^{+}}$red compound\n\nThe normal functioning of any one of these genes is sufficient to convert colourless compound to the red compound. The mutant alleles of $\\mathrm{a}^{+}, \\mathrm{b}^{+} \\& \\mathrm{c}^{+}$are $\\mathrm{a}, \\mathrm{b}$ and $\\mathrm{c}$ respectively.\n\nAnother step involved in the pathway is as follows:\n\n[figure1]\n\nA red $(a+/ a, b+/ b, c+/ c, d+/ d)$ plants are selfed, what proportion of $F 2$ corn is colourless? (Express your answer as a fraction.)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-24.jpg?height=109&width=1242&top_left_y=691&top_left_x=558"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_599",
        "problem": "DNA 测序时, 先将待测单链 DNA 模板、引物、有关酶、 4 种脱氧核苷三磷酸 （dNTP， N 代表 A、T、C、G，能提供能量并作为 DNA 复制的原料）均加入 4 个试管中, 再分别加入一定量的含放射性同位素标记的 ddNTP (ddGTP、ddATP、ddCTP、 ddTTP)。不同于 dNTP 的是, ddNTP 的五碳糖的 3 位不含羟基。在 DNA 合成时 ddNTP 随机与 dNTP 竞争核苷酸链延长位点, 并终止 DNA 的延伸, 从而形成不同长度的一系列终止处带有放射性同位素标记的 DNA 链。下图是 DNA 测序时得到含放射性标记的子代 DNA 的电泳图谱。下列相关说法错误的是（）\n\n[图1]\nA: 反应体系中的引物为单链, DNA 聚合酶从引物的 3'端连接脱氧核苷酸\nB: ddNTP 终止子链延伸, 其原理可能是其五碳糖的 3 '位不含羟基, 无法连接新的脱氧核苷酸\nC: 图中加入 $\\mathrm{ddCTP}$ 的一组中, 可以形成 3 种长度的子链\nD: 图中子代 DNA 的碱基序列是 5'-GATCCGAAT-3'\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nDNA 测序时, 先将待测单链 DNA 模板、引物、有关酶、 4 种脱氧核苷三磷酸 （dNTP， N 代表 A、T、C、G，能提供能量并作为 DNA 复制的原料）均加入 4 个试管中, 再分别加入一定量的含放射性同位素标记的 ddNTP (ddGTP、ddATP、ddCTP、 ddTTP)。不同于 dNTP 的是, ddNTP 的五碳糖的 3 位不含羟基。在 DNA 合成时 ddNTP 随机与 dNTP 竞争核苷酸链延长位点, 并终止 DNA 的延伸, 从而形成不同长度的一系列终止处带有放射性同位素标记的 DNA 链。下图是 DNA 测序时得到含放射性标记的子代 DNA 的电泳图谱。下列相关说法错误的是（）\n\n[图1]\n\nA: 反应体系中的引物为单链, DNA 聚合酶从引物的 3'端连接脱氧核苷酸\nB: ddNTP 终止子链延伸, 其原理可能是其五碳糖的 3 '位不含羟基, 无法连接新的脱氧核苷酸\nC: 图中加入 $\\mathrm{ddCTP}$ 的一组中, 可以形成 3 种长度的子链\nD: 图中子代 DNA 的碱基序列是 5'-GATCCGAAT-3'\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-61.jpg?height=737&width=808&top_left_y=160&top_left_x=318"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_406",
        "problem": "果蝇翅形有圆形、粗圆形、镰刀形 3 种, 分别由复等位基因(同源染色体的相同位\n\n点上, 可以存在两种以上的等位基因)R、O、 $\\mathrm{S}$ 控制。让纯合的镰刀形翅雌果蝇与圆形翅雄果蝇杂交, $\\mathrm{F}_{1}$ 雌雄个体均为镰刀形翅。取 $\\mathrm{F}_{1}$ 镰刀形翅雄果蝇与纯合的椭圆形翅雌果蝇杂交, $\\mathrm{F}_{2}$ 雌雄个体均为椭圆形翅。不考虑变异和 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体同源区段的情况，下列分析错误的是（ ）\nA: 根据上述实验结果推知: $O$ 对 $\\mathrm{S}$ 为显性, $\\mathrm{S}$ 对 $\\mathrm{R}$ 为显性\nB: 根据上述实验结果，能判断 $\\mathrm{R} 、 \\mathrm{O} 、 \\mathrm{~S}$ 基因位于常染色体上还是 $\\mathrm{X}$ 染色体上\nC: 根据上述实验结果不能判断 $\\mathrm{R} 、 \\mathrm{O} 、 \\mathrm{~S}$ 基因位于常染色体上还是 $\\mathrm{X}$ 染色体上\nD: 欲通过一次杂交鉴定一椭圆形翅雌果蝇的基因型, 最好选择圆形翅的雄果蝇\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇翅形有圆形、粗圆形、镰刀形 3 种, 分别由复等位基因(同源染色体的相同位\n\n点上, 可以存在两种以上的等位基因)R、O、 $\\mathrm{S}$ 控制。让纯合的镰刀形翅雌果蝇与圆形翅雄果蝇杂交, $\\mathrm{F}_{1}$ 雌雄个体均为镰刀形翅。取 $\\mathrm{F}_{1}$ 镰刀形翅雄果蝇与纯合的椭圆形翅雌果蝇杂交, $\\mathrm{F}_{2}$ 雌雄个体均为椭圆形翅。不考虑变异和 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体同源区段的情况，下列分析错误的是（ ）\n\nA: 根据上述实验结果推知: $O$ 对 $\\mathrm{S}$ 为显性, $\\mathrm{S}$ 对 $\\mathrm{R}$ 为显性\nB: 根据上述实验结果，能判断 $\\mathrm{R} 、 \\mathrm{O} 、 \\mathrm{~S}$ 基因位于常染色体上还是 $\\mathrm{X}$ 染色体上\nC: 根据上述实验结果不能判断 $\\mathrm{R} 、 \\mathrm{O} 、 \\mathrm{~S}$ 基因位于常染色体上还是 $\\mathrm{X}$ 染色体上\nD: 欲通过一次杂交鉴定一椭圆形翅雌果蝇的基因型, 最好选择圆形翅的雄果蝇\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_144",
        "problem": "During last 500 million years, a great amount of trait variation has been created in land plants in the course of evolution. However, there are some constraints over the possible combinations of traits which indicate evolutionary limitations and trade-offs. To investigate trait combinations in more than 46,000 extant vascular plants, a Principal component analysis was conducted (Díaz et al. 2016). Principal component analysis (PCA) is a statistical procedure converts a set of observations of possibly correlated variables into a set of values of linearly uncorrelated variables called principal components. Figure 1-a shows distribution of investigated vascular plants (including angiosperms, gymnosperms and pteridophytes) in a space conducted by the PCA analysis. Each trait changes along its corresponding axis in this space. Areas with high density of points indicate functional hotspots. There are two major hotspots corresponding to woody and non-woody plants (LA: leaf area, SM: diaspore (fruit or spore) mass, H: plant height, SSD: stem specific density, LMA: leaf mass per area, $\\mathrm{N}_{\\text {mass: }}$ leaf nitrogen content per mass.)\n\n[figure1]\nA: Woody plants tend to have higher $\\mathrm{N}_{\\text {mass }}$ than non-woody plants\nB: Plants inhabiting temperate grasslands tend to place near points 4 and 5 .\nC: Dominant plants of boreal forests are mostly to locate near point 1 .\nD: Plants inhabiting tropical forests tend to have higher LMA than plants inhabiting temperate forests.\nE: $\\mathrm{N}_{\\text {mass }}$ correlates with leaf lifespan negatively.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nDuring last 500 million years, a great amount of trait variation has been created in land plants in the course of evolution. However, there are some constraints over the possible combinations of traits which indicate evolutionary limitations and trade-offs. To investigate trait combinations in more than 46,000 extant vascular plants, a Principal component analysis was conducted (Díaz et al. 2016). Principal component analysis (PCA) is a statistical procedure converts a set of observations of possibly correlated variables into a set of values of linearly uncorrelated variables called principal components. Figure 1-a shows distribution of investigated vascular plants (including angiosperms, gymnosperms and pteridophytes) in a space conducted by the PCA analysis. Each trait changes along its corresponding axis in this space. Areas with high density of points indicate functional hotspots. There are two major hotspots corresponding to woody and non-woody plants (LA: leaf area, SM: diaspore (fruit or spore) mass, H: plant height, SSD: stem specific density, LMA: leaf mass per area, $\\mathrm{N}_{\\text {mass: }}$ leaf nitrogen content per mass.)\n\n[figure1]\n\nA: Woody plants tend to have higher $\\mathrm{N}_{\\text {mass }}$ than non-woody plants\nB: Plants inhabiting temperate grasslands tend to place near points 4 and 5 .\nC: Dominant plants of boreal forests are mostly to locate near point 1 .\nD: Plants inhabiting tropical forests tend to have higher LMA than plants inhabiting temperate forests.\nE: $\\mathrm{N}_{\\text {mass }}$ correlates with leaf lifespan negatively.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-49.jpg?height=1063&width=1402&top_left_y=939&top_left_x=316"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_557",
        "problem": "我国科学家成功地用 iPS 细胞克隆出了活体小鼠, 部分流程如下图所示, 其中 $\\mathrm{Kdm} 4 \\mathrm{~d}$为组蛋白去甲基化酶, TSA 为组蛋白脱乙酰酶抑制剂。下列说法正确的是（）\n\n[图1]\nA: 组蛋白脱乙酰化和去甲基化有利于重构胚后续的胚胎发育过程\nB: 用电刺激、 $\\mathrm{Ca}^{2+}$ 载体等方法激活重构胚, 使其完成细胞分裂和发育进程\nC: (3)过程中使用有活性的病毒处理的目的是诱导细胞融合\nD: 图示流程运用了重组 DNA、体细胞核移植、肧胎移植等技术\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n我国科学家成功地用 iPS 细胞克隆出了活体小鼠, 部分流程如下图所示, 其中 $\\mathrm{Kdm} 4 \\mathrm{~d}$为组蛋白去甲基化酶, TSA 为组蛋白脱乙酰酶抑制剂。下列说法正确的是（）\n\n[图1]\n\nA: 组蛋白脱乙酰化和去甲基化有利于重构胚后续的胚胎发育过程\nB: 用电刺激、 $\\mathrm{Ca}^{2+}$ 载体等方法激活重构胚, 使其完成细胞分裂和发育进程\nC: (3)过程中使用有活性的病毒处理的目的是诱导细胞融合\nD: 图示流程运用了重组 DNA、体细胞核移植、肧胎移植等技术\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-080.jpg?height=404&width=1468&top_left_y=1363&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1148",
        "problem": "## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAASquid possess hardened beaks used for gripping and ripping apart their prey. These beaks are extremely hard and indigestible. They are commonly found in the stomachs of seabirds and marine mammals that prey on the squid. Typically, these organisms regurgitate the indigestible beaks rather than allowing them to pass through the gut to be excreted with the faeces.\n\n[figure3]\n\nCount the number of \"Extra large plastic fragments\" in Bolus 1 and record your answer.",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n## NEW ZEALAND \\& BEYOND - MARINE PLASTIC POLLUTION, A GLOBAL ISSUE\n\nCanterbury Museum ornithologist Paul Scofield, who does autopsies on 400 muttonbirds caught accidentally by fishing boats every year, has shown that most New Zealand seabirds have plastics in their stomachs. He has also found red Coke bottle tops, cigarette lighters, pieces of fishing buoys and other plastic material in Albatross colonies on Campbell Island, $700 \\mathrm{~km}$ south of Bluff. He has also seen albatross chicks that have died because they had so much plastic in their stomachs there was no room for food.\n\nMarine plastic pollution is a major threat to seabirds and of growing concern worldwide. Seabirds that feed on the surface of the ocean by dipping or scavenging, such as albatross, are at greatest risk. They can mistake pieces of plastic for their normal food of squid, crustaceans such as krill, fish eggs (typically attached to floating pumice and seeds) and fish larvae. Types of plastic ingested by albatross include single-use \"user\" plastic e.g. bottle caps, plastic toys, cigarette lighters, light sticks, industrial pellets known as nurdles, and fishing floats. Seabirds such as the albatrosses also eat fishing line.\n\nAlbatross feed their chicks by regurgitating food into the chick's mouth. Plastics ingested in error by the adults are also fed to their chicks in this way. Albatross chicks regurgitate a bolus of indigestible remains just before they leave the nest to begin their ocean-going adult life. This bolus should contain the indigestible remains of fish (50\\%), squid $(32 \\%)$, crustacea $(5 \\%)$ and stomach oil $(10 \\%)$. In recent years studies have shown albatross bolus' to contain natural indigestible materials, primarily squid beaks, and un-natural indigestible materials such as plastics. If the parents are feeding lots of plastics to the chicks the chicks grow more slowly as they become easily satiated (full feeling). Chicks can become so full of plastic that they are unable to regurgitate a bolus and die. (Information in this section is from: Oikonos, Ecosystem Knowledge. http://www.oikonos.org/projects/oceanstewardship projects.htm and approved for educational use)\n\n[figure1]\n\nA recently dead Laysan Albatross chick with its belly full of plastic.\n\nPhoto: Claire Johnson/NOAA\n\n[figure2]\n\nRinsing 306 pieces of plastic debris from the stomach of the albatross chick.\n\nPhoto: Claire Johnson/NOAA\n\nproblem:\nSquid possess hardened beaks used for gripping and ripping apart their prey. These beaks are extremely hard and indigestible. They are commonly found in the stomachs of seabirds and marine mammals that prey on the squid. Typically, these organisms regurgitate the indigestible beaks rather than allowing them to pass through the gut to be excreted with the faeces.\n\n[figure3]\n\nCount the number of \"Extra large plastic fragments\" in Bolus 1 and record your answer.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=800&width=617&top_left_y=1299&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-34.jpg?height=817&width=1097&top_left_y=1299&top_left_x=845",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-35.jpg?height=343&width=511&top_left_y=251&top_left_x=1087"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1146",
        "problem": "Understanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.The introduced plant species consumed most frequently by the orange-fronted parakeet on Maud Island was?\nA: Sycamore\nB: Makomako\nC: Tree lucerne\nD: Manuka\nE: Mahoe\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nUnderstanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.\n\nproblem:\nThe introduced plant species consumed most frequently by the orange-fronted parakeet on Maud Island was?\n\nA: Sycamore\nB: Makomako\nC: Tree lucerne\nD: Manuka\nE: Mahoe\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1065",
        "problem": "Synanthropic species lives near and benefit from the association with humans and the habitats that humans create around them. Changes in bird community structure and composition along a wilderness to urbanized landscape gradient are provided below.\n\n[figure1]\n\nBased on the information provided in the graph, which of the following is correct?\nA: The graph indicates a increased predation pressure on native bird species in urban ecosystem.\nB: The increase in total bird abundance is indicative of greater food availability in urban ecosystem as compared to wildland.\nC: Habitat fragmentation, forest destruction has resulted in reduced number of species in urban region.\nD: Bird species that cannot exist in unpredictable or resource poor environments are likely to flourish in urban ecosystem.\nE: Dominance of adopted species over resource patches in urban ecosystem can outcompete less fit native species evenness in urban ecosystem.\nF: Urbanization creates a novel ecosystem which attracts and stabilizes a large number of bird species from nearby ecosystems.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nSynanthropic species lives near and benefit from the association with humans and the habitats that humans create around them. Changes in bird community structure and composition along a wilderness to urbanized landscape gradient are provided below.\n\n[figure1]\n\nBased on the information provided in the graph, which of the following is correct?\n\nA: The graph indicates a increased predation pressure on native bird species in urban ecosystem.\nB: The increase in total bird abundance is indicative of greater food availability in urban ecosystem as compared to wildland.\nC: Habitat fragmentation, forest destruction has resulted in reduced number of species in urban region.\nD: Bird species that cannot exist in unpredictable or resource poor environments are likely to flourish in urban ecosystem.\nE: Dominance of adopted species over resource patches in urban ecosystem can outcompete less fit native species evenness in urban ecosystem.\nF: Urbanization creates a novel ecosystem which attracts and stabilizes a large number of bird species from nearby ecosystems.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E, F].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-34.jpg?height=867&width=1632&top_left_y=209&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_367",
        "problem": "研究人员为了研究半保留复制的具体过程, 以 $\\mathrm{T}_{4}$ 噬菌体和大肠杆菌为实验对象进行了实验探究: $20^{\\circ} \\mathrm{C}$ 条件下, 用 $\\mathrm{T}_{4}$ 噬菌体侵染大肠杆菌, 进入 $\\mathrm{T}_{4}$ 噬菌体 DNA 活跃复制期时, 在培养基中添加含 ${ }^{3} \\mathrm{H}$ 标记的胸腺嘧啶脱氧核苷酸, 培养不同时间后, 阻断 DNA 复制, 将 DNA 变性处理为单链后, 离心分离不同长度的 $\\mathrm{T}_{4}$ 噬菌体的 DNA 片段, 检测离心管不同位置的放射性强度, 结果如图所示 (DNA 片段越短, 与离心管顶部距离越近)。下列叙述正确的是（ ）\n\n[图1]\nA: 可在培养基中添加秋水仙素，阻断 DNA 复制\nB: 为了节省实验时间, DNA 不变性处理成单链直接离心也可\nC: 实验结果说明 DNA 复制时子链合成的过程, 短时间内首先合成的是长片段 DNA\nD: 如果实验中抑制 DNA 连接酶的活性, 则大部分放射性会出现在离心管顶部\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究人员为了研究半保留复制的具体过程, 以 $\\mathrm{T}_{4}$ 噬菌体和大肠杆菌为实验对象进行了实验探究: $20^{\\circ} \\mathrm{C}$ 条件下, 用 $\\mathrm{T}_{4}$ 噬菌体侵染大肠杆菌, 进入 $\\mathrm{T}_{4}$ 噬菌体 DNA 活跃复制期时, 在培养基中添加含 ${ }^{3} \\mathrm{H}$ 标记的胸腺嘧啶脱氧核苷酸, 培养不同时间后, 阻断 DNA 复制, 将 DNA 变性处理为单链后, 离心分离不同长度的 $\\mathrm{T}_{4}$ 噬菌体的 DNA 片段, 检测离心管不同位置的放射性强度, 结果如图所示 (DNA 片段越短, 与离心管顶部距离越近)。下列叙述正确的是（ ）\n\n[图1]\n\nA: 可在培养基中添加秋水仙素，阻断 DNA 复制\nB: 为了节省实验时间, DNA 不变性处理成单链直接离心也可\nC: 实验结果说明 DNA 复制时子链合成的过程, 短时间内首先合成的是长片段 DNA\nD: 如果实验中抑制 DNA 连接酶的活性, 则大部分放射性会出现在离心管顶部\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-69.jpg?height=768&width=597&top_left_y=181&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_398",
        "problem": "图 1 为某家系中某种单基因遗传病的遗传系谱图（相关基因用 $\\mathrm{A} / \\mathrm{a}$ 表示）, 图 2 表示对该家系的部分成员中和该遗传病有关的正常基因和异常基因片段的电泳结果,且 12 试卷第 86 页，共 93 页\n号个体是三体患者, 他的三条同源染色体在减数分裂过程中任意两条配对, 另外一条随机进入子细胞中，下列说法错误的是（ ）\n\n[图1]\n\n正常男性\n\n正常女性患病男性\n\n[图2]\n\n图2\nA: 图 2 中表示 $\\mathrm{A}$ 基因的是条带 1\nB: $3 、 4$ 号生一个患病孩子的概率是 $1 / 8$\nC: 若仅考虑这三条染色体, 12 号个体最多可产生 4 种精子\nD: 若不考虑基因突变, 12 号个体多出的一条染色体来自 6 或 7 号\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图 1 为某家系中某种单基因遗传病的遗传系谱图（相关基因用 $\\mathrm{A} / \\mathrm{a}$ 表示）, 图 2 表示对该家系的部分成员中和该遗传病有关的正常基因和异常基因片段的电泳结果,且 12 试卷第 86 页，共 93 页\n号个体是三体患者, 他的三条同源染色体在减数分裂过程中任意两条配对, 另外一条随机进入子细胞中，下列说法错误的是（ ）\n\n[图1]\n\n正常男性\n\n正常女性患病男性\n\n[图2]\n\n图2\n\nA: 图 2 中表示 $\\mathrm{A}$ 基因的是条带 1\nB: $3 、 4$ 号生一个患病孩子的概率是 $1 / 8$\nC: 若仅考虑这三条染色体, 12 号个体最多可产生 4 种精子\nD: 若不考虑基因突变, 12 号个体多出的一条染色体来自 6 或 7 号\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-87.jpg?height=302&width=691&top_left_y=320&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-87.jpg?height=240&width=426&top_left_y=371&top_left_x=1363"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_617",
        "problem": "无尾猫是一种观赏猫。猫的无尾, 有尾是一对相对性状, 按基因的分离定律遗传。为了选育纯种的无尾猫,让无尾猫自交多代, 但发现每一代中总会出现约 $1 / 3$ 的有尾猫, 其余均为无尾猫。由此推断正确的是（）\nA: 猫的有尾性状是由显性基因控制的\nB: 自交后代出现有尾猫是基因突变所致\nC: 自交后代无尾猫中既有杂合子又有纯合子\nD: 无尾猫与有尾猫杂交后代中无尾猫约占 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n无尾猫是一种观赏猫。猫的无尾, 有尾是一对相对性状, 按基因的分离定律遗传。为了选育纯种的无尾猫,让无尾猫自交多代, 但发现每一代中总会出现约 $1 / 3$ 的有尾猫, 其余均为无尾猫。由此推断正确的是（）\n\nA: 猫的有尾性状是由显性基因控制的\nB: 自交后代出现有尾猫是基因突变所致\nC: 自交后代无尾猫中既有杂合子又有纯合子\nD: 无尾猫与有尾猫杂交后代中无尾猫约占 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_731",
        "problem": "人类 $(2 n=46) 14$ 号与 21 号染色体二者的长臂在着丝粒处融合形成 $14 / 21$ 平衡易位染色体，该染色体携带者具有正常的表型，但在产生生殖细胞的过该染色体携带者具有正常的表型, 但在产生生殖细胞的过程中, 其细胞中形成复杂的联会复合物（如图）,在进行减数分裂时, 若该联会复合物的染色体遵循正常的染色体位行为规律 (不考虑染色体互换），下列关于平衡易位的叙述正确的是（ ）\n\n[图1]\nA: 观察平衡易位染色体可选择 MII中期细胞\nB: 男性携带者的初级精母细胞含有 46 条染色体\nC: 女性携带者的卵细胞最多含 24 种形态不同的染色体\nD: 若只考虑图示染色体，女性携带者的卵细胞可能有 6 种类型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人类 $(2 n=46) 14$ 号与 21 号染色体二者的长臂在着丝粒处融合形成 $14 / 21$ 平衡易位染色体，该染色体携带者具有正常的表型，但在产生生殖细胞的过该染色体携带者具有正常的表型, 但在产生生殖细胞的过程中, 其细胞中形成复杂的联会复合物（如图）,在进行减数分裂时, 若该联会复合物的染色体遵循正常的染色体位行为规律 (不考虑染色体互换），下列关于平衡易位的叙述正确的是（ ）\n\n[图1]\n\nA: 观察平衡易位染色体可选择 MII中期细胞\nB: 男性携带者的初级精母细胞含有 46 条染色体\nC: 女性携带者的卵细胞最多含 24 种形态不同的染色体\nD: 若只考虑图示染色体，女性携带者的卵细胞可能有 6 种类型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-32.jpg?height=448&width=780&top_left_y=1312&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1259",
        "problem": "Contractile vacuoles are found in a wide variety of unicellular organisms, but not in the cells of mammals. This is because mammalian cells\nA: have membranes that are not selectively permeable\nB: are generally surrounded by solutions with the same solute potential\nC: have ionic pumps\nD: expend energy to maintain cell volume\nE: are more osmotically resistant to osmotic stress\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nContractile vacuoles are found in a wide variety of unicellular organisms, but not in the cells of mammals. This is because mammalian cells\n\nA: have membranes that are not selectively permeable\nB: are generally surrounded by solutions with the same solute potential\nC: have ionic pumps\nD: expend energy to maintain cell volume\nE: are more osmotically resistant to osmotic stress\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_936",
        "problem": "下列关于“核酸是遗传物质的证据”的相关实验的叙述, 正确的是( )\nA: 噬菌体侵染细菌实验中, 用 ${ }^{32} \\mathrm{P}$ 标记的噬菌体侵染细菌后的子代噬菌体多数具有放射性\nB: 肺炎双球菌活体细菌转化实验中, $\\mathrm{R}$ 型肺炎双球菌转化为 $\\mathrm{S}$ 型菌是基因突变的结果\nC: 肺炎双球菌离体细菌转化实验中, $\\mathrm{S}$ 型菌的 DNA 使 $\\mathrm{R}$ 型菌转化为 $\\mathrm{S}$ 型菌, 说明 DNA 是遗传物质, 蛋白质不是遗传物质\nD: 烟草花叶病毒感染和重建实验中, 用 TMV A 的 RNA 和 TMV B 的蛋白质重建的病毒感染烟草叶片细胞后,可检测到 A 型病毒, 说明 RNA 是 TMV A 的遗传物质\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列关于“核酸是遗传物质的证据”的相关实验的叙述, 正确的是( )\n\nA: 噬菌体侵染细菌实验中, 用 ${ }^{32} \\mathrm{P}$ 标记的噬菌体侵染细菌后的子代噬菌体多数具有放射性\nB: 肺炎双球菌活体细菌转化实验中, $\\mathrm{R}$ 型肺炎双球菌转化为 $\\mathrm{S}$ 型菌是基因突变的结果\nC: 肺炎双球菌离体细菌转化实验中, $\\mathrm{S}$ 型菌的 DNA 使 $\\mathrm{R}$ 型菌转化为 $\\mathrm{S}$ 型菌, 说明 DNA 是遗传物质, 蛋白质不是遗传物质\nD: 烟草花叶病毒感染和重建实验中, 用 TMV A 的 RNA 和 TMV B 的蛋白质重建的病毒感染烟草叶片细胞后,可检测到 A 型病毒, 说明 RNA 是 TMV A 的遗传物质\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_537",
        "problem": "下图表示某家族中两种遗传病的患病情况, 其中 $\\mathrm{I}_{1} 、 \\mathrm{III}_{2}$ 患甲病, $\\mathrm{IV}_{2} 、 \\mathrm{IV}_{3}$ 患乙病。已知甲病在人群中的发病率为 $1 / 100$, 下列相关叙述错误的是（）\n\n[图1]\n\n## 图例\n\n$\\square$ 正常男性\n\n正常女性\n\n$\\square$ 甲遗传病男\n\n甲遗传病女\n\n乙遗传病男\nA: 甲病为常染色体隐性遗传病, 乙病是隐性遗传病\nB: 若乙病致病基因在 XY 染色体的同源区段, $\\mathrm{III}_{3}$ 与 $\\mathrm{III}_{4}$ 再生一个同时患甲、乙病男孩的概率为 $1 / 264$\nC: 若 $\\mathrm{III}_{3}$ 与 $\\mathrm{III}_{4}$ 又生一个女儿为乙病患者, 则 $\\mathrm{III}_{3}$ 与 $\\mathrm{III}_{4}$ 再生另一个孩子患病的概率是 $3 / 11$\nD: 若 $\\mathrm{III}_{4}$ 不携带乙病的致病基因, 且 $\\mathrm{IV}_{3}$ 染色体组成为 $\\mathrm{XXY}$, 则 $\\mathrm{III}_{3}$ 在减数分裂 $\\mathrm{II}$ 后期出错\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图表示某家族中两种遗传病的患病情况, 其中 $\\mathrm{I}_{1} 、 \\mathrm{III}_{2}$ 患甲病, $\\mathrm{IV}_{2} 、 \\mathrm{IV}_{3}$ 患乙病。已知甲病在人群中的发病率为 $1 / 100$, 下列相关叙述错误的是（）\n\n[图1]\n\n## 图例\n\n$\\square$ 正常男性\n\n正常女性\n\n$\\square$ 甲遗传病男\n\n甲遗传病女\n\n乙遗传病男\n\nA: 甲病为常染色体隐性遗传病, 乙病是隐性遗传病\nB: 若乙病致病基因在 XY 染色体的同源区段, $\\mathrm{III}_{3}$ 与 $\\mathrm{III}_{4}$ 再生一个同时患甲、乙病男孩的概率为 $1 / 264$\nC: 若 $\\mathrm{III}_{3}$ 与 $\\mathrm{III}_{4}$ 又生一个女儿为乙病患者, 则 $\\mathrm{III}_{3}$ 与 $\\mathrm{III}_{4}$ 再生另一个孩子患病的概率是 $3 / 11$\nD: 若 $\\mathrm{III}_{4}$ 不携带乙病的致病基因, 且 $\\mathrm{IV}_{3}$ 染色体组成为 $\\mathrm{XXY}$, 则 $\\mathrm{III}_{3}$ 在减数分裂 $\\mathrm{II}$ 后期出错\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-36.jpg?height=580&width=679&top_left_y=1943&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1167",
        "problem": "Two similar plots of cabbage plants were used in an investigation to determine the effectiveness of an insecticide. One plot was sprayed with the insecticide on 26th December. The second plot was left unsprayed as a control. The graph shows the number of caterpillars that were found on the plots during the following January, February and March.\n\n[figure1]\n\nWhich one of the following statements is the best explanation of these data?\nA: The caterpillars on the cabbage sprayed with insecticide were resistant to the insecticide.\nB: In the sprayed plot the insecticide killed many of the natural predators of the caterpillars.\nC: The effect of the insecticide had worn off by the time the caterpillars had hatched from their eggs.\nD: The insecticide was applied too late in the life cycle and had the effect of delaying pupation.\nE: The insecticide had no effect on the caterpillars.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTwo similar plots of cabbage plants were used in an investigation to determine the effectiveness of an insecticide. One plot was sprayed with the insecticide on 26th December. The second plot was left unsprayed as a control. The graph shows the number of caterpillars that were found on the plots during the following January, February and March.\n\n[figure1]\n\nWhich one of the following statements is the best explanation of these data?\n\nA: The caterpillars on the cabbage sprayed with insecticide were resistant to the insecticide.\nB: In the sprayed plot the insecticide killed many of the natural predators of the caterpillars.\nC: The effect of the insecticide had worn off by the time the caterpillars had hatched from their eggs.\nD: The insecticide was applied too late in the life cycle and had the effect of delaying pupation.\nE: The insecticide had no effect on the caterpillars.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-13.jpg?height=472&width=899&top_left_y=415&top_left_x=116"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1369",
        "problem": "The following graph shows relative oxygen concentrations in an airtight container containing one spinach plant exposed to varying light intensities. The y axis indicates oxygen concentration and the $x$ axis indicates increasing light intensity from left to right. The units of both axes are arbitrary; they do not matter.\n\n[figure1]\n\nConsider the points on the graph labelled $\\mathrm{W}, \\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$. Each point has been paired up with a description of what is happening at that point. Select the INCORRECT pairing.\nA: At point $\\mathrm{W}$, the rate of $\\mathrm{CO}_{2}$ consumption balances the rate of $\\mathrm{CO}_{2}$ production.\nB: At point $X$, the rate of photosynthesis exceeds the rate of respiration.\nC: At point $\\mathrm{Y}, \\mathrm{CO}_{2}$ concentration is a possible limiting factor.\nD: At point $\\mathrm{Z}, \\mathrm{O}_{2}$ concentration is decreasing and therefore the rate of respiration is higher than the rate of photosynthesis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following graph shows relative oxygen concentrations in an airtight container containing one spinach plant exposed to varying light intensities. The y axis indicates oxygen concentration and the $x$ axis indicates increasing light intensity from left to right. The units of both axes are arbitrary; they do not matter.\n\n[figure1]\n\nConsider the points on the graph labelled $\\mathrm{W}, \\mathrm{X}, \\mathrm{Y}$ and $\\mathrm{Z}$. Each point has been paired up with a description of what is happening at that point. Select the INCORRECT pairing.\n\nA: At point $\\mathrm{W}$, the rate of $\\mathrm{CO}_{2}$ consumption balances the rate of $\\mathrm{CO}_{2}$ production.\nB: At point $X$, the rate of photosynthesis exceeds the rate of respiration.\nC: At point $\\mathrm{Y}, \\mathrm{CO}_{2}$ concentration is a possible limiting factor.\nD: At point $\\mathrm{Z}, \\mathrm{O}_{2}$ concentration is decreasing and therefore the rate of respiration is higher than the rate of photosynthesis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-18.jpg?height=765&width=1354&top_left_y=563&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_382",
        "problem": "下图为某家族的遗传系谱图, 其中有些家庭成员血型已经清楚 (见标注)。已知该地区人群中 $\\mathrm{I}^{\\mathrm{A}}$ 基因频率为 $0.1, \\mathrm{I}^{\\mathrm{B}}$ 基因频率为 $0.1, \\mathrm{i}$ 基因频率为 0.8 。甲遗传病在男性群体中患病率为 20\\%,4 号个体不携带甲病致病基因, 甲病基因不位于 X、Y 同源区段上。下列叙述正确的是（）\n\n[图1]\nA: 甲病为伴 X 染色体隐性遗传病, 6 号个体的次级精母细胞含有 2 个甲病致病基因\nB: 10 号个体为 $\\mathrm{O}$ 型血且为甲病患者的概率为 $1 / 17$\nC: $\\mathrm{AB}$ 血型的个体红细胞上既有 $\\mathrm{A}$ 抗原又有 $\\mathrm{B}$ 抗原, 人群中 $\\mathrm{AB}$ 血型个体占 $1 \\%$\nD: 11 号个体为 A 型血且为甲病患者的概率为 $1 / 25$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为某家族的遗传系谱图, 其中有些家庭成员血型已经清楚 (见标注)。已知该地区人群中 $\\mathrm{I}^{\\mathrm{A}}$ 基因频率为 $0.1, \\mathrm{I}^{\\mathrm{B}}$ 基因频率为 $0.1, \\mathrm{i}$ 基因频率为 0.8 。甲遗传病在男性群体中患病率为 20\\%,4 号个体不携带甲病致病基因, 甲病基因不位于 X、Y 同源区段上。下列叙述正确的是（）\n\n[图1]\n\nA: 甲病为伴 X 染色体隐性遗传病, 6 号个体的次级精母细胞含有 2 个甲病致病基因\nB: 10 号个体为 $\\mathrm{O}$ 型血且为甲病患者的概率为 $1 / 17$\nC: $\\mathrm{AB}$ 血型的个体红细胞上既有 $\\mathrm{A}$ 抗原又有 $\\mathrm{B}$ 抗原, 人群中 $\\mathrm{AB}$ 血型个体占 $1 \\%$\nD: 11 号个体为 A 型血且为甲病患者的概率为 $1 / 25$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-033.jpg?height=409&width=1445&top_left_y=498&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_132",
        "problem": "Ethanol inhibits microbial growth. Nevertheless, some strains of the yeast Saccharomyces cerevisiae can adapt to high concentrations of ethanol. Many studies have documented the alteration of cellular lipid composition in response to ethanol exposure.\n\nIn this investigation, we systematically altered the fatty acid composition in $S$. cerevisiae by knocking out $O L E 1$ gene coding for integral membrane desaturase, responsible for the formation of mono-unsaturated palmitoleic acid $\\left(\\Delta^{9}-\\mathrm{C}_{16: 1}\\right)$ and oleic acid $\\left(\\Delta^{9}-\\mathrm{C}_{18: 1}\\right)$. The knockout strain was then: (1) reconstituted with $O L E 1$ gene by transformation with YEpOLE1 plasmid; (2) transformed with YEpOLE1- $\\Delta^{9} \\mathrm{~Hz}$, YEpOLE1- $\\Delta^{9} T n$, YEpOLE1$\\triangle^{\\prime \\prime} \\mathrm{Hz}$ and YEpOLE1- $\\Delta^{\\prime \\prime} \\mathrm{Tn}$ plasmid containing the open reading frame of OLE1 ligated to $\\Delta^{9}$ or $\\Delta^{11}$ desaturases of two lepidopteran insect (moths) Helicoverpa zea $(\\mathrm{Hz})$ or Trichoplusia ni (Tni) via a four codon linker. Fatty acid component and growth curve of each mutant were investigated and shown in table and figure below.\n\nTable Q.6. Composition of major fatty acids of S. cerevisiae transformants at mid-log phase\n\n| Transformant | Fatty acid content (\\%) |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | Saturated |  | Monounsaturated |  |\n|  | $C_{16: 0}$ | $C_{18: 0}$ | $C_{16: 1}$ | $C_{18: 1}$ |\n| $\\Delta^{9}$ |  |  |  |  |\n| OLE1 | $45.5 \\pm 2.2$ | $4.7 \\pm 2.4$ | $34.9 \\pm 0.8$ | $14.9 \\pm 1.0$ |\n| $O L E 1-\\Delta^{9} H z$ | $45.5 \\pm 5.5$ | $7.9 \\pm 2.2$ | $31.7 \\pm 5.6$ | $11.0 \\pm 2.0$ |\n| $O L E 1-\\Delta^{9} T n$ | $46.9 \\pm 4.0$ | $8.6 \\pm 3.9$ | $12.8 \\pm 1.9$ | $31.7 \\pm 5.8$ |\n| $\\Delta^{\\prime \\prime}$ |  |  |  |  |\n| $O L E 1-\\Delta^{11} H z$ | $45.6 \\pm 3.6$ | $11.9 \\pm 2.8$ | $42.6 \\pm 6.3$ | 0 |\n| $O L E 1-\\Delta^{\\prime \\prime} T n$ | $49.7 \\pm 4.8$ | $12.5 \\pm 0.1$ | $41.8 \\pm 11.8$ | $11.2 \\pm 1.5$ |\n\n[figure1]\n\nFig.Q6. Growth curves of $S$. cerevisiae strains transformed with plasmid containing OLEI (), OLE1 $-\\Delta^{9} \\mathrm{~Hz}(\\bullet), O L E 1-\\Delta^{\\prime \\prime} \\mathrm{Hz}(\\boxtimes), O L E 1-\\Delta^{9} \\mathrm{Tn}$ () and OLE1- $\\Delta^{\\prime \\prime} \\mathrm{Tn}$ () in YPD medium $(A)$ and in YPD medium containing $5 \\%$ ethanol $(B)$\nA: The lag phase of transformant $O L E 1$ in YPD medium is shorter than those of OLE1 $-\\Delta^{9} \\mathrm{~Hz}, O L E 1-\\Delta^{9} \\mathrm{Tn}, O L E 1-\\Delta^{\\prime \\prime} \\mathrm{Hz}$ and $O L E 1-\\Delta^{\\prime \\prime} \\mathrm{Tn}$ due to the presence of native desaturase in yeast cells.\nB: OLEl was expressed well in all transformants.\nC: The content of mono-unsaturated fatty acids is a good indicator of the ethanol tolerance in $S$. cerevisiae.\nD: Higher ratio of $\\Delta^{9}-C_{18: 1}$ to $\\Delta^{9}-C_{16: I}$ causes higher ethanol tolerance in $S$. cerevisiae Answer key:\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nEthanol inhibits microbial growth. Nevertheless, some strains of the yeast Saccharomyces cerevisiae can adapt to high concentrations of ethanol. Many studies have documented the alteration of cellular lipid composition in response to ethanol exposure.\n\nIn this investigation, we systematically altered the fatty acid composition in $S$. cerevisiae by knocking out $O L E 1$ gene coding for integral membrane desaturase, responsible for the formation of mono-unsaturated palmitoleic acid $\\left(\\Delta^{9}-\\mathrm{C}_{16: 1}\\right)$ and oleic acid $\\left(\\Delta^{9}-\\mathrm{C}_{18: 1}\\right)$. The knockout strain was then: (1) reconstituted with $O L E 1$ gene by transformation with YEpOLE1 plasmid; (2) transformed with YEpOLE1- $\\Delta^{9} \\mathrm{~Hz}$, YEpOLE1- $\\Delta^{9} T n$, YEpOLE1$\\triangle^{\\prime \\prime} \\mathrm{Hz}$ and YEpOLE1- $\\Delta^{\\prime \\prime} \\mathrm{Tn}$ plasmid containing the open reading frame of OLE1 ligated to $\\Delta^{9}$ or $\\Delta^{11}$ desaturases of two lepidopteran insect (moths) Helicoverpa zea $(\\mathrm{Hz})$ or Trichoplusia ni (Tni) via a four codon linker. Fatty acid component and growth curve of each mutant were investigated and shown in table and figure below.\n\nTable Q.6. Composition of major fatty acids of S. cerevisiae transformants at mid-log phase\n\n| Transformant | Fatty acid content (\\%) |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | Saturated |  | Monounsaturated |  |\n|  | $C_{16: 0}$ | $C_{18: 0}$ | $C_{16: 1}$ | $C_{18: 1}$ |\n| $\\Delta^{9}$ |  |  |  |  |\n| OLE1 | $45.5 \\pm 2.2$ | $4.7 \\pm 2.4$ | $34.9 \\pm 0.8$ | $14.9 \\pm 1.0$ |\n| $O L E 1-\\Delta^{9} H z$ | $45.5 \\pm 5.5$ | $7.9 \\pm 2.2$ | $31.7 \\pm 5.6$ | $11.0 \\pm 2.0$ |\n| $O L E 1-\\Delta^{9} T n$ | $46.9 \\pm 4.0$ | $8.6 \\pm 3.9$ | $12.8 \\pm 1.9$ | $31.7 \\pm 5.8$ |\n| $\\Delta^{\\prime \\prime}$ |  |  |  |  |\n| $O L E 1-\\Delta^{11} H z$ | $45.6 \\pm 3.6$ | $11.9 \\pm 2.8$ | $42.6 \\pm 6.3$ | 0 |\n| $O L E 1-\\Delta^{\\prime \\prime} T n$ | $49.7 \\pm 4.8$ | $12.5 \\pm 0.1$ | $41.8 \\pm 11.8$ | $11.2 \\pm 1.5$ |\n\n[figure1]\n\nFig.Q6. Growth curves of $S$. cerevisiae strains transformed with plasmid containing OLEI (), OLE1 $-\\Delta^{9} \\mathrm{~Hz}(\\bullet), O L E 1-\\Delta^{\\prime \\prime} \\mathrm{Hz}(\\boxtimes), O L E 1-\\Delta^{9} \\mathrm{Tn}$ () and OLE1- $\\Delta^{\\prime \\prime} \\mathrm{Tn}$ () in YPD medium $(A)$ and in YPD medium containing $5 \\%$ ethanol $(B)$\n\nA: The lag phase of transformant $O L E 1$ in YPD medium is shorter than those of OLE1 $-\\Delta^{9} \\mathrm{~Hz}, O L E 1-\\Delta^{9} \\mathrm{Tn}, O L E 1-\\Delta^{\\prime \\prime} \\mathrm{Hz}$ and $O L E 1-\\Delta^{\\prime \\prime} \\mathrm{Tn}$ due to the presence of native desaturase in yeast cells.\nB: OLEl was expressed well in all transformants.\nC: The content of mono-unsaturated fatty acids is a good indicator of the ethanol tolerance in $S$. cerevisiae.\nD: Higher ratio of $\\Delta^{9}-C_{18: 1}$ to $\\Delta^{9}-C_{16: I}$ causes higher ethanol tolerance in $S$. cerevisiae Answer key:\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-016.jpg?height=740&width=1430&top_left_y=336&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_311",
        "problem": "大鼠的毛色由独立遗传的两对等位基因控制。用黄色大鼠与黑色大鼠进行杂交实验, 结果如图。据图判断,下列叙述正确的是 ( )\n\n[图1]\nA: 黄色为显性性状，黑色为隐性性状\nB: $F_{1}$ 与黄色亲本杂交, 后代有两种表现型\nC: $F_{1}$ 和 $F_{2}$ 中灰色大鼠均为杂合体\nD: $F_{2}$ 黑色大鼠与米色大鼠杂交, 其后代中出现米色大鼠的概率为 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n大鼠的毛色由独立遗传的两对等位基因控制。用黄色大鼠与黑色大鼠进行杂交实验, 结果如图。据图判断,下列叙述正确的是 ( )\n\n[图1]\n\nA: 黄色为显性性状，黑色为隐性性状\nB: $F_{1}$ 与黄色亲本杂交, 后代有两种表现型\nC: $F_{1}$ 和 $F_{2}$ 中灰色大鼠均为杂合体\nD: $F_{2}$ 黑色大鼠与米色大鼠杂交, 其后代中出现米色大鼠的概率为 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_ea324bee4b9236ad5b87g-06.jpg?height=292&width=506&top_left_y=665&top_left_x=181"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_55",
        "problem": "Understanding how plant species richness affects community biomass production is important for biodiversity and ecosystem conservation. In a grassland, scientists created 72 experimental plots ( $1 \\mathrm{~m}^{2}$ each) with different numbers of vascular plant species (from 1 to 10 species), with species combinations assembled randomly. Both local light and soil conditions were similar among the plots before establishing vegetation. After three years of this experiment, they harvested aboveground vegetation to measure aboveground biomass in each plot. The figure shows the relationship between species richness (number of species) and the dry weight of aboveground biomass $\\left(\\mathrm{kg} \\mathrm{m}^{-2}\\right)$ of plant communities in each plot. The line indicates the linear relationship obtained from the least square regression.\n\n[figure1]\nA: Niche difference among species is one reason for producing a positive association between species richness and aboveground biomass. 199\nB: The plot showing the largest aboveground biomass also has the highest species richness.\nC: On average, increasing aboveground biomass of $0.1 \\mathrm{~kg} \\mathrm{~m}^{-2}$ in a plot requires an additional eight species.\nD: The greater chance of including more productive species in species-rich plots is one reason for producing the positive association between species richness and aboveground biomass.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nUnderstanding how plant species richness affects community biomass production is important for biodiversity and ecosystem conservation. In a grassland, scientists created 72 experimental plots ( $1 \\mathrm{~m}^{2}$ each) with different numbers of vascular plant species (from 1 to 10 species), with species combinations assembled randomly. Both local light and soil conditions were similar among the plots before establishing vegetation. After three years of this experiment, they harvested aboveground vegetation to measure aboveground biomass in each plot. The figure shows the relationship between species richness (number of species) and the dry weight of aboveground biomass $\\left(\\mathrm{kg} \\mathrm{m}^{-2}\\right)$ of plant communities in each plot. The line indicates the linear relationship obtained from the least square regression.\n\n[figure1]\n\nA: Niche difference among species is one reason for producing a positive association between species richness and aboveground biomass. 199\nB: The plot showing the largest aboveground biomass also has the highest species richness.\nC: On average, increasing aboveground biomass of $0.1 \\mathrm{~kg} \\mathrm{~m}^{-2}$ in a plot requires an additional eight species.\nD: The greater chance of including more productive species in species-rich plots is one reason for producing the positive association between species richness and aboveground biomass.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-65.jpg?height=865&width=922&top_left_y=1072&top_left_x=521"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_881",
        "problem": "某家族有甲、乙两种单基因遗传病, 其中一种病的致病基因位于 $\\mathrm{X}$ 染色体上。控制甲病的基因用 $\\mathrm{A} / \\mathrm{a}$ 表示, 某遗传标记与 $\\mathrm{A} / \\mathrm{a}$ 基因紧密连锁，可特异性标记并追踪 $\\mathrm{A} / \\mathrm{a}$基因在家系中的遗传。图 1 是该家族的遗传系谱图, 图 2 是家族中部分个体 DNA 标记\n成分（N1、N2）的体外复制产物电泳结果图，不考虑基因突变和 $X 、 Y$ 同源区段情况。下列说法正确的是（ ）\n\n[图1]\n\n图1图2\nA: 甲病为常染色体隐性遗传病, 乙病与血友病遗传方式相同\nB: I1 和II3 基因型一定相同, I3 与III1 基因可能相同\nC: $\\mathrm{N} 1$ 和 $\\mathrm{A}$ 连锁, $\\mathrm{N} 2$ 和 $\\mathrm{a}$ 连锁\nD: II2 和II 3 再生一个孩子患甲病的概率为 $1 / 6$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某家族有甲、乙两种单基因遗传病, 其中一种病的致病基因位于 $\\mathrm{X}$ 染色体上。控制甲病的基因用 $\\mathrm{A} / \\mathrm{a}$ 表示, 某遗传标记与 $\\mathrm{A} / \\mathrm{a}$ 基因紧密连锁，可特异性标记并追踪 $\\mathrm{A} / \\mathrm{a}$基因在家系中的遗传。图 1 是该家族的遗传系谱图, 图 2 是家族中部分个体 DNA 标记\n成分（N1、N2）的体外复制产物电泳结果图，不考虑基因突变和 $X 、 Y$ 同源区段情况。下列说法正确的是（ ）\n\n[图1]\n\n图1图2\n\nA: 甲病为常染色体隐性遗传病, 乙病与血友病遗传方式相同\nB: I1 和II3 基因型一定相同, I3 与III1 基因可能相同\nC: $\\mathrm{N} 1$ 和 $\\mathrm{A}$ 连锁, $\\mathrm{N} 2$ 和 $\\mathrm{a}$ 连锁\nD: II2 和II 3 再生一个孩子患甲病的概率为 $1 / 6$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-066.jpg?height=283&width=1239&top_left_y=361&top_left_x=383"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1355",
        "problem": "n an experiment to measure the speed of conduction of a nerve impulse along a giant axon, the distance between the stimulating and recording electrodes was varied and the delay between stimulus and response was recorded for each distance. The results are shown in the graph below.\n\n[figure1]\n\nFrom these results the mean speed of conduction was found to be:\nA: $2.0 \\mathrm{~cm} \\mathrm{~s}^{-1}$\nB: $2.5 \\mathrm{~cm} \\mathrm{~s}^{-1}$\nC: $0.5 \\mathrm{~m} \\mathrm{~s}^{-1}$\nD: $4.0 \\mathrm{~m} \\mathrm{~s}^{-1}$\nE: $6.0 \\mathrm{~m} \\mathrm{~s}^{-1}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nn an experiment to measure the speed of conduction of a nerve impulse along a giant axon, the distance between the stimulating and recording electrodes was varied and the delay between stimulus and response was recorded for each distance. The results are shown in the graph below.\n\n[figure1]\n\nFrom these results the mean speed of conduction was found to be:\n\nA: $2.0 \\mathrm{~cm} \\mathrm{~s}^{-1}$\nB: $2.5 \\mathrm{~cm} \\mathrm{~s}^{-1}$\nC: $0.5 \\mathrm{~m} \\mathrm{~s}^{-1}$\nD: $4.0 \\mathrm{~m} \\mathrm{~s}^{-1}$\nE: $6.0 \\mathrm{~m} \\mathrm{~s}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-03.jpg?height=726&width=1202&top_left_y=585&top_left_x=427"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1283",
        "problem": "The phloem and xylem of an experimental shoot were carefully separated along part of their lengths, and a cylinder of waxed paper was placed between them. A control shoot was prepared using non-waxed paper. The base of each shoot was then placed in a solution containing a radioactive isotope of potassium. After five hours the radioactivity of different shoot regions was determined. The figures in the diagram give the location and concentration of radioactive potassium in parts per million.\n\n[figure1]\n\nThe best interpretation of these results is that the transport of potassium ions:\nA: Is upward through the phloem.\nB: Is prevented by the insertion of paper cylinders around the xylem.\nC: Occurs longitudinally through both the xylem and the phloem.\nD: Occurs longitudinally through the phloem only.\nE: Occurs laterally from the xylem to the phloem.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe phloem and xylem of an experimental shoot were carefully separated along part of their lengths, and a cylinder of waxed paper was placed between them. A control shoot was prepared using non-waxed paper. The base of each shoot was then placed in a solution containing a radioactive isotope of potassium. After five hours the radioactivity of different shoot regions was determined. The figures in the diagram give the location and concentration of radioactive potassium in parts per million.\n\n[figure1]\n\nThe best interpretation of these results is that the transport of potassium ions:\n\nA: Is upward through the phloem.\nB: Is prevented by the insertion of paper cylinders around the xylem.\nC: Occurs longitudinally through both the xylem and the phloem.\nD: Occurs longitudinally through the phloem only.\nE: Occurs laterally from the xylem to the phloem.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-12.jpg?height=714&width=874&top_left_y=1613&top_left_x=591"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_774",
        "problem": "下图为某家族 $\\mathrm{ABO}$ 血型与某遗传疾病的遗传系谱图, 其中有些家庭的成员血型已经清楚, (见标注)。已知该地区人群中 $\\mathrm{I}^{\\mathrm{A}}$ 基因频率为 $0.1, \\mathrm{I}^{\\mathrm{B}}$ 基因频率为 $0.1, \\mathrm{i}$ 基因频率为 0.8 ; 人群中患病的频率为 $1 / 10000$ 。已知控制血型与该病的基因是独立分配的。据\n图分析, 下列叙述错误的是\n\n[图1]\nA: 该遗传病为常染色体隐性遗传病\nB: 9 号为 $\\mathrm{AB}$ 型的几率为 $3 / 40$\nC: 3 号为 $\\mathrm{O}$ 型血的概率为 $4 / 5$\nD: 3 号与 4 号再生一个孩子是 $\\mathrm{AB}$ 型同时患病的几率是 1/7272\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为某家族 $\\mathrm{ABO}$ 血型与某遗传疾病的遗传系谱图, 其中有些家庭的成员血型已经清楚, (见标注)。已知该地区人群中 $\\mathrm{I}^{\\mathrm{A}}$ 基因频率为 $0.1, \\mathrm{I}^{\\mathrm{B}}$ 基因频率为 $0.1, \\mathrm{i}$ 基因频率为 0.8 ; 人群中患病的频率为 $1 / 10000$ 。已知控制血型与该病的基因是独立分配的。据\n图分析, 下列叙述错误的是\n\n[图1]\n\nA: 该遗传病为常染色体隐性遗传病\nB: 9 号为 $\\mathrm{AB}$ 型的几率为 $3 / 40$\nC: 3 号为 $\\mathrm{O}$ 型血的概率为 $4 / 5$\nD: 3 号与 4 号再生一个孩子是 $\\mathrm{AB}$ 型同时患病的几率是 1/7272\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-035.jpg?height=349&width=948&top_left_y=248&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1438",
        "problem": "Which of the following is found in RNA but not in DNA?\nA: Thymine\nB: Guanine\nC: Uracil\nD: Phosphate group\nE: Cytosine\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is found in RNA but not in DNA?\n\nA: Thymine\nB: Guanine\nC: Uracil\nD: Phosphate group\nE: Cytosine\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_844",
        "problem": "某二倍体植物的花色由三对独立遗传的等位基因（A/a、B/b、 $\\mathrm{D} / \\mathrm{d}$ ）控制, 体细胞中 $\\mathrm{d}$ 基因数多于 $\\mathrm{D}$ 基因时, $\\mathrm{D}$ 基因不能表达, 其花色遗传如图 1 所示。为了确定 aaBbDdd\n植株属于图 2 中的哪一种突变体, 让该突变体与基因型为 aaBBDD 的植株杂交, 观察并统计子代的表现型与比例（各种配子正常存活）。下列叙述错误的是（ ）\n[图1]\n\n图1\n\n[图2]\n\n(1)\n\n(基因型为 $a a B b D d d$ 种可能的突变体类型)图2\nA: 在没有突变的情况下, 橙花性状的基因型有 4 种\nB: 突变体 1 可以产生 8 种不同基因型的配子\nC: 若子代中黄色:橙色 $=1: 4$, 则其为突变体(2) D．等位基因 $\\mathrm{A}$ 和 $\\mathrm{a}$ 的本质区别是碱基排列顺序不同\nD: 等位基因 $\\mathrm{A}$ 和 $\\mathrm{a}$ 的本质区别是碱基排列顺序不同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二倍体植物的花色由三对独立遗传的等位基因（A/a、B/b、 $\\mathrm{D} / \\mathrm{d}$ ）控制, 体细胞中 $\\mathrm{d}$ 基因数多于 $\\mathrm{D}$ 基因时, $\\mathrm{D}$ 基因不能表达, 其花色遗传如图 1 所示。为了确定 aaBbDdd\n植株属于图 2 中的哪一种突变体, 让该突变体与基因型为 aaBBDD 的植株杂交, 观察并统计子代的表现型与比例（各种配子正常存活）。下列叙述错误的是（ ）\n[图1]\n\n图1\n\n[图2]\n\n(1)\n\n(基因型为 $a a B b D d d$ 种可能的突变体类型)图2\n\nA: 在没有突变的情况下, 橙花性状的基因型有 4 种\nB: 突变体 1 可以产生 8 种不同基因型的配子\nC: 若子代中黄色:橙色 $=1: 4$, 则其为突变体(2) D．等位基因 $\\mathrm{A}$ 和 $\\mathrm{a}$ 的本质区别是碱基排列顺序不同\nD: 等位基因 $\\mathrm{A}$ 和 $\\mathrm{a}$ 的本质区别是碱基排列顺序不同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-006.jpg?height=296&width=630&top_left_y=370&top_left_x=352",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-006.jpg?height=220&width=214&top_left_y=347&top_left_x=1092"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_63",
        "problem": "MicroRNAs are short non-coding RNAs expressed in various tissues and cell types that suppress the expression of target genes. MicroRNAs involved in cancer are called oncomiRs. Inhibition of oncomiRs using antisense oligomers (that is, antimiRs) is an evolving therapeutic strategy. However, the in vivo efficacy of current antimiR technologies is hindered by physiological and cellular barriers for delivery into targeted cells.\n\nA novel antimiR delivery platform that specifically targets tumour microenvironment makes use of synthetic molecules called peptide nucleic acid (PNA) antimiRs attached to a peptide called pHLIP. PNA antimiRs are antimiRs whose nucleotides are connected by peptide bonds instead of the normal phosphodiester bonds. The structure of pHLIP is $\\mathrm{pH}$ dependent. At low $\\mathrm{pH}$, a transmembrane structure is induced in pHLIP that facilitates transport of attached PNA into tumour cells (Fig. 1). pHLIP mediated transport of antimiR-155 effectively inhibited the miR-155 oncomiR in cultured cells (Fig. 2).\n\n## a\n\n[figure1]\n\nSubcutaneous tumour\n\n[figure2]\n\nFree antimiR-155\n\nFig. 1: Targeting miR-155 in a mouse lymphoma model using PNA- antimiR-155-pHLIP a: Distribution of pHLIP (yellow and red zones) $36 \\mathrm{~h}$ after injection into tail. b: Schematic presentation of pHLIP-mediated PNA antimiR delivery.\na\n\n[figure3]\n\nb\n\n[figure4]\n\nFig. 2: pH dependent transport and activity of PNA- antimiR-155-pHLIP. a: Microscopy images of A549 cells incubated with labelled PNA- antimiR-155-pHLIP (red fluorescence) at two pHs; the label for the nucleus is blue. b: Effects of pHLIP-antimiR-155 on cell viability at two different $\\mathrm{pHs}$.\nA: Based on the data, you can conclude the liver does not function in clearing the pHLIP-antimiRs from the body.\nB: At pH less than 7, pHLIP inserts into the lipid bilayer and thus facilitates delivery of attached antimiR-155.\nC: The acidic microenvironment of tumours is responsible for intracellular release of antimiR-155.\nD: The antimiR cargo will be trapped within endosomes.\nE: Transition from random coil to helix of pHLIP enhances cell death.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nMicroRNAs are short non-coding RNAs expressed in various tissues and cell types that suppress the expression of target genes. MicroRNAs involved in cancer are called oncomiRs. Inhibition of oncomiRs using antisense oligomers (that is, antimiRs) is an evolving therapeutic strategy. However, the in vivo efficacy of current antimiR technologies is hindered by physiological and cellular barriers for delivery into targeted cells.\n\nA novel antimiR delivery platform that specifically targets tumour microenvironment makes use of synthetic molecules called peptide nucleic acid (PNA) antimiRs attached to a peptide called pHLIP. PNA antimiRs are antimiRs whose nucleotides are connected by peptide bonds instead of the normal phosphodiester bonds. The structure of pHLIP is $\\mathrm{pH}$ dependent. At low $\\mathrm{pH}$, a transmembrane structure is induced in pHLIP that facilitates transport of attached PNA into tumour cells (Fig. 1). pHLIP mediated transport of antimiR-155 effectively inhibited the miR-155 oncomiR in cultured cells (Fig. 2).\n\n## a\n\n[figure1]\n\nSubcutaneous tumour\n\n[figure2]\n\nFree antimiR-155\n\nFig. 1: Targeting miR-155 in a mouse lymphoma model using PNA- antimiR-155-pHLIP a: Distribution of pHLIP (yellow and red zones) $36 \\mathrm{~h}$ after injection into tail. b: Schematic presentation of pHLIP-mediated PNA antimiR delivery.\na\n\n[figure3]\n\nb\n\n[figure4]\n\nFig. 2: pH dependent transport and activity of PNA- antimiR-155-pHLIP. a: Microscopy images of A549 cells incubated with labelled PNA- antimiR-155-pHLIP (red fluorescence) at two pHs; the label for the nucleus is blue. b: Effects of pHLIP-antimiR-155 on cell viability at two different $\\mathrm{pHs}$.\n\nA: Based on the data, you can conclude the liver does not function in clearing the pHLIP-antimiRs from the body.\nB: At pH less than 7, pHLIP inserts into the lipid bilayer and thus facilitates delivery of attached antimiR-155.\nC: The acidic microenvironment of tumours is responsible for intracellular release of antimiR-155.\nD: The antimiR cargo will be trapped within endosomes.\nE: Transition from random coil to helix of pHLIP enhances cell death.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-08.jpg?height=924&width=624&top_left_y=1009&top_left_x=264",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-08.jpg?height=913&width=797&top_left_y=954&top_left_x=989",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-09.jpg?height=442&width=414&top_left_y=330&top_left_x=276",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-09.jpg?height=674&width=671&top_left_y=291&top_left_x=1132"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1280",
        "problem": "Three views of the same cube are shown below. Which symbol is opposite the $X$ ?\n[figure1]\n\nAnswers\n\n[figure2]\n\nhttp://www.psychometric-success.com/aptitude-tests/spatial-ability-tests-cubes.htm\nA: A\nB: B\nC: C\nD: D\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThree views of the same cube are shown below. Which symbol is opposite the $X$ ?\n[figure1]\n\nAnswers\n\n[figure2]\n\nhttp://www.psychometric-success.com/aptitude-tests/spatial-ability-tests-cubes.htm\n\nA: A\nB: B\nC: C\nD: D\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-10.jpg?height=286&width=822&top_left_y=2124&top_left_x=134",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-10.jpg?height=168&width=733&top_left_y=2229&top_left_x=1164"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1260",
        "problem": "The genetic relatedness between various sets of relatives is shown in the table below. For relatedness to your brother, half-brother, first cousin, and so forth, simply read across the \"self \" row. For relatedness to your father's brother (your uncle), your father's half-brother, and so forth, read across the second row.\n\n| Generation | Brother | Half-brother | First cousin | Half- <br> first cousin | Second <br> cousin | Half- <br> second <br> cousin |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: |\n| Self | $1 / 2$ | $1 / 4$ | $1 / 8$ | $1 / 16$ | $1 / 32$ | $1 / 64$ |\n| Father's | $1 / 4$ | $1 / 8$ | $1 / 16$ | $1 / 32$ | $1 / 64$ | $1 / 128$ |\n| Grandfather's | $1 / 8$ | $1 / 16$ | $1 / 32$ | $1 / 64$ | $1 / 128$ | $1 / 256$ |\n| Great-grandfather's | $1 / 16$ | $1 / 32$ | $1 / 64$ | $1 / 128$ | $1 / 256$ | $1 / 512$ |\n| Great-great-grandfather's | $1 / 32$ | $1 / 64$ | $1 / 128$ | $1 / 256$ | $1 / 512$ | $1 / 1,024$ |\n\nTable modified from The Altruism Equation: Seven Scientists Search for the Origins of Goodness\" by Lee Alan Dugatkin (2006)\n\nWhat is your genetic relatedness to your great half-uncle?\nA: $1 / 2$\nB: $1 / 4$\nC: $1 / 8$\nD: $1 / 16$\nE: $1 / 32$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe genetic relatedness between various sets of relatives is shown in the table below. For relatedness to your brother, half-brother, first cousin, and so forth, simply read across the \"self \" row. For relatedness to your father's brother (your uncle), your father's half-brother, and so forth, read across the second row.\n\n| Generation | Brother | Half-brother | First cousin | Half- <br> first cousin | Second <br> cousin | Half- <br> second <br> cousin |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: |\n| Self | $1 / 2$ | $1 / 4$ | $1 / 8$ | $1 / 16$ | $1 / 32$ | $1 / 64$ |\n| Father's | $1 / 4$ | $1 / 8$ | $1 / 16$ | $1 / 32$ | $1 / 64$ | $1 / 128$ |\n| Grandfather's | $1 / 8$ | $1 / 16$ | $1 / 32$ | $1 / 64$ | $1 / 128$ | $1 / 256$ |\n| Great-grandfather's | $1 / 16$ | $1 / 32$ | $1 / 64$ | $1 / 128$ | $1 / 256$ | $1 / 512$ |\n| Great-great-grandfather's | $1 / 32$ | $1 / 64$ | $1 / 128$ | $1 / 256$ | $1 / 512$ | $1 / 1,024$ |\n\nTable modified from The Altruism Equation: Seven Scientists Search for the Origins of Goodness\" by Lee Alan Dugatkin (2006)\n\nWhat is your genetic relatedness to your great half-uncle?\n\nA: $1 / 2$\nB: $1 / 4$\nC: $1 / 8$\nD: $1 / 16$\nE: $1 / 32$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1076",
        "problem": "Consider the following pedigree.\n\n[figure1]\n\nAssume that one of the grandparents has the genotype A1/A2 and the other A3/A4. The probability that the individual $\\mathrm{R}$ will be homozygous for any of the given alleles is:\nA: $1 / 2$\nB: $1 / 4$\nC: $1 / 8$\nD: $1 / 16$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nConsider the following pedigree.\n\n[figure1]\n\nAssume that one of the grandparents has the genotype A1/A2 and the other A3/A4. The probability that the individual $\\mathrm{R}$ will be homozygous for any of the given alleles is:\n\nA: $1 / 2$\nB: $1 / 4$\nC: $1 / 8$\nD: $1 / 16$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_50609c20e4e5afbc647ag-09.jpg?height=523&width=832&top_left_y=828&top_left_x=668"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1472",
        "problem": "On the left is corn yield, on the right is seal reproduction. Interpret these and answer the questions below:\n[figure1]\n\nIn the corn figure, the declining line indicates:\nA: the influence of global warming on corn crops.\nB: an increase in available resources as population density increases.\nC: increasing competition as population density increases\nD: decreasing competition as population density increases\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOn the left is corn yield, on the right is seal reproduction. Interpret these and answer the questions below:\n[figure1]\n\nIn the corn figure, the declining line indicates:\n\nA: the influence of global warming on corn crops.\nB: an increase in available resources as population density increases.\nC: increasing competition as population density increases\nD: decreasing competition as population density increases\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-06.jpg?height=520&width=1200&top_left_y=1599&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1341",
        "problem": "In 1976, minced beef was tested in Cork City, Ireland, for the level of bacterial contamination using processes called the total aerobic plate count (APC) and the coliform count. In some areas, meat is classed as unsafe to eat if it contains more than 10 million per gram total viable count (APC method) or 50 per gram of coliform count (total coliform method). Results from 5 different markets were sampled on two different days and the results (count $/ \\mathrm{g}$ ) are shown below.\n\n| Market | Sample 1 |  | Sample 2 |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | APC | Total Coliform | APC | Total Coliform |\n| A | $2.4 \\times 10^{7}$ | $3 \\times 10^{5}$ | $4.1 \\times 10^{7}$ | $2.9 \\times 10^{5}$ |\n| B | $3.5 \\times 10^{5}$ | $8 \\times 10^{4}$ | $3.0 \\times 10^{6}$ | $3.0 \\times 10^{6}$ |\n| C | $3.7 \\times 10^{8}$ | $4.8 \\times 10^{5}$ | $1.0 \\times 10^{8}$ | $4.0 \\times 10^{5}$ |\n| D | $3.2 \\times 10^{6}$ | $4.2 \\times 10^{5}$ | $1.9 \\times 10^{7}$ | $1.3 \\times 10^{5}$ |\n| E | $2.0 \\times 10^{7}$ | $5.3 \\times 10^{5}$ | $2.1 \\times 10^{7}$ | $2.2 \\times 10^{5}$ |\n\nhttp://www.jstor.org/stable/i25555828\n\nWhich market(s) sells minced beef that always meets the safety guidelines given?\nA: Market A.\nB: Markets B.\nC: Markets B and D.\nD: None of the markets, due to coliform count.\nE: None of the markets, due to coliform count and/or APC count.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn 1976, minced beef was tested in Cork City, Ireland, for the level of bacterial contamination using processes called the total aerobic plate count (APC) and the coliform count. In some areas, meat is classed as unsafe to eat if it contains more than 10 million per gram total viable count (APC method) or 50 per gram of coliform count (total coliform method). Results from 5 different markets were sampled on two different days and the results (count $/ \\mathrm{g}$ ) are shown below.\n\n| Market | Sample 1 |  | Sample 2 |  |\n| :---: | :---: | :---: | :---: | :---: |\n|  | APC | Total Coliform | APC | Total Coliform |\n| A | $2.4 \\times 10^{7}$ | $3 \\times 10^{5}$ | $4.1 \\times 10^{7}$ | $2.9 \\times 10^{5}$ |\n| B | $3.5 \\times 10^{5}$ | $8 \\times 10^{4}$ | $3.0 \\times 10^{6}$ | $3.0 \\times 10^{6}$ |\n| C | $3.7 \\times 10^{8}$ | $4.8 \\times 10^{5}$ | $1.0 \\times 10^{8}$ | $4.0 \\times 10^{5}$ |\n| D | $3.2 \\times 10^{6}$ | $4.2 \\times 10^{5}$ | $1.9 \\times 10^{7}$ | $1.3 \\times 10^{5}$ |\n| E | $2.0 \\times 10^{7}$ | $5.3 \\times 10^{5}$ | $2.1 \\times 10^{7}$ | $2.2 \\times 10^{5}$ |\n\nhttp://www.jstor.org/stable/i25555828\n\nWhich market(s) sells minced beef that always meets the safety guidelines given?\n\nA: Market A.\nB: Markets B.\nC: Markets B and D.\nD: None of the markets, due to coliform count.\nE: None of the markets, due to coliform count and/or APC count.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1335",
        "problem": "The flatworm (Dendrocoelum lacteum) increases its rate of turning as light intensity increases and, as a consequence, is found aggregated in dark places. This is an example of:\nA: Negative klinokinesis\nB: Klinokinesis\nC: Negative phototaxis\nD: Orthokinesis\nE: Negative orthokinesis\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe flatworm (Dendrocoelum lacteum) increases its rate of turning as light intensity increases and, as a consequence, is found aggregated in dark places. This is an example of:\n\nA: Negative klinokinesis\nB: Klinokinesis\nC: Negative phototaxis\nD: Orthokinesis\nE: Negative orthokinesis\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1246",
        "problem": "In an experiment to measure the speed of conduction of a nerve impulse along a giant axon, the distance between the stimulating and recording electrodes was varied and the delay between stimulus and response was recorded for each distance. The results are shown in the graph below.\n\n[figure1]\nA: A\nB: B\nC: C\nD: D\nE: E\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn an experiment to measure the speed of conduction of a nerve impulse along a giant axon, the distance between the stimulating and recording electrodes was varied and the delay between stimulus and response was recorded for each distance. The results are shown in the graph below.\n\n[figure1]\n\nA: A\nB: B\nC: C\nD: D\nE: E\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-08.jpg?height=617&width=1331&top_left_y=1342&top_left_x=131"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_283",
        "problem": "Which of the following pairs does not show a monophyletic group - paraphyletic group relationship?\nA: Monocots - Dicots\nB: Tetrapods - Bony fishes\nC: Echinoderms - Chordata\nD: Birds - Reptiles\nE: Vascular plants - Nonvascular plants\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following pairs does not show a monophyletic group - paraphyletic group relationship?\n\nA: Monocots - Dicots\nB: Tetrapods - Bony fishes\nC: Echinoderms - Chordata\nD: Birds - Reptiles\nE: Vascular plants - Nonvascular plants\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1007",
        "problem": "You observe an animal in the early stages of development. The zygote divides radially into eight cells. If you separate these early cells, each one can successfully develop into an animal on its own and the coelom forms from an infolding pocket of the archenteron. Which of the following is a valid candidate for the animal you are observing?\nA: Horseshoe crab (Chelicerata)\nB: Fruit fly (Insecta)\nC: Sea urchin (Echinoidea)\nD: Land snail (Gastropoda)\nE: Earthworm (Oligochaeta)\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nYou observe an animal in the early stages of development. The zygote divides radially into eight cells. If you separate these early cells, each one can successfully develop into an animal on its own and the coelom forms from an infolding pocket of the archenteron. Which of the following is a valid candidate for the animal you are observing?\n\nA: Horseshoe crab (Chelicerata)\nB: Fruit fly (Insecta)\nC: Sea urchin (Echinoidea)\nD: Land snail (Gastropoda)\nE: Earthworm (Oligochaeta)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_996",
        "problem": "Which of the following populations is most likely to be close to HardyWeinberg equilibrium?\nA: The human population of Toronto, Canada.\nB: A population of 100 fruit flies living in a habitat with little environmental fluctuation that has no other populations of fruit flies nearby.\nC: A population of 1 million fruit flies living in a habitat with little environmental fluctuation that has many other populations of fruit flies nearby.\nD: A population of 100 fruit flies living in a habitat with little environmental fluctuation that has many other populations of fruit flies nearby.\nE: A population of 1 million fruit flies living in a habitat with little environmental fluctuation that has no other populations of fruit flies nearby.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following populations is most likely to be close to HardyWeinberg equilibrium?\n\nA: The human population of Toronto, Canada.\nB: A population of 100 fruit flies living in a habitat with little environmental fluctuation that has no other populations of fruit flies nearby.\nC: A population of 1 million fruit flies living in a habitat with little environmental fluctuation that has many other populations of fruit flies nearby.\nD: A population of 100 fruit flies living in a habitat with little environmental fluctuation that has many other populations of fruit flies nearby.\nE: A population of 1 million fruit flies living in a habitat with little environmental fluctuation that has no other populations of fruit flies nearby.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1374",
        "problem": "Sucrose is produced in leaves and translocated short and long distances through veins to non-photosynthetic organs such as roots, stems, flowers and fruits. Sucrose transport is essential for the distribution of carbohydrates in plants. Two principal pathways include symplast and apoplast by which sucrose molecules are transported in the phloem of leaves as shown in the Figure. In the symplastic pathway, sucrose moves from cell to cell via plasmodesmata, which are continuous junctions between neighbouring plant cells. The apoplastic pathway involves a step where sucrose is momentarily transported outside of a plant cell plasma membrane.\n\nA\n\n[figure1]\n\nB\n[figure2]\n\nFigure. Diagram of the whole plant phloem network. M - Mesophyll, BS - Bundle sheath, MS - Mestome sheath, PP - Phloem parenchyma, VP - Vascular parenchyma, CC - Companion cell, TST - Thick walled sieve element, ST - Sieve element.\n\nWhich of the following is correct?\nA: Sucrose is synthesised in leaves and transported long distance through phloem to sinks under hydrostatic pressure gradient.\nB: Loading sucrose in the symplastic pathway requires energy at several steps due to the movement across the secondary walls of living cells.\nC: In the apoplastic pathway, sucrose molecules are passively loaded through plasmodemata.\nD: Unloading sucrose molecules at the sinks requires no energy release because of movement against a gradient concentration of sucrose.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSucrose is produced in leaves and translocated short and long distances through veins to non-photosynthetic organs such as roots, stems, flowers and fruits. Sucrose transport is essential for the distribution of carbohydrates in plants. Two principal pathways include symplast and apoplast by which sucrose molecules are transported in the phloem of leaves as shown in the Figure. In the symplastic pathway, sucrose moves from cell to cell via plasmodesmata, which are continuous junctions between neighbouring plant cells. The apoplastic pathway involves a step where sucrose is momentarily transported outside of a plant cell plasma membrane.\n\nA\n\n[figure1]\n\nB\n[figure2]\n\nFigure. Diagram of the whole plant phloem network. M - Mesophyll, BS - Bundle sheath, MS - Mestome sheath, PP - Phloem parenchyma, VP - Vascular parenchyma, CC - Companion cell, TST - Thick walled sieve element, ST - Sieve element.\n\nWhich of the following is correct?\n\nA: Sucrose is synthesised in leaves and transported long distance through phloem to sinks under hydrostatic pressure gradient.\nB: Loading sucrose in the symplastic pathway requires energy at several steps due to the movement across the secondary walls of living cells.\nC: In the apoplastic pathway, sucrose molecules are passively loaded through plasmodemata.\nD: Unloading sucrose molecules at the sinks requires no energy release because of movement against a gradient concentration of sucrose.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-11.jpg?height=582&width=597&top_left_y=760&top_left_x=250",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-11.jpg?height=598&width=946&top_left_y=751&top_left_x=840"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_867",
        "problem": "在某纯合的黑体直刚毛果蝇种群中, 偶然发现一只灰体直刚毛雌果蝇（甲）和一只\n黑体焦刚毛雄果蝇（乙）, 为探究其遗传方式，研究小组进行如下实验：(1)黑体直刚毛雄果蝇和甲杂交, $F_{1}$ 的雌雄子代均为灰体直刚毛: 黑体直刚毛-1: 1; (2)黑体直刚毛雌果蝇和乙杂交, $F_{1}$ 雌雄均为黑体直刚毛, $F_{1}$ 随机交配, $F_{2}$ 中雌性均为黑体直刚毛, 雄性中黑体直刚毛：黑体焦刚毛为 1：1。下列叙述错误的是（）\nA: 甲果蝇发生了显性突变，乙发生隐性突变\nB: (1)实验说明发生了性状分离, 分离比为 1:1\nC: (1)(2)实验可以判断这两对性状遵循自由组合定律\nD: 若让甲乙相交得 $F_{1}, F_{1}$ 随机交配, $F_{2}$ 中灰体焦刚毛占 $7 / 64$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在某纯合的黑体直刚毛果蝇种群中, 偶然发现一只灰体直刚毛雌果蝇（甲）和一只\n黑体焦刚毛雄果蝇（乙）, 为探究其遗传方式，研究小组进行如下实验：(1)黑体直刚毛雄果蝇和甲杂交, $F_{1}$ 的雌雄子代均为灰体直刚毛: 黑体直刚毛-1: 1; (2)黑体直刚毛雌果蝇和乙杂交, $F_{1}$ 雌雄均为黑体直刚毛, $F_{1}$ 随机交配, $F_{2}$ 中雌性均为黑体直刚毛, 雄性中黑体直刚毛：黑体焦刚毛为 1：1。下列叙述错误的是（）\n\nA: 甲果蝇发生了显性突变，乙发生隐性突变\nB: (1)实验说明发生了性状分离, 分离比为 1:1\nC: (1)(2)实验可以判断这两对性状遵循自由组合定律\nD: 若让甲乙相交得 $F_{1}, F_{1}$ 随机交配, $F_{2}$ 中灰体焦刚毛占 $7 / 64$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1334",
        "problem": "Some clover plants are cyanogenic because they produce hydrogen cyanide when the leaves are mechanically damaged. Acyanogenic plants do not produce cyanide. When two acyanogenic plants were crossed, the following results were obtained.\n\n| Offspring | Cyanogenic | Acyanogenic |\n| :---: | :---: | :---: |\n| $F_{1}$ | All | None |\n| $F_{2}$ | 151 | 41 |\n\nThese results could be explained if the parent plants\nA: both lacked the allele for an enzyme to convert the substrate to hydrogen cyanide\nB: both lacked the allele for the synthesis of the substrate from a precursor\nC: had different genotypes, one carrying the allele for the enzyme and the other for the synthesis of the substrate\nD: had the same genotype, being heterozygous for both substrate and enzyme genes\nE: had different genotypes, one being heterozygous for two loci, the other homozygous recessive for both\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSome clover plants are cyanogenic because they produce hydrogen cyanide when the leaves are mechanically damaged. Acyanogenic plants do not produce cyanide. When two acyanogenic plants were crossed, the following results were obtained.\n\n| Offspring | Cyanogenic | Acyanogenic |\n| :---: | :---: | :---: |\n| $F_{1}$ | All | None |\n| $F_{2}$ | 151 | 41 |\n\nThese results could be explained if the parent plants\n\nA: both lacked the allele for an enzyme to convert the substrate to hydrogen cyanide\nB: both lacked the allele for the synthesis of the substrate from a precursor\nC: had different genotypes, one carrying the allele for the enzyme and the other for the synthesis of the substrate\nD: had the same genotype, being heterozygous for both substrate and enzyme genes\nE: had different genotypes, one being heterozygous for two loci, the other homozygous recessive for both\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_890",
        "problem": "蛋白质可以与多种物质结合, 赋予蛋白质复合体多种功能, 下列叙述错误的是\nA: 冬小麦细胞内蛋白质与水结合，有利于其抵抗寒冷环境\nB: 肝细胞膜上的蛋白质与糖类结合，可以参与细胞间的信息交流\nC: 大肠杆菌细胞内蛋白质与 RNA 结合，可以参与蛋白质的合成\nD: 洋葱根尖分生区细胞核中的蛋白质与 DNA 结合, 作为其遗传物质\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n蛋白质可以与多种物质结合, 赋予蛋白质复合体多种功能, 下列叙述错误的是\n\nA: 冬小麦细胞内蛋白质与水结合，有利于其抵抗寒冷环境\nB: 肝细胞膜上的蛋白质与糖类结合，可以参与细胞间的信息交流\nC: 大肠杆菌细胞内蛋白质与 RNA 结合，可以参与蛋白质的合成\nD: 洋葱根尖分生区细胞核中的蛋白质与 DNA 结合, 作为其遗传物质\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1400",
        "problem": "Turmeric is a flowering plant, Curcuma longa of the ginger family, Zingiberaceae. It has a very long history of medicinal use, dating back nearly 4000 years. In the last 25 years, over 3000 new publications have been released regarding the health effects of turmeric. The primary compound in turmeric is curcumin, which has been purported to have antiinflammatory and antioxidant effects. Scientists conducted a study investigating the effect of varying concentrations and length of exposure of curcumin on adipocytes from a human cell line. Figure A shows the effect of curcumin concentration on the viability of a population of adipocytes.\n\n[figure1]\n\nFurther investigation showed that the effect on viability was due to changes in the regulation of apoptosis. Caspases are known to play essential roles in apoptosis and are an important regulator of this process is the Bax protein, which causes the release of cytochrome $c$.\n\nApoptosis is a process known as:\nA: programmed cell death\nB: mitosis\nC: transcription\nD: cellular injury\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTurmeric is a flowering plant, Curcuma longa of the ginger family, Zingiberaceae. It has a very long history of medicinal use, dating back nearly 4000 years. In the last 25 years, over 3000 new publications have been released regarding the health effects of turmeric. The primary compound in turmeric is curcumin, which has been purported to have antiinflammatory and antioxidant effects. Scientists conducted a study investigating the effect of varying concentrations and length of exposure of curcumin on adipocytes from a human cell line. Figure A shows the effect of curcumin concentration on the viability of a population of adipocytes.\n\n[figure1]\n\nFurther investigation showed that the effect on viability was due to changes in the regulation of apoptosis. Caspases are known to play essential roles in apoptosis and are an important regulator of this process is the Bax protein, which causes the release of cytochrome $c$.\n\nApoptosis is a process known as:\n\nA: programmed cell death\nB: mitosis\nC: transcription\nD: cellular injury\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-08.jpg?height=560&width=797&top_left_y=774&top_left_x=618"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_446",
        "problem": "人群中甲病（由基因 $\\mathrm{A} 、 \\mathrm{a}$ 控制）和乙病（由基因 $\\mathrm{B} 、 \\mathrm{~b}$ 控制）均为单基因遗传病，其中有一种病为伴性遗传病。已知人群中每 100 人中有一个甲病患者。下图为某家族的系谱图， $\\mathrm{II}_{3}$ 无甲病致病基因。下列关于甲、乙两病的叙述错误的是（ ）\n\n[图1]\nA: 乙病的致病基因位于 X 染色体上\nB: $\\mathrm{II}_{6}$ 与 $\\mathrm{I}_{1}$ 基因型相同的概率为 $\\frac{2}{11}$\nC: $\\mathrm{II}_{4}$ 产生同时含甲、乙两病致病基因配子的概率为 $\\frac{1}{6}$\nD: 若 $\\mathrm{III}_{7}$ 与 $\\mathrm{III}_{8}$ 婚配, 他们生一个两病兼患孩子的概率为 $\\frac{1}{24}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人群中甲病（由基因 $\\mathrm{A} 、 \\mathrm{a}$ 控制）和乙病（由基因 $\\mathrm{B} 、 \\mathrm{~b}$ 控制）均为单基因遗传病，其中有一种病为伴性遗传病。已知人群中每 100 人中有一个甲病患者。下图为某家族的系谱图， $\\mathrm{II}_{3}$ 无甲病致病基因。下列关于甲、乙两病的叙述错误的是（ ）\n\n[图1]\n\nA: 乙病的致病基因位于 X 染色体上\nB: $\\mathrm{II}_{6}$ 与 $\\mathrm{I}_{1}$ 基因型相同的概率为 $\\frac{2}{11}$\nC: $\\mathrm{II}_{4}$ 产生同时含甲、乙两病致病基因配子的概率为 $\\frac{1}{6}$\nD: 若 $\\mathrm{III}_{7}$ 与 $\\mathrm{III}_{8}$ 婚配, 他们生一个两病兼患孩子的概率为 $\\frac{1}{24}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-05.jpg?height=357&width=757&top_left_y=2112&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_979",
        "problem": "Which of these is NOT true about allopatric speciation?\nA: It describes formation of a new species due to geographic isolation\nB: It is more rare than sympatric speciation\nC: It can be caused by sexual selection\nD: It can be caused by natural selection under different environmental conditions\nE: It can prevent interbreeding with the parent population\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of these is NOT true about allopatric speciation?\n\nA: It describes formation of a new species due to geographic isolation\nB: It is more rare than sympatric speciation\nC: It can be caused by sexual selection\nD: It can be caused by natural selection under different environmental conditions\nE: It can prevent interbreeding with the parent population\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_909",
        "problem": "人类的 GJB6 基因表达为耳蜗细胞膜的离子通道蛋白, 该基因突变可导致耳狵。下表是对小李(患者)及部分听力正常的家庭成员 GJB6 基因检测的结果(不考虑基因突变和染色体畸变)。下列叙述正确的是（ ）\n\n小李家族 GJB6 基因检测结果\n\n| 检测对象 | 小李 | 父亲 | 母亲 | 祖父 | 祖母 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| GJB6 基 <br> 因 | - | + | + | + | + |\n| 突变基因 | + | $?$ | + | + | - |\n\n注：“+”表示个体有相应基因，“-”表示个体无相应基因, “? ”表示未确定基因是否存在\nA: 仅通过祖父或祖母的基因组成就可判断致病基因的显隐性\nB: 若小李只有一个突变基因, 则父亲携带致病基因\nC: 若小李有两个致病基因且为男性, 则父亲基因型为 $X^{A} Y^{a}$\nD: 若小李有两个致病基因且为女性，则小李可能有两种基因型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n人类的 GJB6 基因表达为耳蜗细胞膜的离子通道蛋白, 该基因突变可导致耳狵。下表是对小李(患者)及部分听力正常的家庭成员 GJB6 基因检测的结果(不考虑基因突变和染色体畸变)。下列叙述正确的是（ ）\n\n小李家族 GJB6 基因检测结果\n\n| 检测对象 | 小李 | 父亲 | 母亲 | 祖父 | 祖母 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| GJB6 基 <br> 因 | - | + | + | + | + |\n| 突变基因 | + | $?$ | + | + | - |\n\n注：“+”表示个体有相应基因，“-”表示个体无相应基因, “? ”表示未确定基因是否存在\n\nA: 仅通过祖父或祖母的基因组成就可判断致病基因的显隐性\nB: 若小李只有一个突变基因, 则父亲携带致病基因\nC: 若小李有两个致病基因且为男性, 则父亲基因型为 $X^{A} Y^{a}$\nD: 若小李有两个致病基因且为女性，则小李可能有两种基因型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_461",
        "problem": "果蝇的性别是由早期胚胎的性指数 (X 染色体数目与常染色体组数之比, 即 $\\mathrm{x}: \\mathrm{A}$ )所决定的。 $\\mathrm{X}: \\mathrm{A}=1$ 时, 激活性别相关基因 $\\mathrm{M}$ 进而发育为雌性, 若基因 $\\mathrm{M}$ 发生突变,则发育为雄性; $X: A=0.5$ 时, 无法激活基因 $M$ 而发育为雄性。已知 $Y$ 染色体只决定雄蝇的可育性, $\\mathrm{M}$ 基因仅位于 $\\mathrm{X}$ 染色体上, 不考虑其他变异。下列说法正确的是（ ）\nA: 对果蝇基因组测序, 应测定 4 条染色体上 DNA 的碱基序列\nB: 染色体组成为 XXY 的个体, 一定发育为雌性\nC: $X^{M} X^{m}$ 和 $X^{m} Y$ 的果蝇杂交子代雌雄之比为 $1: 3$\nD: $\\mathrm{X}^{\\mathrm{M}} \\mathrm{X}^{\\mathrm{M}}$ 和 $\\mathrm{X}^{\\mathrm{M}} \\mathrm{O}$ 的果蝇杂交子代雌雄之比为 $1: 1$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的性别是由早期胚胎的性指数 (X 染色体数目与常染色体组数之比, 即 $\\mathrm{x}: \\mathrm{A}$ )所决定的。 $\\mathrm{X}: \\mathrm{A}=1$ 时, 激活性别相关基因 $\\mathrm{M}$ 进而发育为雌性, 若基因 $\\mathrm{M}$ 发生突变,则发育为雄性; $X: A=0.5$ 时, 无法激活基因 $M$ 而发育为雄性。已知 $Y$ 染色体只决定雄蝇的可育性, $\\mathrm{M}$ 基因仅位于 $\\mathrm{X}$ 染色体上, 不考虑其他变异。下列说法正确的是（ ）\n\nA: 对果蝇基因组测序, 应测定 4 条染色体上 DNA 的碱基序列\nB: 染色体组成为 XXY 的个体, 一定发育为雌性\nC: $X^{M} X^{m}$ 和 $X^{m} Y$ 的果蝇杂交子代雌雄之比为 $1: 3$\nD: $\\mathrm{X}^{\\mathrm{M}} \\mathrm{X}^{\\mathrm{M}}$ 和 $\\mathrm{X}^{\\mathrm{M}} \\mathrm{O}$ 的果蝇杂交子代雌雄之比为 $1: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1166",
        "problem": "Geological analysis suggests the initial separation of New Zealand from Gondwana took place approximately 80 million years ago. However, it is hypothesised that full separation was not complete until 60 million years ago. Some scientists propose that a common ancestor of the Moa and the Kiwi (a proto-Moa) floated away from Gondwana during this time. Others propose that Moa and Kiwi arrived in New Zealand at two different times. Four possible phylogenetic trees, describing different possible evolutionary relationships within the ratites, are shown below. Use these trees to answer the following questions.\n[figure1]Which tree would support the hypothesis of a protoMoa ancestor flying to New Zealand?\nA: Tree 1\nB: Tree 2\nC: Tree 3\nD: Tree 4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nGeological analysis suggests the initial separation of New Zealand from Gondwana took place approximately 80 million years ago. However, it is hypothesised that full separation was not complete until 60 million years ago. Some scientists propose that a common ancestor of the Moa and the Kiwi (a proto-Moa) floated away from Gondwana during this time. Others propose that Moa and Kiwi arrived in New Zealand at two different times. Four possible phylogenetic trees, describing different possible evolutionary relationships within the ratites, are shown below. Use these trees to answer the following questions.\n[figure1]\n\nproblem:\nWhich tree would support the hypothesis of a protoMoa ancestor flying to New Zealand?\n\nA: Tree 1\nB: Tree 2\nC: Tree 3\nD: Tree 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-26.jpg?height=1120&width=1812&top_left_y=505&top_left_x=150"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_799",
        "problem": "葫芦科植物喷瓜的性别由染色体上一组复等位基因决定: 雄性基因 $\\mathrm{a}^{\\mathrm{D}}$ 、雌雄同体基因 $\\mathrm{a}^{+}$和雌性基因 $\\mathrm{a}^{\\mathrm{d}}$, 其中 $\\mathrm{a}^{\\mathrm{D}}$ 对 $\\mathrm{a}^{+}$和 $\\mathrm{a}^{\\mathrm{d}}$ 为显性, $\\mathrm{a}^{+}$对 $\\mathrm{a}^{\\mathrm{d}}$ 为显性。喷瓜的叶色由另外一对染色体上的一对等位基因 ( $\\mathrm{B} / \\mathrm{b})$ 控制, $\\mathrm{BB}$ 表现深绿; $\\mathrm{Bb}$ 表现浅绿; $\\mathrm{bb}$ 呈黄色, 幼苗阶段死亡。下列叙述正确的是 ( )\nA: 基因 $\\mathrm{a}^{\\mathrm{D}} 、 \\mathrm{a}^{+}$和 $\\mathrm{a}^{\\mathrm{d}}$ 位于性染色体上\nB: 基因 $\\mathrm{a}^{\\mathrm{D}} 、 \\mathrm{a}^{+}$和 $\\mathrm{a}^{\\mathrm{d}}$ 的遗传遵循自由组合定律\nC: 不考虑基因突变, 雄性喷瓜的基因型有 2 种\nD: 若模拟 $a^{D} a^{+}$和 $a^{+} a^{d}$ 杂交, 则需要 3 个容器\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n葫芦科植物喷瓜的性别由染色体上一组复等位基因决定: 雄性基因 $\\mathrm{a}^{\\mathrm{D}}$ 、雌雄同体基因 $\\mathrm{a}^{+}$和雌性基因 $\\mathrm{a}^{\\mathrm{d}}$, 其中 $\\mathrm{a}^{\\mathrm{D}}$ 对 $\\mathrm{a}^{+}$和 $\\mathrm{a}^{\\mathrm{d}}$ 为显性, $\\mathrm{a}^{+}$对 $\\mathrm{a}^{\\mathrm{d}}$ 为显性。喷瓜的叶色由另外一对染色体上的一对等位基因 ( $\\mathrm{B} / \\mathrm{b})$ 控制, $\\mathrm{BB}$ 表现深绿; $\\mathrm{Bb}$ 表现浅绿; $\\mathrm{bb}$ 呈黄色, 幼苗阶段死亡。下列叙述正确的是 ( )\n\nA: 基因 $\\mathrm{a}^{\\mathrm{D}} 、 \\mathrm{a}^{+}$和 $\\mathrm{a}^{\\mathrm{d}}$ 位于性染色体上\nB: 基因 $\\mathrm{a}^{\\mathrm{D}} 、 \\mathrm{a}^{+}$和 $\\mathrm{a}^{\\mathrm{d}}$ 的遗传遵循自由组合定律\nC: 不考虑基因突变, 雄性喷瓜的基因型有 2 种\nD: 若模拟 $a^{D} a^{+}$和 $a^{+} a^{d}$ 杂交, 则需要 3 个容器\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-03.jpg?height=58&width=1356&top_left_y=1630&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1216",
        "problem": "A new medication, named \"Beclopa\" is currently undergoing trials to determine its effectiveness in lowering the blood pressure in patients presenting with hypertension (which is the increase in blood pressure above the normal level). The trials were done on two groups of patients (children and the elderly), where each group was divided into half (without the patient's knowing) and given either a pill containing Beclopa or a pill containing water (the control). The table below shows the percentage of patients who showed a significant decrease in blood pressure.\n\n|  | Patients given pills with Beclopa | Patients given pills with water |\n| :--- | :---: | :---: |\n| Children | $10 \\%$ | $11 \\%$ |\n| The elderly | $80 \\%$ | $10 \\%$ |\n\nWhich of the following is true?\nA: Beclopa was more effective for the children than the elderly patients.\nB: Beclopa causes a decrease in blood pressure for all patients.\nC: Beclopa raises blood pressure in children.\nD: Beclopa has no effect on blood pressure in children.\nE: None of the above.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA new medication, named \"Beclopa\" is currently undergoing trials to determine its effectiveness in lowering the blood pressure in patients presenting with hypertension (which is the increase in blood pressure above the normal level). The trials were done on two groups of patients (children and the elderly), where each group was divided into half (without the patient's knowing) and given either a pill containing Beclopa or a pill containing water (the control). The table below shows the percentage of patients who showed a significant decrease in blood pressure.\n\n|  | Patients given pills with Beclopa | Patients given pills with water |\n| :--- | :---: | :---: |\n| Children | $10 \\%$ | $11 \\%$ |\n| The elderly | $80 \\%$ | $10 \\%$ |\n\nWhich of the following is true?\n\nA: Beclopa was more effective for the children than the elderly patients.\nB: Beclopa causes a decrease in blood pressure for all patients.\nC: Beclopa raises blood pressure in children.\nD: Beclopa has no effect on blood pressure in children.\nE: None of the above.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_61",
        "problem": "Density dependence is the fundamental process governing the population dynamics of organisms. The graph below describes per capita (per-individual) birth rate and death rate as a function of population density in two types of species (I and II).\n\n[figure1]\n\nPopulation density\n\n[figure2]\n\nPopulation density\nA: Asexually reproducing species are more likely to be type I than sexually reproducing ones.\nB: Population density is kept constant around all points of $A, B$, and $D$ with a density-dependent manner. \nC: The aggregation of individuals is advantageous, rather than detrimental, below the density threshold of $\\mathrm{C}$.\nD: Type I species are more likely to go extinct when the population is severely decreased, than type II species. \n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nDensity dependence is the fundamental process governing the population dynamics of organisms. The graph below describes per capita (per-individual) birth rate and death rate as a function of population density in two types of species (I and II).\n\n[figure1]\n\nPopulation density\n\n[figure2]\n\nPopulation density\n\nA: Asexually reproducing species are more likely to be type I than sexually reproducing ones.\nB: Population density is kept constant around all points of $A, B$, and $D$ with a density-dependent manner. \nC: The aggregation of individuals is advantageous, rather than detrimental, below the density threshold of $\\mathrm{C}$.\nD: Type I species are more likely to go extinct when the population is severely decreased, than type II species. \n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-60.jpg?height=551&width=626&top_left_y=798&top_left_x=201",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-60.jpg?height=554&width=626&top_left_y=797&top_left_x=955"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_484",
        "problem": "鱼鳞病是一种单基因遗传所致的皮肤疾病, 该疾病在人群中的发病率男性大于女性。某家庭中父亲和母亲都不患鱼鳞病, 却生了一个患鱼鳞病的女儿, 经过细胞学鉴定该女\n儿少了一条 X 染色体。若不考虑基因突变, 下列分析错误的是（ ）\nA: 人体内控制鱼鳞病的致病基因为隐性基因\nB: 该女儿细胞内的致病基因来自于她的母亲\nC: 该女儿少一条染色体与父亲的异常精子有关\nD: 该夫妇再生一个女孩患鱼鳞病的概率为 $1 / 2$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n鱼鳞病是一种单基因遗传所致的皮肤疾病, 该疾病在人群中的发病率男性大于女性。某家庭中父亲和母亲都不患鱼鳞病, 却生了一个患鱼鳞病的女儿, 经过细胞学鉴定该女\n儿少了一条 X 染色体。若不考虑基因突变, 下列分析错误的是（ ）\n\nA: 人体内控制鱼鳞病的致病基因为隐性基因\nB: 该女儿细胞内的致病基因来自于她的母亲\nC: 该女儿少一条染色体与父亲的异常精子有关\nD: 该夫妇再生一个女孩患鱼鳞病的概率为 $1 / 2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1323",
        "problem": "In chloroplast lamellae, the photosystems comprise proteins and are embedded in a phospholipid bilayer. To extract the two photosystems in a functional state the best treatment would be to\nA: use a mild detergent\nB: boil and cool the chloroplast\nC: denature the membrane proteins with acid\nD: reduce the chlorophyll molecules by illumination\nE: separate by differential centrifugation\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn chloroplast lamellae, the photosystems comprise proteins and are embedded in a phospholipid bilayer. To extract the two photosystems in a functional state the best treatment would be to\n\nA: use a mild detergent\nB: boil and cool the chloroplast\nC: denature the membrane proteins with acid\nD: reduce the chlorophyll molecules by illumination\nE: separate by differential centrifugation\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1169",
        "problem": "Understanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.The most important food type for orange-fronted parakeets in the South Island in spring was?\nA: Leaves.\nB: Flowers.\nC: Fruit.\nD: Beech trees.\nE: Invertebrates\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nUnderstanding the foraging ecology of a species is crucial when conservation management involves translocation of the species. Work by Kearvell et al. (2002) had described the foods consumed by orange-fronted parakeets in the South Island and noted that for most of the year the species fed almost exclusively on Nothofagus spp (beech trees) and that invertebrates made up nearly $70 \\%$ of the food items consumed in spring.\n\nLuis' research group recorded data on the diet of translocated orange-fronted parakeets on Maud Island, in the Marlborough Sounds, South Island. They recorded the foraging of the parakeets on each research visit to Maud Island from March 2007 to January 2009, visiting approximately every two months (17 visits in total).\n\n132 feeding bouts were recorded with a total of 124 observations (81\\%) consisting of dietary items and 29 (19\\%) of non-dietary items such as bark, sticks and grit. Orange-fronted parakeets were observed to consume fruits 94 times, leaves 19 times, flowers six times and invertebrates five times.\n\nThe table below gives the plant species and food types ingested by translocated orange-fronted parakeets on Maud Island.\n\n| Species | Type | Proportion of diet (feeding <br> bouts in brackets) |\n| :---: | :---: | :---: |\n| Sycamore (Acer pseudoplatanus ${ }^{*}$ ) | Fruits | $3.36(4)$ |\n| Titoki (Alectryon excelsus) | Fruits | $1.68(2)$ |\n| Makomako (Aristotelia serrata) | Fruits, leaves | $13.44(16)$ |\n| Putaputaweta (Carpodeus serratus) | Fruits, leaves | $5.88(7)$ |\n| Karamu (Coprosma robusta) | Fruits | $8.40(10)$ |\n| Tree lucerne (Cytisus palmensis* $)$ | Flowers, leaves | $5.04(6)$ |\n| Akeake (Dodonea viscosa) | Leaves | $0.84(1)$ |\n| Kohekohe (Dysoxilum spectabile) | Flowers | $0.84(1)$ |\n| Koromiko (Hebe stricta) | Flowers | $1.68(2)$ |\n| Manuka (Leptospermum scoparium) | Fruits | $7.56(9)$ |\n| Mahoe (Melycitus ramiflorus) | Fruits, leaves, flowers | $43.70(52)$ |\n| Whauwhaupaku (Pseudopanax arboreus) | Fruits | $5.04(6)$ |\n| Pine (Pinus radiata ${ }^{*}$ ) | Leaves | $1.68(2)$ |\n| Karo (Pittosporum sp.) | Fruits | $0.84(1)$ |\n\n## ${ }^{*}$ Introduced species.\n\nproblem:\nThe most important food type for orange-fronted parakeets in the South Island in spring was?\n\nA: Leaves.\nB: Flowers.\nC: Fruit.\nD: Beech trees.\nE: Invertebrates\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_791",
        "problem": "正确运用科学方法是取得研究成果的重要前提, 下列科学方法运用正确的是 ( )\nA: 科学家用同位素标记小鼠和人细胞膜证明了细胞膜具有流动性\nB: 鲁宾、卡门运用苂光标记追踪二氧化碳中的碳转移途径\nC: 恩格尔曼用建构模型法证明叶绿体是光合作用的场所\nD: 孟德尔和摩尔根都运用假说-演绎法发现了遗传规律\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n正确运用科学方法是取得研究成果的重要前提, 下列科学方法运用正确的是 ( )\n\nA: 科学家用同位素标记小鼠和人细胞膜证明了细胞膜具有流动性\nB: 鲁宾、卡门运用苂光标记追踪二氧化碳中的碳转移途径\nC: 恩格尔曼用建构模型法证明叶绿体是光合作用的场所\nD: 孟德尔和摩尔根都运用假说-演绎法发现了遗传规律\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_134",
        "problem": "In a recognition memory task, subjects (human being) viewed a series of different visual objects that occurred at different times within the breathing cycle. The interval between displaying images was 3-6 seconds. After a 20 min break, subjects were presented with the old pictures from the encoding session plus an equal number of new pictures; and the subjects pressed the \"Yes\" button if they had previously viewed the object's image, and pressed the \"No\" button if the image was new. Subjects' memory performance was examined in the episodes of inspiration and expiration, and in mouth breathing and nasal breathing (In figures \"*\" means $\\mathrm{p}<0.05$ which indicates significant difference between groups).\n\nA\n[figure1]\n\nC\n\n| Encode | Retrieve |\n| :--- | :--- |\n| $\\square$ insp | insp |\n| $\\square$ exp | insp |\n| $\\square$ insp | $\\exp$ |\n| $\\square$ exp | $\\exp$ |\n\n[figure2]\nA: The inspiration and expiration phase does not affect memory performance during oral breathing.\nB: Memory function during nasal breathing is significantly more accurate than during oral breathing.\nC: Regardless of the respiratory phase in which the image is encoded, if the object's images are depicted in the inspiration phase during the retrieval, memory performance is significantly greater.\nD: In breathing through the nose, unlike encoding, the accuracy of retrieval in the inspiration phase is significantly higher than in the expiration phase.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn a recognition memory task, subjects (human being) viewed a series of different visual objects that occurred at different times within the breathing cycle. The interval between displaying images was 3-6 seconds. After a 20 min break, subjects were presented with the old pictures from the encoding session plus an equal number of new pictures; and the subjects pressed the \"Yes\" button if they had previously viewed the object's image, and pressed the \"No\" button if the image was new. Subjects' memory performance was examined in the episodes of inspiration and expiration, and in mouth breathing and nasal breathing (In figures \"*\" means $\\mathrm{p}<0.05$ which indicates significant difference between groups).\n\nA\n[figure1]\n\nC\n\n| Encode | Retrieve |\n| :--- | :--- |\n| $\\square$ insp | insp |\n| $\\square$ exp | insp |\n| $\\square$ insp | $\\exp$ |\n| $\\square$ exp | $\\exp$ |\n\n[figure2]\n\nA: The inspiration and expiration phase does not affect memory performance during oral breathing.\nB: Memory function during nasal breathing is significantly more accurate than during oral breathing.\nC: Regardless of the respiratory phase in which the image is encoded, if the object's images are depicted in the inspiration phase during the retrieval, memory performance is significantly greater.\nD: In breathing through the nose, unlike encoding, the accuracy of retrieval in the inspiration phase is significantly higher than in the expiration phase.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-27.jpg?height=1170&width=1378&top_left_y=728&top_left_x=310",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-27.jpg?height=397&width=445&top_left_y=1503&top_left_x=1231"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1293",
        "problem": "Which of the following correctly lists the flow of membrane lipids and proteins though the organelles of the endomembranous system of a eukaryotic cell.\nA: Rough ER $\\rightarrow$ smooth ER $\\rightarrow$ transport vesicles $\\rightarrow$ Golgi $\\rightarrow$ plasma membranes\nB: Nuclear envelope $\\rightarrow$ rough ER $\\rightarrow$ smooth ER $\\rightarrow$ transport vesicles $\\rightarrow$ Golgi $\\rightarrow$ transport vesicles $\\rightarrow$ plasma membranes\nC: Nuclear envelope $\\rightarrow$ rough ER $\\rightarrow$ transport vesicles $\\rightarrow$ smooth ER $\\rightarrow$ transport vesicles $\\rightarrow$ Golgi $\\rightarrow$ plasma membranes\nD: Nuclear envelope $\\rightarrow$ rough ER $\\rightarrow$ transport vesicles $\\rightarrow$ smooth ER $\\rightarrow$ transport vesicles $\\rightarrow$ Golgi $\\rightarrow$ transport vesicles $\\rightarrow$ plasma membranes\nE: Rough ER $\\rightarrow$ smooth ER $\\rightarrow$ transport vesicles $\\rightarrow$ Golgi $\\rightarrow$ transport vesicles $\\rightarrow$ plasma membranes\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following correctly lists the flow of membrane lipids and proteins though the organelles of the endomembranous system of a eukaryotic cell.\n\nA: Rough ER $\\rightarrow$ smooth ER $\\rightarrow$ transport vesicles $\\rightarrow$ Golgi $\\rightarrow$ plasma membranes\nB: Nuclear envelope $\\rightarrow$ rough ER $\\rightarrow$ smooth ER $\\rightarrow$ transport vesicles $\\rightarrow$ Golgi $\\rightarrow$ transport vesicles $\\rightarrow$ plasma membranes\nC: Nuclear envelope $\\rightarrow$ rough ER $\\rightarrow$ transport vesicles $\\rightarrow$ smooth ER $\\rightarrow$ transport vesicles $\\rightarrow$ Golgi $\\rightarrow$ plasma membranes\nD: Nuclear envelope $\\rightarrow$ rough ER $\\rightarrow$ transport vesicles $\\rightarrow$ smooth ER $\\rightarrow$ transport vesicles $\\rightarrow$ Golgi $\\rightarrow$ transport vesicles $\\rightarrow$ plasma membranes\nE: Rough ER $\\rightarrow$ smooth ER $\\rightarrow$ transport vesicles $\\rightarrow$ Golgi $\\rightarrow$ transport vesicles $\\rightarrow$ plasma membranes\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1098",
        "problem": "Fluid filtration across capillaries depends on several factors. Which of the following causes net filtration at the arterial end and net reabsorption at the venous end?\nA: Greater osmotic pressure of plasma at the arterial end of capillary than that at the venous end.\nB: Greater interstitial fluid osmotic pressure at venous end of capillary than that at arterial end.\nC: Greater capillary hydrostatic pressure than osmotic pressure of plasma at arterial end and the same being lower than plasma colloidal osmotic pressure at venous end.\nD: Sum total of plasma colloidal osmotic pressure \\& interstitial fluid osmotic pressure at venous end being greater than the sum total at the arterial end capillary.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFluid filtration across capillaries depends on several factors. Which of the following causes net filtration at the arterial end and net reabsorption at the venous end?\n\nA: Greater osmotic pressure of plasma at the arterial end of capillary than that at the venous end.\nB: Greater interstitial fluid osmotic pressure at venous end of capillary than that at arterial end.\nC: Greater capillary hydrostatic pressure than osmotic pressure of plasma at arterial end and the same being lower than plasma colloidal osmotic pressure at venous end.\nD: Sum total of plasma colloidal osmotic pressure \\& interstitial fluid osmotic pressure at venous end being greater than the sum total at the arterial end capillary.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1398",
        "problem": "Julie noticed an introduced species of weed disrupted the growth of native plant seedlings in the local park. She also recalled that symbiotic bacteria in the soil play a role in native plant seedling growth. Julie took two soil samples from an area where the weed species were growing ('Affected' soil samples) and sterilised ONE sample using ultraviolet light. She then obtained another two soil samples from an area without any evidence of the weed species ('Unaffected' soil samples) and sterilised one sample as well. She then planted some native seedlings in each soil sample and let them grow for 3 months. At the end of 3 months, she removed all the roots, stems, and leaves of the seedlings and weighed the biomass of the collected plant material.\n\nShe repeated this experiment 5 times with five different sets of soils and documented the results in a table:\n\n| Soil category | Weighed seedling biomass (g) |  |  |  |  |\n| :--- | :---: | :---: | :---: | :---: | :---: |\n|  | First set | Second set | Third set | Fourth set | Fifth set |\n| Affected | 6 | 11 | 1 | 2 | 4 |\n| Unaffected | 141 | 149 | 139 | 128 | 140 |\n| Sterilised affected | 3 | 7 | 5 | 2 | 5 |\n| Sterilised unaffected | 10 | 21 | 9 | 7 | 14 |\n\nJulie also examined the soil samples with and without the weed species under a light microscope to examine the symbiotic bacteria present in the soil.\n\n| Soil category | Number of symbiotic bacteria viewed per $\\mathbf{m m}^{2}$ under the microscope |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | First set | Second set | Third set | Fourth set | Fifth set |\n| Affected | 6 | 7 | 11 | 6 | 8 |\n| Unaffected | 515 | 534 | 501 | 456 | 466 |\n\nWhich conclusion is most supported by the data results?\nA: The symbiotic bacteria in the soil are a shared resource that the weed species outcompetes the native seedlings for\nB: The presence of symbiotic bacteria supports the growth of the introduced weed species by forming a mutualistic relationship\nC: The native seedlings promote symbiotic bacteria growth in the soil, which then inhibits the growth of the introduced weed species\nD: The data shows there is no relationship between the symbiotic bacteria, introduced weed species, and native seedlings\nE: The weed species suppresses native seedling growth by decreasing the number of symbiotic bacteria in the soil\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nJulie noticed an introduced species of weed disrupted the growth of native plant seedlings in the local park. She also recalled that symbiotic bacteria in the soil play a role in native plant seedling growth. Julie took two soil samples from an area where the weed species were growing ('Affected' soil samples) and sterilised ONE sample using ultraviolet light. She then obtained another two soil samples from an area without any evidence of the weed species ('Unaffected' soil samples) and sterilised one sample as well. She then planted some native seedlings in each soil sample and let them grow for 3 months. At the end of 3 months, she removed all the roots, stems, and leaves of the seedlings and weighed the biomass of the collected plant material.\n\nShe repeated this experiment 5 times with five different sets of soils and documented the results in a table:\n\n| Soil category | Weighed seedling biomass (g) |  |  |  |  |\n| :--- | :---: | :---: | :---: | :---: | :---: |\n|  | First set | Second set | Third set | Fourth set | Fifth set |\n| Affected | 6 | 11 | 1 | 2 | 4 |\n| Unaffected | 141 | 149 | 139 | 128 | 140 |\n| Sterilised affected | 3 | 7 | 5 | 2 | 5 |\n| Sterilised unaffected | 10 | 21 | 9 | 7 | 14 |\n\nJulie also examined the soil samples with and without the weed species under a light microscope to examine the symbiotic bacteria present in the soil.\n\n| Soil category | Number of symbiotic bacteria viewed per $\\mathbf{m m}^{2}$ under the microscope |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | First set | Second set | Third set | Fourth set | Fifth set |\n| Affected | 6 | 7 | 11 | 6 | 8 |\n| Unaffected | 515 | 534 | 501 | 456 | 466 |\n\nWhich conclusion is most supported by the data results?\n\nA: The symbiotic bacteria in the soil are a shared resource that the weed species outcompetes the native seedlings for\nB: The presence of symbiotic bacteria supports the growth of the introduced weed species by forming a mutualistic relationship\nC: The native seedlings promote symbiotic bacteria growth in the soil, which then inhibits the growth of the introduced weed species\nD: The data shows there is no relationship between the symbiotic bacteria, introduced weed species, and native seedlings\nE: The weed species suppresses native seedling growth by decreasing the number of symbiotic bacteria in the soil\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_852",
        "problem": "果蝇的一个精原细胞有 8 条染色体, 某科研人员将完全被 ${ }^{15} \\mathrm{~N}$ 标记的精原细胞转移到普通培养基（仅含 ${ }^{14} \\mathrm{~N}$ 的氮源）中培养。下列有关叙述错误的是（）\nA: 普通培养基上的精原细胞有丝分裂一次，形成的子代精原细胞全部含 ${ }^{15} \\mathrm{~N}$\nB: 普通培养基上的精原细胞第二次有丝分裂后期，有 4 条染色体含 ${ }^{15} \\mathrm{~N}$\nC: 普通培养基上的精原细胞减数分裂, 形成的精细胞可能全部含 ${ }^{15} \\mathrm{~N}$\nD: 普通培养基上的精原细胞减数分裂II后期, 可能有 8 条染色体含 ${ }^{15} \\mathrm{~N}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的一个精原细胞有 8 条染色体, 某科研人员将完全被 ${ }^{15} \\mathrm{~N}$ 标记的精原细胞转移到普通培养基（仅含 ${ }^{14} \\mathrm{~N}$ 的氮源）中培养。下列有关叙述错误的是（）\n\nA: 普通培养基上的精原细胞有丝分裂一次，形成的子代精原细胞全部含 ${ }^{15} \\mathrm{~N}$\nB: 普通培养基上的精原细胞第二次有丝分裂后期，有 4 条染色体含 ${ }^{15} \\mathrm{~N}$\nC: 普通培养基上的精原细胞减数分裂, 形成的精细胞可能全部含 ${ }^{15} \\mathrm{~N}$\nD: 普通培养基上的精原细胞减数分裂II后期, 可能有 8 条染色体含 ${ }^{15} \\mathrm{~N}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_122",
        "problem": "Frequency-dependent selection is an evolutionary process where the fitness of a phenotype depends on its frequency relative to other phenotypes in a given population. In positive frequency-dependent selection, the fitness of a phenotype increases as it becomes more common. And in negative frequency-dependent selection, the fitness of a phenotype decreases as it becomes more common.\nA: Plant self-incompatibility alleles are an example of negative frequency-dependent selection.\nB: Spreading of a newly emerged virus in a human population is controlled by negative frequency-dependent selection.\nC: Prevalence of Papilio memmon whose females resemble unpalatable Papilio coon is an example of positive frequency-dependent selection.\nD: Spread of genes responsible for warning coloration in toxic or distasteful organisms in population is controlled by positive frequency-dependent selection.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFrequency-dependent selection is an evolutionary process where the fitness of a phenotype depends on its frequency relative to other phenotypes in a given population. In positive frequency-dependent selection, the fitness of a phenotype increases as it becomes more common. And in negative frequency-dependent selection, the fitness of a phenotype decreases as it becomes more common.\n\nA: Plant self-incompatibility alleles are an example of negative frequency-dependent selection.\nB: Spreading of a newly emerged virus in a human population is controlled by negative frequency-dependent selection.\nC: Prevalence of Papilio memmon whose females resemble unpalatable Papilio coon is an example of positive frequency-dependent selection.\nD: Spread of genes responsible for warning coloration in toxic or distasteful organisms in population is controlled by positive frequency-dependent selection.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1031",
        "problem": "Which of the following is NOT true about genetic drift?\nA: Can cause harmful alleles to become fixed\nB: Can cause allele frequencies to change at random\nC: Significant in small populations\nD: Can lead to a loss of genetic variation within populations\nE: Prevents allele frequencies from fluctuating over time\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is NOT true about genetic drift?\n\nA: Can cause harmful alleles to become fixed\nB: Can cause allele frequencies to change at random\nC: Significant in small populations\nD: Can lead to a loss of genetic variation within populations\nE: Prevents allele frequencies from fluctuating over time\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1125",
        "problem": "The flat body plan of platyhelminthes helps them to:\nA: attain effective locomotion in any direction.\nB: minimize the loss of body fluid and maximize surface area.\nC: facilitate successful reproduction.\nD: maximize exposure of animal's cells/ tissues to the surrounding medium.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe flat body plan of platyhelminthes helps them to:\n\nA: attain effective locomotion in any direction.\nB: minimize the loss of body fluid and maximize surface area.\nC: facilitate successful reproduction.\nD: maximize exposure of animal's cells/ tissues to the surrounding medium.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1500",
        "problem": "Dynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\nSpecies iii is found to have the sequence GCATGCA.\n\nWhich is true?\nA: Species ii and i are more closely related than species ii and iii.\nB: The best alignment between species ii and iii has a higher score than the best alignment between species ii and i.\nC: There are more high scoring alignments between species ii and iii than between species ii and i.\nD: There are more gaps in the alignment between species ii and iii than between species ii and i.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\nSpecies iii is found to have the sequence GCATGCA.\n\nWhich is true?\n\nA: Species ii and i are more closely related than species ii and iii.\nB: The best alignment between species ii and iii has a higher score than the best alignment between species ii and i.\nC: There are more high scoring alignments between species ii and iii than between species ii and i.\nD: There are more gaps in the alignment between species ii and iii than between species ii and i.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=925&width=928&top_left_y=354&top_left_x=238",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=922&width=928&top_left_y=1458&top_left_x=238"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1208",
        "problem": "## RATITE EVOLUTION\n\nRatites are named for their flat \"raft-like\" sternum that lacks a keel and thus cannot provide sufficient anchorage for flight muscles to effectively generate the power required to fly. Cassowary (Australia and Papua New Guinea), Emu (Australia), Rhea (South America), Ostrich (Africa), Kiwi (New Zealand), and extinct Moa (New Zealand) are all examples of ground dwelling, flightless ratites. Given their large, flightless nature, biologists have proposed that they have all descended from a common ancestor present in ancient Gondwana before it broke apart. The exact timing and evolutionary origin of the Moa and Kiwi in New Zealand have been greatly debated in the scientific literature.\n\n[figure1]\nfrom Antarctica and India\n\n- 130 mya South America breaks free of Africa\n- 80 mya Zealandia breaks free\n- 60 mya New Zealand separates from Australia\n- 40 mya Australia separates from Antarctica\n- 30 mya South America breaks from West Antarctica (Antarctica freezes over)\n\n[figure2]\n\nFuture Zealandia\n\n## Flying Cousins\n\nRatites are one of only two groups of birds belonging to the \"old jaw\" Paleognaths originating in Gondwana, the other comprises the 47 living Tinamou species of Central and South America. Tinamou species are generally ground dwelling, though they do have wings that allow for limited flight.\n\nIt is generally accepted that loss of flight in birds is due to the development of successful foraging behaviours and diminished predation on eggs and nests.\n\n## Fossil Evidence\n\nExamination of the oldest known Moa fossils, dating from 19 million years ago, determined they had thickened leg bones and no wing structures at all. At least two species of flightless Moa were present in New Zealand at this time.\n\nThe oldest Tinamou fossils in South America are 10 million years old.\n\nThe oldest Kiwi fossil dates back 1 million years. Though diminished in size, kiwis have wings and flight feathers. Kiwis are the only known bird to have nostrils at the end of their beaks and the only ratite with two functioning ovaries.Scientists have proposed that Moa and Kiwi share an immediate common ancestor. Which of the following statements would BEST support this hypothesis?\nA: They are both found in New Zealand.\nB: They are both flightless birds.\nC: Kiwi and Moa are not found outside of New Zealand.\nD: Without help, Kiwi are likely to go extinct.\nE: Oldest Kiwi fossils are much younger than the oldest Moa fossils.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## RATITE EVOLUTION\n\nRatites are named for their flat \"raft-like\" sternum that lacks a keel and thus cannot provide sufficient anchorage for flight muscles to effectively generate the power required to fly. Cassowary (Australia and Papua New Guinea), Emu (Australia), Rhea (South America), Ostrich (Africa), Kiwi (New Zealand), and extinct Moa (New Zealand) are all examples of ground dwelling, flightless ratites. Given their large, flightless nature, biologists have proposed that they have all descended from a common ancestor present in ancient Gondwana before it broke apart. The exact timing and evolutionary origin of the Moa and Kiwi in New Zealand have been greatly debated in the scientific literature.\n\n[figure1]\nfrom Antarctica and India\n\n- 130 mya South America breaks free of Africa\n- 80 mya Zealandia breaks free\n- 60 mya New Zealand separates from Australia\n- 40 mya Australia separates from Antarctica\n- 30 mya South America breaks from West Antarctica (Antarctica freezes over)\n\n[figure2]\n\nFuture Zealandia\n\n## Flying Cousins\n\nRatites are one of only two groups of birds belonging to the \"old jaw\" Paleognaths originating in Gondwana, the other comprises the 47 living Tinamou species of Central and South America. Tinamou species are generally ground dwelling, though they do have wings that allow for limited flight.\n\nIt is generally accepted that loss of flight in birds is due to the development of successful foraging behaviours and diminished predation on eggs and nests.\n\n## Fossil Evidence\n\nExamination of the oldest known Moa fossils, dating from 19 million years ago, determined they had thickened leg bones and no wing structures at all. At least two species of flightless Moa were present in New Zealand at this time.\n\nThe oldest Tinamou fossils in South America are 10 million years old.\n\nThe oldest Kiwi fossil dates back 1 million years. Though diminished in size, kiwis have wings and flight feathers. Kiwis are the only known bird to have nostrils at the end of their beaks and the only ratite with two functioning ovaries.\n\nproblem:\nScientists have proposed that Moa and Kiwi share an immediate common ancestor. Which of the following statements would BEST support this hypothesis?\n\nA: They are both found in New Zealand.\nB: They are both flightless birds.\nC: Kiwi and Moa are not found outside of New Zealand.\nD: Without help, Kiwi are likely to go extinct.\nE: Oldest Kiwi fossils are much younger than the oldest Moa fossils.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-24.jpg?height=711&width=585&top_left_y=689&top_left_x=135",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-24.jpg?height=708&width=1259&top_left_y=688&top_left_x=747"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_880",
        "problem": "染色体平衡易位又称为相互易位, 是指两条染色体发生断裂后相互交换, 形成两条\n新的衍生染色体, 基因总数未丢失, 大部分平衡易位对基因表达和个体发育无严重影响,但对雌雄配子的活性有相同的影响。平衡易位（只考虑非同源染色体间片段的互换）杂合体的情况比较复杂, 当易位区段较长时, 在减数分裂I的前期, 由于同源部分 7 的配对, 而出现富有特征性的“+”字形结构, 称为“四射体”, 如图所示。已知“四射体”中的每一条染色体在减数分裂I的后期会随机移向一极, 不考虑突变、交换等其他变异类型。以下分析说法错误的是（ ）\n\n[图1]\nA: 若图中的四射体是在某雌性果蝇体内发现的, 则该体细胞内含有 6 种形态不同的染色体\nB: 观察平衡易位染色体也可选择有丝分裂中期的细胞\nC: 若仅考虑发生相互易位的两对同源染色体及其上基因, 人类平衡易位杂合体理论上产生的配子种类数为 6 种\nD: 人类平衡易位杂合体与正常人婚后生育的后代也可能是平衡易位杂合体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n染色体平衡易位又称为相互易位, 是指两条染色体发生断裂后相互交换, 形成两条\n新的衍生染色体, 基因总数未丢失, 大部分平衡易位对基因表达和个体发育无严重影响,但对雌雄配子的活性有相同的影响。平衡易位（只考虑非同源染色体间片段的互换）杂合体的情况比较复杂, 当易位区段较长时, 在减数分裂I的前期, 由于同源部分 7 的配对, 而出现富有特征性的“+”字形结构, 称为“四射体”, 如图所示。已知“四射体”中的每一条染色体在减数分裂I的后期会随机移向一极, 不考虑突变、交换等其他变异类型。以下分析说法错误的是（ ）\n\n[图1]\n\nA: 若图中的四射体是在某雌性果蝇体内发现的, 则该体细胞内含有 6 种形态不同的染色体\nB: 观察平衡易位染色体也可选择有丝分裂中期的细胞\nC: 若仅考虑发生相互易位的两对同源染色体及其上基因, 人类平衡易位杂合体理论上产生的配子种类数为 6 种\nD: 人类平衡易位杂合体与正常人婚后生育的后代也可能是平衡易位杂合体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-78.jpg?height=380&width=811&top_left_y=644&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_663",
        "problem": "玉米的花色由两对独立遗传的等位基因 (A/a、B/b) 控制, 其花色遗传如下图所示。玉米的性别由（ $\\mathrm{D} / \\mathrm{d} 、 \\mathrm{~T} / \\mathrm{t}$ ）两对独立遗传的等位基因控制。 $\\mathrm{D} 、 \\mathrm{~T}$ 基因同时存在时，雌花雄花都能发育; 当 $\\mathrm{T}$ 基因存在且无 $\\mathrm{D}$ 基因时植株无雌花; 无 $\\mathrm{T}$ 基因时无雄花。现有一株基因型为 AABBDDTT 的植株与基因型为 aabbddtt 的植株交配，得到 $\\mathrm{F}_{1}$, 下列叙述错误的是 ( )\n\n[图1]\nA: 亲本中母本为白花, 父本为紫花\nB: $F_{1}$ 随机传粉, $F_{2}$ 中红色花占 $3 / 16$, 有雌花的植株占 $13 / 16$\nC: 将 $F_{2}$ ( $F_{1}$ 随机传粉所得) 中同时有雌雄花的植株的雄花摘除, 取只有雄花植株的花粉对其授粉, 子代中只有雄花的个体占 $4 / 27$\nD: 将 $F_{2}\\left(F_{1}\\right.$ 随机传粉所得) 中同时有雌雄花的植株摘除, $F_{2}$ 随机传粉, 雌雄花都有、只有雄花和只有雌花所占比值相同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n玉米的花色由两对独立遗传的等位基因 (A/a、B/b) 控制, 其花色遗传如下图所示。玉米的性别由（ $\\mathrm{D} / \\mathrm{d} 、 \\mathrm{~T} / \\mathrm{t}$ ）两对独立遗传的等位基因控制。 $\\mathrm{D} 、 \\mathrm{~T}$ 基因同时存在时，雌花雄花都能发育; 当 $\\mathrm{T}$ 基因存在且无 $\\mathrm{D}$ 基因时植株无雌花; 无 $\\mathrm{T}$ 基因时无雄花。现有一株基因型为 AABBDDTT 的植株与基因型为 aabbddtt 的植株交配，得到 $\\mathrm{F}_{1}$, 下列叙述错误的是 ( )\n\n[图1]\n\nA: 亲本中母本为白花, 父本为紫花\nB: $F_{1}$ 随机传粉, $F_{2}$ 中红色花占 $3 / 16$, 有雌花的植株占 $13 / 16$\nC: 将 $F_{2}$ ( $F_{1}$ 随机传粉所得) 中同时有雌雄花的植株的雄花摘除, 取只有雄花植株的花粉对其授粉, 子代中只有雄花的个体占 $4 / 27$\nD: 将 $F_{2}\\left(F_{1}\\right.$ 随机传粉所得) 中同时有雌雄花的植株摘除, $F_{2}$ 随机传粉, 雌雄花都有、只有雄花和只有雌花所占比值相同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-052.jpg?height=274&width=419&top_left_y=2022&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_216",
        "problem": "Molecular phylogenetics is a powerful tool for inferring phylogenetic relationships among extant species. The following are methodological statements on molecular phylogenetics.\nA: We must choose gene(s) with faster evolutionary rate(s) when inferring a phylogenetic tree of species with older divergence.\nB: In order to infer phylogenetic relationships between species, paralogous gene(s) that were duplicated during the evolution of the subject group should not be analyzed.\nC: To root a phylogenetic tree with an outgroup, we should choose a species, which is as distantly related to the subject species as possible.\nD: Two species (sp. $\\mathbf{X}$ and sp. Y) are described based on morphological characteristics. Here, we sequenced a gene from five individuals of each species. As a result, we found that the gene sequence of an individual of sp. $\\mathbf{Y}$ is more similar to those of five individuals of $\\mathrm{sp.} \\mathbf{X}$ than those of other four individuals of $\\mathrm{sp.} \\mathbf{Y}$. This result contradicts the biological species concept.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMolecular phylogenetics is a powerful tool for inferring phylogenetic relationships among extant species. The following are methodological statements on molecular phylogenetics.\n\nA: We must choose gene(s) with faster evolutionary rate(s) when inferring a phylogenetic tree of species with older divergence.\nB: In order to infer phylogenetic relationships between species, paralogous gene(s) that were duplicated during the evolution of the subject group should not be analyzed.\nC: To root a phylogenetic tree with an outgroup, we should choose a species, which is as distantly related to the subject species as possible.\nD: Two species (sp. $\\mathbf{X}$ and sp. Y) are described based on morphological characteristics. Here, we sequenced a gene from five individuals of each species. As a result, we found that the gene sequence of an individual of sp. $\\mathbf{Y}$ is more similar to those of five individuals of $\\mathrm{sp.} \\mathbf{X}$ than those of other four individuals of $\\mathrm{sp.} \\mathbf{Y}$. This result contradicts the biological species concept.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_110",
        "problem": "Pentose sugars and NADPH are synthetized through the pentose phosphate pathway.\n\n[figure1]\n\nReduced glutathione (GSH) acts to neutralize reactive oxygen species (ROSs) in the body. The reaction for GSH generation is shown below:\n\n## Glutathione reductase\n\n[figure2]\n\nIndividuals with glucose-6-phospho dehydrogenase (G6PD) deficiency have decreased level of NADPH which can cause the favism disorder. Compared to other populations, frequency of G6PD deficiency is higher in Africans, and its prevalence in Africa positively correlates with the prevalence of malaria.\nA: The high prevalence of G6PD deficiency in regions with high incidence of malaria can be explained by high susceptibility of Plasmodium falciparum to oxidative agents.\nB: Increase in levels of cell division is accompanied by increase in the ratio of fructose- 6 phosphate/ ribose- 5 phosphate production by the pentose phosphate pathway.\nC: G6PD deficiency affects catabolism more than anabolism in individuals with the deficiency.\nD: It is expected that the cause of favism in some affected individuals may be glutathione reductase deficiency.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPentose sugars and NADPH are synthetized through the pentose phosphate pathway.\n\n[figure1]\n\nReduced glutathione (GSH) acts to neutralize reactive oxygen species (ROSs) in the body. The reaction for GSH generation is shown below:\n\n## Glutathione reductase\n\n[figure2]\n\nIndividuals with glucose-6-phospho dehydrogenase (G6PD) deficiency have decreased level of NADPH which can cause the favism disorder. Compared to other populations, frequency of G6PD deficiency is higher in Africans, and its prevalence in Africa positively correlates with the prevalence of malaria.\n\nA: The high prevalence of G6PD deficiency in regions with high incidence of malaria can be explained by high susceptibility of Plasmodium falciparum to oxidative agents.\nB: Increase in levels of cell division is accompanied by increase in the ratio of fructose- 6 phosphate/ ribose- 5 phosphate production by the pentose phosphate pathway.\nC: G6PD deficiency affects catabolism more than anabolism in individuals with the deficiency.\nD: It is expected that the cause of favism in some affected individuals may be glutathione reductase deficiency.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-15.jpg?height=1031&width=1014&top_left_y=450&top_left_x=521",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-15.jpg?height=163&width=500&top_left_y=1663&top_left_x=778"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1392",
        "problem": "The left ventricle of the heart pumps blood around the body. The amount of blood pumped out of the heart during one contraction is called the cardiac output. The cardiac output varies according to end diastolic volume (see diagram below). The end diastolic volume, also known as the preload, is the volume of blood in the left ventricle before contraction. The cardiac output is also influenced by contractility, which is the ability of the heart muscle to contract.\n\n[figure1]\n\nWhat would NOT result from an increased end-diastolic volume?\nA: Increased stretching of the left ventricle.\nB: Weaker contraction of the left ventricle.\nC: Increased preload.\nD: Increased cardiac output.\nE: Increased volume of blood in the left ventricle before contraction.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe left ventricle of the heart pumps blood around the body. The amount of blood pumped out of the heart during one contraction is called the cardiac output. The cardiac output varies according to end diastolic volume (see diagram below). The end diastolic volume, also known as the preload, is the volume of blood in the left ventricle before contraction. The cardiac output is also influenced by contractility, which is the ability of the heart muscle to contract.\n\n[figure1]\n\nWhat would NOT result from an increased end-diastolic volume?\n\nA: Increased stretching of the left ventricle.\nB: Weaker contraction of the left ventricle.\nC: Increased preload.\nD: Increased cardiac output.\nE: Increased volume of blood in the left ventricle before contraction.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-19.jpg?height=939&width=1245&top_left_y=661&top_left_x=405"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_916",
        "problem": "某地发现一种人类遗传病，该病受 $\\mathrm{X}$ 染色体上的两对等位基因(A、a 和 B、b)控制,且只有 $A 、 B$ 基因同时存在时个体才表现正常。如图为某家族的遗传系谱图, I-1 的基因型为 $X^{A b} Y$, 不考虑基因突变和染色体变异，下列分析正确的是（ ）\n\n[图1]\nA: I-2 产生的配子肯定有两种类型\nB: III -2 和III -1 再生一个孩子患病的概率为 $1 / 4$\nC: 患病的II -3 和IV- 1 的基因型相同\nD: II -2 的初级卵母细胞含有 2 个显性基因\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某地发现一种人类遗传病，该病受 $\\mathrm{X}$ 染色体上的两对等位基因(A、a 和 B、b)控制,且只有 $A 、 B$ 基因同时存在时个体才表现正常。如图为某家族的遗传系谱图, I-1 的基因型为 $X^{A b} Y$, 不考虑基因突变和染色体变异，下列分析正确的是（ ）\n\n[图1]\n\nA: I-2 产生的配子肯定有两种类型\nB: III -2 和III -1 再生一个孩子患病的概率为 $1 / 4$\nC: 患病的II -3 和IV- 1 的基因型相同\nD: II -2 的初级卵母细胞含有 2 个显性基因\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-24.jpg?height=377&width=834&top_left_y=837&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1539",
        "problem": "Blue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nEstimate the \\% change in whale population from 2010 to 2020 .\nA: 0.9\nB: 9\nC: 90\nD: 900\nE: 4.5\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBlue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nEstimate the \\% change in whale population from 2010 to 2020 .\n\nA: 0.9\nB: 9\nC: 90\nD: 900\nE: 4.5\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-36.jpg?height=948&width=1673&top_left_y=657&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1154",
        "problem": "In still air, the transpiration rate is low because\nA: stomata open more when the diffusion gradient changes.\nB: the concentration gradient of water molecules is reduced\nC: the water molecules move more slowly\nD: root pressure is reduced\nE: water molecules are more strongly attracted to each other\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn still air, the transpiration rate is low because\n\nA: stomata open more when the diffusion gradient changes.\nB: the concentration gradient of water molecules is reduced\nC: the water molecules move more slowly\nD: root pressure is reduced\nE: water molecules are more strongly attracted to each other\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_452",
        "problem": "果蝇红眼和白眼是由一对等位基因控制的。现有用纯种果蝇进行的两组杂交实验,\n结果如下图所示。下列判断错误的是（）\n\n实验 (1)\n\n$\\mathrm{P}$ 우红眼 $\\times$ 全白眼\n\n$\\mathrm{F}_{1}$ 우红眼 吕红眼性染色体 XX XY实验(2)\n\n$\\mathrm{P}$ 우白眼 $\\mathrm{x}$ 食红眼\n\n$\\mathrm{F}_{1}$ 우红眼 合白眼 우白眼 含红眼 XX $\\quad X Y \\quad X X Y \\quad X$\nA: Y 染色体不是果蝇发育成雄性的必要条件\nB: 实验(2)中子代红眼 0 :白眼 $Q+=1: 1$\nC: 实验(2)子代出现白眼우的原因是母本减数分裂 $\\mathrm{I}$ 异常\nD: 实验(1)中子代红眼\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇红眼和白眼是由一对等位基因控制的。现有用纯种果蝇进行的两组杂交实验,\n结果如下图所示。下列判断错误的是（）\n\n实验 (1)\n\n$\\mathrm{P}$ 우红眼 $\\times$ 全白眼\n\n$\\mathrm{F}_{1}$ 우红眼 吕红眼性染色体 XX XY实验(2)\n\n$\\mathrm{P}$ 우白眼 $\\mathrm{x}$ 食红眼\n\n$\\mathrm{F}_{1}$ 우红眼 合白眼 우白眼 含红眼 XX $\\quad X Y \\quad X X Y \\quad X$\n\nA: Y 染色体不是果蝇发育成雄性的必要条件\nB: 实验(2)中子代红眼 0 :白眼 $Q+=1: 1$\nC: 实验(2)子代出现白眼우的原因是母本减数分裂 $\\mathrm{I}$ 异常\nD: 实验(1)中子代红眼\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1029",
        "problem": "All of the following processes involve hydrogen bonding except:\nA: DNA replication\nB: Formation of ice crystals\nC: Binding of enzyme and substrate\nD: Protein folding\nE: All of the above\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAll of the following processes involve hydrogen bonding except:\n\nA: DNA replication\nB: Formation of ice crystals\nC: Binding of enzyme and substrate\nD: Protein folding\nE: All of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_201",
        "problem": "Two different mutant forms of a gene that encodes a cell surface receptor tyrosine kinase (RTK) are separately inserted into vectors. One mutant encodes a protein with a non-functional kinase domain, and the other lacks a functional ligand binding domain. Each vector is separately introduced into normal cells that can express wild type RTK from their endogenous genes.\n\nIt is known that the cells used have a high capacity for RTK receptors on their surface, that ligands bind to monomeric forms of receptor proteins and that heterodimeric receptors are inactive in signalling. The diagrams below depict four cell types identified. Each diagram shows the only receptor forms observed on the respective cell types in the ratio that they were observed.\n\nThe experiments were performed under non-saturating concentrations of ligand.\n\n[figure1]\n\n1\n\n[figure2]\n\n3\n\n[figure3]\n\n2\n\n[figure4]\n\n4\n\nwild type kinase domain\n\nmutant kinase domain\n\nmutant ligand binding domain\n\nwild type ligand binding domain\nA: In type 1 cells, the mutant receptor with the non-functional kinase domain will interfere with signalling by the cells' normal RTK.\nB: In type 2 cells, the mutant RTK lacking functional ligand binding domain will be inactive for signalling, but will not interfere with normal signalling mediated by the cells' own receptor tyrosine kinases.\nC: Equal levels of signalling will be achieved by type 3 and type 4 cells.\nD: The effects of mutant RTKs of cell type 2 and cell type 3 on levels of signalling by the cells' own normal RTKs will be the same.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nTwo different mutant forms of a gene that encodes a cell surface receptor tyrosine kinase (RTK) are separately inserted into vectors. One mutant encodes a protein with a non-functional kinase domain, and the other lacks a functional ligand binding domain. Each vector is separately introduced into normal cells that can express wild type RTK from their endogenous genes.\n\nIt is known that the cells used have a high capacity for RTK receptors on their surface, that ligands bind to monomeric forms of receptor proteins and that heterodimeric receptors are inactive in signalling. The diagrams below depict four cell types identified. Each diagram shows the only receptor forms observed on the respective cell types in the ratio that they were observed.\n\nThe experiments were performed under non-saturating concentrations of ligand.\n\n[figure1]\n\n1\n\n[figure2]\n\n3\n\n[figure3]\n\n2\n\n[figure4]\n\n4\n\nwild type kinase domain\n\nmutant kinase domain\n\nmutant ligand binding domain\n\nwild type ligand binding domain\n\nA: In type 1 cells, the mutant receptor with the non-functional kinase domain will interfere with signalling by the cells' normal RTK.\nB: In type 2 cells, the mutant RTK lacking functional ligand binding domain will be inactive for signalling, but will not interfere with normal signalling mediated by the cells' own receptor tyrosine kinases.\nC: Equal levels of signalling will be achieved by type 3 and type 4 cells.\nD: The effects of mutant RTKs of cell type 2 and cell type 3 on levels of signalling by the cells' own normal RTKs will be the same.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-13.jpg?height=265&width=417&top_left_y=850&top_left_x=594",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-13.jpg?height=311&width=414&top_left_y=1215&top_left_x=615",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-13.jpg?height=275&width=377&top_left_y=865&top_left_x=1088",
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-13.jpg?height=309&width=420&top_left_y=1219&top_left_x=1041"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1271",
        "problem": "The New Zealand crayfish, Jasus edwardsii, has a complex life cycle. The female attaches the fertilized eggs to her abdomen. The eggs hatch as a small spidery creature called a naupliosoma larva and are released by the female. The naupliosoma swims to the surface and moults, transforming into a leaf-like larva, known as a phyllosoma. In this form, the crayfish larva spends an extended period floating in ocean currents that carry it far from shore. At the end of this time, the final moult transforms the larva into a miniature transparent version of the adult called a puerulus larva. The puerulus swims back to the coast and settles, becoming a small coloured adult.\n\nPlace the photos of the crayfish life stages below into the correct order.\n[figure1]\n\nPhotos are from http://www.teara.govt.nz/en/crabs-crayfish-and-other-crustaceans/media and www.wilderness.org.au\nA: $1,2,3,4$\nB: $1,3,2,4$\nC: $2,3,1,4$\nD: $4,3,2,1$\nE: $4,2,3,1$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe New Zealand crayfish, Jasus edwardsii, has a complex life cycle. The female attaches the fertilized eggs to her abdomen. The eggs hatch as a small spidery creature called a naupliosoma larva and are released by the female. The naupliosoma swims to the surface and moults, transforming into a leaf-like larva, known as a phyllosoma. In this form, the crayfish larva spends an extended period floating in ocean currents that carry it far from shore. At the end of this time, the final moult transforms the larva into a miniature transparent version of the adult called a puerulus larva. The puerulus swims back to the coast and settles, becoming a small coloured adult.\n\nPlace the photos of the crayfish life stages below into the correct order.\n[figure1]\n\nPhotos are from http://www.teara.govt.nz/en/crabs-crayfish-and-other-crustaceans/media and www.wilderness.org.au\n\nA: $1,2,3,4$\nB: $1,3,2,4$\nC: $2,3,1,4$\nD: $4,3,2,1$\nE: $4,2,3,1$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-16.jpg?height=266&width=1470&top_left_y=638&top_left_x=113"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_129",
        "problem": "Microorganisms that live at high salt concentration (above $2 \\mathrm{M}$ of $\\mathrm{NaCl}$ ) are exposed to media with low water activity, and must have mechanisms to avoid water loss by osmosis. Analyses of intracellular ionic concentration of Halobacteriales living in salt lakes show that these microorganisms maintain extremely high salt $(\\mathrm{KCl})$ concentration inside their cells. The presence of high intracellular salt concentration requires special adaptations of the proteins and other macromolecules of the cells.\nA: Most intracellular proteins of Halobacteriales contain a large excess of negative charges on their outer surface.\nB: Halobacteriales spend a lot of ATPs to maintain osmotic pressure.\nC: Most intracellular enzymes of Halobacteriales lose their catalytic activity when suspended in solutions containing less than $1 \\mathrm{M} \\mathrm{NaCl}$.\nD: Amino acids can be imported through $\\mathrm{Na}^{+} /$amino acids antiporters.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nMicroorganisms that live at high salt concentration (above $2 \\mathrm{M}$ of $\\mathrm{NaCl}$ ) are exposed to media with low water activity, and must have mechanisms to avoid water loss by osmosis. Analyses of intracellular ionic concentration of Halobacteriales living in salt lakes show that these microorganisms maintain extremely high salt $(\\mathrm{KCl})$ concentration inside their cells. The presence of high intracellular salt concentration requires special adaptations of the proteins and other macromolecules of the cells.\n\nA: Most intracellular proteins of Halobacteriales contain a large excess of negative charges on their outer surface.\nB: Halobacteriales spend a lot of ATPs to maintain osmotic pressure.\nC: Most intracellular enzymes of Halobacteriales lose their catalytic activity when suspended in solutions containing less than $1 \\mathrm{M} \\mathrm{NaCl}$.\nD: Amino acids can be imported through $\\mathrm{Na}^{+} /$amino acids antiporters.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_593",
        "problem": "研究发现单基因突变导致卵母细胞死亡是女性无法生育的原因之一。图甲为某不孕女性家族系谱图, 图乙为家族成员一对基因的模板链部分测序结果(注：箭头处 GA 表示一个基因的模板链该位点为 $\\mathrm{G}$, 另一个模板链该位点为 $\\mathrm{A} ; \\mathrm{G}$ 表示此位点两条模板链都为 $\\mathrm{G})$\n\n[图1]\n\n甲\n[图2]\n\n乙\n\n下列相关叙述正确的是（ ）\nA: 由图乙可知, II-1 的致病基因来自 I-2\nB: 卵母细胞死亡导致的不育为隐性性状\nC: II-1 不育的根本原因是基因碱基对发生了替换\nD: II-5 和正常女性婚配, 子代患病的概率是 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究发现单基因突变导致卵母细胞死亡是女性无法生育的原因之一。图甲为某不孕女性家族系谱图, 图乙为家族成员一对基因的模板链部分测序结果(注：箭头处 GA 表示一个基因的模板链该位点为 $\\mathrm{G}$, 另一个模板链该位点为 $\\mathrm{A} ; \\mathrm{G}$ 表示此位点两条模板链都为 $\\mathrm{G})$\n\n[图1]\n\n甲\n[图2]\n\n乙\n\n下列相关叙述正确的是（ ）\n\nA: 由图乙可知, II-1 的致病基因来自 I-2\nB: 卵母细胞死亡导致的不育为隐性性状\nC: II-1 不育的根本原因是基因碱基对发生了替换\nD: II-5 和正常女性婚配, 子代患病的概率是 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-40.jpg?height=417&width=648&top_left_y=1573&top_left_x=384",
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-40.jpg?height=430&width=592&top_left_y=1558&top_left_x=1094"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_681",
        "problem": "二倍体牵牛花的花色有红色、粉色和白色三个品种。将两株基因型相同的牵牛花杂交, 后代出现红色: 粉色: 白色=4: 4：1。若不考虑基因突变和染色体变异, 涉及的基因用 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 表示。以下假设和分析正确的是（）\n\n| 项 | 假设等位基 <br> 因数 | 假设基因 <br> 位置 | 花色基因型 | 原因分析 |\n| :---: | :---: | :---: | :---: | :---: |\n| A | 1 对 | 1 对同源 <br> 染色体 | AA 红色、Aa 粉色、aa 白 <br> 色 | 含基因 $\\mathrm{a}$ 的雌配子或雄配 <br> 子致死 |\n| B | 1 对 | 1 对同源 <br> 染色体 | AA 红色、Aa 粉色、aa 白 <br> 色 | 含基因 $\\mathrm{A}$ 的雌配子或雄配 <br> 子 $50 \\%$ 致死 |\n| $\\mathrm{C}$ | 2 对 | 2 对同源 <br> 染色体 | $\\mathrm{AaBb}$ 红色、 $\\mathrm{Aabb}$ 和 $\\mathrm{aaBb}$ <br> 粉色、 $\\mathrm{aabb}$ 白色 | $\\mathrm{AA}$ 或 $\\mathrm{BB}$ 纯合均致死 |\n| $\\mathrm{D}$ | 2 对 | 2 对同源 <br> 染色体 | $\\mathrm{AaBb}$ 红色、 $\\mathrm{Aabb}$ 和 aabb <br> 粉色、 $\\mathrm{aaBb}$ 白色 | $\\mathrm{AA}$ 或 BB 纯合均致死 |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n二倍体牵牛花的花色有红色、粉色和白色三个品种。将两株基因型相同的牵牛花杂交, 后代出现红色: 粉色: 白色=4: 4：1。若不考虑基因突变和染色体变异, 涉及的基因用 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 表示。以下假设和分析正确的是（）\n\n| 项 | 假设等位基 <br> 因数 | 假设基因 <br> 位置 | 花色基因型 | 原因分析 |\n| :---: | :---: | :---: | :---: | :---: |\n| A | 1 对 | 1 对同源 <br> 染色体 | AA 红色、Aa 粉色、aa 白 <br> 色 | 含基因 $\\mathrm{a}$ 的雌配子或雄配 <br> 子致死 |\n| B | 1 对 | 1 对同源 <br> 染色体 | AA 红色、Aa 粉色、aa 白 <br> 色 | 含基因 $\\mathrm{A}$ 的雌配子或雄配 <br> 子 $50 \\%$ 致死 |\n| $\\mathrm{C}$ | 2 对 | 2 对同源 <br> 染色体 | $\\mathrm{AaBb}$ 红色、 $\\mathrm{Aabb}$ 和 $\\mathrm{aaBb}$ <br> 粉色、 $\\mathrm{aabb}$ 白色 | $\\mathrm{AA}$ 或 $\\mathrm{BB}$ 纯合均致死 |\n| $\\mathrm{D}$ | 2 对 | 2 对同源 <br> 染色体 | $\\mathrm{AaBb}$ 红色、 $\\mathrm{Aabb}$ 和 aabb <br> 粉色、 $\\mathrm{aaBb}$ 白色 | $\\mathrm{AA}$ 或 BB 纯合均致死 |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_882",
        "problem": "二倍体生物细胞正在进行着丝点分裂时, 下列有关叙述正确的是 ( )\nA: 细胞中一定不存在同源染色体\nB: 着丝点分裂一定导致 DNA 数目加倍\nC: 染色体 DNA 一定由母链和子链组成\nD: 细胞中染色体数目一定是其体细胞的 2 倍\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n二倍体生物细胞正在进行着丝点分裂时, 下列有关叙述正确的是 ( )\n\nA: 细胞中一定不存在同源染色体\nB: 着丝点分裂一定导致 DNA 数目加倍\nC: 染色体 DNA 一定由母链和子链组成\nD: 细胞中染色体数目一定是其体细胞的 2 倍\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_714",
        "problem": "苯内酮尿症（PKU）患者体内, 因产生过多的苯丙酮酸, 导致神经系统受到伤害。 PHA 基因控制合成苯丙氨酸羟化酶, 对某家系成员的 PHA 相关基因进行扩增, 用 Rsa1 酶切后并电泳，结果如图所示。推测下列说法错误的是（）\n\n[图1]\nA: 导致 PKU 的根本原因是 PHA 基因碱基对发生了替换\nB: PHA 基因是位于常染色体上的隐性基因\nC: 若在形成精子的过程中, PHA 基因与等位基因发生了互换, 则生一个 PKU 患者的概率会增大\nD: 以上体现了基因通过控制酶的合成控制代谢，进而控制性状\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n苯内酮尿症（PKU）患者体内, 因产生过多的苯丙酮酸, 导致神经系统受到伤害。 PHA 基因控制合成苯丙氨酸羟化酶, 对某家系成员的 PHA 相关基因进行扩增, 用 Rsa1 酶切后并电泳，结果如图所示。推测下列说法错误的是（）\n\n[图1]\n\nA: 导致 PKU 的根本原因是 PHA 基因碱基对发生了替换\nB: PHA 基因是位于常染色体上的隐性基因\nC: 若在形成精子的过程中, PHA 基因与等位基因发生了互换, 则生一个 PKU 患者的概率会增大\nD: 以上体现了基因通过控制酶的合成控制代谢，进而控制性状\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-025.jpg?height=968&width=1110&top_left_y=641&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_981",
        "problem": "While preparing to teach an organic chemistry class, Professor Dacb has managed to mix up several beakers. He has taken samples from each beaker and enlisted your help in identifying them. Below are the results of a series of reagent tests.\n\n| Beaker | Biuret | Benedict's | Ninhydrin | Iodine | Sudan |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | Blue | Blue | Blue | Orange | Rather pale |\n| 2 | Blue | Blue | Pale yellow | Red-orange | Red-orange |\n| 3 | Purple | Blue | Pale yellow | Orange | Rather pale |\n| 4 | Blue | Red precipitate | Pale yellow | Orange-yellow | Rather pale |\n| 5 | Blue | Blue | Pale yellow | Orange | Rather pale |\n\nWhich of the following correctly matches each beaker to its possible contents?\nA: 1: Gelatin, 2: Corn oil, 3: A solution of asparagine, 4: A solution of glucose, 5: A solution of sucrose\nB: 1: Gelatin, 2: Triglycerides, 3: A solution of asparagine, 4: A solution of sucrose, 5: A solution of glucose\nC: 1: A solution of asparagine, 2: Triglycerides, 3: Gelatin, 4: A solution of glucose, 5: A solution of sucrose\nD: 1: A solution of sucrose, 2: Corn oil, 3: A solution of glucose, 4: A solution of asparagine, 5: Gelatin\nE: 1: A solution of glucose, 2: Triglycerides, 3: A solution of sucrose, 4: A solution of asparagine, 5: Gelatin\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhile preparing to teach an organic chemistry class, Professor Dacb has managed to mix up several beakers. He has taken samples from each beaker and enlisted your help in identifying them. Below are the results of a series of reagent tests.\n\n| Beaker | Biuret | Benedict's | Ninhydrin | Iodine | Sudan |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | Blue | Blue | Blue | Orange | Rather pale |\n| 2 | Blue | Blue | Pale yellow | Red-orange | Red-orange |\n| 3 | Purple | Blue | Pale yellow | Orange | Rather pale |\n| 4 | Blue | Red precipitate | Pale yellow | Orange-yellow | Rather pale |\n| 5 | Blue | Blue | Pale yellow | Orange | Rather pale |\n\nWhich of the following correctly matches each beaker to its possible contents?\n\nA: 1: Gelatin, 2: Corn oil, 3: A solution of asparagine, 4: A solution of glucose, 5: A solution of sucrose\nB: 1: Gelatin, 2: Triglycerides, 3: A solution of asparagine, 4: A solution of sucrose, 5: A solution of glucose\nC: 1: A solution of asparagine, 2: Triglycerides, 3: Gelatin, 4: A solution of glucose, 5: A solution of sucrose\nD: 1: A solution of sucrose, 2: Corn oil, 3: A solution of glucose, 4: A solution of asparagine, 5: Gelatin\nE: 1: A solution of glucose, 2: Triglycerides, 3: A solution of sucrose, 4: A solution of asparagine, 5: Gelatin\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1524",
        "problem": "Blue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nAnother project uses a mark-recapture method to count whales, but it takes several years to find enough whales. *Note whales can be visited by a boat without causing distress* In the capture period, 300 whales were identified. In the recapture period, 450 whales were found but only 5 were individuals identified in the capture period.\n\nGive another estimate of the whale population.\nA: 2000\nB: 3000\nC: 20000\nD: 30000\nE: 40000\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBlue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nAnother project uses a mark-recapture method to count whales, but it takes several years to find enough whales. *Note whales can be visited by a boat without causing distress* In the capture period, 300 whales were identified. In the recapture period, 450 whales were found but only 5 were individuals identified in the capture period.\n\nGive another estimate of the whale population.\n\nA: 2000\nB: 3000\nC: 20000\nD: 30000\nE: 40000\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-36.jpg?height=948&width=1673&top_left_y=657&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1170",
        "problem": "The diagram represents a pyramid of biomass based on samples taken from marine plankton.\n\nDry mass $\\left(\\mathrm{g} \\mathrm{m}^{-2}\\right)$\n\n[figure1]\n\nWhich one of the following statements best explains this inverted pyramid?\nA: The sampling technique was incorrect\nB: The whole of the phytoplankton material is not eaten by the zooplankton\nC: The productivity of the zooplankton must be higher than that of the phytoplankton\nD: The productivity of the phytoplankton must be higher than that of the zooplankton\nE: The zooplankton must be recycling energy\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe diagram represents a pyramid of biomass based on samples taken from marine plankton.\n\nDry mass $\\left(\\mathrm{g} \\mathrm{m}^{-2}\\right)$\n\n[figure1]\n\nWhich one of the following statements best explains this inverted pyramid?\n\nA: The sampling technique was incorrect\nB: The whole of the phytoplankton material is not eaten by the zooplankton\nC: The productivity of the zooplankton must be higher than that of the phytoplankton\nD: The productivity of the phytoplankton must be higher than that of the zooplankton\nE: The zooplankton must be recycling energy\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-04.jpg?height=209&width=783&top_left_y=1809&top_left_x=671"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1130",
        "problem": "Breeding in birds is mostly of the non-cooperative type, in which the male and female rear the young ones. As they mature into adults, they leave the nest in search of a new nesting site. However, in cooperative breeding, individuals other than the genetic\nparents help raise the young. They are called \"helpers\". Very often these helpers are male offspring from the previous clutch. The average brood size of a pair with helpers has always been found to be larger than the one without helpers.\n\nIn the co-operative breeding system, why would helpers help raise young ones which are not their own offspring? which of the following is correct?\nA: This act will ensure the survival of the species.\nB: It can increase probability of the future breeding in helpers.\nC: It can increase production of non-descendent kin.\nD: It can increase the helper's fecundity by the act of helping.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBreeding in birds is mostly of the non-cooperative type, in which the male and female rear the young ones. As they mature into adults, they leave the nest in search of a new nesting site. However, in cooperative breeding, individuals other than the genetic\nparents help raise the young. They are called \"helpers\". Very often these helpers are male offspring from the previous clutch. The average brood size of a pair with helpers has always been found to be larger than the one without helpers.\n\nIn the co-operative breeding system, why would helpers help raise young ones which are not their own offspring? which of the following is correct?\n\nA: This act will ensure the survival of the species.\nB: It can increase probability of the future breeding in helpers.\nC: It can increase production of non-descendent kin.\nD: It can increase the helper's fecundity by the act of helping.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_968",
        "problem": "What is the primary source of food for developing zebrafish larvae up to 5 days post fertilization (dpf)?\nA: Yolk Sac\nB: Larval fish food flakes\nC: No food required\nD: Paramecium\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the primary source of food for developing zebrafish larvae up to 5 days post fertilization (dpf)?\n\nA: Yolk Sac\nB: Larval fish food flakes\nC: No food required\nD: Paramecium\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1371",
        "problem": "[figure1]\n\nFig 2A: Regulation of PFK activity\n\n[figure2]\n\nFig 2B: Graph between the relative velocity and fructose- 6 phosphate\n\nFigure 2A and 2B: Enzyme activity under different conditions. (PFK = Phosphofructokinase)\n\nFigure $2 A$ shows that for this enzyme ATP is a/an:\nA: allosteric activator.\nB: allosteric inhibitor.\nC: product.\nD: substrate.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nFig 2A: Regulation of PFK activity\n\n[figure2]\n\nFig 2B: Graph between the relative velocity and fructose- 6 phosphate\n\nFigure 2A and 2B: Enzyme activity under different conditions. (PFK = Phosphofructokinase)\n\nFigure $2 A$ shows that for this enzyme ATP is a/an:\n\nA: allosteric activator.\nB: allosteric inhibitor.\nC: product.\nD: substrate.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-33.jpg?height=508&width=671&top_left_y=186&top_left_x=201",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-33.jpg?height=548&width=574&top_left_y=157&top_left_x=1072"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_169",
        "problem": "Isoetes howelli is an amphibious plant that can live in both aerial and submerged conditions. In a completely submerged condition in shallow fresh water, Isoetes howelli shows characteristic metabolism; $\\mathrm{CO}_{2}$ is fixed to malate in a certain time period and released in another period to be used in photosynthetic carbon assimilation. This metabolism is not seen in the aerial condition. There shall be a strong photosynthetic competition in daytime between Isoetes howelli and other photosynthetic organisms.\nA: The malate concentration in the leaves is the highest just before sunrise.\nB: The characteristic metabolism is adaptive because it reduces water loss.\nC: This species has characteristic bundle sheath cells with well-developed chloroplasts.\nD: In the submerged condition to which this species is adapted, it is more difficult to use $\\mathrm{CO}_{2}$ in daytime than in nighttime.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIsoetes howelli is an amphibious plant that can live in both aerial and submerged conditions. In a completely submerged condition in shallow fresh water, Isoetes howelli shows characteristic metabolism; $\\mathrm{CO}_{2}$ is fixed to malate in a certain time period and released in another period to be used in photosynthetic carbon assimilation. This metabolism is not seen in the aerial condition. There shall be a strong photosynthetic competition in daytime between Isoetes howelli and other photosynthetic organisms.\n\nA: The malate concentration in the leaves is the highest just before sunrise.\nB: The characteristic metabolism is adaptive because it reduces water loss.\nC: This species has characteristic bundle sheath cells with well-developed chloroplasts.\nD: In the submerged condition to which this species is adapted, it is more difficult to use $\\mathrm{CO}_{2}$ in daytime than in nighttime.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_187",
        "problem": "When a person born and brought-up in a region at sea level moves to a village at an altitude of 3000 metres above sea level by helicopter, some adaptations of his body occur to compensate for the decreased oxygen pressure at high altitude.\nA: The moment the person arrives at the high altitude, oxyhemoglobin dissociation curve shifts to the left (indicating greater affinity of hemoglobin for oxygen).\nB: After several days living at the high altitude, the person's blood viscosity is decreased, enabling his blood to deliver more oxygen to his tissues.\nC: After several weeks living at the high altitude, the person's lung cells of this person produce more nitric oxide (NO).\nD: Many people who ascend rapidly to high altitude experience some degree of acute mountain sickness (e.g., headache, malaise, and nausea). Which may be treated with a drug that causes bicarbonate to be excreted in the urine.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nWhen a person born and brought-up in a region at sea level moves to a village at an altitude of 3000 metres above sea level by helicopter, some adaptations of his body occur to compensate for the decreased oxygen pressure at high altitude.\n\nA: The moment the person arrives at the high altitude, oxyhemoglobin dissociation curve shifts to the left (indicating greater affinity of hemoglobin for oxygen).\nB: After several days living at the high altitude, the person's blood viscosity is decreased, enabling his blood to deliver more oxygen to his tissues.\nC: After several weeks living at the high altitude, the person's lung cells of this person produce more nitric oxide (NO).\nD: Many people who ascend rapidly to high altitude experience some degree of acute mountain sickness (e.g., headache, malaise, and nausea). Which may be treated with a drug that causes bicarbonate to be excreted in the urine.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1551",
        "problem": "What is the underlying reason trees have evolved to grow so tall?\n\n[figure1]\nA: There is brighter light for photosynthesis further from the ground.\nB: They are able to trap more water from rainfall.\nC: They are in competition with other individuals for light.\nD: To disperse seeds further.\nE: To escape herbivores.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the underlying reason trees have evolved to grow so tall?\n\n[figure1]\n\nA: There is brighter light for photosynthesis further from the ground.\nB: They are able to trap more water from rainfall.\nC: They are in competition with other individuals for light.\nD: To disperse seeds further.\nE: To escape herbivores.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-04.jpg?height=1313&width=897&top_left_y=403&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_671",
        "problem": "某科研小组对蝗虫精巢切片进行显微观察, 测定不同细胞中的染色体数目和核 DNA 数目，结果如图。下列分析正确的是（ ）\n\n[图1]\nA: 细胞 $\\mathrm{c}$ 和细胞 $\\mathrm{g}$ 都可能发生了染色单体分离\nB: 细胞 $d 、 e 、 f$ 中都可能发生同源染色体联会\nC: 细胞 $\\mathrm{b}$ 和细胞 $\\mathrm{c}$ 中都含有同源染色体\nD: 细胞 a 经过间期可变成细胞 b\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某科研小组对蝗虫精巢切片进行显微观察, 测定不同细胞中的染色体数目和核 DNA 数目，结果如图。下列分析正确的是（ ）\n\n[图1]\n\nA: 细胞 $\\mathrm{c}$ 和细胞 $\\mathrm{g}$ 都可能发生了染色单体分离\nB: 细胞 $d 、 e 、 f$ 中都可能发生同源染色体联会\nC: 细胞 $\\mathrm{b}$ 和细胞 $\\mathrm{c}$ 中都含有同源染色体\nD: 细胞 a 经过间期可变成细胞 b\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-62.jpg?height=454&width=854&top_left_y=658&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1498",
        "problem": "A plant is grown in a tank of water under a lamp.\n\nThe rate at which bubbles are given off will be increased four-fold if:\nA: The lamp is placed at half the distance.\nB: The lamp is placed at a quarter of the distance.\nC: The temperature is increased by $40^{\\circ} \\mathrm{C}$.\nD: The carbon dioxide supply is increased two-fold.\nE: The oxygen supply is increased four-fold.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA plant is grown in a tank of water under a lamp.\n\nThe rate at which bubbles are given off will be increased four-fold if:\n\nA: The lamp is placed at half the distance.\nB: The lamp is placed at a quarter of the distance.\nC: The temperature is increased by $40^{\\circ} \\mathrm{C}$.\nD: The carbon dioxide supply is increased two-fold.\nE: The oxygen supply is increased four-fold.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_498",
        "problem": "果蝇的正常刚毛和截刚毛是一对相对性状。为了研究其遗传规律, 现用纯合体果蝇进行下表所示的杂交实验。相关叙述正确的是（ ）\n\n| 杂交组 <br> 合 | 一 | 二 |\n| :---: | :---: | :---: |\n| $\\mathrm{P}$ | 正常刚毛 $+\\times$ 截刚毛 | 截刚毛 $+\\times$ 正常刚毛 |\n| $\\mathrm{F}_{1}$ | 雌雄均为正常刚毛 | 雌雄均为正常刚毛 |\n| $\\mathrm{F}_{2}$ | 正常刚毛 + : 正常刚毛 $\\hat{\\delta}$ : 截刚毛 <br> $\\hat{\\delta}=2: 1: 1$ | 正常刚毛 $q$ : 截刚毛 $q$ : 正常刚毛 <br> $\\hat{\\delta}=1: 1: 2$ |\nA: 这对相对性状受两对等位基因控制, 遵循自由组合定律\nB: 截刚毛的遗传属于伴 X 染色体隐性遗传, 雄性截刚毛多于雌性\nC: 杂交一的 $F_{2}$ 果蝇自由交配, $F_{3}$ 中正常刚毛雌性果蝇所占比例为 $7 / 16$\nD: 杂交二的 $F_{2}$ 果蝇自由交配, $F_{3}$ 中截刚毛果蝇所占比例为 $1 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的正常刚毛和截刚毛是一对相对性状。为了研究其遗传规律, 现用纯合体果蝇进行下表所示的杂交实验。相关叙述正确的是（ ）\n\n| 杂交组 <br> 合 | 一 | 二 |\n| :---: | :---: | :---: |\n| $\\mathrm{P}$ | 正常刚毛 $+\\times$ 截刚毛 | 截刚毛 $+\\times$ 正常刚毛 |\n| $\\mathrm{F}_{1}$ | 雌雄均为正常刚毛 | 雌雄均为正常刚毛 |\n| $\\mathrm{F}_{2}$ | 正常刚毛 + : 正常刚毛 $\\hat{\\delta}$ : 截刚毛 <br> $\\hat{\\delta}=2: 1: 1$ | 正常刚毛 $q$ : 截刚毛 $q$ : 正常刚毛 <br> $\\hat{\\delta}=1: 1: 2$ |\n\nA: 这对相对性状受两对等位基因控制, 遵循自由组合定律\nB: 截刚毛的遗传属于伴 X 染色体隐性遗传, 雄性截刚毛多于雌性\nC: 杂交一的 $F_{2}$ 果蝇自由交配, $F_{3}$ 中正常刚毛雌性果蝇所占比例为 $7 / 16$\nD: 杂交二的 $F_{2}$ 果蝇自由交配, $F_{3}$ 中截刚毛果蝇所占比例为 $1 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1222",
        "problem": "Colour blindness results from the presence of a sex-linked allele which is recessive in females. A colour blind woman and a man with normal vision produced a child with normal vision and a sex chromosome complement $X X Y$. This suggests that chromosomes failed to separate during the formation of the\nA: sperm\nB: egg\nC: sperm and egg\nD: sperm or egg\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nColour blindness results from the presence of a sex-linked allele which is recessive in females. A colour blind woman and a man with normal vision produced a child with normal vision and a sex chromosome complement $X X Y$. This suggests that chromosomes failed to separate during the formation of the\n\nA: sperm\nB: egg\nC: sperm and egg\nD: sperm or egg\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_674",
        "problem": "在孟德尔两对相对性状杂交实验中, $F_{1}$ 黄色圆粒踠豆（YyRr）自交产生 $F_{2}$ 。下列表述正确的是 ( )\nA: $\\mathrm{F}_{1}$ 产生 4 个配子, 比例为 $1: 1: 1: 1$\nB: $\\mathrm{F}_{1}$ 产生基因型 YR 的卵和基因型 YR 的精子数量之比为 1: 1\nC: 基因自由组合定律是指 F1 产生的 4 种类型的精子和卵可以自由组合\nD: $\\mathrm{F}_{1}$ 产生的精子中, 基因型为 $\\mathrm{YR}$ 和基因型为 $\\mathrm{yr}$ 的比例为 1: 1\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在孟德尔两对相对性状杂交实验中, $F_{1}$ 黄色圆粒踠豆（YyRr）自交产生 $F_{2}$ 。下列表述正确的是 ( )\n\nA: $\\mathrm{F}_{1}$ 产生 4 个配子, 比例为 $1: 1: 1: 1$\nB: $\\mathrm{F}_{1}$ 产生基因型 YR 的卵和基因型 YR 的精子数量之比为 1: 1\nC: 基因自由组合定律是指 F1 产生的 4 种类型的精子和卵可以自由组合\nD: $\\mathrm{F}_{1}$ 产生的精子中, 基因型为 $\\mathrm{YR}$ 和基因型为 $\\mathrm{yr}$ 的比例为 1: 1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_261",
        "problem": "Phosphorylation is a major post-translational modification widely used in the regulation of many cellular processes. A method to determine the phosphorylation status of proteins is to run an electrophoresis in a modified gel with a chemical group containing metal ions (M) that can reversibly bind phosphates and thus affects migration of phosphorylated proteins.\n\n[figure1]\n\nFig.Q59A. Phospho-tag polyacrylamide gel\n\nThis technique was used to study the phosphorylation of protein $\\mathrm{p} 35$. Three mutant forms of this protein were generated: a serine to alanine substitution in position $8(\\mathrm{~S} 8 \\mathrm{~A})$; a threonine to alanine substitution in position 138 (T138A) and both amino acid substitutions (2A). Note that serine and threonine can be phosphorylated while alanine cannot. Then two yeast strains with normal (wt) or inactive cyclin-dependent kinase 5 (Cdk5) (kn) were transformed with either the wild type version of p35 gene (wt) or one of the three mutant forms. Cell lysate of the eight resulting strains was loaded on a Phospho-tag gel. The proteins from the gel were transferred by western-blot to a membrane that was treated with anti-p35 antibodies. The result is shown below.\n\n[figure2]\n\nFig.Q59B. Immunoblotting with anti-p35. The arrow indicates the direction of migration p35 bands are named $\\mathrm{M} 1, \\mathrm{M} 2, \\mathrm{~L} 1, \\mathrm{~L} 2, \\mathrm{~L} 3$, and $\\mathrm{L} 4$. 4 band corresponds to the completely non-phosphorylated form of $\\mathrm{p} 35$.\nA: Protein 35 has only two phosphorylation sites: serine 8 and threonine 138.\nB: Protein $\\mathrm{p} 35$ can be phosphorylated by a protein kinase different from Cdk5.\nC: In strain Cdk5-wt p35-S8A only a few p35 molecules are phosphorylated at T138.\nD: Phosphate groups attached to $\\$ 8$ are more accessible to phosphate binding groups of the Phospho-tag gel than phosphate groups attached to T138.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPhosphorylation is a major post-translational modification widely used in the regulation of many cellular processes. A method to determine the phosphorylation status of proteins is to run an electrophoresis in a modified gel with a chemical group containing metal ions (M) that can reversibly bind phosphates and thus affects migration of phosphorylated proteins.\n\n[figure1]\n\nFig.Q59A. Phospho-tag polyacrylamide gel\n\nThis technique was used to study the phosphorylation of protein $\\mathrm{p} 35$. Three mutant forms of this protein were generated: a serine to alanine substitution in position $8(\\mathrm{~S} 8 \\mathrm{~A})$; a threonine to alanine substitution in position 138 (T138A) and both amino acid substitutions (2A). Note that serine and threonine can be phosphorylated while alanine cannot. Then two yeast strains with normal (wt) or inactive cyclin-dependent kinase 5 (Cdk5) (kn) were transformed with either the wild type version of p35 gene (wt) or one of the three mutant forms. Cell lysate of the eight resulting strains was loaded on a Phospho-tag gel. The proteins from the gel were transferred by western-blot to a membrane that was treated with anti-p35 antibodies. The result is shown below.\n\n[figure2]\n\nFig.Q59B. Immunoblotting with anti-p35. The arrow indicates the direction of migration p35 bands are named $\\mathrm{M} 1, \\mathrm{M} 2, \\mathrm{~L} 1, \\mathrm{~L} 2, \\mathrm{~L} 3$, and $\\mathrm{L} 4$. 4 band corresponds to the completely non-phosphorylated form of $\\mathrm{p} 35$.\n\nA: Protein 35 has only two phosphorylation sites: serine 8 and threonine 138.\nB: Protein $\\mathrm{p} 35$ can be phosphorylated by a protein kinase different from Cdk5.\nC: In strain Cdk5-wt p35-S8A only a few p35 molecules are phosphorylated at T138.\nD: Phosphate groups attached to $\\$ 8$ are more accessible to phosphate binding groups of the Phospho-tag gel than phosphate groups attached to T138.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-020.jpg?height=435&width=919&top_left_y=676&top_left_x=565",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-020.jpg?height=438&width=819&top_left_y=1921&top_left_x=615"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_518",
        "problem": "某鳞翅目昆虫的性别决定方式为 ZW 型, 触角羽状和棒状为一对相对性状, 由基因 $\\mathrm{A} 、 \\mathrm{a}$ 控制，体型扁平形和圆形为一对相对性状，由基因 B、 $\\mathrm{b}$ 控制。现将纯合的羽状扁平形个体与棒状圆形个体正反交, 得到的 $\\mathrm{F}_{1}$ 个体随机交配产生 $\\mathrm{F}_{1}$, 表型及比例如表所示 (不考虑 ZW 染色体同源区段), 下列说法正确的是（ ）\n\n| 杂 | 亲本 | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :--- | :--- | :--- | :--- |\n\n\n| 交 |  |  |  |\n| :---: | :---: | :---: | :---: |\n| 方 |  |  |  |\n| 式 |  |  |  |\n| 正 | 羽状扁平形 $($ Q $) \\times$ | 雌雄均为羽状圆形 |  |\n| 交 | 棒状圆形 $\\left(\\right.$ Ō $\\left.^{2}\\right)$ |  | 棒状扁平形 $=15: 5: 3: 1$ |\nA: 亲本中羽状扁平形昆虫的基因型为 $\\mathrm{bbZ}^{\\mathrm{x}} \\mathrm{Z}^{\\mathrm{X}}$ 或 $\\mathrm{bbZ}^{\\mathrm{x}} \\mathrm{W}$\nB: 正交 $F_{2}$ 的四种表型比例的出现一定与基因型的致死有关\nC: 正交 $F_{2}$ 中羽状圆形雄性个体中, 纯合子的比例为 $1 / 6$\nD: 反交 $\\mathrm{F}_{2}$ 中羽状圆形: 羽状扁平形: 棒状圆形: 棒状扁平形可能等于 5: 5: 1: 1\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某鳞翅目昆虫的性别决定方式为 ZW 型, 触角羽状和棒状为一对相对性状, 由基因 $\\mathrm{A} 、 \\mathrm{a}$ 控制，体型扁平形和圆形为一对相对性状，由基因 B、 $\\mathrm{b}$ 控制。现将纯合的羽状扁平形个体与棒状圆形个体正反交, 得到的 $\\mathrm{F}_{1}$ 个体随机交配产生 $\\mathrm{F}_{1}$, 表型及比例如表所示 (不考虑 ZW 染色体同源区段), 下列说法正确的是（ ）\n\n| 杂 | 亲本 | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :--- | :--- | :--- | :--- |\n\n\n| 交 |  |  |  |\n| :---: | :---: | :---: | :---: |\n| 方 |  |  |  |\n| 式 |  |  |  |\n| 正 | 羽状扁平形 $($ Q $) \\times$ | 雌雄均为羽状圆形 |  |\n| 交 | 棒状圆形 $\\left(\\right.$ Ō $\\left.^{2}\\right)$ |  | 棒状扁平形 $=15: 5: 3: 1$ |\n\nA: 亲本中羽状扁平形昆虫的基因型为 $\\mathrm{bbZ}^{\\mathrm{x}} \\mathrm{Z}^{\\mathrm{X}}$ 或 $\\mathrm{bbZ}^{\\mathrm{x}} \\mathrm{W}$\nB: 正交 $F_{2}$ 的四种表型比例的出现一定与基因型的致死有关\nC: 正交 $F_{2}$ 中羽状圆形雄性个体中, 纯合子的比例为 $1 / 6$\nD: 反交 $\\mathrm{F}_{2}$ 中羽状圆形: 羽状扁平形: 棒状圆形: 棒状扁平形可能等于 5: 5: 1: 1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-84.jpg?height=57&width=1282&top_left_y=2530&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1312",
        "problem": "Conservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.The maximum distance known to have been travelled by a Maui's dolphin within a single year is?\nA: $17.88 \\mathrm{~km}$\nB: $26.44 \\mathrm{~km}$\nC: $46.30 \\mathrm{~km}$\nD: $78.62 \\mathrm{~km}$\nE: $80.43 \\mathrm{~km}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nConservation of the critically endangered Maui's dolphin requires a good understanding of the trends in abundance and effective population sizes so that conservation actions can be planned and evaluated. DOC (2012) reported on the genetic monitoring of the Maui's dolphin using DNA profiles to estimate the current abundance and effective population size, as well as to document movements of individuals. The study collected of 37 dart-biopsy samples in summer 2010 and a further 36 in summer 2011. DNA profiles were completed for each sample and the sex was identified. These profiles were used to identify individual Maui's dolphins and Hector's dolphin migrants, to describe individual movements, and to estimate the abundance, population trend and effective population size of Maui's dolphins for 2010-11, including comparison with data from a previous set of samples collected in 2001-07.\n\nBased on the microsatellite genotyping, 26 individuals were identified from the 37 samples collected in 2010 (16 females, 10 males) and 27 individuals from 36 samples collected in 2011 (16 females, 11 males). Twelve individuals were sampled in both 2010 and 2011, and with the addition of one unique male washed up on a beach in 2010, this provided a minimum census of 42 individuals ( 25 females, 17 males) alive at some point during the two years of the survey. Of this total, two females were identified as West Coast South Island Hector's dolphin ( $C$. h. hectori) migrants based on distinct mtDNA haplotypes and genotype-based population assignment procedures.\n\nPopulation size can be estimated by the Lincoln-Petersen estimator:\n\n$N=\\left[\\left(n_{1}+1\\right)\\left(n_{2}+1\\right) /\\left(m_{2}+1\\right)\\right]-1$\n\nWhere $N=$ abundance\n\n$n 1=$ number of individuals sampled in occasion 1\n\n$n 2=$ number of individuals sampled in occasion 2\n\n$m 2=$ number of individuals sampled in both occasions 1 and 2\n\nIndividual movements of Maui's dolphins and a Hector's dolphin migrant $\\left({ }^{\\wedge}\\right)$ that were sampled more than once during 2010-11 are given in Table 2 in the Resource Pack.\n\nAn Expert Panel of New Zealand and international scientists, convened by the New Zealand government in 2012, estimated that five Maui's dolphins were killed in fishing gear each year - one in trawl fisheries and four in gillnet fisheries. The number of gillnet mortalities per year is estimated to have decreased from four to at best two per year as a result of the $350 \\mathrm{sq}$. $\\mathrm{km}$ set net restriction extension in the WCNIMMS. The Ministry of Primary Industries (MPI) and the Department of Conservation (DOC) reviewed the Maui's dolphin portion of the Threat Management Plan (TMP) in 2013. During the TMP review process it was highlighted that non-fishing-related threats such as seismic surveying, oil and gas exploration, vessel strikes, and disease also pose a serious risk to the long-term viability of Maui's dolphins. These threats represented $4.5 \\%$ of the estimated dolphin mortalities. The importance of both the fishing and nonfishing risks needs to be assessed relative to the Potential Biological Removal (PBR) level which is defined as the number of human-caused deaths the stock can withstand annually. The PBR for Maui's dolphins is one dolphin in $10-$ 23 years or $0.044-0.1$ per year.\n\nFor Maui's dolphin the impact of seismic surveying was combined within all mining and oil activities. This was estimated to contribute to the equivalent of 0.10 deaths per year ( $95 \\%$ confidence interval $0.01-0.46$ ), with a $61.3 \\%$ likelihood of exceeding the PBR. In terms of seismic surveying, the greatest concern is noise in the marine environment. Noise leading to trauma was scored at 0.01 deaths per year $(95 \\% \\mathrm{Cl}:<0.01-0.13)$ and a likelihood of exceeding the PBR of $8.8 \\%$, while non-trauma noise effects was scored at 0.03 deaths per year ( $95 \\% \\mathrm{Cl}:<0.01-0.23$ ) and a likelihood of exceeding the PBR of $28.6 \\%$.\n\n(http://www.fish.govt.nz/en-nz/Environmental/Hectors+Dolphins/default.htm)\n\nThe Minister of Conservation has proposed the following measures:\n\n- Making the Code of Conduct for Minimising Acoustic Disturbance to Marine Mammals from Seismic Survey Operations a mandatory standard by reference under section 28 of the Marine Mammal Protection Act.\n- Developing a voluntary code of conduct with the inshore boat racing community to minimise the potential for vessel strike, and\n- Ensuring that disease investigations are a priority in the Maui's dolphin Research Advisory Group.\n\nThere has been no recommendation to prevent set net and trawl fishing throughout the range of Maui's dolphins as to do so would need to extend fishing restrictions as far south as Whanganui, out to the $100 \\mathrm{~m}$ depth contour, and include all harbours.\n\nproblem:\nThe maximum distance known to have been travelled by a Maui's dolphin within a single year is?\n\nA: $17.88 \\mathrm{~km}$\nB: $26.44 \\mathrm{~km}$\nC: $46.30 \\mathrm{~km}$\nD: $78.62 \\mathrm{~km}$\nE: $80.43 \\mathrm{~km}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1209",
        "problem": "In a series of breeding experiments using Drosophila, a linkage group composed of five genes was found to show the following recombination frequencies per 100 fertilised eggs.\n\n|  | Singed <br> Bristles <br> $(s n)$ | Vermilion <br> eye $(v)$ | White <br> Eye $(w)$ | Ruby <br> Eye $(r b)$ | Tan <br> Body $(t)$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| Singed <br> Bristles <br> $(s n)$ | 0.0 | 12.0 | 19.5 | 13.5 | 6.5 |\n| Vermilion <br> eye $(v)$ | 12.0 | 0.0 | 31.5 | 25.5 | 5.5 |\n| White <br> eye $(w)$ | 19.5 | 31.5 | 0.0 | 6.0 | 26.0 |\n| Ruby <br> eye $(r b)$ | 13.5 | 25.5 | 6.0 | 0.0 | 20.0 |\n| Tan <br> body $(t)$ | 6.5 | 5.5 | 26.0 | 20.0 | 0.0 |\n\nFrom this table of results it can be deduced that the correct sequence of the genes on the chromosomes is\nA: $s n-t-v-r b-w$\nB: $s n-v-w-r b-t$\nC: $w-t-s n-r b-v$\nD: $w-r b-s n-t-v$\nE: $w-r b-t-s n-v$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn a series of breeding experiments using Drosophila, a linkage group composed of five genes was found to show the following recombination frequencies per 100 fertilised eggs.\n\n|  | Singed <br> Bristles <br> $(s n)$ | Vermilion <br> eye $(v)$ | White <br> Eye $(w)$ | Ruby <br> Eye $(r b)$ | Tan <br> Body $(t)$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| Singed <br> Bristles <br> $(s n)$ | 0.0 | 12.0 | 19.5 | 13.5 | 6.5 |\n| Vermilion <br> eye $(v)$ | 12.0 | 0.0 | 31.5 | 25.5 | 5.5 |\n| White <br> eye $(w)$ | 19.5 | 31.5 | 0.0 | 6.0 | 26.0 |\n| Ruby <br> eye $(r b)$ | 13.5 | 25.5 | 6.0 | 0.0 | 20.0 |\n| Tan <br> body $(t)$ | 6.5 | 5.5 | 26.0 | 20.0 | 0.0 |\n\nFrom this table of results it can be deduced that the correct sequence of the genes on the chromosomes is\n\nA: $s n-t-v-r b-w$\nB: $s n-v-w-r b-t$\nC: $w-t-s n-r b-v$\nD: $w-r b-s n-t-v$\nE: $w-r b-t-s n-v$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1558",
        "problem": "The complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nWhich pathway links the complement system to the adaptive immune system?\nA: Pathway A\nB: Pathway B\nC: Pathway C\nD: Pathway D\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe complement system is so called because it complements the immune system by tagging and destroying foreign materials. Individual reactions in this system were discovered separately. The reactions are shown.\n[figure1]\n\n\n\n[figure2]\n\nComplement factors circulate freely in extracellular fluid, but C3B has a highly reactive group which binds any surfaces it touches. The membrane attack complex punches large holes in membranes, whilst C3B attracts immune cells and stimulates them to attack. The complement system can cause a variety of diseases so scientists need to guess which pathways to target with drugs. ${ }^{* *}$ Come up with a hypothesis about what each pathway does.**\n\nWhich pathway links the complement system to the adaptive immune system?\n\nA: Pathway A\nB: Pathway B\nC: Pathway C\nD: Pathway D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=790&width=1714&top_left_y=486&top_left_x=228",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-40.jpg?height=930&width=1806&top_left_y=1335&top_left_x=228"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_536",
        "problem": "豌豆花的顶生和腋生是一对相对性状, 有多对遗传因子共同控制并且各自独立遗传\n\n(用基因 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b} \\ldots \\ldots$. 表示), 用纯合花的顶生和纯合腋生踠豆作为亲本杂交得 $\\mathrm{F}_{1}, \\mathrm{~F}_{1}$自交得 $F_{2}, F_{2}$ 中顶生: 腋生=63: 1。下列相关叙述错误的是 $(\\quad)$\nA: 该相对性状至少由 3 对等位基因共同控制\nB: 将 $F_{1}$ 进行测交, 后代中腋生植株所占比例为 $1 / 8$\nC: 将 $\\mathrm{F}_{2}$ 中顶生个体进行测交, 后代中腋生植株所占比例为 $5 / 6$\nD: 让 $\\mathrm{F}_{2}$ 植株中顶生个体进行自交不发生性状分离所占比例为 $37 / 63$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n豌豆花的顶生和腋生是一对相对性状, 有多对遗传因子共同控制并且各自独立遗传\n\n(用基因 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b} \\ldots \\ldots$. 表示), 用纯合花的顶生和纯合腋生踠豆作为亲本杂交得 $\\mathrm{F}_{1}, \\mathrm{~F}_{1}$自交得 $F_{2}, F_{2}$ 中顶生: 腋生=63: 1。下列相关叙述错误的是 $(\\quad)$\n\nA: 该相对性状至少由 3 对等位基因共同控制\nB: 将 $F_{1}$ 进行测交, 后代中腋生植株所占比例为 $1 / 8$\nC: 将 $\\mathrm{F}_{2}$ 中顶生个体进行测交, 后代中腋生植株所占比例为 $5 / 6$\nD: 让 $\\mathrm{F}_{2}$ 植株中顶生个体进行自交不发生性状分离所占比例为 $37 / 63$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1513",
        "problem": "The way groups of organisms live depends on fundamental aspects of their biology. For each of these characteristics, state whether they belong to plants, animals, fungi and/or bacteria.\n\nHave mitochondria.\nA: Plants, animals, fungi\nB: Plants, bacteria, fungi\nC: Animals, fungi, bacteria\nD: Plants, animals, bacteria\nE: Plants, animals, bacteria, fungi\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe way groups of organisms live depends on fundamental aspects of their biology. For each of these characteristics, state whether they belong to plants, animals, fungi and/or bacteria.\n\nHave mitochondria.\n\nA: Plants, animals, fungi\nB: Plants, bacteria, fungi\nC: Animals, fungi, bacteria\nD: Plants, animals, bacteria\nE: Plants, animals, bacteria, fungi\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1401",
        "problem": "If a length of DNA comprises 10,000 nucleotides of which $26 \\%$ is adenine, what is the predicted number of cytosine bases?\nA: 48\nB: 96\nC: 1200\nD: 2400\nE: 4800\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf a length of DNA comprises 10,000 nucleotides of which $26 \\%$ is adenine, what is the predicted number of cytosine bases?\n\nA: 48\nB: 96\nC: 1200\nD: 2400\nE: 4800\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1406",
        "problem": "When eukaryotic cells divide, they pass through the 'cell cycle', which can be subdivided into a series of phases. This is depicted in the picture below. Cells that begin to prepare for cell division enter the G1 phase ('Gap 1' phase). In the G1 phase, cellular and cytoplasmic contents are produced in preparation for cell division (such as proteins, organelles, and the cytoplasm itself). After this phase, the cell progresses into the $S$ phase ('Synthesis' phase) where DNA replication occurs, and then into the G2 phase ('Gap 2' phase) where the cell grows further by producing and assembling the cytoplasmic materials for mitosis and cytokinesis. Finally, the cell enters the mitotic phase (labelled 'M') and cytokinesis (labelled 'C') and completes cell division.\n\n[figure1]\n\nLiam is using $x$-ray radiation to induce mutations in a strain of animal cells in his laboratory. At which phase of the cell cycle is a cell most likely to mutate when exposed to radiation?\nA: $\\mathrm{G} 1$\nB: $\\mathrm{S}$\nC: $\\mathrm{G} 2$\nD: $\\mathrm{M}$\nE: $C$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen eukaryotic cells divide, they pass through the 'cell cycle', which can be subdivided into a series of phases. This is depicted in the picture below. Cells that begin to prepare for cell division enter the G1 phase ('Gap 1' phase). In the G1 phase, cellular and cytoplasmic contents are produced in preparation for cell division (such as proteins, organelles, and the cytoplasm itself). After this phase, the cell progresses into the $S$ phase ('Synthesis' phase) where DNA replication occurs, and then into the G2 phase ('Gap 2' phase) where the cell grows further by producing and assembling the cytoplasmic materials for mitosis and cytokinesis. Finally, the cell enters the mitotic phase (labelled 'M') and cytokinesis (labelled 'C') and completes cell division.\n\n[figure1]\n\nLiam is using $x$-ray radiation to induce mutations in a strain of animal cells in his laboratory. At which phase of the cell cycle is a cell most likely to mutate when exposed to radiation?\n\nA: $\\mathrm{G} 1$\nB: $\\mathrm{S}$\nC: $\\mathrm{G} 2$\nD: $\\mathrm{M}$\nE: $C$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-33.jpg?height=545&width=462&top_left_y=1589&top_left_x=794"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1062",
        "problem": "A tuberculosis patient died within a few minutes of second intravenous administration of streptomycin and the post-mortem report gave anaphylactic shock as the cause of death. This is explained as:\nA: a sudden surge in IgM synthesis leading to complement activation causing fatal drop in blood pressure.\nB: a vigorous precipitation of soluble antigens in body fluids triggered by the IgA insurgence leading to blockage of capillaries to vital organs.\nC: IgE induced mast cell degranulation triggering abrupt dilation of peripheral blood vessels resulting in a precipitous drop in blood pressure.\nD: IgD induced agglutination of antigen-bearing cells in vital organs leading to multi-organ failure.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA tuberculosis patient died within a few minutes of second intravenous administration of streptomycin and the post-mortem report gave anaphylactic shock as the cause of death. This is explained as:\n\nA: a sudden surge in IgM synthesis leading to complement activation causing fatal drop in blood pressure.\nB: a vigorous precipitation of soluble antigens in body fluids triggered by the IgA insurgence leading to blockage of capillaries to vital organs.\nC: IgE induced mast cell degranulation triggering abrupt dilation of peripheral blood vessels resulting in a precipitous drop in blood pressure.\nD: IgD induced agglutination of antigen-bearing cells in vital organs leading to multi-organ failure.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_595",
        "problem": "一个班级分组进行减数分裂实验, 下列操作及表述正确的是( )\nA: 换用高倍镜时，从侧面观察，防止物镜与装片碰擦\nB: 因观察材料较少, 性母细胞较小, 显微镜视野应适当调亮\nC: 为观察染色体不同层面的精细结构，用粗准焦螺旋调节焦距\nD: 一个视野中，用 $10 \\times$ 物镜看到 8 个细胞，用 $40 \\times$ 物镜则可看到 32 个细胞\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一个班级分组进行减数分裂实验, 下列操作及表述正确的是( )\n\nA: 换用高倍镜时，从侧面观察，防止物镜与装片碰擦\nB: 因观察材料较少, 性母细胞较小, 显微镜视野应适当调亮\nC: 为观察染色体不同层面的精细结构，用粗准焦螺旋调节焦距\nD: 一个视野中，用 $10 \\times$ 物镜看到 8 个细胞，用 $40 \\times$ 物镜则可看到 32 个细胞\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1068",
        "problem": "A common method used to compare the patterns of species richness and abundance between communities involves plotting the relative abundance of each species against rank (where rank is defined by the order of species from the most to the least abundant). The most abundant species is plotted first along the $x$-axis, with the corresponding value on the $y$-axis being the value of relative abundance. The following diagram depicts the rank-abundance curve for two forest communities 1 and 2 .\n\n[figure1]\n\nA few statements regarding the forest communities are made below.\n\ni. Community 1 has a higher number of species.\n\nii. Community 2 shows a higher degree of evenness in terms of abundance.\n\niii. The most dominant member of community 1 has the highest relative abundance among all the species depicted in the graph.\n\nBased on the graph, the true statement/s is/are:\nA: i , ii and iii\nB: iii only\nC: i and iii only\nD: i only\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA common method used to compare the patterns of species richness and abundance between communities involves plotting the relative abundance of each species against rank (where rank is defined by the order of species from the most to the least abundant). The most abundant species is plotted first along the $x$-axis, with the corresponding value on the $y$-axis being the value of relative abundance. The following diagram depicts the rank-abundance curve for two forest communities 1 and 2 .\n\n[figure1]\n\nA few statements regarding the forest communities are made below.\n\ni. Community 1 has a higher number of species.\n\nii. Community 2 shows a higher degree of evenness in terms of abundance.\n\niii. The most dominant member of community 1 has the highest relative abundance among all the species depicted in the graph.\n\nBased on the graph, the true statement/s is/are:\n\nA: i , ii and iii\nB: iii only\nC: i and iii only\nD: i only\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-10.jpg?height=759&width=1280&top_left_y=1583&top_left_x=474"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1066",
        "problem": "In an experiment, E. coli cells growing at $37^{\\circ} \\mathrm{C}$ were shifted to $20^{\\circ} \\mathrm{C}$ and grown for a few generations. Which of the following changes in the membrane would help the $E$. coli cells adapt to the new environment?\nA: Increase in the unsaturated fatty acid content.\nB: Increase in the number of integral membrane proteins.\nC: Increase in the phospholipid content.\nD: Increase in the length of the hydrophobic tail.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn an experiment, E. coli cells growing at $37^{\\circ} \\mathrm{C}$ were shifted to $20^{\\circ} \\mathrm{C}$ and grown for a few generations. Which of the following changes in the membrane would help the $E$. coli cells adapt to the new environment?\n\nA: Increase in the unsaturated fatty acid content.\nB: Increase in the number of integral membrane proteins.\nC: Increase in the phospholipid content.\nD: Increase in the length of the hydrophobic tail.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_15",
        "problem": "As shown in the picture below, microarray was used to find genes whose expression is regulated when a plant is treated with the ABA hormone.\n\n[figure1]\n\nWhich of the following explanations is not correct about the microarray experiment?\nA: All cDNAs of the expressed mRNA from both the experimental group and the control group hybridizes competitively with the corresponding genes on the DNA chip.\nB: Genes whose expressions are induced by ABA appear red after hybridization.\nC: Because we used different colored probes with each sample, we can measure the relative amount of genes which are expressed differentially.\nD: We can only know the expression profile of genes which are included on the microarray.\nE: This process includes reverse transcription and hybridization.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAs shown in the picture below, microarray was used to find genes whose expression is regulated when a plant is treated with the ABA hormone.\n\n[figure1]\n\nWhich of the following explanations is not correct about the microarray experiment?\n\nA: All cDNAs of the expressed mRNA from both the experimental group and the control group hybridizes competitively with the corresponding genes on the DNA chip.\nB: Genes whose expressions are induced by ABA appear red after hybridization.\nC: Because we used different colored probes with each sample, we can measure the relative amount of genes which are expressed differentially.\nD: We can only know the expression profile of genes which are included on the microarray.\nE: This process includes reverse transcription and hybridization.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-13.jpg?height=1071&width=1465&top_left_y=487&top_left_x=273"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_824",
        "problem": "如图表示某种单基因显性遗传病（相关基因用 $\\mathrm{T} 、 \\mathrm{t}$ 表示）的家系图和家庭成员基因检测的结果，且这种遗传病只有在成年后才发病，因此不确定小孩是否携带致病基因。由于采样时将样本弄混, 无法对应甲、乙、丙、丁 4 份检测结果。下列分析错误的是\n[图1]\nA: 致病基因也可能位于 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段\nB: 2 号个体不携带致病基因, 3 号和 4 号个体成年后都会患病\nC: 若致病基因位于常染色体上, 则 3 号和 4 号的基因型相同\nD: 可以确定 1 号个体的基因检测结果即为样本乙所示结果\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图表示某种单基因显性遗传病（相关基因用 $\\mathrm{T} 、 \\mathrm{t}$ 表示）的家系图和家庭成员基因检测的结果，且这种遗传病只有在成年后才发病，因此不确定小孩是否携带致病基因。由于采样时将样本弄混, 无法对应甲、乙、丙、丁 4 份检测结果。下列分析错误的是\n[图1]\n\nA: 致病基因也可能位于 $\\mathrm{X} 、 \\mathrm{Y}$ 染色体的同源区段\nB: 2 号个体不携带致病基因, 3 号和 4 号个体成年后都会患病\nC: 若致病基因位于常染色体上, 则 3 号和 4 号的基因型相同\nD: 可以确定 1 号个体的基因检测结果即为样本乙所示结果\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-57.jpg?height=326&width=1462&top_left_y=154&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1370",
        "problem": "A Poikilotherm is an organism whose internal temperature varies considerably with their environment. In comparison, a homeotherm is able to maintain a stable internal body temperature regardless of external influence.\n\n[figure1]\n\nHomeotherms and poikilotherms generally have different energy requirements per body mass. Which response is correct?\nA: A given food source can support a greater density of poikilothermic animals than homeothermic animals\nB: In niches with limited food, homeotherms tend to out compete poikilotherms\nC: Poikilotherms require more energy than homeotherms to reproduce\nD: A given food source can support the same density of poikilothermic and homeothermic animals\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA Poikilotherm is an organism whose internal temperature varies considerably with their environment. In comparison, a homeotherm is able to maintain a stable internal body temperature regardless of external influence.\n\n[figure1]\n\nHomeotherms and poikilotherms generally have different energy requirements per body mass. Which response is correct?\n\nA: A given food source can support a greater density of poikilothermic animals than homeothermic animals\nB: In niches with limited food, homeotherms tend to out compete poikilotherms\nC: Poikilotherms require more energy than homeotherms to reproduce\nD: A given food source can support the same density of poikilothermic and homeothermic animals\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-06.jpg?height=683&width=763&top_left_y=1466&top_left_x=638"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1090",
        "problem": "In an experiment of mitotically dividing animal cells, nuclei of cells in G1 and G2 phases were removed. In the subsequent step, the G2 phase nuclei were introduced in enucleated cells of $\\mathrm{G} 1$ phase. If these cells are cultured, what will be the consequence?\nA: Cells will abort cell cycle and enter G0 phase.\nB: Cells will shift directly from G1 to G2 phase.\nC: Cells will continue to stay in G1 phase.\nD: Cells will proceed from G1 to $S$ phase.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn an experiment of mitotically dividing animal cells, nuclei of cells in G1 and G2 phases were removed. In the subsequent step, the G2 phase nuclei were introduced in enucleated cells of $\\mathrm{G} 1$ phase. If these cells are cultured, what will be the consequence?\n\nA: Cells will abort cell cycle and enter G0 phase.\nB: Cells will shift directly from G1 to G2 phase.\nC: Cells will continue to stay in G1 phase.\nD: Cells will proceed from G1 to $S$ phase.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1008",
        "problem": "The virus Ijustmadethisup causes the following symptoms: overly fragile bones, fatigue, bone pain, an abnormal EKG with a short $Q T$ interval and a long $T$ wave, kidney stones, constipation, and vomiting. You theorize that the virus is infecting endocrine cells and causing the overproduction of a certain hormone. Which hormone should you test for first?\nA: Prolactin\nB: Parathyroid hormone (PTH)\nC: Triiodothyronine\nD: Melatonin\nE: Oxytocin\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe virus Ijustmadethisup causes the following symptoms: overly fragile bones, fatigue, bone pain, an abnormal EKG with a short $Q T$ interval and a long $T$ wave, kidney stones, constipation, and vomiting. You theorize that the virus is infecting endocrine cells and causing the overproduction of a certain hormone. Which hormone should you test for first?\n\nA: Prolactin\nB: Parathyroid hormone (PTH)\nC: Triiodothyronine\nD: Melatonin\nE: Oxytocin\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1252",
        "problem": "10 different Australian honeyeaters are found in the Mount Lofty Ranges near Adelaide, South Australia. The species differ in beak length with shorter beaked species feeding chiefly on insects and longer -beaked species feeding more often on flowers. Insect eating species may glean insects from leaves and bark or capture them in the air. The co-existance of these species in the same habitat is an example of\nA: Competition\nB: Resource partitioning\nC: Character displacement\nD: Competitive exclusion\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n10 different Australian honeyeaters are found in the Mount Lofty Ranges near Adelaide, South Australia. The species differ in beak length with shorter beaked species feeding chiefly on insects and longer -beaked species feeding more often on flowers. Insect eating species may glean insects from leaves and bark or capture them in the air. The co-existance of these species in the same habitat is an example of\n\nA: Competition\nB: Resource partitioning\nC: Character displacement\nD: Competitive exclusion\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1036",
        "problem": "Natural selection is effective in the evolutionary processes because it:\nA: Causes evolution.\nB: Changes allele frequencies.\nC: Changes genotype frequencies.\nD: Leads to fixation or loss of particular alleles.\nE: Increases the mean fitness of a population.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nNatural selection is effective in the evolutionary processes because it:\n\nA: Causes evolution.\nB: Changes allele frequencies.\nC: Changes genotype frequencies.\nD: Leads to fixation or loss of particular alleles.\nE: Increases the mean fitness of a population.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1043",
        "problem": "In the absence of active transport, the passive sodium and potassium ion fluxes across the plasma membrane are still coupled. What makes these two passive ion fluxes dependent on each other?\nA: The potassium channels\nB: The $\\mathrm{Na}^{+}: \\mathrm{K}^{+}$pumping ratio\nC: The cholesterol: phospholipid ratio in the plasma membrane\nD: The membrane potential, $\\mathbf{V}_{\\mathrm{m}}$\nE: The relative chemical potentials of sodium ions and potassium ions to the chemical potential of chloride ion\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the absence of active transport, the passive sodium and potassium ion fluxes across the plasma membrane are still coupled. What makes these two passive ion fluxes dependent on each other?\n\nA: The potassium channels\nB: The $\\mathrm{Na}^{+}: \\mathrm{K}^{+}$pumping ratio\nC: The cholesterol: phospholipid ratio in the plasma membrane\nD: The membrane potential, $\\mathbf{V}_{\\mathrm{m}}$\nE: The relative chemical potentials of sodium ions and potassium ions to the chemical potential of chloride ion\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_477",
        "problem": "某植物的花色由两对独立遗传的基因控制, 这两对基因与花色的关系如图所示. 此外， $a$ 基因存在时抑制 B 基因的表达. 现将基因型为 AABB 的个体与基因型为 $a a b b$ 的个体杂交得到 $F_{1}$, 下列分析正确的是（）\n\n[图1]\nA: $F_{1}$ 全部开红色花\nB: $F_{1}$ 测交后代中花色的表现型及比例是白: 粉: 红 $=1: 2: 1$\nC: $F_{1}$ 自交后代中花色的表现型及比例是白：粉：红=4: 9: 3\nD: 图示过程体现基因对性状的直接控制\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某植物的花色由两对独立遗传的基因控制, 这两对基因与花色的关系如图所示. 此外， $a$ 基因存在时抑制 B 基因的表达. 现将基因型为 AABB 的个体与基因型为 $a a b b$ 的个体杂交得到 $F_{1}$, 下列分析正确的是（）\n\n[图1]\n\nA: $F_{1}$ 全部开红色花\nB: $F_{1}$ 测交后代中花色的表现型及比例是白: 粉: 红 $=1: 2: 1$\nC: $F_{1}$ 自交后代中花色的表现型及比例是白：粉：红=4: 9: 3\nD: 图示过程体现基因对性状的直接控制\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-038.jpg?height=191&width=774&top_left_y=167&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1160",
        "problem": "A series of experiments were conducted with one-day and three-day old chicks. The one-day old chicks were removed from the parents at hatching whereas the three-day old chicks had been reared with their parents until the time of the experiment. Both sets of chicks were presented with models of the parent head and the following responses obtained:\n\n[figure1]\n\nWhich is the best interpretation of these results?\nA: Pecking behaviour is a fixed action pattern where any long pointed object acts as an equally effective stimulus.\nB: The pecking rate of laughing gull chicks increases with age\nC: The response of one-day old chicks is more pronounced when the model is closer to that of the parent\nD: The act of pecking is an innate behaviour whilst the discriminatory capacity of the chick is a result of learning\nE: Pecking behaviour is triggered by a sign stimulus, the head colour of the adult gull\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA series of experiments were conducted with one-day and three-day old chicks. The one-day old chicks were removed from the parents at hatching whereas the three-day old chicks had been reared with their parents until the time of the experiment. Both sets of chicks were presented with models of the parent head and the following responses obtained:\n\n[figure1]\n\nWhich is the best interpretation of these results?\n\nA: Pecking behaviour is a fixed action pattern where any long pointed object acts as an equally effective stimulus.\nB: The pecking rate of laughing gull chicks increases with age\nC: The response of one-day old chicks is more pronounced when the model is closer to that of the parent\nD: The act of pecking is an innate behaviour whilst the discriminatory capacity of the chick is a result of learning\nE: Pecking behaviour is triggered by a sign stimulus, the head colour of the adult gull\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-12.jpg?height=723&width=659&top_left_y=777&top_left_x=733"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_973",
        "problem": "A species of insect was found to have a resistance to a commonly used insecticide. Which of the following is the most likely explanation?\nA: Stabilizing selection caused development of resistance in the insect population\nB: The original gene pool included genes that conferred resistance to the insecticide\nC: The insecticide stimulated development of resistance in certain individuals and this was inherited\nD: The insecticide caused a mutation that was favorable to resistance and this was inherited\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA species of insect was found to have a resistance to a commonly used insecticide. Which of the following is the most likely explanation?\n\nA: Stabilizing selection caused development of resistance in the insect population\nB: The original gene pool included genes that conferred resistance to the insecticide\nC: The insecticide stimulated development of resistance in certain individuals and this was inherited\nD: The insecticide caused a mutation that was favorable to resistance and this was inherited\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_214",
        "problem": "Endotherm animals can only tolerate narrow range of body temperature $\\left(30-45^{\\circ} \\mathrm{C}\\right)$, but their ability to generate heat internally allowed them to expand their distributions. On the other hand, the cost to be endotherm is high because of their constant demand for food to support the energy needed for heat production.\n\n[figure1]\nA: In a drastically-changing environment optimal body temperature of an endotherm animal will be largely maintained with minor behavioural\nB: When body temperature of endotherm animal starts dropping, it triggers the increase of metabolic heat generation, independent of animal habitat.\nC: Basal metabolic rate of an endotherm animal stays constant throughout a limited range of environmental temperatures.\nD: One way that some endotherm animal species tolerate extreme cold conditions is to alter their lower critical temperature.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nEndotherm animals can only tolerate narrow range of body temperature $\\left(30-45^{\\circ} \\mathrm{C}\\right)$, but their ability to generate heat internally allowed them to expand their distributions. On the other hand, the cost to be endotherm is high because of their constant demand for food to support the energy needed for heat production.\n\n[figure1]\n\nA: In a drastically-changing environment optimal body temperature of an endotherm animal will be largely maintained with minor behavioural\nB: When body temperature of endotherm animal starts dropping, it triggers the increase of metabolic heat generation, independent of animal habitat.\nC: Basal metabolic rate of an endotherm animal stays constant throughout a limited range of environmental temperatures.\nD: One way that some endotherm animal species tolerate extreme cold conditions is to alter their lower critical temperature.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3c69afdd8d3df34c0783g-58.jpg?height=591&width=1031&top_left_y=578&top_left_x=518"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_168",
        "problem": "Fabian MA et al. (2005) A small molecule-kinase interaction map for clinical kinase inhibitors. Nat. Biotechnol. 23, 329-336.\n\nYou identified a gene in fission yeast, homologous to a telomerase subunit from a protozoan. You then make a targeted deletion of one copy of the gene in a diploid strain of the yeast and then induce sporulation to produce haploid organisms. All four spores germinate perfectly, and you are able to grow colonies on nutrient agar plates. Every 3 days, you re-streak colonies onto fresh plates. After four such serial transfers, the descendants of two of the original four spores grow poorly, if at all. You take cells from the 3-, 6-, and 9-day master plates, prepare DNA from them, and cleave the samples at a chromosomal site about 35 nucleotides away from the start of the telomere repeats. You separate the fragments by gel electrophoresis, and hybridize them to a radioactive telomere-specific probe (Fig.Q.52). Assume that generation time is 6 hours.\n\n[figure1]\n\nFig.Q.52. Analysis of telomeres from four fission-yeast spores.\n\nWT is the normal diploid yeast\nA: The average length of telomere in fission yeast is 280 nucleotides.\nB: Spores 2 and 4 appear to lack telomerase.\nC: Fission yeast telomeres lose less than 10 nucleotides per replication.\nD: The fission yeasts that lose their telomeres will have normal size.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFabian MA et al. (2005) A small molecule-kinase interaction map for clinical kinase inhibitors. Nat. Biotechnol. 23, 329-336.\n\nYou identified a gene in fission yeast, homologous to a telomerase subunit from a protozoan. You then make a targeted deletion of one copy of the gene in a diploid strain of the yeast and then induce sporulation to produce haploid organisms. All four spores germinate perfectly, and you are able to grow colonies on nutrient agar plates. Every 3 days, you re-streak colonies onto fresh plates. After four such serial transfers, the descendants of two of the original four spores grow poorly, if at all. You take cells from the 3-, 6-, and 9-day master plates, prepare DNA from them, and cleave the samples at a chromosomal site about 35 nucleotides away from the start of the telomere repeats. You separate the fragments by gel electrophoresis, and hybridize them to a radioactive telomere-specific probe (Fig.Q.52). Assume that generation time is 6 hours.\n\n[figure1]\n\nFig.Q.52. Analysis of telomeres from four fission-yeast spores.\n\nWT is the normal diploid yeast\n\nA: The average length of telomere in fission yeast is 280 nucleotides.\nB: Spores 2 and 4 appear to lack telomerase.\nC: Fission yeast telomeres lose less than 10 nucleotides per replication.\nD: The fission yeasts that lose their telomeres will have normal size.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-007.jpg?height=652&width=1063&top_left_y=1110&top_left_x=514"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1106",
        "problem": "A few statements regarding the sexual and asexual modes of reproduction in plants are given.\n1. In sexual reproduction, progeny are genetically different from each other.\n2. In asexual mode of reproduction, progeny are genetically identical to each other but different from the parent.\n3. Sexual reproduction is more conducive for driving evolution.\n4. A minor change in the habitat may adversely affect all offspring derived by asexual reproduction.\n5. A bisexual plant grown in isolation is always incapable of sexual reproduction.\n\nWhich of these statements are correct?\nA: 1,3 and 4\nB: 1,2 and 5\nC: 3,4 and 5\nD: 2, 3 and 4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA few statements regarding the sexual and asexual modes of reproduction in plants are given.\n1. In sexual reproduction, progeny are genetically different from each other.\n2. In asexual mode of reproduction, progeny are genetically identical to each other but different from the parent.\n3. Sexual reproduction is more conducive for driving evolution.\n4. A minor change in the habitat may adversely affect all offspring derived by asexual reproduction.\n5. A bisexual plant grown in isolation is always incapable of sexual reproduction.\n\nWhich of these statements are correct?\n\nA: 1,3 and 4\nB: 1,2 and 5\nC: 3,4 and 5\nD: 2, 3 and 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1258",
        "problem": "Which one of the following graphs best represents the changes in substrate concentration during the course of an enzyme reaction in vitro?\n[figure1]\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following graphs best represents the changes in substrate concentration during the course of an enzyme reaction in vitro?\n[figure1]\n\nA: A\nB: B\nC: C\nD: D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-02.jpg?height=368&width=1604&top_left_y=320&top_left_x=258"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1480",
        "problem": "For the three different diseases, pedigree analysis was performed on family trees.\n\n[figure1]\n\nSecondly, if the frequency of the recessive allele in the population is 0.1 , calculate the probability that a new child of 3 and 4 gets the disease.\nA: 0.05\nB: 0.1\nC: 0.25\nD: 0.5\nE: 1\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor the three different diseases, pedigree analysis was performed on family trees.\n\n[figure1]\n\nSecondly, if the frequency of the recessive allele in the population is 0.1 , calculate the probability that a new child of 3 and 4 gets the disease.\n\nA: 0.05\nB: 0.1\nC: 0.25\nD: 0.5\nE: 1\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-43.jpg?height=1110&width=1636&top_left_y=500&top_left_x=240"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1126",
        "problem": "To obtain a clear and systematic picture of mortality and survival within a population, ecologists use an approach involving the construction of life tables. The life table is an age-specific account of mortality. The life table of a Gypsy Moth population is represented below.\n\n$l_{z}$ represents the probability of the given stage surviving relative to the first stage.\n\n$q_{x}$ represents the rate of mortality for the given stage of life cycle.\n\n| $x$ | $l_{x}$ | $q_{x}$ |\n| :--- | :--- | :--- |\n| Eggs | 1.000 | 0.300 |\n| Instars I-III | 0.700 | 0.819 |\n| Instars IV-VII | 0.127 | 0.582 |\n| Prepupae | 0.053 | 0.038 |\n| Pupae | 0.051 | 0.29 |\n| Adults | 0.036 | 1.000 |\n\nIf the number of eggs were 450 , what would be the number of pupae that do not survive to become adult Gypsy Moths?",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\n\nproblem:\nTo obtain a clear and systematic picture of mortality and survival within a population, ecologists use an approach involving the construction of life tables. The life table is an age-specific account of mortality. The life table of a Gypsy Moth population is represented below.\n\n$l_{z}$ represents the probability of the given stage surviving relative to the first stage.\n\n$q_{x}$ represents the rate of mortality for the given stage of life cycle.\n\n| $x$ | $l_{x}$ | $q_{x}$ |\n| :--- | :--- | :--- |\n| Eggs | 1.000 | 0.300 |\n| Instars I-III | 0.700 | 0.819 |\n| Instars IV-VII | 0.127 | 0.582 |\n| Prepupae | 0.053 | 0.038 |\n| Pupae | 0.051 | 0.29 |\n| Adults | 0.036 | 1.000 |\n\nIf the number of eggs were 450 , what would be the number of pupae that do not survive to become adult Gypsy Moths?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            null,
            null
        ],
        "answer_sequence": null,
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_247",
        "problem": "Huntington's disease (HD) is a genetic disorder characterized by devastating degeneration of nerve tissues that progresses with age. Huntingtin (HTT) is known to be the causative protein of HD. Near the transcriptional initiation point of HTT gene, there is a sequence containing repeated CAG (corresponding to glutamine), which are usually between 9 to 35 repeats in healthy individuals. These repeats are 35 to 75 in HD-population. The symptoms of HD tend to appear at a younger age and are more severe when there are an increased number of CAG repeats.\n\nRecently, scientists in France have revealed that HTT plays an important role in maintaining neuronal fast axonal transport (FAT, Figure 1). By careful observations with fluorescence microscopy, they first showed that HTT was co-localized with motor proteins (kinesin and dynein) that are involved in FAT. HTT was also shown to be co-localized with synaptic vesicles, as well as with glyceraldehyde-3-phosphate dehydrogenase (GAPDH). Interestingly, HTT was not found with mitochondria that were transported by FAT. Next, using cultured neurons, they investigated the effects of oligomycin, an inhibitor of ATP production in mitochondria, and iodoacetate acid, an inhibitor of GAPDH activity (Table 1). Furthermore, when HTT expression was suppressed by RNAi treatment, only the FAT of synaptic vesicles, not that of mitochondria, was significantly reduced. These results indicate that HTT was solely involved in FAT of synaptic vesicles.\n\n[figure1]\n\nFigure 1 Fast axonal transport (FAT) in nerve cells. a, active transportation of synaptic vesicles and mitochondria outward to the nerve ends is called anterograde FAT (aFAT). Transportation in the opposite inward direction is called retrograde FAT (rFAT). Measured velocity and its distribution (\\%) is shown in $\\mathbf{b}$.\n\n|  | Synaptic vesicles |  |  | Mitochondria |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | Control | Oligomycin | lodoacetate | Control | Oligomycin | lodoacetate |\n| aFAT | $2.3 \\pm 0.1$ | $2.2 \\pm 0.2$ | $0.3 \\pm 0.1$ | $0.9 \\pm 0.1$ | $0.3 \\pm 0.1$ | $1.0 \\pm 0.1$ |\n| rFAT | $-1.9 \\pm 0.1$ | $-1.9 \\pm 0.2$ | $-0.2 \\pm 0.1$ | $-1.2 \\pm 0.1$ | $-0.4 \\pm 0.2$ | $-1.0 \\pm 0.1$ |\n\nTable 1 Effects of oligomycin and iodoacetate on the velocity [ $\\mu \\mathrm{m} / \\mathrm{s}]$ of anterograde (aFAT) and retrograde ( $r F A T)$ transportation. In the experiments to determine FAT velocity with iodoacetate, pyruvate was included to maintain ATP production by mitochondria. Control experiments were carried out in a buffer medium without inhibitors. Under all experimental conditions, ATP/ADP ratio in axons was maintained $>80 \\%$.\nA: Near the N-terminal end of HTT molecules in HD patients, there is a larger number of glutamine repeats compared to that in healthy individuals. 80\nB: It is possible that HTT helps to anchor GAPDH and motor proteins to synaptic vesicles.\nC: ATP produced by mitochondria is not efficiently used for FAT of synaptic vesicles, even though it can maintain a sufficiently high ATP concentration within axons.\nD: ATP produced by glycolysis is crucial for the FAT of mitochondria.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nHuntington's disease (HD) is a genetic disorder characterized by devastating degeneration of nerve tissues that progresses with age. Huntingtin (HTT) is known to be the causative protein of HD. Near the transcriptional initiation point of HTT gene, there is a sequence containing repeated CAG (corresponding to glutamine), which are usually between 9 to 35 repeats in healthy individuals. These repeats are 35 to 75 in HD-population. The symptoms of HD tend to appear at a younger age and are more severe when there are an increased number of CAG repeats.\n\nRecently, scientists in France have revealed that HTT plays an important role in maintaining neuronal fast axonal transport (FAT, Figure 1). By careful observations with fluorescence microscopy, they first showed that HTT was co-localized with motor proteins (kinesin and dynein) that are involved in FAT. HTT was also shown to be co-localized with synaptic vesicles, as well as with glyceraldehyde-3-phosphate dehydrogenase (GAPDH). Interestingly, HTT was not found with mitochondria that were transported by FAT. Next, using cultured neurons, they investigated the effects of oligomycin, an inhibitor of ATP production in mitochondria, and iodoacetate acid, an inhibitor of GAPDH activity (Table 1). Furthermore, when HTT expression was suppressed by RNAi treatment, only the FAT of synaptic vesicles, not that of mitochondria, was significantly reduced. These results indicate that HTT was solely involved in FAT of synaptic vesicles.\n\n[figure1]\n\nFigure 1 Fast axonal transport (FAT) in nerve cells. a, active transportation of synaptic vesicles and mitochondria outward to the nerve ends is called anterograde FAT (aFAT). Transportation in the opposite inward direction is called retrograde FAT (rFAT). Measured velocity and its distribution (\\%) is shown in $\\mathbf{b}$.\n\n|  | Synaptic vesicles |  |  | Mitochondria |  |  |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | Control | Oligomycin | lodoacetate | Control | Oligomycin | lodoacetate |\n| aFAT | $2.3 \\pm 0.1$ | $2.2 \\pm 0.2$ | $0.3 \\pm 0.1$ | $0.9 \\pm 0.1$ | $0.3 \\pm 0.1$ | $1.0 \\pm 0.1$ |\n| rFAT | $-1.9 \\pm 0.1$ | $-1.9 \\pm 0.2$ | $-0.2 \\pm 0.1$ | $-1.2 \\pm 0.1$ | $-0.4 \\pm 0.2$ | $-1.0 \\pm 0.1$ |\n\nTable 1 Effects of oligomycin and iodoacetate on the velocity [ $\\mu \\mathrm{m} / \\mathrm{s}]$ of anterograde (aFAT) and retrograde ( $r F A T)$ transportation. In the experiments to determine FAT velocity with iodoacetate, pyruvate was included to maintain ATP production by mitochondria. Control experiments were carried out in a buffer medium without inhibitors. Under all experimental conditions, ATP/ADP ratio in axons was maintained $>80 \\%$.\n\nA: Near the N-terminal end of HTT molecules in HD patients, there is a larger number of glutamine repeats compared to that in healthy individuals. 80\nB: It is possible that HTT helps to anchor GAPDH and motor proteins to synaptic vesicles.\nC: ATP produced by mitochondria is not efficiently used for FAT of synaptic vesicles, even though it can maintain a sufficiently high ATP concentration within axons.\nD: ATP produced by glycolysis is crucial for the FAT of mitochondria.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-29.jpg?height=682&width=813&top_left_y=1561&top_left_x=610"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_854",
        "problem": "细菌 $\\operatorname{glg}$ 基因编码的 UDPG 焦磷酸化酶在糖原合成中起关键作用。细菌糖原合成的平衡受到 CsrAB 系统的调节。CsrA 蛋白可以结合 glg mRNA 分子, 也可结合非编码 RNA 分子 $\\mathrm{CsrB}$ ，如图所示。下列叙述不正确的是（）\n\n[图1]\nA: 细菌 $\\operatorname{glg}$ 基因转录时, RNA 聚合酶识别和结合 $\\operatorname{glg}$ 基因的启动子并驱动转录\nB: 细菌合成 UDPG 焦磷酸化酶的肽链时, 核糖体沿 glgmRNA 从 5'端向 3'端移动\nC: $\\mathrm{CsrB}$ 是 $\\mathrm{CsrB}$ 基因的转录产物, 抑制 $\\mathrm{CsrB}$ 基因的转录能促进细菌糖原合成\nD: CsrA 蛋白都结合到 CsrB 上, 有利于细菌糖原合成\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n细菌 $\\operatorname{glg}$ 基因编码的 UDPG 焦磷酸化酶在糖原合成中起关键作用。细菌糖原合成的平衡受到 CsrAB 系统的调节。CsrA 蛋白可以结合 glg mRNA 分子, 也可结合非编码 RNA 分子 $\\mathrm{CsrB}$ ，如图所示。下列叙述不正确的是（）\n\n[图1]\n\nA: 细菌 $\\operatorname{glg}$ 基因转录时, RNA 聚合酶识别和结合 $\\operatorname{glg}$ 基因的启动子并驱动转录\nB: 细菌合成 UDPG 焦磷酸化酶的肽链时, 核糖体沿 glgmRNA 从 5'端向 3'端移动\nC: $\\mathrm{CsrB}$ 是 $\\mathrm{CsrB}$ 基因的转录产物, 抑制 $\\mathrm{CsrB}$ 基因的转录能促进细菌糖原合成\nD: CsrA 蛋白都结合到 CsrB 上, 有利于细菌糖原合成\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-70.jpg?height=476&width=691&top_left_y=156&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_60",
        "problem": "An animal's territory is an exclusive area defended by an individual to keep resources, such as food, and mates. Territory is different from home range, because home range simply represents an area over which an animal regularly moves and may overlap with those of neighboring animals of the same species. The size of the territory is determined by the cost and benefit obtained from the area, in a way that maximizes the net benefit gain of individual animals. The graph below shows how cost and benefit change with the size of the territory.\n\n[figure1]\n\nTerritory size\nA: The benefit curve shows saturation at a large territory size due to the depletion of resources.\nB: The optimal territory size is the intersection point between the cost line and the benefit curve\nC: When resources become scarce while the cost line is unchanged, the optimal territory size becomes larger\nD: When the population density increases and intraspecific competition becomes intense, territorial behavior could disappear.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn animal's territory is an exclusive area defended by an individual to keep resources, such as food, and mates. Territory is different from home range, because home range simply represents an area over which an animal regularly moves and may overlap with those of neighboring animals of the same species. The size of the territory is determined by the cost and benefit obtained from the area, in a way that maximizes the net benefit gain of individual animals. The graph below shows how cost and benefit change with the size of the territory.\n\n[figure1]\n\nTerritory size\n\nA: The benefit curve shows saturation at a large territory size due to the depletion of resources.\nB: The optimal territory size is the intersection point between the cost line and the benefit curve\nC: When resources become scarce while the cost line is unchanged, the optimal territory size becomes larger\nD: When the population density increases and intraspecific competition becomes intense, territorial behavior could disappear.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-61.jpg?height=494&width=826&top_left_y=861&top_left_x=181",
            "https://cdn.mathpix.com/cropped/2024_03_14_6d440fcb8a6b50c185f5g-30.jpg?height=537&width=674&top_left_y=491&top_left_x=294"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_239",
        "problem": "To study the effect of phytohormone on fruit maturation, reseachers used abscisic acid (ABA) and ethephon to treat of sweet cherry fruits and afterward to evaluate the expression of PacNCED1 gene which encodes 9-cis-epoxycarotenoid dioxygenase, a key enzyme in ABA biosynthesis. They also checked the expression of PacACO1 gene encoding 1-aminocyclopropane-1-carboxylic acid oxidase enzyme that involves in ethylene biosynthesis. The transcript of PacACT1 (one $\\beta$-actin cDNA fragment was cloned and designated as PacACT1, accession number FJ560908) was used to standardize for all expression (Fig.Q17-D).\n\n| Treetment | Firmness of <br> pulp | Soluble solids content/ <br> titratable acidity | Anthocyanin(U.g $\\left.\\mathbf{~}^{\\mathbf{1}}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| Control | $20.3 \\mathrm{a}$ | $14.4 \\mathrm{a}$ | $13.4 \\mathrm{a}$ |\n| Ethephon | $19.6 \\mathrm{a}$ | $15.3 \\mathrm{a}$ | $14.4 \\mathrm{a}$ |\n| ABA | $11.9 \\mathrm{~b}$ | $16.8 \\mathrm{~b}$ | $23.8 \\mathrm{~b}$ |\n\n( $a$ and $b$ show values that are significantly different).\n\n[figure1]\n\n(C)\n[figure2]\n\n(D)\n\n[figure3]\n\nFig. Q17. Effects of ABA and ethephon application on ABA content (A), accumulation of PacNCED1 (B), ethylene production during ripening (C) and PacACO1(D).\nA: Both $\\mathrm{ABA}$ and Ethephon stimulate the expression of PacACO1 and PacNCED genes in sweet cherry fruit.\nB: The expression of PacNCED1 and ABA accumulation in pulp are higher than in the seed in the treatments of $\\mathrm{ABA}$ and ethephon.\nC: ABA induces the maturation of sweet cherry fruit via stimulation of ethylene production.\nD: Ethephon shows lower effect on anthocyanin and indogenous ABA production than exogenous ABA does.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nTo study the effect of phytohormone on fruit maturation, reseachers used abscisic acid (ABA) and ethephon to treat of sweet cherry fruits and afterward to evaluate the expression of PacNCED1 gene which encodes 9-cis-epoxycarotenoid dioxygenase, a key enzyme in ABA biosynthesis. They also checked the expression of PacACO1 gene encoding 1-aminocyclopropane-1-carboxylic acid oxidase enzyme that involves in ethylene biosynthesis. The transcript of PacACT1 (one $\\beta$-actin cDNA fragment was cloned and designated as PacACT1, accession number FJ560908) was used to standardize for all expression (Fig.Q17-D).\n\n| Treetment | Firmness of <br> pulp | Soluble solids content/ <br> titratable acidity | Anthocyanin(U.g $\\left.\\mathbf{~}^{\\mathbf{1}}\\right)$ |\n| :---: | :---: | :---: | :---: |\n| Control | $20.3 \\mathrm{a}$ | $14.4 \\mathrm{a}$ | $13.4 \\mathrm{a}$ |\n| Ethephon | $19.6 \\mathrm{a}$ | $15.3 \\mathrm{a}$ | $14.4 \\mathrm{a}$ |\n| ABA | $11.9 \\mathrm{~b}$ | $16.8 \\mathrm{~b}$ | $23.8 \\mathrm{~b}$ |\n\n( $a$ and $b$ show values that are significantly different).\n\n[figure1]\n\n(C)\n[figure2]\n\n(D)\n\n[figure3]\n\nFig. Q17. Effects of ABA and ethephon application on ABA content (A), accumulation of PacNCED1 (B), ethylene production during ripening (C) and PacACO1(D).\n\nA: Both $\\mathrm{ABA}$ and Ethephon stimulate the expression of PacACO1 and PacNCED genes in sweet cherry fruit.\nB: The expression of PacNCED1 and ABA accumulation in pulp are higher than in the seed in the treatments of $\\mathrm{ABA}$ and ethephon.\nC: ABA induces the maturation of sweet cherry fruit via stimulation of ethylene production.\nD: Ethephon shows lower effect on anthocyanin and indogenous ABA production than exogenous ABA does.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-040.jpg?height=476&width=684&top_left_y=839&top_left_x=342",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-040.jpg?height=912&width=1342&top_left_y=910&top_left_x=342",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-040.jpg?height=208&width=368&top_left_y=1337&top_left_x=1246"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1367",
        "problem": "Hemoglobin and myoglobin are proteins that have oxygen-carrying capacity. Hemoglobin is found in red blood cells and myoglobin is found in muscles. Hemoglobin can bind to four oxygen molecules at any point in time. In the lungs, where there is a high oxygen concentration, hemoglobin binds to oxygen forming oxyhemoglobin. Oxyhemoglobin then circulates the body in blood and unloads oxygen to tissues with a low oxygen concentration, reforming hemoglobin. Myoglobin stores oxygen in the muscle tissue and only releases oxygen when the partial pressure of oxygen has fallen drastically. Myoglobin is only found in the bloodstream after muscle injury.\n\nThe oxygen dissociation curve, depicted below, describes the relationship between the partial pressure of oxygen ( $x$ axis) and the oxygen saturation of hemoglobin or myoglobin ( $y$ axis).\n\n[figure1]\n\nAs hemoglobin unloads oxygen:\nA: It becomes easier to unload more oxygen\nB: It becomes harder to unload more oxygen\nC: Unloading oxygen continues at the same rate\nD: Carbon dioxide is exchanged for oxygen on hemoglobin\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHemoglobin and myoglobin are proteins that have oxygen-carrying capacity. Hemoglobin is found in red blood cells and myoglobin is found in muscles. Hemoglobin can bind to four oxygen molecules at any point in time. In the lungs, where there is a high oxygen concentration, hemoglobin binds to oxygen forming oxyhemoglobin. Oxyhemoglobin then circulates the body in blood and unloads oxygen to tissues with a low oxygen concentration, reforming hemoglobin. Myoglobin stores oxygen in the muscle tissue and only releases oxygen when the partial pressure of oxygen has fallen drastically. Myoglobin is only found in the bloodstream after muscle injury.\n\nThe oxygen dissociation curve, depicted below, describes the relationship between the partial pressure of oxygen ( $x$ axis) and the oxygen saturation of hemoglobin or myoglobin ( $y$ axis).\n\n[figure1]\n\nAs hemoglobin unloads oxygen:\n\nA: It becomes easier to unload more oxygen\nB: It becomes harder to unload more oxygen\nC: Unloading oxygen continues at the same rate\nD: Carbon dioxide is exchanged for oxygen on hemoglobin\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-12.jpg?height=820&width=1014&top_left_y=892&top_left_x=538"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1473",
        "problem": "The control of thyroid hormone secretion follows the pathway depicted in the figure. Thyroid releasing hormone (TRH) is secreted by the hypothalamus, which stimulates the production of thyroid stimulating hormone (TSH) from the anterior pituitary. TSH then stimulates the production of the hormones thyroxine (T4) and triiodothyronine (T3) from the thyroid gland.\n\nThe thyroid gland requires dietary iodine to form T4 and T3. T4 contains four iodine atoms while T3 contains three.\n\n[figure1]\n\nHypothyroidism can be caused by iodine deficiency. What is the likely consequence of hypothyroidism?\nA: High TSH, low T4 and low T3 levels.\nB: Low TSH, low T4 and low T3 levels.\nC: High TSH, low T4 and high T3 levels.\nD: Low TSH, high T4 and high T3 levels.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe control of thyroid hormone secretion follows the pathway depicted in the figure. Thyroid releasing hormone (TRH) is secreted by the hypothalamus, which stimulates the production of thyroid stimulating hormone (TSH) from the anterior pituitary. TSH then stimulates the production of the hormones thyroxine (T4) and triiodothyronine (T3) from the thyroid gland.\n\nThe thyroid gland requires dietary iodine to form T4 and T3. T4 contains four iodine atoms while T3 contains three.\n\n[figure1]\n\nHypothyroidism can be caused by iodine deficiency. What is the likely consequence of hypothyroidism?\n\nA: High TSH, low T4 and low T3 levels.\nB: Low TSH, low T4 and low T3 levels.\nC: High TSH, low T4 and high T3 levels.\nD: Low TSH, high T4 and high T3 levels.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-27.jpg?height=1714&width=1017&top_left_y=725&top_left_x=631"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_49",
        "problem": "To study the effects of cadmium (Cd) on root development, two experiments on maize seedlings with 6-cm-long root were conducted. In the first experiment, seedlings were grown either in media supplemented with $5 \\mu \\mathrm{M} \\mathrm{Cd}(\\mathrm{Cd} 5)$ or without $\\mathrm{Cd}(\\mathrm{Cd} 0)$. In the second experiment, seedlings were grown either in two layers of agar without $\\mathrm{Cd}$ ( $\\mathrm{Cd0}$ $\\mathrm{Cd} 0)$ or unilaterally to $100 \\mu \\mathrm{M} \\mathrm{Cd}(\\mathrm{Cd} 0-\\mathrm{Cd} 100)$. Four days later, root growth was recorded and cross-sections of roots were stained to visualize suberin lamillae in endodermis.\n\n[figure1]\n\nFig.Q61-1. Distance (\\%) from root tip to root base is shown on the left. The presence of suberin lamillae in the roots are shown as solid and dashed lines in the center figures. White arrows in $A$ and $B$ indicate suberin lamellae in the endodermis.\n[figure2]\n\nFig.Q61-2. Distance (mm) from root apex to root base is shown on the center. The presence of suberin lamillae in the roots are shown as solid and dashed lines. White arrows in A, B, D, E and F indicate suberin lamellae in the endodermis.\nA: The treatment of $\\mathrm{Cd}$ resulted in the reduction of elongation zone of the root, leading to decreased root length.\nB: Endodermal cells with suberin lamellae were already present at a distance of approximately $0.5 \\mathrm{~cm}$ from the root apex in tissues adjacent to agar containing $\\mathrm{Cd} 100$, however, suberinized cells were found much later in the other side.\nC: In roots exposed unilaterally to $\\mathrm{Cd}(\\mathrm{Cd} 0-\\mathrm{Cd} 100)$, the development of the endodermis was accelerated and asymmetrical.\nD: In high Cd containing media, suberin lamellae in endodermal cells were not present in older parts of the root likely due to the restriction of $\\mathrm{Cd}$ in younger part.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nTo study the effects of cadmium (Cd) on root development, two experiments on maize seedlings with 6-cm-long root were conducted. In the first experiment, seedlings were grown either in media supplemented with $5 \\mu \\mathrm{M} \\mathrm{Cd}(\\mathrm{Cd} 5)$ or without $\\mathrm{Cd}(\\mathrm{Cd} 0)$. In the second experiment, seedlings were grown either in two layers of agar without $\\mathrm{Cd}$ ( $\\mathrm{Cd0}$ $\\mathrm{Cd} 0)$ or unilaterally to $100 \\mu \\mathrm{M} \\mathrm{Cd}(\\mathrm{Cd} 0-\\mathrm{Cd} 100)$. Four days later, root growth was recorded and cross-sections of roots were stained to visualize suberin lamillae in endodermis.\n\n[figure1]\n\nFig.Q61-1. Distance (\\%) from root tip to root base is shown on the left. The presence of suberin lamillae in the roots are shown as solid and dashed lines in the center figures. White arrows in $A$ and $B$ indicate suberin lamellae in the endodermis.\n[figure2]\n\nFig.Q61-2. Distance (mm) from root apex to root base is shown on the center. The presence of suberin lamillae in the roots are shown as solid and dashed lines. White arrows in A, B, D, E and F indicate suberin lamellae in the endodermis.\n\nA: The treatment of $\\mathrm{Cd}$ resulted in the reduction of elongation zone of the root, leading to decreased root length.\nB: Endodermal cells with suberin lamellae were already present at a distance of approximately $0.5 \\mathrm{~cm}$ from the root apex in tissues adjacent to agar containing $\\mathrm{Cd} 100$, however, suberinized cells were found much later in the other side.\nC: In roots exposed unilaterally to $\\mathrm{Cd}(\\mathrm{Cd} 0-\\mathrm{Cd} 100)$, the development of the endodermis was accelerated and asymmetrical.\nD: In high Cd containing media, suberin lamellae in endodermal cells were not present in older parts of the root likely due to the restriction of $\\mathrm{Cd}$ in younger part.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-025.jpg?height=801&width=1302&top_left_y=1024&top_left_x=380",
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-026.jpg?height=792&width=1070&top_left_y=185&top_left_x=476"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_391",
        "problem": "已知某环境条件下某种动物的 $\\mathrm{AA}$ 和 $\\mathrm{Aa}$ 个体全部存活, aa 个体在出生前会全部死亡。现有该动物的一个大群体，只有 $\\mathrm{AA} 、 \\mathrm{Aa}$ 两种基因型，其比例为 $1: 2$ 。假设每对亲本只交配一次且成功受孕, 均为单胎, 在上述环境条件下，理论上该群体随机交配产生的第一代中 $\\mathrm{AA}$ 和 Aa 的比例是(\nA: $1: 1$\nB: $1: 2$\nC: 2: 1\nD: 3: 1\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知某环境条件下某种动物的 $\\mathrm{AA}$ 和 $\\mathrm{Aa}$ 个体全部存活, aa 个体在出生前会全部死亡。现有该动物的一个大群体，只有 $\\mathrm{AA} 、 \\mathrm{Aa}$ 两种基因型，其比例为 $1: 2$ 。假设每对亲本只交配一次且成功受孕, 均为单胎, 在上述环境条件下，理论上该群体随机交配产生的第一代中 $\\mathrm{AA}$ 和 Aa 的比例是(\n\nA: $1: 1$\nB: $1: 2$\nC: 2: 1\nD: 3: 1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1339",
        "problem": "[figure1]\n\nhttp://www5.pbrc.hawaii.edu/allen/ch19/\nVorticella is a genus of protozoa found mainly in freshwater streams. On the left is a transmission electron micrograph showing food vacuoles (fv), a stalk (stalk) and a macronucleus (mac). The bar on the bottom right is $2 \\mu \\mathrm{m}$.\n\nWhat is the maximum diameter of the largest food vacuole shown?\nA: $8 \\mu \\mathrm{m}$\nB: $18 \\mu \\mathrm{m}$\nC: $180 \\mu \\mathrm{m}$\nD: $0.18 \\mathrm{~mm}$\nE: $0.8 \\mathrm{~mm}$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nhttp://www5.pbrc.hawaii.edu/allen/ch19/\nVorticella is a genus of protozoa found mainly in freshwater streams. On the left is a transmission electron micrograph showing food vacuoles (fv), a stalk (stalk) and a macronucleus (mac). The bar on the bottom right is $2 \\mu \\mathrm{m}$.\n\nWhat is the maximum diameter of the largest food vacuole shown?\n\nA: $8 \\mu \\mathrm{m}$\nB: $18 \\mu \\mathrm{m}$\nC: $180 \\mu \\mathrm{m}$\nD: $0.18 \\mathrm{~mm}$\nE: $0.8 \\mathrm{~mm}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-07.jpg?height=677&width=617&top_left_y=1763&top_left_x=114"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1427",
        "problem": "What is the term that best describes the net movement of uncharged molecules from an area of high concentration to an area of low concentration?\nA: diffusion\nB: active transport\nC: passive transport\nD: osmosis\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the term that best describes the net movement of uncharged molecules from an area of high concentration to an area of low concentration?\n\nA: diffusion\nB: active transport\nC: passive transport\nD: osmosis\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_586",
        "problem": "果蝇眼色这一性状有红眼、伊红眼和奶油眼等表型, 已知这一性状由 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 两对等位基因控制。为研究其遗传规律, 某科研小组选取了两只红眼果蝇作为亲本进行杂交, 子代中雌果蝇全为红眼, 雄果蝇中, 红眼: 伊红眼: 奶油眼 $=4: 3: 1$. 下列说法错误的是 ( )\nA: $\\mathrm{A} / \\mathrm{a}$ 与 $\\mathrm{B} / \\mathrm{b}$ 遵循自由组合定律\nB: $\\mathrm{A} / \\mathrm{a}$ 与 $\\mathrm{B} / \\mathrm{b}$ 中有一对位于 $\\mathrm{X}$ 染色体上\nC: 子代果蝇的基因型共 9 种\nD: 子代个体中, 纯合红眼雌性个体和纯合红眼雄性个体所占比例相同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇眼色这一性状有红眼、伊红眼和奶油眼等表型, 已知这一性状由 $\\mathrm{A} / \\mathrm{a} 、 \\mathrm{~B} / \\mathrm{b}$ 两对等位基因控制。为研究其遗传规律, 某科研小组选取了两只红眼果蝇作为亲本进行杂交, 子代中雌果蝇全为红眼, 雄果蝇中, 红眼: 伊红眼: 奶油眼 $=4: 3: 1$. 下列说法错误的是 ( )\n\nA: $\\mathrm{A} / \\mathrm{a}$ 与 $\\mathrm{B} / \\mathrm{b}$ 遵循自由组合定律\nB: $\\mathrm{A} / \\mathrm{a}$ 与 $\\mathrm{B} / \\mathrm{b}$ 中有一对位于 $\\mathrm{X}$ 染色体上\nC: 子代果蝇的基因型共 9 种\nD: 子代个体中, 纯合红眼雌性个体和纯合红眼雄性个体所占比例相同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1266",
        "problem": "The following table shows the interaction between pairs of species\n\n| Pair | Species 1 | Species 2 |\n| :---: | :---: | :---: |\n| 1 | 0 | + |\n| 2 | - | + |\n| 3 | + | + |\n| 4 | + | - |\n| 5 | + | - |\n\nWhich one of the following gives the correct type of interactions for species $\\mathbf{1} \\mathbf{- 5}$\nA: mutualism, parasitism, commensalism, herbivory, predation\nB: mutualism, parasitism, commensalism, predation, herbivory\nC: commensalism, parasitism, mutualism, herbivory, predation\nD: commensalism, herbivory, parasitism, mutualism, predation\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following table shows the interaction between pairs of species\n\n| Pair | Species 1 | Species 2 |\n| :---: | :---: | :---: |\n| 1 | 0 | + |\n| 2 | - | + |\n| 3 | + | + |\n| 4 | + | - |\n| 5 | + | - |\n\nWhich one of the following gives the correct type of interactions for species $\\mathbf{1} \\mathbf{- 5}$\n\nA: mutualism, parasitism, commensalism, herbivory, predation\nB: mutualism, parasitism, commensalism, predation, herbivory\nC: commensalism, parasitism, mutualism, herbivory, predation\nD: commensalism, herbivory, parasitism, mutualism, predation\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1042",
        "problem": "Which of the following statements about carnivorous plants is FALSE?\nA: All carnivorous plants are angiosperms.\nB: All carnivorous plants are evolved for nutrient-poor environments.\nC: All carnivorous plants are limited to small arthropods or, in the case of bladderwort, zooplankton as prey.\nD: All carnivorous plants have mechanisms to attract and capture prey.\nE: All carnivorous plants derive energy from trapping and consuming arthropods and/or zooplankton\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following statements about carnivorous plants is FALSE?\n\nA: All carnivorous plants are angiosperms.\nB: All carnivorous plants are evolved for nutrient-poor environments.\nC: All carnivorous plants are limited to small arthropods or, in the case of bladderwort, zooplankton as prey.\nD: All carnivorous plants have mechanisms to attract and capture prey.\nE: All carnivorous plants derive energy from trapping and consuming arthropods and/or zooplankton\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1188",
        "problem": "Geological analysis suggests the initial separation of New Zealand from Gondwana took place approximately 80 million years ago. However, it is hypothesised that full separation was not complete until 60 million years ago. Some scientists propose that a common ancestor of the Moa and the Kiwi (a proto-Moa) floated away from Gondwana during this time. Others propose that Moa and Kiwi arrived in New Zealand at two different times. Four possible phylogenetic trees, describing different possible evolutionary relationships within the ratites, are shown below. Use these trees to answer the following questions.\n[figure1]Which tree is most consistent with present day locations of Paleognath species?\nA: Tree 1\nB: Tree 2\nC: Tree 3\nD: Tree 4\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nGeological analysis suggests the initial separation of New Zealand from Gondwana took place approximately 80 million years ago. However, it is hypothesised that full separation was not complete until 60 million years ago. Some scientists propose that a common ancestor of the Moa and the Kiwi (a proto-Moa) floated away from Gondwana during this time. Others propose that Moa and Kiwi arrived in New Zealand at two different times. Four possible phylogenetic trees, describing different possible evolutionary relationships within the ratites, are shown below. Use these trees to answer the following questions.\n[figure1]\n\nproblem:\nWhich tree is most consistent with present day locations of Paleognath species?\n\nA: Tree 1\nB: Tree 2\nC: Tree 3\nD: Tree 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-26.jpg?height=1120&width=1812&top_left_y=505&top_left_x=150"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1439",
        "problem": "Turmeric is a flowering plant, Curcuma longa of the ginger family, Zingiberaceae. It has a very long history of medicinal use, dating back nearly 4000 years. In the last 25 years, over 3000 new publications have been released regarding the health effects of turmeric. The primary compound in turmeric is curcumin, which has been purported to have antiinflammatory and antioxidant effects. Scientists conducted a study investigating the effect of varying concentrations and length of exposure of curcumin on adipocytes from a human cell line. Figure A shows the effect of curcumin concentration on the viability of a population of adipocytes.\n\n[figure1]\n\nFurther investigation showed that the effect on viability was due to changes in the regulation of apoptosis. Caspases are known to play essential roles in apoptosis and are an important regulator of this process is the Bax protein, which causes the release of cytochrome $c$.\n\n$\\mathrm{Bcl}-2$ is a protein that prevents the release of cytochrome $\\mathrm{c}$. The Bax/Bcl-2 ratio determines the balance of apoptosis. Figure B shows the effect of $0,10,20$ and $40 \\mu \\mathrm{M}$ curcumin for a $24 \\mathrm{~h}$ period on the Bax/Bcl- 2 ratio. An asterisk ( ${ }^{*}$ ) denotes a statistically significant result. Figure $\\mathrm{C}$ shows the result of a western blot.\n\nB\n\n[figure2]\n\n[figure3]\n\n$\\mathrm{Bcl}-2$\n\n[figure4]\n\nResults of the present study suggest that curcumin may be a promising therapeutic agent for:\nA: Heart disease, by increasing adipocyte numbers surrounding the heart\nB: Obesity, by increasing adipocyte numbers through the induction of adipocyte apoptosis\nC: Obesity, by decreasing adipocyte numbers through the induction of adipocyte apoptosis\nD: Cardiovascular health, by decreasing a person's fat free mass\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTurmeric is a flowering plant, Curcuma longa of the ginger family, Zingiberaceae. It has a very long history of medicinal use, dating back nearly 4000 years. In the last 25 years, over 3000 new publications have been released regarding the health effects of turmeric. The primary compound in turmeric is curcumin, which has been purported to have antiinflammatory and antioxidant effects. Scientists conducted a study investigating the effect of varying concentrations and length of exposure of curcumin on adipocytes from a human cell line. Figure A shows the effect of curcumin concentration on the viability of a population of adipocytes.\n\n[figure1]\n\nFurther investigation showed that the effect on viability was due to changes in the regulation of apoptosis. Caspases are known to play essential roles in apoptosis and are an important regulator of this process is the Bax protein, which causes the release of cytochrome $c$.\n\n$\\mathrm{Bcl}-2$ is a protein that prevents the release of cytochrome $\\mathrm{c}$. The Bax/Bcl-2 ratio determines the balance of apoptosis. Figure B shows the effect of $0,10,20$ and $40 \\mu \\mathrm{M}$ curcumin for a $24 \\mathrm{~h}$ period on the Bax/Bcl- 2 ratio. An asterisk ( ${ }^{*}$ ) denotes a statistically significant result. Figure $\\mathrm{C}$ shows the result of a western blot.\n\nB\n\n[figure2]\n\n[figure3]\n\n$\\mathrm{Bcl}-2$\n\n[figure4]\n\nResults of the present study suggest that curcumin may be a promising therapeutic agent for:\n\nA: Heart disease, by increasing adipocyte numbers surrounding the heart\nB: Obesity, by increasing adipocyte numbers through the induction of adipocyte apoptosis\nC: Obesity, by decreasing adipocyte numbers through the induction of adipocyte apoptosis\nD: Cardiovascular health, by decreasing a person's fat free mass\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-08.jpg?height=560&width=797&top_left_y=774&top_left_x=618",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-09.jpg?height=411&width=557&top_left_y=1128&top_left_x=407",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-09.jpg?height=168&width=640&top_left_y=1138&top_left_x=1048",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-09.jpg?height=140&width=572&top_left_y=1346&top_left_x=1116"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1368",
        "problem": "Turmeric is a flowering plant, Curcuma longa of the ginger family, Zingiberaceae. It has a very long history of medicinal use, dating back nearly 4000 years. In the last 25 years, over 3000 new publications have been released regarding the health effects of turmeric. The primary compound in turmeric is curcumin, which has been purported to have antiinflammatory and antioxidant effects. Scientists conducted a study investigating the effect of varying concentrations and length of exposure of curcumin on adipocytes from a human cell line. Figure A shows the effect of curcumin concentration on the viability of a population of adipocytes.\n\n[figure1]\n\nFurther investigation showed that the effect on viability was due to changes in the regulation of apoptosis. Caspases are known to play essential roles in apoptosis and are an important regulator of this process is the Bax protein, which causes the release of cytochrome $c$.\n\nCytochrome $c$ :\nA: Initiates the release of caspases from the mitochondria\nB: Interacts with DNA, leading to nuclear fragmentation\nC: Activates the first caspase of an enzymatic cascade\nD: Is an integral protein\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTurmeric is a flowering plant, Curcuma longa of the ginger family, Zingiberaceae. It has a very long history of medicinal use, dating back nearly 4000 years. In the last 25 years, over 3000 new publications have been released regarding the health effects of turmeric. The primary compound in turmeric is curcumin, which has been purported to have antiinflammatory and antioxidant effects. Scientists conducted a study investigating the effect of varying concentrations and length of exposure of curcumin on adipocytes from a human cell line. Figure A shows the effect of curcumin concentration on the viability of a population of adipocytes.\n\n[figure1]\n\nFurther investigation showed that the effect on viability was due to changes in the regulation of apoptosis. Caspases are known to play essential roles in apoptosis and are an important regulator of this process is the Bax protein, which causes the release of cytochrome $c$.\n\nCytochrome $c$ :\n\nA: Initiates the release of caspases from the mitochondria\nB: Interacts with DNA, leading to nuclear fragmentation\nC: Activates the first caspase of an enzymatic cascade\nD: Is an integral protein\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-08.jpg?height=560&width=797&top_left_y=774&top_left_x=618"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_818",
        "problem": "家蚕是二倍体生物 $(2 n=56)$, 雌、雄个体性染色体组成分别是 $\\mathrm{ZW} 、 \\mathrm{ZZ}$ 。幼蚕不抗浓核病与抗浓核病由常染色体上的等位基因 D 和 $\\mathrm{d}$ 控制; 幼蚕的体色黑色（E）对巧克力色 (e) 为显性。不考虑基因突变和染色体变异, 下列相关叙述正确的是（）\nA: 雌家蚕一个染色体组有 28 条染色体，处于减数分裂II后期的细胞含有 2 条 Z 染色体\nB: $\\mathrm{DdZ} \\mathrm{Z}^{\\mathrm{e} W}$ 个体测交, 若含 $\\mathrm{dW}$ 的雌配子 $1 / 2$ 致死, 则子代中 $\\mathrm{ddZ}^{e} Z^{\\mathrm{e}}$ 的个体所占比例为 $2 / 7$\nC: 不抗浓核病的家蚕与抗浓核病的家蚕正反交可以确定这对基因的显隐性\nD: 巧克力色雌蚕和黑色雄蚕杂交, 若子代表型与性别关联, 则说明该基因位于 Z 染色体上\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n家蚕是二倍体生物 $(2 n=56)$, 雌、雄个体性染色体组成分别是 $\\mathrm{ZW} 、 \\mathrm{ZZ}$ 。幼蚕不抗浓核病与抗浓核病由常染色体上的等位基因 D 和 $\\mathrm{d}$ 控制; 幼蚕的体色黑色（E）对巧克力色 (e) 为显性。不考虑基因突变和染色体变异, 下列相关叙述正确的是（）\n\nA: 雌家蚕一个染色体组有 28 条染色体，处于减数分裂II后期的细胞含有 2 条 Z 染色体\nB: $\\mathrm{DdZ} \\mathrm{Z}^{\\mathrm{e} W}$ 个体测交, 若含 $\\mathrm{dW}$ 的雌配子 $1 / 2$ 致死, 则子代中 $\\mathrm{ddZ}^{e} Z^{\\mathrm{e}}$ 的个体所占比例为 $2 / 7$\nC: 不抗浓核病的家蚕与抗浓核病的家蚕正反交可以确定这对基因的显隐性\nD: 巧克力色雌蚕和黑色雄蚕杂交, 若子代表型与性别关联, 则说明该基因位于 Z 染色体上\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_183",
        "problem": "Female vampire bats live in colonies made up of unrelated females and their offspring, and they feed exclusively on blood of herbivores. The fully-fed bats often share some of the blood that they collected with bats that are starving, and are more likely to receive blood from these individuals when they themselves starve.\n\nIndicate in the answer sheet if each of the following statements about this behavior of bats is mostly true or false.\nA: Kin selection has an important role to play in the evolution of sharing behavior.\nB: Vampire bats are likely to live in colonies only for short periods of time.\nC: The bats showing this behavior have higher indirect fitness than if they would not do so.\nD: Vampire bats are able to recognize and remember other individual bats.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFemale vampire bats live in colonies made up of unrelated females and their offspring, and they feed exclusively on blood of herbivores. The fully-fed bats often share some of the blood that they collected with bats that are starving, and are more likely to receive blood from these individuals when they themselves starve.\n\nIndicate in the answer sheet if each of the following statements about this behavior of bats is mostly true or false.\n\nA: Kin selection has an important role to play in the evolution of sharing behavior.\nB: Vampire bats are likely to live in colonies only for short periods of time.\nC: The bats showing this behavior have higher indirect fitness than if they would not do so.\nD: Vampire bats are able to recognize and remember other individual bats.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1556",
        "problem": "Dynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\nWhat is the highest score of your alignments?\nA: -1\nB: 0\nC: 1\nD: 2\nE: 3\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDynamic programming was invented to align DNA sequences, but now underpins countless processes, such as language-translation. It breaks a problem into small steps to find the optimal solution. In this question, you will use dynamic programming to align two DNA sequences. First you need a system to score an alignment. In this algorithm:\n\n- $\\quad$ Matching bases will be scored +1\n- $\\quad$ Miss-matched bases will be scored -1\n- Gaps in the alignment will be scored -1\n- The highest score 'wins'\n\nNext, the two sequences are arranged in a grid.\n\n|  |  | $\\mathbf{G}$ | $\\mathbf{C}$ | $\\mathbf{A}$ | $\\mathbf{T}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 |  |  |  |  |\n| $\\mathbf{G}$ |  |  |  |  |  |\n| $\\mathbf{A}$ |  |  |  |  |  |\n| $\\mathbf{T}$ |  |  |  |  |  |\n| $\\boldsymbol{T}$ |  |  |  |  |  |\n\nYou then generate a score for each cell. Moving horizontally or vertically indicates you are skipping bases, creating gaps in the alignment, so you add -1 to the previous score.\n\n[figure1]\n\nMoving diagonally indicates you are aligning matching or miss-matching bases. You add +1 to the previous score if they match, or -1 if they miss-match.\n\n[figure2]\n\nCells in the middle could be scored based on horizontal, vertical or diagonal movements. The cell should be given the highest score possible.\n\nYou then mark with an arrow which movement this score came from.\n\n|  |  | G | C | A | T |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 |\n| G | -1 | 1 | 0 | -1 | -2 |\n| A | -2 | 0 | 0 | 1 | 0 |\n| T | -3 | -1 | -1 | 0 | 2 |\n| T | -4 | -2 | -2 | -1 | 1 |\n\nNote that some cells can gain an equal score from more than one movement, so both arrows are included.\n\nYou then trace backwards along arrows from the bottom right to generate your highest scoring alignments.\n\n|  |  | $G$ | $C$ | $A$ | $T$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n|  | 0 | -1 | -2 | -3 | -4 |\n| $G$ | -1 |  |  | -1 | -2 |\n| $\\mathrm{A}$ | -2 | 0 | 0 |  | 0 |\n| $\\mathrm{T}$ | -3 | -1 | -1 | 0 | $?$ |\n| $\\mathrm{T}$ | -4 | -2 | -2 | -1 | 1 |\n\nIn this example, the two possible alignments are:\n\n- G-ATT\n\n1 GCA-T\n\n- G-ATT\n\n2 GCAT\n\nNow extend the above example to align these two sequences from species i and ii.\n\ni) GCATGCT\n\nii) GATTACA\n\nA part filled table is provided for you. You will probably need to do this on a sheet of rough paper.\n\n|  |  | $G$ | $C$ | $A$ | $T$ | $G$ | $C$ | $T$ |\n| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|  | 0 | -1 | -2 | -3 | -4 | -5 | -6 | -7 |\n| $G$ | -1 | 1 | 0 | -1 | -2 | -3 | -4 | -5 |\n| A | -2 | 0 | 0 | 1 | 0 | -1 | -2 | -3 |\n| T | -3 | -1 | -1 | 0 | 2 | 1 | 0 | -1 |\n| T | -4 | -2 | -2 | -1 | 1 |  |  |  |\n| A | -5 | -3 | -3 | -1 |  |  |  |  |\n| C | -6 | -4 | -2 | -2 |  |  |  |  |\n| A | -7 | -5 | -3 | -1 |  |  |  |  |\n\nWhat is the highest score of your alignments?\n\nA: -1\nB: 0\nC: 1\nD: 2\nE: 3\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=925&width=928&top_left_y=354&top_left_x=238",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-62.jpg?height=922&width=928&top_left_y=1458&top_left_x=238"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_140",
        "problem": "Gall aphids (Pemphigus betae) live in poplar plants. Adult females produce galls on poplar leaves. Some fraction of these galls will emerge and survive to adulthood. Female aphids complete their life cycle after laying eggs in the leaves. All the progeny of a single female aphid are contained in one gall. A student recorded observation on several aphid populations, shown in the table below. All environmental parameters are assumed constant.\n\n| Population | Number of <br> aborted galls | Number of <br> successful galls | Female/Male ratio in <br> adult stage |\n| :---: | :---: | :---: | :---: |\n| 1 | 35 | 70 | $1 / 1$ |\n| 2 | 25 | 75 | $1 / 2$ |\n| 3 | 21 | 63 | Not given |\n| 4 | 16 | 32 | $1 / 1$ |\n\nAn equation representing number of female aphids in $t^{\\text {th }}$ generation is established as below:\n\n$$\nN_{t}=[\\mathrm{f} \\times \\mathrm{r} \\times(1-\\mathrm{m})]^{\\mathrm{t}} \\times \\mathrm{N}_{\\mathrm{o}}\n$$\n\nWhereas:\n\n$\\mathrm{N}_{\\mathrm{t}}$ - number of adult female aphids in the $\\mathrm{t}^{\\text {th }}$ generation\n\n$\\mathrm{N}_{\\mathrm{o}}$ - number of adult female aphids in the initial generation\n\n$\\mathrm{m}$ - fraction mortality of the young aphids\n\n$\\mathrm{f}$ - number of progeny per female aphid\n\n$r$ - ratio of female aphids to total adult aphids.\n\nTheoretically $\\mathrm{f}, \\mathrm{m}$ and $\\mathrm{r}$ are constant.\nA: Population 1 has a constant number of adult females across generations when each female produces 4 progeny.\nB: When every female in population 2 produces 3 progeny, this population will have a constant number of adult females across generations.\nC: When population 3 has a constant number of adult females across generations and each female aphid produces 4 progeny, the female/male ratio of the population in adult stage is $1 / 2$.\nD: Given that each female in population 4 produces 6 aphids and taking the offspring of population 4 to be in the first generation, the number of adult females in the third generation will be 384 .\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nGall aphids (Pemphigus betae) live in poplar plants. Adult females produce galls on poplar leaves. Some fraction of these galls will emerge and survive to adulthood. Female aphids complete their life cycle after laying eggs in the leaves. All the progeny of a single female aphid are contained in one gall. A student recorded observation on several aphid populations, shown in the table below. All environmental parameters are assumed constant.\n\n| Population | Number of <br> aborted galls | Number of <br> successful galls | Female/Male ratio in <br> adult stage |\n| :---: | :---: | :---: | :---: |\n| 1 | 35 | 70 | $1 / 1$ |\n| 2 | 25 | 75 | $1 / 2$ |\n| 3 | 21 | 63 | Not given |\n| 4 | 16 | 32 | $1 / 1$ |\n\nAn equation representing number of female aphids in $t^{\\text {th }}$ generation is established as below:\n\n$$\nN_{t}=[\\mathrm{f} \\times \\mathrm{r} \\times(1-\\mathrm{m})]^{\\mathrm{t}} \\times \\mathrm{N}_{\\mathrm{o}}\n$$\n\nWhereas:\n\n$\\mathrm{N}_{\\mathrm{t}}$ - number of adult female aphids in the $\\mathrm{t}^{\\text {th }}$ generation\n\n$\\mathrm{N}_{\\mathrm{o}}$ - number of adult female aphids in the initial generation\n\n$\\mathrm{m}$ - fraction mortality of the young aphids\n\n$\\mathrm{f}$ - number of progeny per female aphid\n\n$r$ - ratio of female aphids to total adult aphids.\n\nTheoretically $\\mathrm{f}, \\mathrm{m}$ and $\\mathrm{r}$ are constant.\n\nA: Population 1 has a constant number of adult females across generations when each female produces 4 progeny.\nB: When every female in population 2 produces 3 progeny, this population will have a constant number of adult females across generations.\nC: When population 3 has a constant number of adult females across generations and each female aphid produces 4 progeny, the female/male ratio of the population in adult stage is $1 / 2$.\nD: Given that each female in population 4 produces 6 aphids and taking the offspring of population 4 to be in the first generation, the number of adult females in the third generation will be 384 .\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_123",
        "problem": "Scientists monitored the number of individuals for ant species in a 5 hectare plot of land for 50 years. The below figures represent the dominance rank of observed species in terms of their abundance, that is, the number of individuals for each species. Each open circle represents the value for each species. Note that the most abundant species is given rank 1.\n[figure1]\n\nSpecies dominance rank\nA: The total number of species does not change in the three periods observed.\nB: Species A has gradually outcompeted other species over time.\nC: The top three species account for more than $75 \\%$ of the total number of individuals in the first year.\nD: During the 50 years, evenness among species has decrease\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nScientists monitored the number of individuals for ant species in a 5 hectare plot of land for 50 years. The below figures represent the dominance rank of observed species in terms of their abundance, that is, the number of individuals for each species. Each open circle represents the value for each species. Note that the most abundant species is given rank 1.\n[figure1]\n\nSpecies dominance rank\n\nA: The total number of species does not change in the three periods observed.\nB: Species A has gradually outcompeted other species over time.\nC: The top three species account for more than $75 \\%$ of the total number of individuals in the first year.\nD: During the 50 years, evenness among species has decrease\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-64.jpg?height=526&width=1556&top_left_y=820&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1156",
        "problem": "In mice the allele $C$ must be present before colour pigments are produced in the fur. Individuals lacking this allele are albino. Another pair of alleles determines whether the fur will be agouti $(A)$ or black (a).\n\nWhich one of the rows $A$ to $E$ shows, as far as these two pairs of alleles are concerned, the number of different genotypes which could result in albino, agouti and black mice?\nA: albino: 3, agouti: 4, black: 2\nB: albino: 4, agouti: 2, black: 3\nC: albino: 2, agouti: 3, black: 4\nD: albino: 3, agouti: 1, black: 2\nE: albino: 4, agouti: 4, black: 2\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn mice the allele $C$ must be present before colour pigments are produced in the fur. Individuals lacking this allele are albino. Another pair of alleles determines whether the fur will be agouti $(A)$ or black (a).\n\nWhich one of the rows $A$ to $E$ shows, as far as these two pairs of alleles are concerned, the number of different genotypes which could result in albino, agouti and black mice?\n\nA: albino: 3, agouti: 4, black: 2\nB: albino: 4, agouti: 2, black: 3\nC: albino: 2, agouti: 3, black: 4\nD: albino: 3, agouti: 1, black: 2\nE: albino: 4, agouti: 4, black: 2\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1533",
        "problem": "The animals below are found in the same food web, but fit into different food chains. One chain takes place close to the ocean surface, and one begins with dead animals resting on the seabed.\n\nWhich animals go into the ocean surface food chain? The order does not matter.\nA: Phytoplankton / algae, zooplankton, shrimp, whale, small fish\nB: Phytoplankton / algae, zooplankton, scavenging crab, octopus, small fish\nC: Phytoplankton / algae, zooplankton, scavenging crab, small fish, hammerhead shark\nD: Whale, hammerhead shark, scavenging crab, shrimp\nE: Phytoplankton / algae, zooplankton, shrimp, hammerhead shark\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe animals below are found in the same food web, but fit into different food chains. One chain takes place close to the ocean surface, and one begins with dead animals resting on the seabed.\n\nWhich animals go into the ocean surface food chain? The order does not matter.\n\nA: Phytoplankton / algae, zooplankton, shrimp, whale, small fish\nB: Phytoplankton / algae, zooplankton, scavenging crab, octopus, small fish\nC: Phytoplankton / algae, zooplankton, scavenging crab, small fish, hammerhead shark\nD: Whale, hammerhead shark, scavenging crab, shrimp\nE: Phytoplankton / algae, zooplankton, shrimp, hammerhead shark\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_373",
        "problem": "研究发现在某些物质的诱导下, 基因的工作状态会发生改变, 这一过程称为可诱导调节，相关机制如图所示。为验证这一机制，研究人员将大肠杆菌放在乳糖培养液中,发现细菌在早期不生长, 一段时间后开始生长, 随后分离出 $\\beta$-半乳糖苷酶(催化乳糖水解)。根据以上资料推测，下列说法错误的是（ ）\n\n[图1]\nA: 当无诱导物时, 阻遏物通过与操纵基因的结合来影响结构基因的转录过程\nB: 一段时间后细菌开始生长, 表明基因通过控制酶的合成来控制生物体的生命活动\nC: 乳糖通过诱导大肠杆菌合成 $\\beta$-半乳糖苷酶基因来实现对乳糖的水解利用\nD: $\\beta$-半乳糖苷酶基因的转录起点可能位于阻遏基因与操纵基因之间\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n研究发现在某些物质的诱导下, 基因的工作状态会发生改变, 这一过程称为可诱导调节，相关机制如图所示。为验证这一机制，研究人员将大肠杆菌放在乳糖培养液中,发现细菌在早期不生长, 一段时间后开始生长, 随后分离出 $\\beta$-半乳糖苷酶(催化乳糖水解)。根据以上资料推测，下列说法错误的是（ ）\n\n[图1]\n\nA: 当无诱导物时, 阻遏物通过与操纵基因的结合来影响结构基因的转录过程\nB: 一段时间后细菌开始生长, 表明基因通过控制酶的合成来控制生物体的生命活动\nC: 乳糖通过诱导大肠杆菌合成 $\\beta$-半乳糖苷酶基因来实现对乳糖的水解利用\nD: $\\beta$-半乳糖苷酶基因的转录起点可能位于阻遏基因与操纵基因之间\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-05.jpg?height=648&width=988&top_left_y=173&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_430",
        "problem": "某 $X Y$ 型二倍体高等动物（2n=6）雄性个体的基因型为 $\\mathrm{AaBb}$, 其体内某细胞处于细胞分裂某时期的示意图如下。该生物精原细胞中的全部 DNA 分子双链均被 ${ }^{15} \\mathrm{~N}$ 标记,在 ${ }^{14} \\mathrm{~N}$ 的环境中增殖。下列叙述正确的是（）\n\n[图1]\nA: 形成该细胞过程中发生了基因突变和染色体畸变\nB: 若进行两次细胞质分裂后, 被标记的子细胞所占的比例为 1 或 $1 / 2$\nC: 该细胞内含三个四分体，并能观察到染色体的片段交换现象\nD: 该细胞分裂形成的配子的基因型可能为 $a B X 、 a B X^{A} 、 A b Y 、 b Y$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某 $X Y$ 型二倍体高等动物（2n=6）雄性个体的基因型为 $\\mathrm{AaBb}$, 其体内某细胞处于细胞分裂某时期的示意图如下。该生物精原细胞中的全部 DNA 分子双链均被 ${ }^{15} \\mathrm{~N}$ 标记,在 ${ }^{14} \\mathrm{~N}$ 的环境中增殖。下列叙述正确的是（）\n\n[图1]\n\nA: 形成该细胞过程中发生了基因突变和染色体畸变\nB: 若进行两次细胞质分裂后, 被标记的子细胞所占的比例为 1 或 $1 / 2$\nC: 该细胞内含三个四分体，并能观察到染色体的片段交换现象\nD: 该细胞分裂形成的配子的基因型可能为 $a B X 、 a B X^{A} 、 A b Y 、 b Y$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-30.jpg?height=611&width=639&top_left_y=171&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_209",
        "problem": "Bustamante et al. (2005) analysed the pattern of polymorphism in the human genome by sequencing over 11,624 genes from 39 humans. The table below shows the distributions of synonymous (changes that does not result in amino acid replacement) and non-synonymous (changes that result in amino acid replacement) single nucleotide polymorphism (SNPs - variation within species):\n\n|  | Divergence | SNPs |\n| :--- | :---: | :---: |\n| Synonymous | 34,099 | 15,750 |\n| Non-synonymous | 20,467 | 14,311 |\n\nIn the table, divergence refers to the fixed differences between the humans in this study and the chimpanzee genome.\nA: Greater ratio of non-synonymous to synonymous SNPs relative to ratio of nonsynonymous to synonymous divergent sites suggest the effect of negative selection in human.\nB: The ratio of synonymous/non-synonymous SNPs is higher in genes involved in chromatin structure.\nC: Arms races caused by host-specific pathogen would decrease the ratio of synonymous/non-synonymous substitutions in divergent sites between human\nD: SNPs present within chimpanzee population cannot result in any level of reproductive isolation in chimpanzee population.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBustamante et al. (2005) analysed the pattern of polymorphism in the human genome by sequencing over 11,624 genes from 39 humans. The table below shows the distributions of synonymous (changes that does not result in amino acid replacement) and non-synonymous (changes that result in amino acid replacement) single nucleotide polymorphism (SNPs - variation within species):\n\n|  | Divergence | SNPs |\n| :--- | :---: | :---: |\n| Synonymous | 34,099 | 15,750 |\n| Non-synonymous | 20,467 | 14,311 |\n\nIn the table, divergence refers to the fixed differences between the humans in this study and the chimpanzee genome.\n\nA: Greater ratio of non-synonymous to synonymous SNPs relative to ratio of nonsynonymous to synonymous divergent sites suggest the effect of negative selection in human.\nB: The ratio of synonymous/non-synonymous SNPs is higher in genes involved in chromatin structure.\nC: Arms races caused by host-specific pathogen would decrease the ratio of synonymous/non-synonymous substitutions in divergent sites between human\nD: SNPs present within chimpanzee population cannot result in any level of reproductive isolation in chimpanzee population.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1028",
        "problem": "Darwin's Finches are a prime example of adaptive radiation. Which of the following best describes this adaptive radiation correctly?\nA: The genetic variability that can be found among individuals of the same species\nB: The evolutionary process by which different forms, adapted to different niches arose from a common ancestor\nC: A sudden diversification of a group of organisms from closely related species\nD: The evolutionary process that allows for the changes that occur within the same lineage\nE: The evolutionary process of adaptation of species through a form of polymorphism\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDarwin's Finches are a prime example of adaptive radiation. Which of the following best describes this adaptive radiation correctly?\n\nA: The genetic variability that can be found among individuals of the same species\nB: The evolutionary process by which different forms, adapted to different niches arose from a common ancestor\nC: A sudden diversification of a group of organisms from closely related species\nD: The evolutionary process that allows for the changes that occur within the same lineage\nE: The evolutionary process of adaptation of species through a form of polymorphism\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_147",
        "problem": "Four mutant strains of bacteria (1 4) all require substance $S$ to grow (each strain is blocked at one step in the S-biosynthesis pathway). Four plates were prepared with minimal medium and a trace of substance $S$, to allow a small amount of growth of mutant cells. On plate $a$, mutant cells of strain 1 were spread over entire surface of the agar to form a thin lawn of bacteria. On plate $b$, the lawn was composed of mutant cells of strain 2, and so on. On each plate, cells of each of the four mutant types were inoculated over the lawn, as indicated in the figure by the circles. Dark circles indicate excellent growth. A strain blocked at a later step in the $\\mathrm{S}$ substance metabolic pathway accumulates intermediates that can 'feed' a strain blocked at an earlier step.\n\na\n\n[figure1]\n\n$b$\n\n[figure2]\n\n$c$\n\n[figure3]\n\n$d$\n\n[figure4]\n\nWhat is the order of genes (1 4) in the metabolic pathway for synthesis of substance S?\nA: $2 \\rightarrow 4 \\rightarrow 3 \\rightarrow 1$\nB: $2 \\rightarrow 1 \\rightarrow 3 \\rightarrow 4$\nC: $1 \\rightarrow 3 \\rightarrow 4 \\rightarrow 2$\nD: $1 \\rightarrow 2 \\rightarrow 4 \\rightarrow 3$\nE: $3 \\rightarrow 4 \\rightarrow 2 \\rightarrow 1$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFour mutant strains of bacteria (1 4) all require substance $S$ to grow (each strain is blocked at one step in the S-biosynthesis pathway). Four plates were prepared with minimal medium and a trace of substance $S$, to allow a small amount of growth of mutant cells. On plate $a$, mutant cells of strain 1 were spread over entire surface of the agar to form a thin lawn of bacteria. On plate $b$, the lawn was composed of mutant cells of strain 2, and so on. On each plate, cells of each of the four mutant types were inoculated over the lawn, as indicated in the figure by the circles. Dark circles indicate excellent growth. A strain blocked at a later step in the $\\mathrm{S}$ substance metabolic pathway accumulates intermediates that can 'feed' a strain blocked at an earlier step.\n\na\n\n[figure1]\n\n$b$\n\n[figure2]\n\n$c$\n\n[figure3]\n\n$d$\n\n[figure4]\n\nWhat is the order of genes (1 4) in the metabolic pathway for synthesis of substance S?\n\nA: $2 \\rightarrow 4 \\rightarrow 3 \\rightarrow 1$\nB: $2 \\rightarrow 1 \\rightarrow 3 \\rightarrow 4$\nC: $1 \\rightarrow 3 \\rightarrow 4 \\rightarrow 2$\nD: $1 \\rightarrow 2 \\rightarrow 4 \\rightarrow 3$\nE: $3 \\rightarrow 4 \\rightarrow 2 \\rightarrow 1$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-45.jpg?height=360&width=357&top_left_y=1168&top_left_x=381",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-45.jpg?height=343&width=352&top_left_y=1185&top_left_x=772",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-45.jpg?height=343&width=351&top_left_y=1185&top_left_x=1161",
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-45.jpg?height=343&width=368&top_left_y=1185&top_left_x=1549"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_545",
        "problem": "下图是某同学观察玉米 $(2 \\mathrm{~N}=20)$ 细胞分裂的实验中, 在显微镜下看到的几个细胞分裂时期。相关叙述正确的是（ ）\n\n[图1]\n\n(1)\n\n[图2]\n\n(2)\n\n[图3]\n\n(3)\n\n[图4]\n\n(4)\nA: 该同学选择的实验材料是玉米雌荵的子房\nB: 装片制作过程中用碱性染色剂进行染色前需充分漂洗\nC: 细胞(3)中可能存在姐妹染色单体之间的互换\nD: 正在进行等位基因分离的细胞是(2)和(5)\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图是某同学观察玉米 $(2 \\mathrm{~N}=20)$ 细胞分裂的实验中, 在显微镜下看到的几个细胞分裂时期。相关叙述正确的是（ ）\n\n[图1]\n\n(1)\n\n[图2]\n\n(2)\n\n[图3]\n\n(3)\n\n[图4]\n\n(4)\n\nA: 该同学选择的实验材料是玉米雌荵的子房\nB: 装片制作过程中用碱性染色剂进行染色前需充分漂洗\nC: 细胞(3)中可能存在姐妹染色单体之间的互换\nD: 正在进行等位基因分离的细胞是(2)和(5)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-14.jpg?height=254&width=271&top_left_y=1141&top_left_x=344",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-14.jpg?height=257&width=274&top_left_y=1134&top_left_x=640",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-14.jpg?height=254&width=263&top_left_y=1135&top_left_x=928",
            "https://cdn.mathpix.com/cropped/2024_03_31_457e30042f7c293ebe80g-14.jpg?height=252&width=548&top_left_y=1139&top_left_x=1208"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_696",
        "problem": "某家系中有甲、乙两种单基因遗传病（如图）, 其中一种是伴性遗传病。相关分析不正确的是（ ）\n\n[图1]\nA: 甲病是常染色体隐性遗传, 乙病是伴 X 染色体显性遗传\nB: 若III-4 与III-5 结婚, 生育一患两种病孩子的概率是 $5 / 12$\nC: II-2 有一种基因型, III-8 基因型有四种可能\nD: II-3 的致病基因均来自于 I-2\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某家系中有甲、乙两种单基因遗传病（如图）, 其中一种是伴性遗传病。相关分析不正确的是（ ）\n\n[图1]\n\nA: 甲病是常染色体隐性遗传, 乙病是伴 X 染色体显性遗传\nB: 若III-4 与III-5 结婚, 生育一患两种病孩子的概率是 $5 / 12$\nC: II-2 有一种基因型, III-8 基因型有四种可能\nD: II-3 的致病基因均来自于 I-2\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-52.jpg?height=348&width=1213&top_left_y=820&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1220",
        "problem": "[figure1]\n\nSide view\n\n[figure2]\n\nFrontal view\n\nWhich of the following shows a cross section through the dashed line?\nA: ![]([figure3])\nB: ![]([figure4])\nC: ![]([figure5])\nD: ![]([figure6])\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nSide view\n\n[figure2]\n\nFrontal view\n\nWhich of the following shows a cross section through the dashed line?\n\nA: ![]([figure3])\nB: ![]([figure4])\nC: ![]([figure5])\nD: ![]([figure6])\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-12.jpg?height=372&width=579&top_left_y=268&top_left_x=133",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-12.jpg?height=383&width=279&top_left_y=254&top_left_x=820",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-12.jpg?height=431&width=186&top_left_y=790&top_left_x=204",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-12.jpg?height=408&width=206&top_left_y=790&top_left_x=614",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-12.jpg?height=437&width=212&top_left_y=787&top_left_x=1002",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-12.jpg?height=451&width=217&top_left_y=780&top_left_x=1408"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1105",
        "problem": "Muscle contraction is a result of cross-bridging between actin and myosin filaments. Uncoupling of the two can occur only when fresh ATP molecule binds to crossbridged myosin molecule.\n\nStiffening of muscles known as 'rigor mortis' occurs approx 2-6 hrs after death (phase I). It persists for about $48 \\mathrm{hrs}$ (phase II) after which the muscles return to 'relaxed' (flexible) state (phase III).\n\nWhich of the following is correct?\nA: The onset of \"rigor mortis\" (phase I) is due to depletion of ATP storage of the cell.\nB: The long lasting phase II is mainly due to supply of ATP from creatinine phosphate of muscle.\nC: \"Rigor mortis\" is a reversible process in which supply of remaining ATP from degrading tissues leads to uncoupling of actin-myosin bridges.\nD: Muscle with small amount of stored glycogen is likely to go to \"rigor mortis\" earlier than the one that has more amount of glycogen.\nE: Phase III of rigor mortis is a result of protein degradation hence lesser the environmental temperature, longer will be the phase II of rigor mortis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nMuscle contraction is a result of cross-bridging between actin and myosin filaments. Uncoupling of the two can occur only when fresh ATP molecule binds to crossbridged myosin molecule.\n\nStiffening of muscles known as 'rigor mortis' occurs approx 2-6 hrs after death (phase I). It persists for about $48 \\mathrm{hrs}$ (phase II) after which the muscles return to 'relaxed' (flexible) state (phase III).\n\nWhich of the following is correct?\n\nA: The onset of \"rigor mortis\" (phase I) is due to depletion of ATP storage of the cell.\nB: The long lasting phase II is mainly due to supply of ATP from creatinine phosphate of muscle.\nC: \"Rigor mortis\" is a reversible process in which supply of remaining ATP from degrading tissues leads to uncoupling of actin-myosin bridges.\nD: Muscle with small amount of stored glycogen is likely to go to \"rigor mortis\" earlier than the one that has more amount of glycogen.\nE: Phase III of rigor mortis is a result of protein degradation hence lesser the environmental temperature, longer will be the phase II of rigor mortis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1360",
        "problem": "When eukaryotic cells divide, they pass through the 'cell cycle', which can be subdivided into a series of phases. This is depicted in the picture below. Cells that begin to prepare for cell division enter the G1 phase ('Gap 1' phase). In the G1 phase, cellular and cytoplasmic contents are produced in preparation for cell division (such as proteins, organelles, and the cytoplasm itself). After this phase, the cell progresses into the $S$ phase ('Synthesis' phase) where DNA replication occurs, and then into the G2 phase ('Gap 2' phase) where the cell grows further by producing and assembling the cytoplasmic materials for mitosis and cytokinesis. Finally, the cell enters the mitotic phase (labelled 'M') and cytokinesis (labelled 'C') and completes cell division.\n\n[figure1]\n\nAnna is a scientist who looked through the microscope at a dividing animal embryo. She noticed that in the early stages of embryo development, the cells undergo cell division in quick succession and the daughter cells become smaller with each division, as depicted in the image below. Unlike regular cell division, embryo cells divide through a modified cell cycle known as 'cleavage' where some phases of the cell cycle are shortened or skipped entirely.\n\n[figure2]\n\nWhich phase of the cell cycle is shortened the most in 'cleavage' division?\nA: S and G1 phases\nB: G1 and G2 phases\nC: $M$ and $C$ phases\nD: $\\mathrm{G} 2$ and $M$ phases\nE: $M$ and $S$ phases\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen eukaryotic cells divide, they pass through the 'cell cycle', which can be subdivided into a series of phases. This is depicted in the picture below. Cells that begin to prepare for cell division enter the G1 phase ('Gap 1' phase). In the G1 phase, cellular and cytoplasmic contents are produced in preparation for cell division (such as proteins, organelles, and the cytoplasm itself). After this phase, the cell progresses into the $S$ phase ('Synthesis' phase) where DNA replication occurs, and then into the G2 phase ('Gap 2' phase) where the cell grows further by producing and assembling the cytoplasmic materials for mitosis and cytokinesis. Finally, the cell enters the mitotic phase (labelled 'M') and cytokinesis (labelled 'C') and completes cell division.\n\n[figure1]\n\nAnna is a scientist who looked through the microscope at a dividing animal embryo. She noticed that in the early stages of embryo development, the cells undergo cell division in quick succession and the daughter cells become smaller with each division, as depicted in the image below. Unlike regular cell division, embryo cells divide through a modified cell cycle known as 'cleavage' where some phases of the cell cycle are shortened or skipped entirely.\n\n[figure2]\n\nWhich phase of the cell cycle is shortened the most in 'cleavage' division?\n\nA: S and G1 phases\nB: G1 and G2 phases\nC: $M$ and $C$ phases\nD: $\\mathrm{G} 2$ and $M$ phases\nE: $M$ and $S$ phases\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-33.jpg?height=545&width=462&top_left_y=1589&top_left_x=794",
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-34.jpg?height=208&width=717&top_left_y=1221&top_left_x=681"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1342",
        "problem": "The geographic regions with the most recorded species were Australia and Japan, each reporting over 32,000 species. However, South Korea, China, South Africa, Baltic Sea and Gulf of Mexico had the most species per unit area. The graph shows the relationship between total number of recorded species in each region to sea volume (solid dots, dashed line, millions $\\mathrm{km}^{3}$ ), and seabed area (squares, solid line, millions of $\\mathrm{km}^{2}$ ) with linear trend lines shown. The Southern Ocean, Antarctica, is the outlier at top left of graph.\n\n[figure1]\n\nFrom this information you can conclude:\nA: There are always more marine species in regions with a larger seabed area.\nB: Antarctica has an extremely low number of marine species given the sea volume and seabed area in this region.\nC: Australia and Japan had approximately the same number of species per seabed area.\nD: South Korea is represented by the square furthermost to the right on the graph.\nE: There are always more marine species in regions with a larger sea volume.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe geographic regions with the most recorded species were Australia and Japan, each reporting over 32,000 species. However, South Korea, China, South Africa, Baltic Sea and Gulf of Mexico had the most species per unit area. The graph shows the relationship between total number of recorded species in each region to sea volume (solid dots, dashed line, millions $\\mathrm{km}^{3}$ ), and seabed area (squares, solid line, millions of $\\mathrm{km}^{2}$ ) with linear trend lines shown. The Southern Ocean, Antarctica, is the outlier at top left of graph.\n\n[figure1]\n\nFrom this information you can conclude:\n\nA: There are always more marine species in regions with a larger seabed area.\nB: Antarctica has an extremely low number of marine species given the sea volume and seabed area in this region.\nC: Australia and Japan had approximately the same number of species per seabed area.\nD: South Korea is represented by the square furthermost to the right on the graph.\nE: There are always more marine species in regions with a larger sea volume.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-33.jpg?height=797&width=943&top_left_y=338&top_left_x=1042"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1542",
        "problem": "Humans eyes have cone and rod cells with different sensitivities to different wavelengths of light. The minimum brightness cells can detect has been plotted against wavelength.\n\n[figure1]\n\nBright light hitting the cells causes them to reduce signalling to the brain, to maintain contrast in different conditions. The saturation point of rods is plotted below.\n\n[figure2]\n\n\n\nWhich is true?\nA: Green light displays are the best colour for preserving night vision.\nB: Rods are able to detect detail under red light well.\nC: Colour vision requires dimmer lights than monochrome vision.\nD: The colour blue can be identified correctly in dimmer light than the colour yellow.\nE: Bright green light stimulates all cones equally.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHumans eyes have cone and rod cells with different sensitivities to different wavelengths of light. The minimum brightness cells can detect has been plotted against wavelength.\n\n[figure1]\n\nBright light hitting the cells causes them to reduce signalling to the brain, to maintain contrast in different conditions. The saturation point of rods is plotted below.\n\n[figure2]\n\n\n\nWhich is true?\n\nA: Green light displays are the best colour for preserving night vision.\nB: Rods are able to detect detail under red light well.\nC: Colour vision requires dimmer lights than monochrome vision.\nD: The colour blue can be identified correctly in dimmer light than the colour yellow.\nE: Bright green light stimulates all cones equally.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-28.jpg?height=959&width=1385&top_left_y=474&top_left_x=238",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-28.jpg?height=1025&width=1511&top_left_y=1595&top_left_x=227"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_128",
        "problem": "An important question in the evolution of angiosperms is the influence of life form and habit on speciation and extinction. In a phylogenetic study of the morphologically highly diverse plant order Saxifragales, the effect of habit and life history on speciation and extinction was investigated. Figures A-F illustrates estimates of speciation $(\\lambda)$ and extinction $(\\mu)$ rates in annual, perennial, herbaceous, and woody lineages (Soltis et al., 2013). The results are given as posterior probability density, which reflect the uncertainty associated with parameter estimates. Dashed lines represent the median values of each distribution. Net diversification rate $r=\\lambda-\\mu$. Posterior probability density is a function that use to specify the probability of the random variable falling within a particular range of values. Range of value, as oppose to taking on any one value.\n[figure1]\n\n- Woody\n- Herbaceous\n[figure2]\n\nExtinction rate\n\n- Woody perennials\n- Herbaceous perennials\n- Herbaceous annuals\n[figure3]\nA: Among different lineages, woody lineages have higher speciation $(\\lambda)$ and extinction $(\\mu)$ rates.\nB: Results are consistent with the idea that shorter generation time causes higher speciation rate.\nC: As compared to the other groups, herbaceous perennial lineages have higher rates of net diversification.\nD: Generation time and breeding systems affect rates of diversification.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn important question in the evolution of angiosperms is the influence of life form and habit on speciation and extinction. In a phylogenetic study of the morphologically highly diverse plant order Saxifragales, the effect of habit and life history on speciation and extinction was investigated. Figures A-F illustrates estimates of speciation $(\\lambda)$ and extinction $(\\mu)$ rates in annual, perennial, herbaceous, and woody lineages (Soltis et al., 2013). The results are given as posterior probability density, which reflect the uncertainty associated with parameter estimates. Dashed lines represent the median values of each distribution. Net diversification rate $r=\\lambda-\\mu$. Posterior probability density is a function that use to specify the probability of the random variable falling within a particular range of values. Range of value, as oppose to taking on any one value.\n[figure1]\n\n- Woody\n- Herbaceous\n[figure2]\n\nExtinction rate\n\n- Woody perennials\n- Herbaceous perennials\n- Herbaceous annuals\n[figure3]\n\nA: Among different lineages, woody lineages have higher speciation $(\\lambda)$ and extinction $(\\mu)$ rates.\nB: Results are consistent with the idea that shorter generation time causes higher speciation rate.\nC: As compared to the other groups, herbaceous perennial lineages have higher rates of net diversification.\nD: Generation time and breeding systems affect rates of diversification.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-36.jpg?height=1278&width=448&top_left_y=800&top_left_x=313",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-36.jpg?height=1126&width=460&top_left_y=950&top_left_x=775",
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-36.jpg?height=1132&width=462&top_left_y=947&top_left_x=1231"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_680",
        "problem": "在胚胎发育的早期染色体失活中心（XIC）负责 X 染色体计数, 并随机只允许一条 $\\mathrm{X}$ 染色体保持活性, 其余的 X 染色体高度浓缩化后失活, 形成巴氏小体。若某一个早期胚胎细胞的一条 X 染色体失活，则这个祖先细胞分裂的所有子细胞均失活同一条 X 染色体。根据题中信息, 下列有关叙述错误的是（）\n\n[图1]\nA: 巴氏小体可以应用于鉴定性别, 通常情况下，细胞有巴氏小体的个体为雌性\nB: 该机制有利于维持雌、雄个体的 X 染色体上的基因编码的蛋白质在数量上达到平衡\nC: 人类抗维生素 $\\mathrm{D}$ 佝偻病, 基因型为 $\\mathrm{X}^{\\mathrm{D}} \\mathrm{X}^{\\mathrm{d}}$ 的个体比 $\\mathrm{X}^{\\mathrm{D} Y}$ 的个体发病程度轻\nD: 性染色体组成为 XXY 的个体, 其细胞核具有 2 个巴氏小体\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在胚胎发育的早期染色体失活中心（XIC）负责 X 染色体计数, 并随机只允许一条 $\\mathrm{X}$ 染色体保持活性, 其余的 X 染色体高度浓缩化后失活, 形成巴氏小体。若某一个早期胚胎细胞的一条 X 染色体失活，则这个祖先细胞分裂的所有子细胞均失活同一条 X 染色体。根据题中信息, 下列有关叙述错误的是（）\n\n[图1]\n\nA: 巴氏小体可以应用于鉴定性别, 通常情况下，细胞有巴氏小体的个体为雌性\nB: 该机制有利于维持雌、雄个体的 X 染色体上的基因编码的蛋白质在数量上达到平衡\nC: 人类抗维生素 $\\mathrm{D}$ 佝偻病, 基因型为 $\\mathrm{X}^{\\mathrm{D}} \\mathrm{X}^{\\mathrm{d}}$ 的个体比 $\\mathrm{X}^{\\mathrm{D} Y}$ 的个体发病程度轻\nD: 性染色体组成为 XXY 的个体, 其细胞核具有 2 个巴氏小体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-003.jpg?height=434&width=1397&top_left_y=2213&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_422",
        "problem": "如图表示某家族两种单基因遗传病的系谱图。其中一种遗传病为伴性遗传病（致病基因不位于 $\\mathrm{X}$ 和 $\\mathrm{Y}$ 染色体的同源区段上)。已知乙病在人群中的发病率为 1/10000。下列叙述错误的是（ ）\n[图1]\n\n## 2甲甲病男性 $\\bigcirc$ 病女性\nA: 甲病在男性的发病率与甲病的致病基因频率相同\nB: 若 $\\mathrm{II}_{6}$ 不携带致病基因, $\\mathrm{III}_{8}$ 和 $\\mathrm{III}_{9}$ 近亲结婚, 生一个健康孩子的概率是 9/16\nC: 若 $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生一个表型正常的女儿, 女儿和母亲基因型相同的概率是 $1 / 6$\nD: 若 $\\mathrm{II}_{6}$ 不携带致病基因, $\\mathrm{III}_{9}$ 和一个表型正常的女性结婚, 孩子患乙病的概率为 $1 / 404$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图表示某家族两种单基因遗传病的系谱图。其中一种遗传病为伴性遗传病（致病基因不位于 $\\mathrm{X}$ 和 $\\mathrm{Y}$ 染色体的同源区段上)。已知乙病在人群中的发病率为 1/10000。下列叙述错误的是（ ）\n[图1]\n\n## 2甲甲病男性 $\\bigcirc$ 病女性\n\nA: 甲病在男性的发病率与甲病的致病基因频率相同\nB: 若 $\\mathrm{II}_{6}$ 不携带致病基因, $\\mathrm{III}_{8}$ 和 $\\mathrm{III}_{9}$ 近亲结婚, 生一个健康孩子的概率是 9/16\nC: 若 $\\mathrm{II}_{3}$ 和 $\\mathrm{II}_{4}$ 再生一个表型正常的女儿, 女儿和母亲基因型相同的概率是 $1 / 6$\nD: 若 $\\mathrm{II}_{6}$ 不携带致病基因, $\\mathrm{III}_{9}$ 和一个表型正常的女性结婚, 孩子患乙病的概率为 $1 / 404$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-93.jpg?height=410&width=818&top_left_y=1234&top_left_x=391",
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-94.jpg?height=52&width=1391&top_left_y=171&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_490",
        "problem": "下图是基因型为 $\\mathrm{AaBb}$ 的某动物的一个精原细胞经减数分裂过程产生的一个细胞示意图。据图分析相关叙述不正确的是（）\n\n[图1]\nA: 图中细胞处于减数第二次分裂, 内含 8 条染色单体\nB: 此精原细胞可能在四分体时期发生了非等位基因之间的互换, 此变异属于基因重组\nC: 此精原细胞在减数第一次分裂后期, 移向细胞一极的基因可能是 $\\mathrm{A} 、 \\mathrm{~A} 、 \\mathrm{~b} 、 \\mathrm{~b}$\nD: 此图中含有一个染色体组和两套遗传信息\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图是基因型为 $\\mathrm{AaBb}$ 的某动物的一个精原细胞经减数分裂过程产生的一个细胞示意图。据图分析相关叙述不正确的是（）\n\n[图1]\n\nA: 图中细胞处于减数第二次分裂, 内含 8 条染色单体\nB: 此精原细胞可能在四分体时期发生了非等位基因之间的互换, 此变异属于基因重组\nC: 此精原细胞在减数第一次分裂后期, 移向细胞一极的基因可能是 $\\mathrm{A} 、 \\mathrm{~A} 、 \\mathrm{~b} 、 \\mathrm{~b}$\nD: 此图中含有一个染色体组和两套遗传信息\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-90.jpg?height=485&width=489&top_left_y=1942&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1136",
        "problem": "Eukaryotes are organisms with their DNA organised in chromosomes, found in a distinct nucleus. Eukaryotes include algae, fungi, plants, and animals. Prokaryotes are single-celled organisms that have neither a nucleus nor other organelles. Prokaryotes include bacteria and cyanobacteria. There are approximately 8.7 million species of eukaryotes on earth - the relative size of each eukaryote group is listed below.\n\n[figure1]\n\nhttp://www.sciencedaily.com/releases/2011/08/110823180459.htm\n\nFrom the information provided, the best conclusion is?\nA: Fungi are not well described as a taxonomic group.\nB: Animals have gone through the most evolution.\nC: Eukaryotes are better described than prokaryotes.\nD: There are not many prokaryotes on Earth.\nE: Simple life is dependent on photosynthesis.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEukaryotes are organisms with their DNA organised in chromosomes, found in a distinct nucleus. Eukaryotes include algae, fungi, plants, and animals. Prokaryotes are single-celled organisms that have neither a nucleus nor other organelles. Prokaryotes include bacteria and cyanobacteria. There are approximately 8.7 million species of eukaryotes on earth - the relative size of each eukaryote group is listed below.\n\n[figure1]\n\nhttp://www.sciencedaily.com/releases/2011/08/110823180459.htm\n\nFrom the information provided, the best conclusion is?\n\nA: Fungi are not well described as a taxonomic group.\nB: Animals have gone through the most evolution.\nC: Eukaryotes are better described than prokaryotes.\nD: There are not many prokaryotes on Earth.\nE: Simple life is dependent on photosynthesis.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-05.jpg?height=799&width=1097&top_left_y=1828&top_left_x=111"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_549",
        "problem": "家蚕的茧色白色与黄色受到常染色体上两对独立遗传的等位基因 $\\mathrm{I} / \\mathrm{i} 、 \\mathrm{Y} / \\mathrm{y}$ 控制, $\\mathrm{Y}$基因表达产物参与色素的合成， I 基因抑制 $\\mathrm{Y}$ 基因的表达，i 与 y 基因均无该功能。野生型家蚕的茧色为白色, 某种黄茧突变体的产生与 $\\mathrm{Y}$ 基因的表达量异常有关。研究人员比较了纯合野生型家蚕与纯合突变体家蚕的 Y 基因表达量, 结果如图所示。下列分析错误的是 ( )\n\n[图1]\nA: 诱导突变体的 $\\mathrm{Y}$ 基因突变, 培育的后代结白茧\nB: 纯合白茧家蚕的基因型为 IIYY、Iiyy 和 iiyy\nC: 上述野生型家蚕与突变体家蚕杂交, 子一代均结白茧\nD: 提高野生型家蚕 Y 基因的表达量可能获得结黄茧表型\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n家蚕的茧色白色与黄色受到常染色体上两对独立遗传的等位基因 $\\mathrm{I} / \\mathrm{i} 、 \\mathrm{Y} / \\mathrm{y}$ 控制, $\\mathrm{Y}$基因表达产物参与色素的合成， I 基因抑制 $\\mathrm{Y}$ 基因的表达，i 与 y 基因均无该功能。野生型家蚕的茧色为白色, 某种黄茧突变体的产生与 $\\mathrm{Y}$ 基因的表达量异常有关。研究人员比较了纯合野生型家蚕与纯合突变体家蚕的 Y 基因表达量, 结果如图所示。下列分析错误的是 ( )\n\n[图1]\n\nA: 诱导突变体的 $\\mathrm{Y}$ 基因突变, 培育的后代结白茧\nB: 纯合白茧家蚕的基因型为 IIYY、Iiyy 和 iiyy\nC: 上述野生型家蚕与突变体家蚕杂交, 子一代均结白茧\nD: 提高野生型家蚕 Y 基因的表达量可能获得结黄茧表型\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_a4a53182e6105aab5ff8g-88.jpg?height=425&width=448&top_left_y=724&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_910",
        "problem": "鹤鹑的性别决定方式为 ZW 型, 其羽色由两对基因控制。控制栗羽的基因 A, 黄羽的基因 $\\mathrm{a}$ 位于 $\\mathrm{Z}$ 染色体上, 控制黑羽的基因 $\\mathrm{B} / \\mathrm{b}$ 位于常染色体上, 其中基因 $\\mathrm{B}$ 对基因 $\\mathrm{b}$为不完全显性。羽色与基因型的关系如下表所示。下列相关分析错误的是（）\n\n| 表现型 | 基因型 |\n| :---: | :---: |\n| 栗羽 | $\\mathrm{BBZ}^{\\mathrm{A}}-$ |\n\n\n| 不完全黑羽 | $\\mathrm{BbZ}_{-}^{\\mathrm{A}}$ |\n| :---: | :---: |\n| 黑羽 | $\\mathrm{bbZ}^{\\mathrm{A}}$ |\n| 黄羽 | B_Z $Z^{a} Z^{a} 、 B \\_Z^{a} W$ |\n| 灰羽 | $b^{2} Z^{a} Z^{a} 、 b^{2} Z^{a} W$ |\nA: 纯合栗羽和纯合黑羽鹤鹑交配, 不能得到黄羽或灰羽子代\nB: 不完全黑羽雌雄鹤䴔相互交配, 子代的羽色最多有 4 种\nC: 多组纯合栗羽和纯合黄羽鹤鹑交配, 子代不可能出现黄羽鹤鹑\nD: 灰羽雄鹤鹑与纯合栗羽雌鹤鹑交配，子代的羽色与亲本的完全不同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n鹤鹑的性别决定方式为 ZW 型, 其羽色由两对基因控制。控制栗羽的基因 A, 黄羽的基因 $\\mathrm{a}$ 位于 $\\mathrm{Z}$ 染色体上, 控制黑羽的基因 $\\mathrm{B} / \\mathrm{b}$ 位于常染色体上, 其中基因 $\\mathrm{B}$ 对基因 $\\mathrm{b}$为不完全显性。羽色与基因型的关系如下表所示。下列相关分析错误的是（）\n\n| 表现型 | 基因型 |\n| :---: | :---: |\n| 栗羽 | $\\mathrm{BBZ}^{\\mathrm{A}}-$ |\n\n\n| 不完全黑羽 | $\\mathrm{BbZ}_{-}^{\\mathrm{A}}$ |\n| :---: | :---: |\n| 黑羽 | $\\mathrm{bbZ}^{\\mathrm{A}}$ |\n| 黄羽 | B_Z $Z^{a} Z^{a} 、 B \\_Z^{a} W$ |\n| 灰羽 | $b^{2} Z^{a} Z^{a} 、 b^{2} Z^{a} W$ |\n\nA: 纯合栗羽和纯合黑羽鹤鹑交配, 不能得到黄羽或灰羽子代\nB: 不完全黑羽雌雄鹤䴔相互交配, 子代的羽色最多有 4 种\nC: 多组纯合栗羽和纯合黄羽鹤鹑交配, 子代不可能出现黄羽鹤鹑\nD: 灰羽雄鹤鹑与纯合栗羽雌鹤鹑交配，子代的羽色与亲本的完全不同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_643",
        "problem": "某高等动物的毛色由常染色体上的两对等位基因（A、a 和 $B 、 b$ ）控制， A 对 $a$ 、 $\\mathrm{B}$ 对 $\\mathrm{b}$ 完全显性, 其中 $\\mathrm{A}$ 基因控制黑色素的合成， $\\mathrm{B}$ 基因控制黄色素的合成，两种色素均不合成时毛色呈白色。当 $\\mathrm{A} 、 \\mathrm{~B}$ 基因同时存在时, 二者的转录产物会形成双链结构进而无法继续表达。用纯合的黑色和黄色亲本杂交, $F_{1}$ 为白色, $F_{1}$ 雌雄自由交配得到 $F_{2}$ 。\n下列叙述正确的是（ ）\nA: 自然界中白色个体的基因型有 4 种\nB: 含 $\\mathrm{A} 、 \\mathrm{~B}$ 基因的个体毛色是白色的原因是不能翻译出相关蛋白质\nC: 若 $F_{2}$ 中黑色:黄色:白色个体之比接近 3:3:10, 则两对基因独立遗传\nD: 若 $F_{2}$ 中白色个体的比例接近 $1 / 2$, 则 $F_{2}$ 中黑色个体的比例接近 $1 / 4$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某高等动物的毛色由常染色体上的两对等位基因（A、a 和 $B 、 b$ ）控制， A 对 $a$ 、 $\\mathrm{B}$ 对 $\\mathrm{b}$ 完全显性, 其中 $\\mathrm{A}$ 基因控制黑色素的合成， $\\mathrm{B}$ 基因控制黄色素的合成，两种色素均不合成时毛色呈白色。当 $\\mathrm{A} 、 \\mathrm{~B}$ 基因同时存在时, 二者的转录产物会形成双链结构进而无法继续表达。用纯合的黑色和黄色亲本杂交, $F_{1}$ 为白色, $F_{1}$ 雌雄自由交配得到 $F_{2}$ 。\n下列叙述正确的是（ ）\n\nA: 自然界中白色个体的基因型有 4 种\nB: 含 $\\mathrm{A} 、 \\mathrm{~B}$ 基因的个体毛色是白色的原因是不能翻译出相关蛋白质\nC: 若 $F_{2}$ 中黑色:黄色:白色个体之比接近 3:3:10, 则两对基因独立遗传\nD: 若 $F_{2}$ 中白色个体的比例接近 $1 / 2$, 则 $F_{2}$ 中黑色个体的比例接近 $1 / 4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_364",
        "problem": "下图表示基因 C 发生突变的结果图。下列叙述正确的是（ ）\n\n[图1]\nA: 该突变不会改变基因中的氢键数目\nB: 同一个基因可突变出不同基因，体现了基因突变具有随机性\nC: 图中的基因序列作为模板链, 指导合成相应的 mRNA\nD: 与正常基因 C 相比，突变的基因 C 控制合成的蛋白质分子量减小\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图表示基因 C 发生突变的结果图。下列叙述正确的是（ ）\n\n[图1]\n\nA: 该突变不会改变基因中的氢键数目\nB: 同一个基因可突变出不同基因，体现了基因突变具有随机性\nC: 图中的基因序列作为模板链, 指导合成相应的 mRNA\nD: 与正常基因 C 相比，突变的基因 C 控制合成的蛋白质分子量减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-79.jpg?height=728&width=891&top_left_y=407&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_322",
        "problem": "下列关于遗传物质的叙述, 正确的是( )\nA: 烟草的遗传物质可被 RNA 酶水解\nB: 肺炎双球菌的遗传物质主要是 DNA\nC: 劳氏肉瘤病毒的遗传物质可逆转录出单链 DNA\nD: $\\mathrm{T}_{2}$ 噬菌体的遗传物质可被水解成 4 种脱氧核糖核酸\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列关于遗传物质的叙述, 正确的是( )\n\nA: 烟草的遗传物质可被 RNA 酶水解\nB: 肺炎双球菌的遗传物质主要是 DNA\nC: 劳氏肉瘤病毒的遗传物质可逆转录出单链 DNA\nD: $\\mathrm{T}_{2}$ 噬菌体的遗传物质可被水解成 4 种脱氧核糖核酸\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_611",
        "problem": "对下图所表示的生物学意义的描述, 正确的是\n\n[图1]\n\n甲\n\n[图2]\n\nz\n\n[图3]\n\n两\n\n[图4]\n\nT\nA: 若图甲表示雄果蝇精原细胞染色体组成图, 体细胞中最多含有四个染色体组\nB: 对图乙代表的生物测交，其后代中，基因型为 AADD 的个体的概率为 $1 / 4$\nC: 图丙细胞处于有丝分裂后期, 染色单体数、DNA 数均为 8 条\nD: 图丁所示家系中男性患者明显多于女性患者, 该病是伴 X 隐性遗传病\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n对下图所表示的生物学意义的描述, 正确的是\n\n[图1]\n\n甲\n\n[图2]\n\nz\n\n[图3]\n\n两\n\n[图4]\n\nT\n\nA: 若图甲表示雄果蝇精原细胞染色体组成图, 体细胞中最多含有四个染色体组\nB: 对图乙代表的生物测交，其后代中，基因型为 AADD 的个体的概率为 $1 / 4$\nC: 图丙细胞处于有丝分裂后期, 染色单体数、DNA 数均为 8 条\nD: 图丁所示家系中男性患者明显多于女性患者, 该病是伴 X 隐性遗传病\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-065.jpg?height=223&width=302&top_left_y=822&top_left_x=383",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-065.jpg?height=200&width=209&top_left_y=845&top_left_x=752",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-065.jpg?height=183&width=191&top_left_y=862&top_left_x=1047",
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-065.jpg?height=211&width=254&top_left_y=840&top_left_x=1301"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1549",
        "problem": "Blue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nWhich method is better for estimating the total whale population?\nA: Mark-recapture.\nB: Counting from images.\nC: They are equal.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBlue whales are the largest organisms ever to have lived. The British Antarctic Survey is developing technology to monitor them from space.\n\n[figure1]\n\nYou want to estimate the blue whale population. Most satellites do not take photographs of sufficient quality, so you focus on ten patches of ocean which measure 1000 square miles each. The area of the entire oceans where blue whales live is approximately 20 million square miles. In each square of ocean you count the following numbers of whales:\n\nYear 2010\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 3 | 0 | 0 | 0 | 3 | 2 | 1 | 1 | 0 | 0 |\n\nYear 2020\n\n| Image | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| Number <br> of <br> whales | 0 | 1 | 3 | 0 | 4 | 5 | 1 | 4 | 0 | 1 |\n\n\n\nWhich method is better for estimating the total whale population?\n\nA: Mark-recapture.\nB: Counting from images.\nC: They are equal.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-36.jpg?height=948&width=1673&top_left_y=657&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1276",
        "problem": "Interestingly, a recent paper by Evans et al. 2012 in Antarctic Science suggests that both Antarctic icefish and Arctic cods have evolved essentially identical AFGPs, which are synthesized and recycled in similar ways. This is an example of convergent evolution (where distantly related organisms independently evolve similar traits) in response to the identical problem of how to deal with internal ice in freezing environments.\n\nAFGPs are synthesized in the exocrine pancreas in both groups of fish. They are then discharged into the gastrointestinal tract (gut) to inhibit the growth of ingested ice. AFGPs bound to ice are lost with the faeces or if unbound, absorbed from the gut in the rectum. AFGPs circulate in the blood and interstitial fluids where they are available to bind to ice crystals that may form. It is thought that AFGPs are phagocytosed by macrophages (engulfed by the cell) and build up in the macrophages of the spleen where they remain bound to ice crystals until a warming event occurs. AFGPs in the blood are ultimately secreted into the bile and re-enter the gut when bile is secreted for digestion. Arctic cods, unlike the Antarctic icefish, also synthesize AFGP in the liver.\n\n[figure1]Describe one possible pathway of an AFGP molecule that encounters an ice crystal formed in the interstitial fluid of an Antarctic icefish in August (winter).\nA: Binds to the ice crystal in the interstitial fluid, enters the bile, is secreted into the gut, passed into the rectum and lost in the faeces.\nB: Binds to the ice crystal in the interstitial fluid, enters the blood and circulates to the liver, enters the bile, is secreted into the gut with the bile, passed into the rectum and lost in the faeces.\nC: Binds to the ice crystal in the interstitial fluid, enters the blood, is engulfed by macrophages, is stored in the spleen until a warming event occurs, is released from the ice crystal as it melts, circulates in the blood.\nD: Binds to an ice crystal in the blood, circulates to the liver, enters the bile, is secreted into the gut with the bile, passed into the rectum and then absorbed from the gut in the rectum to circulate in the blood.\nE: Binds to an ice crystal in the blood, enters the bile, is secreted into the gut with the bile, passed into the rectum where the ice melts and then absorbed from the gut in the rectum to circulate in the blood.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nInterestingly, a recent paper by Evans et al. 2012 in Antarctic Science suggests that both Antarctic icefish and Arctic cods have evolved essentially identical AFGPs, which are synthesized and recycled in similar ways. This is an example of convergent evolution (where distantly related organisms independently evolve similar traits) in response to the identical problem of how to deal with internal ice in freezing environments.\n\nAFGPs are synthesized in the exocrine pancreas in both groups of fish. They are then discharged into the gastrointestinal tract (gut) to inhibit the growth of ingested ice. AFGPs bound to ice are lost with the faeces or if unbound, absorbed from the gut in the rectum. AFGPs circulate in the blood and interstitial fluids where they are available to bind to ice crystals that may form. It is thought that AFGPs are phagocytosed by macrophages (engulfed by the cell) and build up in the macrophages of the spleen where they remain bound to ice crystals until a warming event occurs. AFGPs in the blood are ultimately secreted into the bile and re-enter the gut when bile is secreted for digestion. Arctic cods, unlike the Antarctic icefish, also synthesize AFGP in the liver.\n\n[figure1]\n\nproblem:\nDescribe one possible pathway of an AFGP molecule that encounters an ice crystal formed in the interstitial fluid of an Antarctic icefish in August (winter).\n\nA: Binds to the ice crystal in the interstitial fluid, enters the bile, is secreted into the gut, passed into the rectum and lost in the faeces.\nB: Binds to the ice crystal in the interstitial fluid, enters the blood and circulates to the liver, enters the bile, is secreted into the gut with the bile, passed into the rectum and lost in the faeces.\nC: Binds to the ice crystal in the interstitial fluid, enters the blood, is engulfed by macrophages, is stored in the spleen until a warming event occurs, is released from the ice crystal as it melts, circulates in the blood.\nD: Binds to an ice crystal in the blood, circulates to the liver, enters the bile, is secreted into the gut with the bile, passed into the rectum and then absorbed from the gut in the rectum to circulate in the blood.\nE: Binds to an ice crystal in the blood, enters the bile, is secreted into the gut with the bile, passed into the rectum where the ice melts and then absorbed from the gut in the rectum to circulate in the blood.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-31.jpg?height=1053&width=780&top_left_y=290&top_left_x=1072"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_418",
        "problem": "科研人员以侏儒小鼠和野生型小鼠作为材料, 进行如下表所示的遗传杂交实验。相关叙述最能解释实验结果的是 ( )\n\n| 杂交组合 | 亲本 | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :---: | :---: | :---: | :---: |\n| 正交 | ð侏儒小鼠×甲野生型 <br> 小鼠 | 侏儒小鼠 | 侏儒小鼠: 野生型小鼠=1:1 |\n| 反交 | ○侏儒小鼠×ð̄野生型 <br> 小鼠 | 野生型小鼠 | 侏儒小鼠: 野生型小鼠=1:1 |\nA: 控制侏儒性状的基因在线粒体 DNA 上\nB: 有两对基因共同控制小鼠的侏儒性状\nC: 含侏儒基因的精子不能参与受精作用\nD: 来源于母本的侏儒和野生型基因不表达\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n科研人员以侏儒小鼠和野生型小鼠作为材料, 进行如下表所示的遗传杂交实验。相关叙述最能解释实验结果的是 ( )\n\n| 杂交组合 | 亲本 | $\\mathrm{F}_{1}$ | $\\mathrm{~F}_{2}$ |\n| :---: | :---: | :---: | :---: |\n| 正交 | ð侏儒小鼠×甲野生型 <br> 小鼠 | 侏儒小鼠 | 侏儒小鼠: 野生型小鼠=1:1 |\n| 反交 | ○侏儒小鼠×ð̄野生型 <br> 小鼠 | 野生型小鼠 | 侏儒小鼠: 野生型小鼠=1:1 |\n\nA: 控制侏儒性状的基因在线粒体 DNA 上\nB: 有两对基因共同控制小鼠的侏儒性状\nC: 含侏儒基因的精子不能参与受精作用\nD: 来源于母本的侏儒和野生型基因不表达\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1487",
        "problem": "The 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nHere are some observations:\n\n1. The average speed with which allele frequency changes\n2. Some alleles takeover very large populations even when they have impacts too small to measure in laboratories\n3. The probability an allele will takeover a population\n4. The speed an allele's frequency changes in any one population at any one time\n5. The frequency at which an allele is initially introduced alters the probability it will takeover a population\n6. More fit alleles sometimes go extinct before less fit alleles\n\n\nWhich observations suggest that the UK-variant is genuinely more transmissible?\nA: Very rapid rise in frequency of UK-variant in the UK.\nB: Rise in frequency of UK-variant in multiple different countries.\nC: Countries near the UK show a higher frequency of UK-variant than distant countries.\nD: Covid cases in the UK were very low around the time the UK-variant is thought to have appeared.\nE: a and b only.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe 'UK variant' of Covid-19 was discovered to be more transmissible using genetic analyses of relative fitness. This involved deciding whether it was taking-over from other variants by chance, or by natural selection. In these simulations, a 'blue' allele appears in a population. The relative fitness of the blue allele, its starting frequency in the population, and the size of the population was modified. The simulations were repeated 5 times for each condition (shown as separate lines).\n\nBlue allele\n\nfrequency\n[figure1]\n[figure2]\n\nHere are some observations:\n\n1. The average speed with which allele frequency changes\n2. Some alleles takeover very large populations even when they have impacts too small to measure in laboratories\n3. The probability an allele will takeover a population\n4. The speed an allele's frequency changes in any one population at any one time\n5. The frequency at which an allele is initially introduced alters the probability it will takeover a population\n6. More fit alleles sometimes go extinct before less fit alleles\n\n\nWhich observations suggest that the UK-variant is genuinely more transmissible?\n\nA: Very rapid rise in frequency of UK-variant in the UK.\nB: Rise in frequency of UK-variant in multiple different countries.\nC: Countries near the UK show a higher frequency of UK-variant than distant countries.\nD: Covid cases in the UK were very low around the time the UK-variant is thought to have appeared.\nE: a and b only.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=970&width=704&top_left_y=778&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-58.jpg?height=966&width=706&top_left_y=778&top_left_x=1118"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_795",
        "problem": "若将处于 $\\mathrm{G}_{1}$ 期的胡萝卜愈伤组织细胞置于含 ${ }^{3} \\mathrm{H}$ 标记的胸腺嘧啶脱氧核苷酸培养液中, 培养至第二次分裂中期。下列叙述正确的是(\nA: 每条染色体中的两条染色单体均含 ${ }^{3} \\mathrm{H}$\nB: 每个 DNA 分子的两条脱氧核苷酸链均含 ${ }^{3} \\mathrm{H}$\nC: 每个 DNA 分子中均只有一条脱氧核苷酸链含 ${ }^{3} \\mathrm{H}$\nD: 每条染色单体均只有一个 DNA 分子的两条脱氧核苷酸链含 ${ }^{3} \\mathrm{H}$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n若将处于 $\\mathrm{G}_{1}$ 期的胡萝卜愈伤组织细胞置于含 ${ }^{3} \\mathrm{H}$ 标记的胸腺嘧啶脱氧核苷酸培养液中, 培养至第二次分裂中期。下列叙述正确的是(\n\nA: 每条染色体中的两条染色单体均含 ${ }^{3} \\mathrm{H}$\nB: 每个 DNA 分子的两条脱氧核苷酸链均含 ${ }^{3} \\mathrm{H}$\nC: 每个 DNA 分子中均只有一条脱氧核苷酸链含 ${ }^{3} \\mathrm{H}$\nD: 每条染色单体均只有一个 DNA 分子的两条脱氧核苷酸链含 ${ }^{3} \\mathrm{H}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_7",
        "problem": "A student measured length and height of rhizophores of Rhizophora mangle plant (Fig.Q15-1). She also made cross sections of rhizophores and observed their anatomical characteristics. The results are shown in Q15-2 and Q15-3.\n\n[figure1]\n\nFig.Q15-1. Rhizophore height and length measurement.\n\n[figure2]\n\nFig.Q15-2. Change in height and in length/height proportion in five sequential orders of rhizophores from Rhizophora mangle plants.\n\n[figure3]\n\nFig.Q15-3. Relative proportions of bark (including aerenchyma), xylem and pith along the length of individual first-order rhizophores (A), and at the base of rhizophores of sequential orders (B) Rhizophora mangle mangrove trees.\nA: There are monotonic decreases in rhizophore height and the length/height proportion in the rhizophores as a function of the rhizophore order.\nB: Within first-order rhizophores, the xylem proportion in the cross-section is larger when closer to the main stem, and decreases progressively as the rhizophore approached the ground while increasing the proportion of bark and pith.\nC: When rhizophore order decreases, bark and pith proportion decreases, while xylem propotion increases.\nD: The supportive function is likely enhanced in the first-order rhizophores, with lower length/height proportion, and higher proportion of xylem compared with bark and pith.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA student measured length and height of rhizophores of Rhizophora mangle plant (Fig.Q15-1). She also made cross sections of rhizophores and observed their anatomical characteristics. The results are shown in Q15-2 and Q15-3.\n\n[figure1]\n\nFig.Q15-1. Rhizophore height and length measurement.\n\n[figure2]\n\nFig.Q15-2. Change in height and in length/height proportion in five sequential orders of rhizophores from Rhizophora mangle plants.\n\n[figure3]\n\nFig.Q15-3. Relative proportions of bark (including aerenchyma), xylem and pith along the length of individual first-order rhizophores (A), and at the base of rhizophores of sequential orders (B) Rhizophora mangle mangrove trees.\n\nA: There are monotonic decreases in rhizophore height and the length/height proportion in the rhizophores as a function of the rhizophore order.\nB: Within first-order rhizophores, the xylem proportion in the cross-section is larger when closer to the main stem, and decreases progressively as the rhizophore approached the ground while increasing the proportion of bark and pith.\nC: When rhizophore order decreases, bark and pith proportion decreases, while xylem propotion increases.\nD: The supportive function is likely enhanced in the first-order rhizophores, with lower length/height proportion, and higher proportion of xylem compared with bark and pith.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-034.jpg?height=683&width=1309&top_left_y=572&top_left_x=365",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-034.jpg?height=954&width=734&top_left_y=1416&top_left_x=658",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-035.jpg?height=714&width=1105&top_left_y=223&top_left_x=470"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_250",
        "problem": "Glucagon is secreted from pancreatic A-cells and works as a signal via receptors (GLR) on the cells of target tissues. The amount of GLR expressed on cell surfaces is important in determining the magnitude of the response to glucagon in each target tissue. Figure 1 shows the amount of GLR mRNA in different rat tissues. In the data shown here, the glucagon receptor is not detected in brain tissue, but recent reports have revealed that it is present even in a very small amount, e.g., in the hypothalamus.\n\n[figure1]\n\nFigure 1 Relative abundance of GLR (glucagon receptor) mRNA in rat tissue. * indicates less than detectable level.\nA: Liver expresses the largest amount of GLR because it is working as one of the major organs that uptake and storage glucose in response to glucagon.\nB: A lack of mRNA detection in brain tissue indicates that neural tissue in the brain does not require much glucose as a nutrient.\nC: Skeletal muscles hold stores of glucose only used in exercise. This is consistent with the absence of GLR from the results of this experiment.\nD: Adipose tissue, which has high levels of expression of GLR, is most important energy sources during starvation.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nGlucagon is secreted from pancreatic A-cells and works as a signal via receptors (GLR) on the cells of target tissues. The amount of GLR expressed on cell surfaces is important in determining the magnitude of the response to glucagon in each target tissue. Figure 1 shows the amount of GLR mRNA in different rat tissues. In the data shown here, the glucagon receptor is not detected in brain tissue, but recent reports have revealed that it is present even in a very small amount, e.g., in the hypothalamus.\n\n[figure1]\n\nFigure 1 Relative abundance of GLR (glucagon receptor) mRNA in rat tissue. * indicates less than detectable level.\n\nA: Liver expresses the largest amount of GLR because it is working as one of the major organs that uptake and storage glucose in response to glucagon.\nB: A lack of mRNA detection in brain tissue indicates that neural tissue in the brain does not require much glucose as a nutrient.\nC: Skeletal muscles hold stores of glucose only used in exercise. This is consistent with the absence of GLR from the results of this experiment.\nD: Adipose tissue, which has high levels of expression of GLR, is most important energy sources during starvation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-27.jpg?height=902&width=1113&top_left_y=908&top_left_x=423"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1187",
        "problem": "There were approximately 1733 tonnes of oil onboard the Rena when she ran aground. Approximately 350 tonnes were not recovered from the wreck and most of this washed ashore on Matakana Island and the coastline from Mt Maunganui to Maketu between 9 and 11 October 2011. Surveys examining the effects of the oil and debris washed onto the surf beaches focused on the northern tuatua (Paphies subtriangulata), as this is one of the most common species found burrowing in the sand on the open coast surf beaches that were most heavily fouled by oil from the Rena, and they are an important kai moana species.\n\nThe figure at right shows the level of total PAH in the tissue of tuatua from Papamoa and Omanu beaches from October 52011 to 30 June 2012. Before impact total PAH levels (background) were about $0.7 \\mu \\mathrm{g} / \\mathrm{kg}$ at Papamoa beach and $0.2 \\mu \\mathrm{g} / \\mathrm{kg}$ at Ōmanu Beach. These values were produced on a wet weight basis.\n\n[figure1]\n\nHeavily oiled\n\nModerately oiled\n\nLightly oiled\n\n[figure2]\n\nThe graph at left shows the total PAH\nlevels in tuatua in winter 2012 from beaches from Waihi - East Cape. They are colour-coded to represent the degree of oiling. No winter PAH data was available for Maketū Spit - Mid (shore level). Results obtained from Waihi and Ōhope beaches are considered background levels and an average between these levels (20.6 $\\mu \\mathrm{g} / \\mathrm{kg}$ ) is plotted as a dashed line. These values were produced on a dry weight basis.From these results it can be concluded:\nA: Total PAH levels in tuatua are strongly related to the degree of oil fouling.\nB: Tuatua at beaches that were fouled by oil from the Rena show significant ongoing contamination with PAH.\nC: Total PAH levels in tuatua from Ōmanu Beach can be directly compared and show an increase of 103-fold.\nD: Tuatua communities on Bay of Plenty open surf beaches do not appear to be catastrophically affected by the Rena oil spill in the long term.\nE: None of the above is a valid conclusion.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nThere were approximately 1733 tonnes of oil onboard the Rena when she ran aground. Approximately 350 tonnes were not recovered from the wreck and most of this washed ashore on Matakana Island and the coastline from Mt Maunganui to Maketu between 9 and 11 October 2011. Surveys examining the effects of the oil and debris washed onto the surf beaches focused on the northern tuatua (Paphies subtriangulata), as this is one of the most common species found burrowing in the sand on the open coast surf beaches that were most heavily fouled by oil from the Rena, and they are an important kai moana species.\n\nThe figure at right shows the level of total PAH in the tissue of tuatua from Papamoa and Omanu beaches from October 52011 to 30 June 2012. Before impact total PAH levels (background) were about $0.7 \\mu \\mathrm{g} / \\mathrm{kg}$ at Papamoa beach and $0.2 \\mu \\mathrm{g} / \\mathrm{kg}$ at Ōmanu Beach. These values were produced on a wet weight basis.\n\n[figure1]\n\nHeavily oiled\n\nModerately oiled\n\nLightly oiled\n\n[figure2]\n\nThe graph at left shows the total PAH\nlevels in tuatua in winter 2012 from beaches from Waihi - East Cape. They are colour-coded to represent the degree of oiling. No winter PAH data was available for Maketū Spit - Mid (shore level). Results obtained from Waihi and Ōhope beaches are considered background levels and an average between these levels (20.6 $\\mu \\mathrm{g} / \\mathrm{kg}$ ) is plotted as a dashed line. These values were produced on a dry weight basis.\n\nproblem:\nFrom these results it can be concluded:\n\nA: Total PAH levels in tuatua are strongly related to the degree of oil fouling.\nB: Tuatua at beaches that were fouled by oil from the Rena show significant ongoing contamination with PAH.\nC: Total PAH levels in tuatua from Ōmanu Beach can be directly compared and show an increase of 103-fold.\nD: Tuatua communities on Bay of Plenty open surf beaches do not appear to be catastrophically affected by the Rena oil spill in the long term.\nE: None of the above is a valid conclusion.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-26.jpg?height=528&width=1105&top_left_y=196&top_left_x=772",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-26.jpg?height=708&width=1125&top_left_y=914&top_left_x=154"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_108",
        "problem": "Information about the relationships among organisms is a useful source of data for scientists investigating a wide variety of biological questions.\nA: Phylogenetic trees can be used to determine how many times a particular trait independently evolved.\nB: Phylogenetic trees can suggest a particular trait is the ancestral state.\nC: Phylogenetic trees can be used to determine the order in which evolutionary lineages split but not the timing of those splits.\nD: Phylogenetic trees can be used to determine the virus's origins in human populations.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nInformation about the relationships among organisms is a useful source of data for scientists investigating a wide variety of biological questions.\n\nA: Phylogenetic trees can be used to determine how many times a particular trait independently evolved.\nB: Phylogenetic trees can suggest a particular trait is the ancestral state.\nC: Phylogenetic trees can be used to determine the order in which evolutionary lineages split but not the timing of those splits.\nD: Phylogenetic trees can be used to determine the virus's origins in human populations.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-109.jpg?height=716&width=1155&top_left_y=178&top_left_x=541"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_74",
        "problem": "The Peromyscus polionotus inhabits the mainland of the Florida peninsula (Figure 1 (4)) and has a dark-colored coat (Figure 1). In contrast, P. polionotus, inhabiting the light-colored sandy coastal dunes (Figure 1 (1)-(3), which are estimated to be 6,000 years old, has a lighter-colored coat (Figure 1). These mice show obvious differences in color patterns according to their habitat. The researchers compared the melanocortin 1 receptor gene (MClR), a key gene for melanogenesis, and revealed the existence of two alleles, of which $65^{\\text {th }}$ amino acid residue is $\\mathrm{R}$ or $\\mathrm{C}$, among these mice populations.\n\n[figure1]\n\n[figure2]\n\nFigure 1\n\n[figure3]\n\n$(\\log \\mathrm{M})$\n\nFigure 2\n\nFigure 1. (left) Four localities of habitat of P. polionotus in the Florida peninsula. (right) Cartoons represent the color patterns of the mice in each locality. Pie charts indicate the frequencies of the $\\mathrm{R}$ allele (black) and $\\mathrm{C}$ allele (white). $\\mathrm{n}$ indicates the number of individuals surveyed.\n\nFigure 2. Plot of the cAMP response to the MSH (melanocyte stimulating hormone) stimulation for MC1R expressing cultured cells. The $\\mathrm{X}$ and $\\mathrm{Y}$ axis indicate the concentration of the MSH and cAMP, respectively. The MC1R proteins encoded by $\\mathrm{R}$ or $\\mathrm{C}$ alleles were examined in this experiment.\nA: In addition to the $M C 1 R$ gene, other genes are likely to be involved in the body color of these subspecies.\nB: The dark color coat population is likely to have emerged from the light color coat population.\nC: It is likely that the $\\mathrm{C}$ alleles ( $65^{\\text {th }}$ amino acid residue is $\\mathrm{C}$ ) of each population (1) to (3) results from an independent mutation.\nD: In Figure 2, the white circle and white triangle represent $\\mathrm{R}$ and $\\mathrm{C}$ alleles, respectively.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe Peromyscus polionotus inhabits the mainland of the Florida peninsula (Figure 1 (4)) and has a dark-colored coat (Figure 1). In contrast, P. polionotus, inhabiting the light-colored sandy coastal dunes (Figure 1 (1)-(3), which are estimated to be 6,000 years old, has a lighter-colored coat (Figure 1). These mice show obvious differences in color patterns according to their habitat. The researchers compared the melanocortin 1 receptor gene (MClR), a key gene for melanogenesis, and revealed the existence of two alleles, of which $65^{\\text {th }}$ amino acid residue is $\\mathrm{R}$ or $\\mathrm{C}$, among these mice populations.\n\n[figure1]\n\n[figure2]\n\nFigure 1\n\n[figure3]\n\n$(\\log \\mathrm{M})$\n\nFigure 2\n\nFigure 1. (left) Four localities of habitat of P. polionotus in the Florida peninsula. (right) Cartoons represent the color patterns of the mice in each locality. Pie charts indicate the frequencies of the $\\mathrm{R}$ allele (black) and $\\mathrm{C}$ allele (white). $\\mathrm{n}$ indicates the number of individuals surveyed.\n\nFigure 2. Plot of the cAMP response to the MSH (melanocyte stimulating hormone) stimulation for MC1R expressing cultured cells. The $\\mathrm{X}$ and $\\mathrm{Y}$ axis indicate the concentration of the MSH and cAMP, respectively. The MC1R proteins encoded by $\\mathrm{R}$ or $\\mathrm{C}$ alleles were examined in this experiment.\n\nA: In addition to the $M C 1 R$ gene, other genes are likely to be involved in the body color of these subspecies.\nB: The dark color coat population is likely to have emerged from the light color coat population.\nC: It is likely that the $\\mathrm{C}$ alleles ( $65^{\\text {th }}$ amino acid residue is $\\mathrm{C}$ ) of each population (1) to (3) results from an independent mutation.\nD: In Figure 2, the white circle and white triangle represent $\\mathrm{R}$ and $\\mathrm{C}$ alleles, respectively.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-52.jpg?height=457&width=397&top_left_y=968&top_left_x=184",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-52.jpg?height=457&width=442&top_left_y=962&top_left_x=590",
            "https://cdn.mathpix.com/cropped/2024_03_14_285df5ea8616a0714b70g-52.jpg?height=531&width=834&top_left_y=934&top_left_x=1045"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1014",
        "problem": "Robert MacArthur and E. O. Wilson studied island biogeography and developed the island equilibrium model. Which of the following scenarios will DEFINITELY result in an island $\\mathrm{X}$ having a greater equilibrium number of species than island $\\mathrm{Y}$ ?\nA: Island $\\mathrm{X}$ is bigger than Island $\\mathrm{Y}$. The two islands are the same distance from the mainland.\nB: Island $\\mathrm{X}$ is farther from the mainland than Island $\\mathrm{Y}$. The two islands are the same size.\nC: Island $\\mathrm{X}$ is bigger than Island $\\mathrm{Y}$ and farther from the mainland.\nD: Island $\\mathrm{X}$ is smaller than Island $\\mathrm{Y}$ and closer to the mainland.\nE: Island $\\mathrm{X}$ and $\\mathrm{Y}$ have the same rate of immigration and extinction.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRobert MacArthur and E. O. Wilson studied island biogeography and developed the island equilibrium model. Which of the following scenarios will DEFINITELY result in an island $\\mathrm{X}$ having a greater equilibrium number of species than island $\\mathrm{Y}$ ?\n\nA: Island $\\mathrm{X}$ is bigger than Island $\\mathrm{Y}$. The two islands are the same distance from the mainland.\nB: Island $\\mathrm{X}$ is farther from the mainland than Island $\\mathrm{Y}$. The two islands are the same size.\nC: Island $\\mathrm{X}$ is bigger than Island $\\mathrm{Y}$ and farther from the mainland.\nD: Island $\\mathrm{X}$ is smaller than Island $\\mathrm{Y}$ and closer to the mainland.\nE: Island $\\mathrm{X}$ and $\\mathrm{Y}$ have the same rate of immigration and extinction.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1550",
        "problem": "ATP is important for maintaining a normal membrane potential in nerve cells. A squid giant axon was injected with radioactive ${ }^{24} \\mathrm{Na}^{+}$. The rate of ${ }^{24} \\mathrm{Na}^{+}$leaving the cell was measured overtime. The external seawater was replaced with seawater containing a mitochondrial poison between 100 and 200 minutes.\n\n## Rate of radioactivity <br> leaving cell (\\%)\n\n[figure1]\n\nWhich is false?\nA: This experiment should have been carried out in the presence of oxygen.\nB: Most of the $\\mathrm{Na}^{+}$leaks out of cells by specific transportation.\nC: The effect of the poison is slightly delayed because there is some ATP stored in axons.\nD: The poison is a reversible inhibitor.\nE: The total amount of $\\mathrm{Na}^{+}$(radioactive and non-radioactive) in the axon decreases overtime.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nATP is important for maintaining a normal membrane potential in nerve cells. A squid giant axon was injected with radioactive ${ }^{24} \\mathrm{Na}^{+}$. The rate of ${ }^{24} \\mathrm{Na}^{+}$leaving the cell was measured overtime. The external seawater was replaced with seawater containing a mitochondrial poison between 100 and 200 minutes.\n\n## Rate of radioactivity <br> leaving cell (\\%)\n\n[figure1]\n\nWhich is false?\n\nA: This experiment should have been carried out in the presence of oxygen.\nB: Most of the $\\mathrm{Na}^{+}$leaks out of cells by specific transportation.\nC: The effect of the poison is slightly delayed because there is some ATP stored in axons.\nD: The poison is a reversible inhibitor.\nE: The total amount of $\\mathrm{Na}^{+}$(radioactive and non-radioactive) in the axon decreases overtime.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_903bc497c8e35b97faffg-31.jpg?height=625&width=828&top_left_y=647&top_left_x=471"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1269",
        "problem": "## DEAR ENEMIES \\& NASTY NEIGHBORS\n\n\"Dear Enemy, I curse you, and hope that something slightly unpleasant happens to you, like an onion falling on your head.\"\n\n## Blackadder I, The Archbishop\n\nMany species show territorial behaviour (actively defending an area containing resources) allowing the species to monopolise resources such as food or mates. Territoriality is costly in terms of energy and time so some species reduce these costs by being less aggressive towards their neighbours than towards unfamiliar strangers, the so called \"dear enemy\" hypothesis. The converse are \"nasty neighbours\" in which species are more, not less aggressive towards their neighbours.\n\nDr Brunton's research group, at the Institute of Natural Sciences at Massey University, has been studying the behaviour of the New Zealand bellbird,\n\n[figure1]\n(Anthornis melanura) on Tiritiri Matangi Island. Both sexes sing prolifically and are known to use vocalisations to recognise individuals. Male and female bellbirds counter-sing in response to the vocalisations of their neighbours in a territorial behaviour that may lead to chasing of an individual bird intruding on a territory.\n\nThe researchers used speakers to play the song of neighbouring females or stranger females at different breeding stages. They recorded the responses of male and female bellbirds, including counter-singing and whether the birds approached the speaker. An asterisk indicates significant differences between the response to neighbours and strangers. Due to small sample sizes, the responses to the neighbours versus strangers were not tested during the incubation stage. Control 1 was silence, control 2 played back the\n\n[figure2]\nsong of a different species.\n[figure3]\n\nb) Counter-singing\n[figure4]\n\nUsing the information and the graphs above answer Questions 54 - 58 on the following page.During courtship, male bellbirds respond how much more frequently to the sound of a stranger female by approaching the speaker than by counter-singing?\nA: $88 \\%$\nB: $68 \\%$\nC: $56 \\%$\nD: $22 \\%$\nE: $20 \\%$\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n## DEAR ENEMIES \\& NASTY NEIGHBORS\n\n\"Dear Enemy, I curse you, and hope that something slightly unpleasant happens to you, like an onion falling on your head.\"\n\n## Blackadder I, The Archbishop\n\nMany species show territorial behaviour (actively defending an area containing resources) allowing the species to monopolise resources such as food or mates. Territoriality is costly in terms of energy and time so some species reduce these costs by being less aggressive towards their neighbours than towards unfamiliar strangers, the so called \"dear enemy\" hypothesis. The converse are \"nasty neighbours\" in which species are more, not less aggressive towards their neighbours.\n\nDr Brunton's research group, at the Institute of Natural Sciences at Massey University, has been studying the behaviour of the New Zealand bellbird,\n\n[figure1]\n(Anthornis melanura) on Tiritiri Matangi Island. Both sexes sing prolifically and are known to use vocalisations to recognise individuals. Male and female bellbirds counter-sing in response to the vocalisations of their neighbours in a territorial behaviour that may lead to chasing of an individual bird intruding on a territory.\n\nThe researchers used speakers to play the song of neighbouring females or stranger females at different breeding stages. They recorded the responses of male and female bellbirds, including counter-singing and whether the birds approached the speaker. An asterisk indicates significant differences between the response to neighbours and strangers. Due to small sample sizes, the responses to the neighbours versus strangers were not tested during the incubation stage. Control 1 was silence, control 2 played back the\n\n[figure2]\nsong of a different species.\n[figure3]\n\nb) Counter-singing\n[figure4]\n\nUsing the information and the graphs above answer Questions 54 - 58 on the following page.\n\nproblem:\nDuring courtship, male bellbirds respond how much more frequently to the sound of a stranger female by approaching the speaker than by counter-singing?\n\nA: $88 \\%$\nB: $68 \\%$\nC: $56 \\%$\nD: $22 \\%$\nE: $20 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=357&width=571&top_left_y=524&top_left_x=1368",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=485&width=311&top_left_y=931&top_left_x=1495",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=486&width=1734&top_left_y=1493&top_left_x=114",
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-28.jpg?height=534&width=1744&top_left_y=2042&top_left_x=108"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_785",
        "problem": "某种昆虫的黑身 (B) 对灰身 (b) 为显性, 眼型的椭圆眼 ( $R$ ) 对圆眼 ( $r$ ) 为显性,基因 $\\mathrm{B} / \\mathrm{b} 、 \\mathrm{R} / \\mathrm{r}$ 均位于常染色体上。让黑身圆眼和灰身椭圆眼昆虫进行杂交得到 $\\mathrm{F}_{1}$, 选择 $F_{1}$ 中黑身粗圆眼昆虫相互杂交, $F_{2}$ 的表现型及比例为黑身椭圆眼：黑身圆眼：灰身\n椭圆眼：灰身圆眼 $=5: 3: 3: 1$ 。下列叙述正确的是（ ）\nA: 雌、雄亲本的基因型均为纯合\nB: $F_{1}$ 中雌、雄黑身椭圆眼昆虫的基因型为 $B b R r$\nC: $F_{2}$ 黑身粗圆眼昆虫中的纯合子所占比例为 $1 / 5$\nD: 该种昆虫产生的基因型为 $\\mathrm{Br}$ 的雄配子具有致死效应\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某种昆虫的黑身 (B) 对灰身 (b) 为显性, 眼型的椭圆眼 ( $R$ ) 对圆眼 ( $r$ ) 为显性,基因 $\\mathrm{B} / \\mathrm{b} 、 \\mathrm{R} / \\mathrm{r}$ 均位于常染色体上。让黑身圆眼和灰身椭圆眼昆虫进行杂交得到 $\\mathrm{F}_{1}$, 选择 $F_{1}$ 中黑身粗圆眼昆虫相互杂交, $F_{2}$ 的表现型及比例为黑身椭圆眼：黑身圆眼：灰身\n椭圆眼：灰身圆眼 $=5: 3: 3: 1$ 。下列叙述正确的是（ ）\n\nA: 雌、雄亲本的基因型均为纯合\nB: $F_{1}$ 中雌、雄黑身椭圆眼昆虫的基因型为 $B b R r$\nC: $F_{2}$ 黑身粗圆眼昆虫中的纯合子所占比例为 $1 / 5$\nD: 该种昆虫产生的基因型为 $\\mathrm{Br}$ 的雄配子具有致死效应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1118",
        "problem": "In which of the following animals is the 'heart' located dorsally?\nA: Flightless birds\nB: Lungfish\nC: Snakes\nD: Crabs\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn which of the following animals is the 'heart' located dorsally?\n\nA: Flightless birds\nB: Lungfish\nC: Snakes\nD: Crabs\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_254",
        "problem": "In a scientific excursion the students found four plant specimens that they had never seen before. They studied the plant organs carefully and made a few sections from them to look for the plants transport systems. The following table was completed based on their observation.\n\n| Plant number | Companion cell | Sieve cell | Sieve tube cell | Tracheid | Vessel elements | Pollen |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | + | - | + | + | + | + |\n| 2 | - | + | - | + | - | + |\n| 3 | + | - | + | + | + | - |\n| 4 | - | + | - | + | - | - |\n\nSupposing that these plants are representatives and all possible conductive elements were found,\nA: Plant 1 has seeds with more than two cotyledons.\nB: Plant 2 has winged pollen grain and double fertilization.\nC: Plants 3 and 1 have smaller female gametophyte as compared to plants 2 and 4 .\nD: Plant 3 does not belong to the flowering plants, because it does not have pollen grain.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn a scientific excursion the students found four plant specimens that they had never seen before. They studied the plant organs carefully and made a few sections from them to look for the plants transport systems. The following table was completed based on their observation.\n\n| Plant number | Companion cell | Sieve cell | Sieve tube cell | Tracheid | Vessel elements | Pollen |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | + | - | + | + | + | + |\n| 2 | - | + | - | + | - | + |\n| 3 | + | - | + | + | + | - |\n| 4 | - | + | - | + | - | - |\n\nSupposing that these plants are representatives and all possible conductive elements were found,\n\nA: Plant 1 has seeds with more than two cotyledons.\nB: Plant 2 has winged pollen grain and double fertilization.\nC: Plants 3 and 1 have smaller female gametophyte as compared to plants 2 and 4 .\nD: Plant 3 does not belong to the flowering plants, because it does not have pollen grain.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1064",
        "problem": "Partial pressures of carbon dioxide $\\left(\\mathrm{PCO}_{2}\\right)$ in body fluids of four animals (I, II, III and IV) when they are at rest are shown.\n\n[figure1]\n\nWhich of the following is correct?\nA: Animals in I and II represent invertebrates while III and IV represent vertebrates.\nB: Mode of gas exchange for animals I and II is most likely to be aquatic while for III and IV to be terrestrial.\nC: The body fluids of animals III and IV must be much more acidic than I and II.\nD: Bicarbonate concentration in body fluids of III and IV will be higher than that in I and II.\nE: Rate of metabolism of animals III and IV is likely to be higher than that of animals I and II.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nPartial pressures of carbon dioxide $\\left(\\mathrm{PCO}_{2}\\right)$ in body fluids of four animals (I, II, III and IV) when they are at rest are shown.\n\n[figure1]\n\nWhich of the following is correct?\n\nA: Animals in I and II represent invertebrates while III and IV represent vertebrates.\nB: Mode of gas exchange for animals I and II is most likely to be aquatic while for III and IV to be terrestrial.\nC: The body fluids of animals III and IV must be much more acidic than I and II.\nD: Bicarbonate concentration in body fluids of III and IV will be higher than that in I and II.\nE: Rate of metabolism of animals III and IV is likely to be higher than that of animals I and II.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_3a69655fece2dd580110g-22.jpg?height=434&width=960&top_left_y=1255&top_left_x=496"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1466",
        "problem": "[figure1]\n\nWhat is the name of the organelle labelled A?\nA: Golgi apparatus\nB: Chloroplast\nC: Ribosome\nD: Mitochondria\nE: Nucleus\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nWhat is the name of the organelle labelled A?\n\nA: Golgi apparatus\nB: Chloroplast\nC: Ribosome\nD: Mitochondria\nE: Nucleus\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-03.jpg?height=434&width=485&top_left_y=314&top_left_x=777"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_628",
        "problem": "如图表示人体细胞正常分裂过程中，核 DNA、染色体或染色体组数量变化的部分图形，下列叙述正确的是（ ）\n\n[图1]\nA: 若图示为有丝分裂过程中染色体的数量变化, 则 $\\mathrm{a}$ 时刻前发生了着丝粒分裂、 $\\mathrm{c}$时刻后细胞中不含同源染色体\nB: 若图示为减数分裂过程中染色体的数量变化, 则 $\\mathrm{a}$ 时刻前发生了着丝粒分裂、 $\\mathrm{c}$时刻后细胞中不含同源染色体\nC: 若图示为核 DNA 的数量变化且 $\\mathrm{a}$ 时刻前已复制, 则 ab 段染色体与核 DNA 的数量相等\nD: 若图示为染色体组的数量变化且 $a$ 时刻前已加倍, 则 $a b$ 段染色体组数为 2 个或 4 个\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图表示人体细胞正常分裂过程中，核 DNA、染色体或染色体组数量变化的部分图形，下列叙述正确的是（ ）\n\n[图1]\n\nA: 若图示为有丝分裂过程中染色体的数量变化, 则 $\\mathrm{a}$ 时刻前发生了着丝粒分裂、 $\\mathrm{c}$时刻后细胞中不含同源染色体\nB: 若图示为减数分裂过程中染色体的数量变化, 则 $\\mathrm{a}$ 时刻前发生了着丝粒分裂、 $\\mathrm{c}$时刻后细胞中不含同源染色体\nC: 若图示为核 DNA 的数量变化且 $\\mathrm{a}$ 时刻前已复制, 则 ab 段染色体与核 DNA 的数量相等\nD: 若图示为染色体组的数量变化且 $a$ 时刻前已加倍, 则 $a b$ 段染色体组数为 2 个或 4 个\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9fd5dd4d4b2128719128g-71.jpg?height=420&width=537&top_left_y=1463&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1243",
        "problem": "Geological analysis suggests the initial separation of New Zealand from Gondwana took place approximately 80 million years ago. However, it is hypothesised that full separation was not complete until 60 million years ago. Some scientists propose that a common ancestor of the Moa and the Kiwi (a proto-Moa) floated away from Gondwana during this time. Others propose that Moa and Kiwi arrived in New Zealand at two different times. Four possible phylogenetic trees, describing different possible evolutionary relationships within the ratites, are shown below. Use these trees to answer the following questions.\n[figure1]Which of the following provides the BEST evidence for two distinct migrations?\nA: Moa fossils are much older than Kiwi fossils.\nB: DNA analysis estimates that Moa and Kiwi each evolved into flightless species approximately 60 mya.\nC: Mitochondrial evidence suggests that Kiwi and Moa split from a common ancestor 80 mya.\nD: Moa have no wings, Kiwi still possess flight feathers.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nGeological analysis suggests the initial separation of New Zealand from Gondwana took place approximately 80 million years ago. However, it is hypothesised that full separation was not complete until 60 million years ago. Some scientists propose that a common ancestor of the Moa and the Kiwi (a proto-Moa) floated away from Gondwana during this time. Others propose that Moa and Kiwi arrived in New Zealand at two different times. Four possible phylogenetic trees, describing different possible evolutionary relationships within the ratites, are shown below. Use these trees to answer the following questions.\n[figure1]\n\nproblem:\nWhich of the following provides the BEST evidence for two distinct migrations?\n\nA: Moa fossils are much older than Kiwi fossils.\nB: DNA analysis estimates that Moa and Kiwi each evolved into flightless species approximately 60 mya.\nC: Mitochondrial evidence suggests that Kiwi and Moa split from a common ancestor 80 mya.\nD: Moa have no wings, Kiwi still possess flight feathers.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-26.jpg?height=1120&width=1812&top_left_y=505&top_left_x=150"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_91",
        "problem": "The competitive exclusion principle states that whenever two or more species compete for same set of limiting resources the superior competitors can drive inferior competitors extinct. On the other hand, some species coexist in naturally nutrient-poor systems while they need same limited resources. For example, pitcher plants (carnivorous plants) are considered inferior competitors. Brewer (2003) investigated an example of interaction between pitcher plants and another species.\n\n[figure1]\n\nEffect of pitchers starving and neighbour removal on production of total dry mass in pitchers' roots. Error bars are $\\pm 1$ standard error (Pitcher open means the pitcher plant can easily feed. after Brewer, 2003).\nA: When using the same resources, inferior competitors fail to coexist with superior competitors.\nB: Inferior competitors can coexist with superior competitors if periodic disturbance remove or reduce population of superior competitors.\nC: Intense competition can result in competing species to diversify so they can coexist while reducing the intensity of competitions.\nD: When pitcher plants are starved, their ability to coexist with their superior competitor is significantly decreased.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nThe competitive exclusion principle states that whenever two or more species compete for same set of limiting resources the superior competitors can drive inferior competitors extinct. On the other hand, some species coexist in naturally nutrient-poor systems while they need same limited resources. For example, pitcher plants (carnivorous plants) are considered inferior competitors. Brewer (2003) investigated an example of interaction between pitcher plants and another species.\n\n[figure1]\n\nEffect of pitchers starving and neighbour removal on production of total dry mass in pitchers' roots. Error bars are $\\pm 1$ standard error (Pitcher open means the pitcher plant can easily feed. after Brewer, 2003).\n\nA: When using the same resources, inferior competitors fail to coexist with superior competitors.\nB: Inferior competitors can coexist with superior competitors if periodic disturbance remove or reduce population of superior competitors.\nC: Intense competition can result in competing species to diversify so they can coexist while reducing the intensity of competitions.\nD: When pitcher plants are starved, their ability to coexist with their superior competitor is significantly decreased.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-47.jpg?height=677&width=822&top_left_y=627&top_left_x=617"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1117",
        "problem": "In nature, summer squash fruit is found in three colours, namely, white, yellow and green. When a white fruit variety (genotype WWYY) was crossed with green variety (genotype wwyy), the progeny obtained in the F2 generation was as follows:\n\nF2 generation - white : yellow: green :: 12:3:1.\n\nWhich of the following is correct?\nA: Gene W shows epistatic effect over gene Y.\nB: $W$ and $Y$ are allelic forms of the same gene.\nC: Gene $\\mathrm{Y}$ is dominant over its allelic form $\\mathrm{y}$ and is responsible for yellow fruit.\nD: Presence of both genes $W$ and $Y$ are essential for the fruit colour.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nIn nature, summer squash fruit is found in three colours, namely, white, yellow and green. When a white fruit variety (genotype WWYY) was crossed with green variety (genotype wwyy), the progeny obtained in the F2 generation was as follows:\n\nF2 generation - white : yellow: green :: 12:3:1.\n\nWhich of the following is correct?\n\nA: Gene W shows epistatic effect over gene Y.\nB: $W$ and $Y$ are allelic forms of the same gene.\nC: Gene $\\mathrm{Y}$ is dominant over its allelic form $\\mathrm{y}$ and is responsible for yellow fruit.\nD: Presence of both genes $W$ and $Y$ are essential for the fruit colour.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1379",
        "problem": "[figure1]\n\nFigure 1. 3D visualisation of the protein backbone structure for an enzyme involved in glycolysis. Please note some of the following questions relate to the circles and rectangles positioned on this visualisation.\n\nThe reaction that this enzyme catalyses is:\n\n[figure2]\n\nThe enzyme is composed of four subunits. There are certain sites on the enzyme that can bind to other molecules, indicated by the rectangles and ovals. Binding sites at the front of the molecule are indicated by solid lines. Binding sites at the back of the molecule is indicated by dotted lines. Oval sites can bind to fructose-6-phosphate. The sites indicated with rectangles bind fructose-2,6-bisphosphate. Interestingly, ATP can bind to both sites (ovals and rectangles), but ADP can only bind to rectangular sites.\n\nATP, ADP, and fructose-6-phosphate are all negative molecules.\n\nIn addition, please note that the term allosteric relates to enzymes that change the shape of the binding site of a molecule.\n\nDeficiency in the muscle isoform of the enzyme depicted in Figure 1 causes Tauri's disease (Type VII glycogen storage disease). The metabolic outcomes in muscle cells are likely to include:\nA: decreased fructose 1,6 bisphosphate, increased levels of fructose-6-phosphate\nB: decreased fructose 1,6 bisphosphate, decreased levels of fructose-6-phosphate\nC: increased fructose 1,6 bisphosphate, increased levels of fructose-6-phosphate\nD: increased fructose 1,6 bisphosphate, decreased levels of fructose-6-phosphate\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nFigure 1. 3D visualisation of the protein backbone structure for an enzyme involved in glycolysis. Please note some of the following questions relate to the circles and rectangles positioned on this visualisation.\n\nThe reaction that this enzyme catalyses is:\n\n[figure2]\n\nThe enzyme is composed of four subunits. There are certain sites on the enzyme that can bind to other molecules, indicated by the rectangles and ovals. Binding sites at the front of the molecule are indicated by solid lines. Binding sites at the back of the molecule is indicated by dotted lines. Oval sites can bind to fructose-6-phosphate. The sites indicated with rectangles bind fructose-2,6-bisphosphate. Interestingly, ATP can bind to both sites (ovals and rectangles), but ADP can only bind to rectangular sites.\n\nATP, ADP, and fructose-6-phosphate are all negative molecules.\n\nIn addition, please note that the term allosteric relates to enzymes that change the shape of the binding site of a molecule.\n\nDeficiency in the muscle isoform of the enzyme depicted in Figure 1 causes Tauri's disease (Type VII glycogen storage disease). The metabolic outcomes in muscle cells are likely to include:\n\nA: decreased fructose 1,6 bisphosphate, increased levels of fructose-6-phosphate\nB: decreased fructose 1,6 bisphosphate, decreased levels of fructose-6-phosphate\nC: increased fructose 1,6 bisphosphate, increased levels of fructose-6-phosphate\nD: increased fructose 1,6 bisphosphate, decreased levels of fructose-6-phosphate\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-31.jpg?height=442&width=400&top_left_y=276&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_06_cdd320068ae134b0f9deg-31.jpg?height=280&width=965&top_left_y=1105&top_left_x=243"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_411",
        "problem": "某闭花传粉植物的花色有红色和白色两种，由基因 $\\mathrm{M} 、 \\mathrm{~m}$ 控制; 花的位置有顶生和\n腋生两种，由基因 $\\mathrm{N} 、 \\mathrm{n}$ 控制。研究人员进行了以下两个实验：\n\n实验一: 用红花腋生植株人工传粉给白花顶生植株, $\\mathrm{F}_{1}$ 表型及其比例为红花腋生: 红花顶生 $=1: 1, \\mathrm{~F}_{1}$ 中红花腋生植株自花受粉, $\\mathrm{F}_{2}$ 表型及其比例为红花腋生：红花顶生：白花腋生: 白花顶生 $=6: 3: 2: 1$ 。\n\n实验二: 从 $\\mathrm{F}_{1}$ 两种表型中各选取一株, 对它们和两个亲本的两对基因（ $\\mathrm{M} 、 \\mathrm{~m}$ 和 $\\mathrm{N}$ 、 $n$ ) 进行 PCR 扩增, 然后进行电泳分离, 结果如下图。已知: (1)条带 1 和 2 是一对等位基因的条带, 条带 3 和 4 是另一对等位基因的条带; (2)图谱二为实验一中亲代红花腋生植株的电泳图谱。下列说法错误的是（）\n\n[图1]\nA: 两对相对性状中，显性性状分别是红花、腋生\nB: 条带 $1 、 4$ 分别代表的基因是 $\\mathrm{m} 、 \\mathrm{~N}$\nC: $\\mathrm{F}_{2}$ 表型出现的原因可能是基因型为 $\\mathrm{NN}$ 的个体完全致死\nD: $\\mathrm{F}_{2}$ 表型出现的原因可能是含有基因 $\\mathrm{N}$ 的花粉 50\\%致死\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某闭花传粉植物的花色有红色和白色两种，由基因 $\\mathrm{M} 、 \\mathrm{~m}$ 控制; 花的位置有顶生和\n腋生两种，由基因 $\\mathrm{N} 、 \\mathrm{n}$ 控制。研究人员进行了以下两个实验：\n\n实验一: 用红花腋生植株人工传粉给白花顶生植株, $\\mathrm{F}_{1}$ 表型及其比例为红花腋生: 红花顶生 $=1: 1, \\mathrm{~F}_{1}$ 中红花腋生植株自花受粉, $\\mathrm{F}_{2}$ 表型及其比例为红花腋生：红花顶生：白花腋生: 白花顶生 $=6: 3: 2: 1$ 。\n\n实验二: 从 $\\mathrm{F}_{1}$ 两种表型中各选取一株, 对它们和两个亲本的两对基因（ $\\mathrm{M} 、 \\mathrm{~m}$ 和 $\\mathrm{N}$ 、 $n$ ) 进行 PCR 扩增, 然后进行电泳分离, 结果如下图。已知: (1)条带 1 和 2 是一对等位基因的条带, 条带 3 和 4 是另一对等位基因的条带; (2)图谱二为实验一中亲代红花腋生植株的电泳图谱。下列说法错误的是（）\n\n[图1]\n\nA: 两对相对性状中，显性性状分别是红花、腋生\nB: 条带 $1 、 4$ 分别代表的基因是 $\\mathrm{m} 、 \\mathrm{~N}$\nC: $\\mathrm{F}_{2}$ 表型出现的原因可能是基因型为 $\\mathrm{NN}$ 的个体完全致死\nD: $\\mathrm{F}_{2}$ 表型出现的原因可能是含有基因 $\\mathrm{N}$ 的花粉 50\\%致死\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_6094917e73c20e0852feg-54.jpg?height=377&width=711&top_left_y=811&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_28",
        "problem": "Which pair of graphs correctly shows the changes in glucose concentrations in the medium and $\\beta$-galactosidase activity within the cells?\nA: [figure1]\nB: [figure2]\nC: [figure3]\nD: [figure4]\nE: [figure5]\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich pair of graphs correctly shows the changes in glucose concentrations in the medium and $\\beta$-galactosidase activity within the cells?\n\nA: [figure1]\nB: [figure2]\nC: [figure3]\nD: [figure4]\nE: [figure5]\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_ee27396ffe2f7352b987g-08.jpg?height=464&width=1006&top_left_y=464&top_left_x=574",
            "https://cdn.mathpix.com/cropped/2024_03_14_ee27396ffe2f7352b987g-08.jpg?height=402&width=898&top_left_y=947&top_left_x=681",
            "https://cdn.mathpix.com/cropped/2024_03_14_ee27396ffe2f7352b987g-08.jpg?height=418&width=898&top_left_y=1358&top_left_x=681",
            "https://cdn.mathpix.com/cropped/2024_03_14_ee27396ffe2f7352b987g-08.jpg?height=402&width=902&top_left_y=1780&top_left_x=677",
            "https://cdn.mathpix.com/cropped/2024_03_14_ee27396ffe2f7352b987g-08.jpg?height=410&width=902&top_left_y=2196&top_left_x=677"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_151",
        "problem": "Scientist has prepared 3 essential components for high-throughput screens of protein kinase inhibitors. First, individual protein kinase genes are fused to the major capsid (head) gene of $\\mathrm{T} 7$ phage. When expressed in bacteria, the fusion proteins are assembled into the phage capsid, with the kinases displayed on the outer surface. Second, an analog of ATP, which can bind to the ATP-binding pocket of the kinases, is attached to magnetic beads. Third, a bank of test compounds is prepared.\n\n[figure1]\n\nFig.Q.51. Screening potential inhibitors of protein kinases\n\nTo measure the ability of a test compound to inhibit a kinase, phage displaying a specific kinase is mixed with the magnetic beads in several wells of a 96-well plate. Then the test compound is added to individual wells over a range of different concentrations. The mixtures are incubated with gentle shaking for 1 hour at $25^{\\circ} \\mathrm{C}$, the beads are pulled to the bottom with a strong magnet, and all the free (unbound) components are washed away. Finally, the remaining, attached phage are dissociated from the beads using an excess of the same ATP analog that is attached to the beads, and counted by measuring the number of plaques they form on a bacterial lawn on a Petri dish (Fig.Q.51).\nA: When the binding process reaches equilibrium, all potential inhibitor molecules will be bound to the kinase.\nB: Test compounds that score well in this assay bind in the ATP-binding cleft of the kinase.\nC: Small differences in evolutionary conserved ATP binding sites on kinases allow targeting specific kinases.\nD: A strong binding test compound will yield a low count in the plaque assay.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nScientist has prepared 3 essential components for high-throughput screens of protein kinase inhibitors. First, individual protein kinase genes are fused to the major capsid (head) gene of $\\mathrm{T} 7$ phage. When expressed in bacteria, the fusion proteins are assembled into the phage capsid, with the kinases displayed on the outer surface. Second, an analog of ATP, which can bind to the ATP-binding pocket of the kinases, is attached to magnetic beads. Third, a bank of test compounds is prepared.\n\n[figure1]\n\nFig.Q.51. Screening potential inhibitors of protein kinases\n\nTo measure the ability of a test compound to inhibit a kinase, phage displaying a specific kinase is mixed with the magnetic beads in several wells of a 96-well plate. Then the test compound is added to individual wells over a range of different concentrations. The mixtures are incubated with gentle shaking for 1 hour at $25^{\\circ} \\mathrm{C}$, the beads are pulled to the bottom with a strong magnet, and all the free (unbound) components are washed away. Finally, the remaining, attached phage are dissociated from the beads using an excess of the same ATP analog that is attached to the beads, and counted by measuring the number of plaques they form on a bacterial lawn on a Petri dish (Fig.Q.51).\n\nA: When the binding process reaches equilibrium, all potential inhibitor molecules will be bound to the kinase.\nB: Test compounds that score well in this assay bind in the ATP-binding cleft of the kinase.\nC: Small differences in evolutionary conserved ATP binding sites on kinases allow targeting specific kinases.\nD: A strong binding test compound will yield a low count in the plaque assay.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_1923091db797f33574ddg-005.jpg?height=823&width=1286&top_left_y=906&top_left_x=421"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_769",
        "problem": "甲病由两对常染色体上的两对等位基因控制(只有基因型为 B_D_才表现正常), 其中 I-1 基因型为 BbDD, 且II-2 与II-3 婚配的子代不会患甲病。乙病为单基因遗传病(由基因 $\\mathrm{G} 、 \\mathrm{~g}$ 控制)，其中 I-2 不携带乙病基因(不考虑突变), 下列叙述正确的是\n[图1]\n\nIII\nA: I-4 和II-1 的基因型都为 $\\mathrm{BbDdX}^{\\mathrm{g}} \\mathrm{Y}$\nB: II-2 和II-3 再生一个正常孩子的概率为 $3 / 8$\nC: III-2 的一个初级精母细胞中可能含有 3 个致病基因\nD: III-1 与基因型为 $\\mathrm{BbDdX}^{\\mathrm{g}} \\mathrm{Y}$ 的男性婚配, 子代患病男孩的概率为 $37 / 128$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n甲病由两对常染色体上的两对等位基因控制(只有基因型为 B_D_才表现正常), 其中 I-1 基因型为 BbDD, 且II-2 与II-3 婚配的子代不会患甲病。乙病为单基因遗传病(由基因 $\\mathrm{G} 、 \\mathrm{~g}$ 控制)，其中 I-2 不携带乙病基因(不考虑突变), 下列叙述正确的是\n[图1]\n\nIII\n\nA: I-4 和II-1 的基因型都为 $\\mathrm{BbDdX}^{\\mathrm{g}} \\mathrm{Y}$\nB: II-2 和II-3 再生一个正常孩子的概率为 $3 / 8$\nC: III-2 的一个初级精母细胞中可能含有 3 个致病基因\nD: III-1 与基因型为 $\\mathrm{BbDdX}^{\\mathrm{g}} \\mathrm{Y}$ 的男性婚配, 子代患病男孩的概率为 $37 / 128$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_73471debdc710d642a8dg-026.jpg?height=244&width=164&top_left_y=414&top_left_x=367"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_204",
        "problem": "Guppies (Poecilia reticulata) show a complex color pattern polymorphism that varies with predation pressure, reflecting a balance between selection for crypsis by predators and selection for conspicuousness by sexual selection. Three experimental ponds were used to study this phenomenon, mimicking the real condition on the native habitat of the guppies and its predators, Rivulus hartii and Crenicichla alta. One pond has the control group while the other two ponds were added with one of the two predators. In the field, C. alta was observed to be more dangerous than $R$. hartii.\n\n[figure1]\n\nFigure Q.45. Changes in the number of spots per fish during the course of the experiment. Line ' $\\mathrm{K}$ ' stands for pond without predator, ' $\\mathrm{R}$ ' for pond with $R$. hartii, and\n\n' $\\mathrm{C}$ ' for pond with $C$. alta. In the $\\mathrm{X}$-axis, ' $\\mathrm{F}$ ' stands for the time when the foundation population was started, ' $S$ ' stands for the time when the predators were added, then ' $\\mathrm{I}$ ' and 'II' stands for the numbering of the following censuses.\nA: The color pattern is responsible for the reduced fitness of $P$. reticulata.\nB: Sexual selection for color pattern of the $P$. reticulata cannot be inferred from the data.\nC: The color pattern of the $P$. reticulata may be advantagous in escaping $R$. hartii.\nD: The two predators possibly use two different mechanisms to detect $P$. reticulata.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGuppies (Poecilia reticulata) show a complex color pattern polymorphism that varies with predation pressure, reflecting a balance between selection for crypsis by predators and selection for conspicuousness by sexual selection. Three experimental ponds were used to study this phenomenon, mimicking the real condition on the native habitat of the guppies and its predators, Rivulus hartii and Crenicichla alta. One pond has the control group while the other two ponds were added with one of the two predators. In the field, C. alta was observed to be more dangerous than $R$. hartii.\n\n[figure1]\n\nFigure Q.45. Changes in the number of spots per fish during the course of the experiment. Line ' $\\mathrm{K}$ ' stands for pond without predator, ' $\\mathrm{R}$ ' for pond with $R$. hartii, and\n\n' $\\mathrm{C}$ ' for pond with $C$. alta. In the $\\mathrm{X}$-axis, ' $\\mathrm{F}$ ' stands for the time when the foundation population was started, ' $S$ ' stands for the time when the predators were added, then ' $\\mathrm{I}$ ' and 'II' stands for the numbering of the following censuses.\n\nA: The color pattern is responsible for the reduced fitness of $P$. reticulata.\nB: Sexual selection for color pattern of the $P$. reticulata cannot be inferred from the data.\nC: The color pattern of the $P$. reticulata may be advantagous in escaping $R$. hartii.\nD: The two predators possibly use two different mechanisms to detect $P$. reticulata.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-098.jpg?height=827&width=944&top_left_y=819&top_left_x=606"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1310",
        "problem": "[figure1]\n\nThe graph above shows the effects of a change from light to dark conditions on the relative concentrations of phosphoglycerate and ribulose bisphosphate in a\n\nWhich one of the following features of the graph does NOT provide support for the hypothesis that ribulose bisphosphate is converted to phosphoglycerate? The concentration of the compounds\nA: Have similar concentrations at the start of the experiment.\nB: Respond immediately to the changed conditions.\nC: Achieve a new steady concentration at the same time.\nD: Have the same rate of response, one negative and the other positive. plant.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n[figure1]\n\nThe graph above shows the effects of a change from light to dark conditions on the relative concentrations of phosphoglycerate and ribulose bisphosphate in a\n\nWhich one of the following features of the graph does NOT provide support for the hypothesis that ribulose bisphosphate is converted to phosphoglycerate? The concentration of the compounds\n\nA: Have similar concentrations at the start of the experiment.\nB: Respond immediately to the changed conditions.\nC: Achieve a new steady concentration at the same time.\nD: Have the same rate of response, one negative and the other positive. plant.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_dea84ffe537dddbf3e65g-15.jpg?height=617&width=714&top_left_y=251&top_left_x=197"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_242",
        "problem": "As shown in the left-hand picture below, neuron $(\\mathrm{N})$ receives signals directly from two separate nerve terminals (a and c). Nerve terminal (b) is synaptically connected to nerve terminal (a). The right-hand graph shows various postsynaptic potentials recorded in neuron $(\\mathrm{N})$ caused by input signals from the three presynaptic terminals.\n[figure1]\n\nWhich of the following statements about the signal transmissions of these synapses are correct?\n\nI. Action potentials would be generated in neuron (N) if nerve terminals (a) and (c) were stimulated simultaneously.\n\nII. The neurotransmitter released from nerve terminal (b) is inhibitory.\n\nIII. When nerve terminal (b) is stimulated alone, an inhibitory postsynaptic potential (IPSP) would be recorded in neuron (N).\n\nIV. When nerve terminals (b) and (c) are stimulated simultaneously, the excitatory postsynaptic potential (EPSP) recorded in neuron (N) is smaller compared to when only nerve terminal (c) is stimulated.\nA: Only I and II\nB: Only I and IV\nC: Only II and III\nD: Only III and IV\nE: I, II and III\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAs shown in the left-hand picture below, neuron $(\\mathrm{N})$ receives signals directly from two separate nerve terminals (a and c). Nerve terminal (b) is synaptically connected to nerve terminal (a). The right-hand graph shows various postsynaptic potentials recorded in neuron $(\\mathrm{N})$ caused by input signals from the three presynaptic terminals.\n[figure1]\n\nWhich of the following statements about the signal transmissions of these synapses are correct?\n\nI. Action potentials would be generated in neuron (N) if nerve terminals (a) and (c) were stimulated simultaneously.\n\nII. The neurotransmitter released from nerve terminal (b) is inhibitory.\n\nIII. When nerve terminal (b) is stimulated alone, an inhibitory postsynaptic potential (IPSP) would be recorded in neuron (N).\n\nIV. When nerve terminals (b) and (c) are stimulated simultaneously, the excitatory postsynaptic potential (EPSP) recorded in neuron (N) is smaller compared to when only nerve terminal (c) is stimulated.\n\nA: Only I and II\nB: Only I and IV\nC: Only II and III\nD: Only III and IV\nE: I, II and III\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-20.jpg?height=532&width=1484&top_left_y=796&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_95",
        "problem": "There are two clinical tests to detect some antibodies against erythrocytes (RBCs), direct coombs test and indirect coombs test.\n\nDuring the direct coombs test a blood sample is taken from a person. RBCs are washed (removing the patient's own plasma) and then incubated with anti-human globulin, which attaches to all IgG antibodies. Coombs test is positive if agglutination reaction occurs.\n\nIn the indirect coombs test, serum is extracted from the blood sample taken from the person. Then, the serum gets incubated with RBCs of known antigenicity and is washed. Finally, anti-human globulin is added. If agglutination occurs, the indirect coombs test is positive.\nA: Direct coombs test will be positive only if autoantibodies are present.\nB: If the result of indirect coombs test using serum of patient 1 and RBCs of patient 2 is positive, so we can use the patient 2 as a blood donor for patient 1.\nC: Anti-human globulin specifically binds to the variable part of human antibodies.\nD: In a positive indirect coombs test, antigen-antibody reaction happens in vivo.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThere are two clinical tests to detect some antibodies against erythrocytes (RBCs), direct coombs test and indirect coombs test.\n\nDuring the direct coombs test a blood sample is taken from a person. RBCs are washed (removing the patient's own plasma) and then incubated with anti-human globulin, which attaches to all IgG antibodies. Coombs test is positive if agglutination reaction occurs.\n\nIn the indirect coombs test, serum is extracted from the blood sample taken from the person. Then, the serum gets incubated with RBCs of known antigenicity and is washed. Finally, anti-human globulin is added. If agglutination occurs, the indirect coombs test is positive.\n\nA: Direct coombs test will be positive only if autoantibodies are present.\nB: If the result of indirect coombs test using serum of patient 1 and RBCs of patient 2 is positive, so we can use the patient 2 as a blood donor for patient 1.\nC: Anti-human globulin specifically binds to the variable part of human antibodies.\nD: In a positive indirect coombs test, antigen-antibody reaction happens in vivo.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1435",
        "problem": "A group of scientists measured the distance covered over 24 hours by two samples of a common periwinkle (Littorina littorea)\n\n| Individual | Sample A (cm) | Sample B (cm) |\n| :---: | :---: | :---: |\n| 1 | 12 | 30 |\n| 2 | 13 | 9 |\n| 3 | 20 | 17 |\n| 4 | 3 | 11 |\n| 5 | 11 | 18 |\n| 6 | 21 | 5 |\n| 7 | 18 | 6 |\n| 8 | 22 | 13 |\n| 9 | 8 | 11 |\n| 10 | 0 | 19 |\n| Standard Deviation | $\\mathbf{7 . 5 8}$ | $\\mathbf{7 . 4 2}$ |\n\nThe mean of a sample is defined as\n\n$$\n\\bar{x}=\\frac{\\sum_{i=1}^{n} x_{i}}{n}\n$$\n\nThe mean of sample $A$ is $12.8 \\mathrm{~cm}$\n\nThe skew of a sample is defined as\n\n$$\n\\text { skew }=\\frac{n}{(n-1)(n-2)} \\sum_{i=1}^{n}\\left(\\frac{x_{i}-\\bar{x}}{s}\\right)^{3}\n$$\n\nwhere s represents standard deviation\n\nWhich of the following statements is true?\nA: The mean of sample B is greater than the mean of sample A\nB: The skew of sample $B$ is greater than the skew of sample $A$\nC: The data in sample A is grouped more closely around the mean than the data in sample $B$\nD: Sample B contains more observations than sample A\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA group of scientists measured the distance covered over 24 hours by two samples of a common periwinkle (Littorina littorea)\n\n| Individual | Sample A (cm) | Sample B (cm) |\n| :---: | :---: | :---: |\n| 1 | 12 | 30 |\n| 2 | 13 | 9 |\n| 3 | 20 | 17 |\n| 4 | 3 | 11 |\n| 5 | 11 | 18 |\n| 6 | 21 | 5 |\n| 7 | 18 | 6 |\n| 8 | 22 | 13 |\n| 9 | 8 | 11 |\n| 10 | 0 | 19 |\n| Standard Deviation | $\\mathbf{7 . 5 8}$ | $\\mathbf{7 . 4 2}$ |\n\nThe mean of a sample is defined as\n\n$$\n\\bar{x}=\\frac{\\sum_{i=1}^{n} x_{i}}{n}\n$$\n\nThe mean of sample $A$ is $12.8 \\mathrm{~cm}$\n\nThe skew of a sample is defined as\n\n$$\n\\text { skew }=\\frac{n}{(n-1)(n-2)} \\sum_{i=1}^{n}\\left(\\frac{x_{i}-\\bar{x}}{s}\\right)^{3}\n$$\n\nwhere s represents standard deviation\n\nWhich of the following statements is true?\n\nA: The mean of sample B is greater than the mean of sample A\nB: The skew of sample $B$ is greater than the skew of sample $A$\nC: The data in sample A is grouped more closely around the mean than the data in sample $B$\nD: Sample B contains more observations than sample A\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_717",
        "problem": "单基因遗传病的遗传方式及再发风险符合孟德尔规律。甲病（控制基因为 $\\mathrm{A} / \\mathrm{a}$ ）和红绿色盲（控制基因为 $\\mathrm{B} / \\mathrm{b}$ ) 均为单基因遗传病, 这两种病在某家系中的遗传情况如图 1 所示。将控制甲病的相关基因用限制酶 MstII处理后进行电泳，结果如图 2 所示。下列叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n未知性别、基因型个体\n\n[图3]\n\n图2\nA: 甲病的遗传方式在人群中取样调查获得\nB: $\\mathrm{III}_{3}$ 若为男性, 其不患色盲的概率为 $1 / 2$\nC: $\\mathrm{II}_{2}$ 的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{B}}$ 或 $\\mathrm{AaX} X^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}, \\mathrm{III}_{2}$ 的 $\\mathrm{a}$ 基因的内部含限制酶 MstII酶切位点\nD: $\\mathrm{III}_{3}$ 若为男性，其与 $\\mathrm{III}_{1}$ 基因型相同的概率是 $3 / 16$\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n单基因遗传病的遗传方式及再发风险符合孟德尔规律。甲病（控制基因为 $\\mathrm{A} / \\mathrm{a}$ ）和红绿色盲（控制基因为 $\\mathrm{B} / \\mathrm{b}$ ) 均为单基因遗传病, 这两种病在某家系中的遗传情况如图 1 所示。将控制甲病的相关基因用限制酶 MstII处理后进行电泳，结果如图 2 所示。下列叙述正确的是（ ）\n\n[图1]\n\n图1\n\n[图2]\n\n未知性别、基因型个体\n\n[图3]\n\n图2\n\nA: 甲病的遗传方式在人群中取样调查获得\nB: $\\mathrm{III}_{3}$ 若为男性, 其不患色盲的概率为 $1 / 2$\nC: $\\mathrm{II}_{2}$ 的基因型为 $\\mathrm{AaX}^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{B}}$ 或 $\\mathrm{AaX} X^{\\mathrm{B}} \\mathrm{X}^{\\mathrm{b}}, \\mathrm{III}_{2}$ 的 $\\mathrm{a}$ 基因的内部含限制酶 MstII酶切位点\nD: $\\mathrm{III}_{3}$ 若为男性，其与 $\\mathrm{III}_{1}$ 基因型相同的概率是 $3 / 16$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-057.jpg?height=312&width=348&top_left_y=161&top_left_x=360",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-057.jpg?height=277&width=423&top_left_y=181&top_left_x=791",
            "https://cdn.mathpix.com/cropped/2024_03_31_dc6560f05174c0bbda68g-057.jpg?height=272&width=471&top_left_y=178&top_left_x=1318"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_192",
        "problem": "Oskar, nanos and hunchback are three major genes that establish the anterior-posterior (A-P) axis during Drosophila embryogenesis. The diagram below shows how the mRNA and protein products of these three genes are distributed in Drosophila eggs (darker shades represent higher concentrations). The table below lists the outcome of disrupting one of these genes, as reflected in protein distribution and anterior-posterior development.\n\n[figure1]\n\n|  | Mutation in <br> nanos | Mutation in <br> hunchback | Mutation in oskar |\n| :---: | :---: | :---: | :---: |\n| Distribution of <br> the Nanos protein | - | normal | abnormal |\n| Distribution of <br> the Hunchback protein | abnormal | - | abnormal |\n| Distribution of <br> the Oskar protein | normal | normal |  |\n| Establishment of normal A-P | abnormal | abnormal |  |\n| polarity |  |  |  |\n\nBased on these data, which of the given statements correctly describes the interaction among the three genes?\nA: The transcription of hunchback gene is suppressed by the Nanos protein.\nB: The Oskar protein activates the translation of nanos mRNA.\nC: The Hunchback protein suppresses the translation of oskar mRNA in the anterior.\nD: The Oskar protein suppresses the transcription of hunchback gene in the posterior.\nE: The Hunchback protein suppresses the transcription of nanos gene in the anterior.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOskar, nanos and hunchback are three major genes that establish the anterior-posterior (A-P) axis during Drosophila embryogenesis. The diagram below shows how the mRNA and protein products of these three genes are distributed in Drosophila eggs (darker shades represent higher concentrations). The table below lists the outcome of disrupting one of these genes, as reflected in protein distribution and anterior-posterior development.\n\n[figure1]\n\n|  | Mutation in <br> nanos | Mutation in <br> hunchback | Mutation in oskar |\n| :---: | :---: | :---: | :---: |\n| Distribution of <br> the Nanos protein | - | normal | abnormal |\n| Distribution of <br> the Hunchback protein | abnormal | - | abnormal |\n| Distribution of <br> the Oskar protein | normal | normal |  |\n| Establishment of normal A-P | abnormal | abnormal |  |\n| polarity |  |  |  |\n\nBased on these data, which of the given statements correctly describes the interaction among the three genes?\n\nA: The transcription of hunchback gene is suppressed by the Nanos protein.\nB: The Oskar protein activates the translation of nanos mRNA.\nC: The Hunchback protein suppresses the translation of oskar mRNA in the anterior.\nD: The Oskar protein suppresses the transcription of hunchback gene in the posterior.\nE: The Hunchback protein suppresses the transcription of nanos gene in the anterior.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-29.jpg?height=571&width=1519&top_left_y=754&top_left_x=331"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_257",
        "problem": "Figure Q. 27 shows the changes in metabolic rates of adult individuals of five animal species in response to temperature changes of the external environment (EM). The individual body weights of the five species were similar (about 30 grams). The data were measured from the animals when they were staying at rest.\n\n[figure1]\n\nTemperature ofEM $\\left({ }^{\\circ} \\mathrm{C}\\right)$\n\n[figure2]\n\nTemperature of $\\mathrm{EM}\\left({ }^{\\circ} \\mathrm{C}\\right)$\n\n[figure3]\n\nTemperature of EM $\\left({ }^{\\circ} \\mathrm{C}\\right)$\n[figure4]\n\nFigure Q. 27\nA: Species IV was an ectothermic animal.\nB: Species II had the highest thermo-insulating ability among the five species.\nC: Species III possessed the highest basal metabolic rate among the five species.\nD: An increase in body temperature in species $V$ was mainly dependent on metabolism.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nFigure Q. 27 shows the changes in metabolic rates of adult individuals of five animal species in response to temperature changes of the external environment (EM). The individual body weights of the five species were similar (about 30 grams). The data were measured from the animals when they were staying at rest.\n\n[figure1]\n\nTemperature ofEM $\\left({ }^{\\circ} \\mathrm{C}\\right)$\n\n[figure2]\n\nTemperature of $\\mathrm{EM}\\left({ }^{\\circ} \\mathrm{C}\\right)$\n\n[figure3]\n\nTemperature of EM $\\left({ }^{\\circ} \\mathrm{C}\\right)$\n[figure4]\n\nFigure Q. 27\n\nA: Species IV was an ectothermic animal.\nB: Species II had the highest thermo-insulating ability among the five species.\nC: Species III possessed the highest basal metabolic rate among the five species.\nD: An increase in body temperature in species $V$ was mainly dependent on metabolism.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-061.jpg?height=410&width=507&top_left_y=778&top_left_x=199",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-061.jpg?height=410&width=507&top_left_y=778&top_left_x=749",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-061.jpg?height=410&width=504&top_left_y=778&top_left_x=1315",
            "https://cdn.mathpix.com/cropped/2024_03_14_6a13c4c20bb973e2d313g-061.jpg?height=476&width=1058&top_left_y=1300&top_left_x=500"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_808",
        "problem": "线粒体 DNA 聚合酶 $\\gamma$ （简称酶 $\\gamma$ ）是由核基因 Polg 负责编码的, 参与线粒体 DNA 复制与错误配对碱基对的切除并修复，使线粒体 DNA 能够不依赖于细胞周期持续进行复制。当 Polg 基因发生某位置的单碱基对替换后会导致酶 $\\gamma$ 修复功能受损, 进而使线粒体损伤。下列叙述正确的是（）\nA: 线粒体 DNA 的复制不受细胞核的控制\nB: Polg 基因在线粒体中完成转录和翻译全过程\nC: 突变后的酶 $\\gamma$ 与正常酶 $\\gamma$ 只有一个氨基酸不同\nD: 酶 $\\gamma$ 既能断开磷酸二酯键也能形成磷酸二酯键\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n线粒体 DNA 聚合酶 $\\gamma$ （简称酶 $\\gamma$ ）是由核基因 Polg 负责编码的, 参与线粒体 DNA 复制与错误配对碱基对的切除并修复，使线粒体 DNA 能够不依赖于细胞周期持续进行复制。当 Polg 基因发生某位置的单碱基对替换后会导致酶 $\\gamma$ 修复功能受损, 进而使线粒体损伤。下列叙述正确的是（）\n\nA: 线粒体 DNA 的复制不受细胞核的控制\nB: Polg 基因在线粒体中完成转录和翻译全过程\nC: 突变后的酶 $\\gamma$ 与正常酶 $\\gamma$ 只有一个氨基酸不同\nD: 酶 $\\gamma$ 既能断开磷酸二酯键也能形成磷酸二酯键\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_328",
        "problem": "一对正常夫妇因妻子的母亲和丈夫的母亲患有相同的单基因遗传病而进行遗传咨询。已知该遗传病在人群中的发病率为 $1 / 10000$ 。不考虑基因突变、XY 同源区段和染色体变异，控制该病的等位基因为 $\\mathrm{A} 、 \\mathrm{a}$ 。下列说法正确的是（ ）\nA: 若该遗传病为隐性遗传病, 则该夫妇生一个患病女儿的概率为 $1 / 8$\nB: 若该夫妇已经生了一个患病儿子, 则该患病儿子的姐姐与正常男性婚配后, 生 一个正常孩子的概率为 $201 / 202$\nC: 这对夫妇的父亲都可能同时含有 A 和 a\nD: 若要研究两位母亲的致病基因序列, 可用 BLAST（基本局部比对搜索工具）对其进行相似性比较\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一对正常夫妇因妻子的母亲和丈夫的母亲患有相同的单基因遗传病而进行遗传咨询。已知该遗传病在人群中的发病率为 $1 / 10000$ 。不考虑基因突变、XY 同源区段和染色体变异，控制该病的等位基因为 $\\mathrm{A} 、 \\mathrm{a}$ 。下列说法正确的是（ ）\n\nA: 若该遗传病为隐性遗传病, 则该夫妇生一个患病女儿的概率为 $1 / 8$\nB: 若该夫妇已经生了一个患病儿子, 则该患病儿子的姐姐与正常男性婚配后, 生 一个正常孩子的概率为 $201 / 202$\nC: 这对夫妇的父亲都可能同时含有 A 和 a\nD: 若要研究两位母亲的致病基因序列, 可用 BLAST（基本局部比对搜索工具）对其进行相似性比较\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_220",
        "problem": "By using modern technology, the gene that determined height in Mendel's pea plants was discovered to be the gene $L e$ that codes for the enzyme involved in biosynthesis of the gibberellin hormone $\\mathrm{GA}_{1}$. The two alleles for this gene, $T$ and $t$, differ in only one nucleotide. The enzyme produced by the recessive $t$ allele has efficiency as low as $1 / 20$ of the normal enzyme. Which of the following statements is correct?\nA: $\\mathrm{GA}_{1}$ is directly involved in auxin biosynthesis in the pea plant.\nB: The product of the $T$ allele is responsible for the normal gibberellin hormone.\nC: A $F_{1}$ plant from a cross between $T T$ and $t t$ plants will have $1 / 20$ enzyme activity of that of normal plant.\nD: Treatment of a tt plant with gibberellin hormone fails to make it grow to become a tall plant.\nE: The mutation is due to a deletion of the gene $L e$.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBy using modern technology, the gene that determined height in Mendel's pea plants was discovered to be the gene $L e$ that codes for the enzyme involved in biosynthesis of the gibberellin hormone $\\mathrm{GA}_{1}$. The two alleles for this gene, $T$ and $t$, differ in only one nucleotide. The enzyme produced by the recessive $t$ allele has efficiency as low as $1 / 20$ of the normal enzyme. Which of the following statements is correct?\n\nA: $\\mathrm{GA}_{1}$ is directly involved in auxin biosynthesis in the pea plant.\nB: The product of the $T$ allele is responsible for the normal gibberellin hormone.\nC: A $F_{1}$ plant from a cross between $T T$ and $t t$ plants will have $1 / 20$ enzyme activity of that of normal plant.\nD: Treatment of a tt plant with gibberellin hormone fails to make it grow to become a tall plant.\nE: The mutation is due to a deletion of the gene $L e$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1424",
        "problem": "The graph shows changes in mass of three individual kangaroo ticks when they were removed and kept for two days at low humidity and then four days at high humidity.\n\n[figure1]\n\nThe best deduction from these data is that:\nA: At high humidity ticks absorb water from the atmosphere\nB: At day 6 the ticks have recovered their body water\nC: The loss of mass during the first two days is due to starvation\nD: Individual ticks contain different amounts of water\nE: Individual ticks lose water at the same rate when held in low humidity conditions\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph shows changes in mass of three individual kangaroo ticks when they were removed and kept for two days at low humidity and then four days at high humidity.\n\n[figure1]\n\nThe best deduction from these data is that:\n\nA: At high humidity ticks absorb water from the atmosphere\nB: At day 6 the ticks have recovered their body water\nC: The loss of mass during the first two days is due to starvation\nD: Individual ticks contain different amounts of water\nE: Individual ticks lose water at the same rate when held in low humidity conditions\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-30.jpg?height=606&width=926&top_left_y=328&top_left_x=565"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_860",
        "problem": "果蝇的性别分化受 X 染色体数与体细胞染色体组数比例的调控。当 X 染色体数与体细胞中染色体组数的比例为 $1: 1$ 时, $\\mathrm{X}$ 染色体上的 $\\mathrm{S}$ 基因表达, 使 $\\mathrm{D}$ 基因表达为 DSXF 蛋白, 个体发育成雌性 当二者比例低于 1: 1 时, $\\mathrm{S}$ 基因不表达, D 基因则表达为 DSXM 蛋白, 个体发育成雄性; 无 $\\mathrm{S}$ 基因时，D 基因则表达为 DSXM 蛋白，个体发育成雄性。若体细胞中 X 染色体超过两条或者低于一条的个体致死。下列说法正确的是（）\nA: 雌性个体中 S 基因促进了 DSXM 蛋白的表达\nB: $X^{s} X^{s}$ 和 $X^{s} Y$ 的果蝇杂交子代雌雄之比为 3:1\nC: $X^{s} X^{s} Y$ 和 $X^{s} Y$ 的果蝇杂交子代雌雄之比为 3: 7\nD: 雌雄个体中 D 基因转录生成的 RNA 一定不同\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n果蝇的性别分化受 X 染色体数与体细胞染色体组数比例的调控。当 X 染色体数与体细胞中染色体组数的比例为 $1: 1$ 时, $\\mathrm{X}$ 染色体上的 $\\mathrm{S}$ 基因表达, 使 $\\mathrm{D}$ 基因表达为 DSXF 蛋白, 个体发育成雌性 当二者比例低于 1: 1 时, $\\mathrm{S}$ 基因不表达, D 基因则表达为 DSXM 蛋白, 个体发育成雄性; 无 $\\mathrm{S}$ 基因时，D 基因则表达为 DSXM 蛋白，个体发育成雄性。若体细胞中 X 染色体超过两条或者低于一条的个体致死。下列说法正确的是（）\n\nA: 雌性个体中 S 基因促进了 DSXM 蛋白的表达\nB: $X^{s} X^{s}$ 和 $X^{s} Y$ 的果蝇杂交子代雌雄之比为 3:1\nC: $X^{s} X^{s} Y$ 和 $X^{s} Y$ 的果蝇杂交子代雌雄之比为 3: 7\nD: 雌雄个体中 D 基因转录生成的 RNA 一定不同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_896",
        "problem": "下列叙述与生物学史实相符的是( )\nA: 孟德尔用山柳菊为实验材料,验证了基因的分离及自由组合规律\nB: 范$\\cdot$海尔蒙特基于柳枝扞插实验,认为植物生长的养料来自土壤、水和空气\nC: 富兰克林和威尔金斯对 DNA 双螺旋结构模型的建立也作出了巨大的贡献\nD: 赫尔希和蔡斯用 ${ }^{35} \\mathrm{~S}$ 和 ${ }^{32} \\mathrm{P}$ 分别标记 $\\mathrm{T}_{2}$ 噬菌体的蛋白质和 DNA,证明了 DNA 的半保留复制\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列叙述与生物学史实相符的是( )\n\nA: 孟德尔用山柳菊为实验材料,验证了基因的分离及自由组合规律\nB: 范$\\cdot$海尔蒙特基于柳枝扞插实验,认为植物生长的养料来自土壤、水和空气\nC: 富兰克林和威尔金斯对 DNA 双螺旋结构模型的建立也作出了巨大的贡献\nD: 赫尔希和蔡斯用 ${ }^{35} \\mathrm{~S}$ 和 ${ }^{32} \\mathrm{P}$ 分别标记 $\\mathrm{T}_{2}$ 噬菌体的蛋白质和 DNA,证明了 DNA 的半保留复制\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1006",
        "problem": "The biodiversity of a community is measured by the following:\nA: Species richness.\nB: Relative abundance.\nC: Total biomass in a given area.\nD: Ratio of plant to animal species.\nE: Both A and B are metrics of biodiversity.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe biodiversity of a community is measured by the following:\n\nA: Species richness.\nB: Relative abundance.\nC: Total biomass in a given area.\nD: Ratio of plant to animal species.\nE: Both A and B are metrics of biodiversity.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1365",
        "problem": "Hemoglobin and myoglobin are proteins that have oxygen-carrying capacity. Hemoglobin is found in red blood cells and myoglobin is found in muscles. Hemoglobin can bind to four oxygen molecules at any point in time. In the lungs, where there is a high oxygen concentration, hemoglobin binds to oxygen forming oxyhemoglobin. Oxyhemoglobin then circulates the body in blood and unloads oxygen to tissues with a low oxygen concentration, reforming hemoglobin. Myoglobin stores oxygen in the muscle tissue and only releases oxygen when the partial pressure of oxygen has fallen drastically. Myoglobin is only found in the bloodstream after muscle injury.\n\nThe oxygen dissociation curve, depicted below, describes the relationship between the partial pressure of oxygen ( $x$ axis) and the oxygen saturation of hemoglobin or myoglobin ( $y$ axis).\n\n[figure1]\n\nIn intense exercise, muscles produce lactic acid. This increase in acidity within the muscle helps hemoglobin unload more oxygen. How would this change the oxyhaemoglobin dissociation curve?\nA: The curve would shift to the left\nB: The curve would shift to the right\nC: The curve would shift up\nD: The curve would shift down\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHemoglobin and myoglobin are proteins that have oxygen-carrying capacity. Hemoglobin is found in red blood cells and myoglobin is found in muscles. Hemoglobin can bind to four oxygen molecules at any point in time. In the lungs, where there is a high oxygen concentration, hemoglobin binds to oxygen forming oxyhemoglobin. Oxyhemoglobin then circulates the body in blood and unloads oxygen to tissues with a low oxygen concentration, reforming hemoglobin. Myoglobin stores oxygen in the muscle tissue and only releases oxygen when the partial pressure of oxygen has fallen drastically. Myoglobin is only found in the bloodstream after muscle injury.\n\nThe oxygen dissociation curve, depicted below, describes the relationship between the partial pressure of oxygen ( $x$ axis) and the oxygen saturation of hemoglobin or myoglobin ( $y$ axis).\n\n[figure1]\n\nIn intense exercise, muscles produce lactic acid. This increase in acidity within the muscle helps hemoglobin unload more oxygen. How would this change the oxyhaemoglobin dissociation curve?\n\nA: The curve would shift to the left\nB: The curve would shift to the right\nC: The curve would shift up\nD: The curve would shift down\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_1e4c53de3699ed10a6e9g-12.jpg?height=820&width=1014&top_left_y=892&top_left_x=538"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_27",
        "problem": "Heritability is a statistical measure of how strongly the phenotype of the offspring resembles the phenotype of the parents. Heritability in broad sense- $\\mathrm{H}^{2}$ is the ratio of total genetic variance $\\left(\\mathrm{V}_{\\mathrm{g}}\\right)$ to total phenotypic variance $\\left(\\mathrm{V}_{\\mathrm{p}}\\right)$ : $\\mathrm{H}^{2}=\\mathrm{V}_{\\mathrm{g}} / \\mathrm{V}_{\\mathrm{p}}$.\n\n$V_{p}=\\sum\\left(X_{i}-X_{\\text {mean }}\\right)^{2} /(n-1)$. The phenotypic variance is often divided into three components: the genetic variance $\\left(V_{g}\\right)$, environmental variance $\\left(V_{e}\\right)$ and the interaction variance $\\left(V_{i}\\right), V_{p}=V_{g}+V_{e}+V_{i}$. Value of $H^{2}$ range from 1 to 0 . The genetic variance is divided into three components: The additive genetic variance $\\left(\\mathrm{V}_{\\mathrm{a}}\\right)$, the dominant genetic variace $\\left(V_{d}\\right)$, and the genetic interaction variance $\\left(V_{i}\\right), V_{g}=V_{a}+V_{d}+V_{i}$. Heritability in narrow sense $\\left(\\mathrm{h}^{2}\\right)=\\mathrm{V}_{\\mathrm{a}} / \\mathrm{V}_{\\mathrm{p}}$.\nA: A low $\\mathrm{H}^{2}$ value of a trait indicates that the trait is determined mainly by the environment.\nB: Genetic variance depends on the environment to which the population is exposed.\nC: The artificial selection for a trait with higher $\\mathrm{H}^{2}$ is more successful than that for a trait with lower $\\mathrm{H}^{2}$.\nD: Two pure strains of been that produce seeds with different weights were crossed. The variances of seeds $\\left(V_{p}\\right)$ from the parental, $F_{1}$, and $F_{2}$ plants are: $\\mathrm{P}_{1}=1.7 ; \\mathrm{P}_{2}=2.1, \\mathrm{~F}_{1}=2.0$ and $\\mathrm{F}_{2}=4.1$. If we ignore interaction effects, $\\mathrm{H}^{2}$ of seeds is 0.54 .\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nHeritability is a statistical measure of how strongly the phenotype of the offspring resembles the phenotype of the parents. Heritability in broad sense- $\\mathrm{H}^{2}$ is the ratio of total genetic variance $\\left(\\mathrm{V}_{\\mathrm{g}}\\right)$ to total phenotypic variance $\\left(\\mathrm{V}_{\\mathrm{p}}\\right)$ : $\\mathrm{H}^{2}=\\mathrm{V}_{\\mathrm{g}} / \\mathrm{V}_{\\mathrm{p}}$.\n\n$V_{p}=\\sum\\left(X_{i}-X_{\\text {mean }}\\right)^{2} /(n-1)$. The phenotypic variance is often divided into three components: the genetic variance $\\left(V_{g}\\right)$, environmental variance $\\left(V_{e}\\right)$ and the interaction variance $\\left(V_{i}\\right), V_{p}=V_{g}+V_{e}+V_{i}$. Value of $H^{2}$ range from 1 to 0 . The genetic variance is divided into three components: The additive genetic variance $\\left(\\mathrm{V}_{\\mathrm{a}}\\right)$, the dominant genetic variace $\\left(V_{d}\\right)$, and the genetic interaction variance $\\left(V_{i}\\right), V_{g}=V_{a}+V_{d}+V_{i}$. Heritability in narrow sense $\\left(\\mathrm{h}^{2}\\right)=\\mathrm{V}_{\\mathrm{a}} / \\mathrm{V}_{\\mathrm{p}}$.\n\nA: A low $\\mathrm{H}^{2}$ value of a trait indicates that the trait is determined mainly by the environment.\nB: Genetic variance depends on the environment to which the population is exposed.\nC: The artificial selection for a trait with higher $\\mathrm{H}^{2}$ is more successful than that for a trait with lower $\\mathrm{H}^{2}$.\nD: Two pure strains of been that produce seeds with different weights were crossed. The variances of seeds $\\left(V_{p}\\right)$ from the parental, $F_{1}$, and $F_{2}$ plants are: $\\mathrm{P}_{1}=1.7 ; \\mathrm{P}_{2}=2.1, \\mathrm{~F}_{1}=2.0$ and $\\mathrm{F}_{2}=4.1$. If we ignore interaction effects, $\\mathrm{H}^{2}$ of seeds is 0.54 .\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1417",
        "problem": "An antibody, also called immunoglobulin ( $\\mathrm{Ig}$ ), is a protective protein produced by the immune system in response to the presence of a foreign substance. Antibodies are specific to one foreign molecule. They help the immune system to target and destroy the foreign material. With modern technology, scientists are now designing antibodies called monoclonal antibodies (mAbs). These are made to target specific proteins, ultimately to treat diseases.\n\nWhich of the following is true:\nA: mAbs provide protection against a wide range of common infections\nB: mAbs can be used as the sole therapy to treat immunodeficiency in patients\nC: mAbs are designed for the treatment of patients with a cancer or disease caused by a specific pathogen\nD: mAbs can be used in place of intravenous IgG (IVIg) as a treatment for people who have antibody deficiencies\nE: mAbs can bind to two different molecules\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn antibody, also called immunoglobulin ( $\\mathrm{Ig}$ ), is a protective protein produced by the immune system in response to the presence of a foreign substance. Antibodies are specific to one foreign molecule. They help the immune system to target and destroy the foreign material. With modern technology, scientists are now designing antibodies called monoclonal antibodies (mAbs). These are made to target specific proteins, ultimately to treat diseases.\n\nWhich of the following is true:\n\nA: mAbs provide protection against a wide range of common infections\nB: mAbs can be used as the sole therapy to treat immunodeficiency in patients\nC: mAbs are designed for the treatment of patients with a cancer or disease caused by a specific pathogen\nD: mAbs can be used in place of intravenous IgG (IVIg) as a treatment for people who have antibody deficiencies\nE: mAbs can bind to two different molecules\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_188",
        "problem": "Figures I III depict the excretory systems of planaria, earthworms, and grasshoppers. Figure IV illustrates the habitat of the salmon life cycle.\n\n[figure1]\n\nWhich of the following statements concerning excretory structures is correct?\nA: In a planarian, the beating of cilia (a) within each flame bulb releases filtrate in the direction of the arrows.\nB: A pair of mesonephridia within each segment of the earthworm collects coelomic fluid from the adjacent anterior segment and excretes that collected fluid.\nC: Structure (b) is called a collecting duct, which collects and excretes concentrated urine that is hyper-osmotic to body fluids.\nD: In a grasshopper, reabsorption of the filtrate occurs mainly in the Malpighian tubules (c), where most solutes are pumped back into the hemolymph, with water following by osmosis.\nE: In freshwater, salmon take up salt from the gills and produce dilute urine; in the ocean, they excrete excess salt through their gills.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFigures I III depict the excretory systems of planaria, earthworms, and grasshoppers. Figure IV illustrates the habitat of the salmon life cycle.\n\n[figure1]\n\nWhich of the following statements concerning excretory structures is correct?\n\nA: In a planarian, the beating of cilia (a) within each flame bulb releases filtrate in the direction of the arrows.\nB: A pair of mesonephridia within each segment of the earthworm collects coelomic fluid from the adjacent anterior segment and excretes that collected fluid.\nC: Structure (b) is called a collecting duct, which collects and excretes concentrated urine that is hyper-osmotic to body fluids.\nD: In a grasshopper, reabsorption of the filtrate occurs mainly in the Malpighian tubules (c), where most solutes are pumped back into the hemolymph, with water following by osmosis.\nE: In freshwater, salmon take up salt from the gills and produce dilute urine; in the ocean, they excrete excess salt through their gills.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_8dfa50632d47d0750540g-35.jpg?height=930&width=1239&top_left_y=483&top_left_x=474"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1238",
        "problem": "The relative diameter of a tree trunk was measured over a period of several days. The results are shown in the graph.\n\n[figure1]\n\nTime (days)\n\nThe best hypothesis to explain these data is that in a transpiring plant\nA: phloem tissue is not involved in transpiration\nB: root pressure reaches a maximum value just after noon\nC: xylem vessels are sometimes under tension\nD: water is transported in the xylem vessels\nE: the rise of sap shows an endogenous rhythm\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe relative diameter of a tree trunk was measured over a period of several days. The results are shown in the graph.\n\n[figure1]\n\nTime (days)\n\nThe best hypothesis to explain these data is that in a transpiring plant\n\nA: phloem tissue is not involved in transpiration\nB: root pressure reaches a maximum value just after noon\nC: xylem vessels are sometimes under tension\nD: water is transported in the xylem vessels\nE: the rise of sap shows an endogenous rhythm\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_070c1fe65d7740f22fd9g-09.jpg?height=388&width=873&top_left_y=332&top_left_x=623"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_358",
        "problem": "DNA 甲基化一般分为从头甲基化和维持甲基化两种方式。前者是指不依赖已有的甲基化 DNA 链, 而在一个新位点产生新的甲基化。后者是指在 DNA 一条链的甲基化位点上经复制, 互补链相关位点也出现甲基化, 即在甲基化 DNA 半保留复制出的新生链相应位置上进行甲基化修饰的过程。下列叙述正确的是（ ）\nA: DNA 甲基化能在修饰碱基序列的过程中改变碱基序列及遗传信息\nB: DNA 从头甲基化会改变基因的表达，不利于基因组的稳定性和细胞分化\nC: 维持甲基化可将母链的甲基化修饰复制到新生链上可能与相关酶有关\nD: 从头甲基化中一般 DNA 两条链相同位置的碱基都会被甲基化\n",
        "prompt": "你正在参加一个国际生物竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\nDNA 甲基化一般分为从头甲基化和维持甲基化两种方式。前者是指不依赖已有的甲基化 DNA 链, 而在一个新位点产生新的甲基化。后者是指在 DNA 一条链的甲基化位点上经复制, 互补链相关位点也出现甲基化, 即在甲基化 DNA 半保留复制出的新生链相应位置上进行甲基化修饰的过程。下列叙述正确的是（ ）\n\nA: DNA 甲基化能在修饰碱基序列的过程中改变碱基序列及遗传信息\nB: DNA 从头甲基化会改变基因的表达，不利于基因组的稳定性和细胞分化\nC: 维持甲基化可将母链的甲基化修饰复制到新生链上可能与相关酶有关\nD: 从头甲基化中一般 DNA 两条链相同位置的碱基都会被甲基化\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Biology_1180",
        "problem": "The global carbon cycle is represented in the diagram at right. Human activities that affect the distribution of carbon include the burning of fossil fuels and cultivation practices such as land clearance. The greatest annual flow of carbon is by:\nA: Dissolving of atmospheric carbon dioxide in water bodies.\nB: Erosion of calcium carbonate in soils and rocks by water.\nC: Release of carbon dioxide during respiration.\nD: Fixation of atmospheric carbon dioxide by plants.\nE: Release of carbon dioxide from volcanoes.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe global carbon cycle is represented in the diagram at right. Human activities that affect the distribution of carbon include the burning of fossil fuels and cultivation practices such as land clearance. The greatest annual flow of carbon is by:\n\nA: Dissolving of atmospheric carbon dioxide in water bodies.\nB: Erosion of calcium carbonate in soils and rocks by water.\nC: Release of carbon dioxide during respiration.\nD: Fixation of atmospheric carbon dioxide by plants.\nE: Release of carbon dioxide from volcanoes.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_1284",
        "problem": "The table shows the amount of plant hormone that passes through short segments of young stems when it is applied either to the end nearest to the shoot-tip or to the end furthest from the shoot tip. Three experiments were carried out under different conditions as indicated below.\n\n| Site of hormone <br> application | Amount of hormone passing through stem (arbitrary units) |  |  |\n| :--- | :---: | :---: | :---: |\n|  | At $3{ }^{\\circ} \\mathrm{C}$ | At $25^{\\circ} \\mathrm{C}$ | At $25^{\\circ} \\mathrm{C}+$ cyanide |\n| basal end | 3.3 | 3.8 | 3.9 |\n| apical end | 3.5 | 15.9 | 4.2 |\n\nThe best interpretation of these results is that\nA: the hormone used was auxin\nB: the hormone shows unidirectional transport\nC: active transport of the hormone occurs\nD: diffusion accounts for the transport of the hormone\nE: diffusion speeds up with rise in temperature\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe table shows the amount of plant hormone that passes through short segments of young stems when it is applied either to the end nearest to the shoot-tip or to the end furthest from the shoot tip. Three experiments were carried out under different conditions as indicated below.\n\n| Site of hormone <br> application | Amount of hormone passing through stem (arbitrary units) |  |  |\n| :--- | :---: | :---: | :---: |\n|  | At $3{ }^{\\circ} \\mathrm{C}$ | At $25^{\\circ} \\mathrm{C}$ | At $25^{\\circ} \\mathrm{C}+$ cyanide |\n| basal end | 3.3 | 3.8 | 3.9 |\n| apical end | 3.5 | 15.9 | 4.2 |\n\nThe best interpretation of these results is that\n\nA: the hormone used was auxin\nB: the hormone shows unidirectional transport\nC: active transport of the hormone occurs\nD: diffusion accounts for the transport of the hormone\nE: diffusion speeds up with rise in temperature\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Biology_4",
        "problem": "Enzyme activity usually correlates with its conformational flexibility, such that higher flexibility (lower rigidity) is usually accompanied with higher activity. Tryptophan residues which emit fluorescence are most commonly located within the nonpolar interior environment of the proteins. An excellent way to experimentally determine the exposure of tryptophan residues to solution is by measuring the quenching (decrease) of their fluorescence. The effect of mutations on the accessibility of tryptophan residues can be studied by measuring amount of fluorescence quenching by potassium iodide (KI). Iodide ions selectively quench fluorescence emitted by exposed tryptophan residues.\n\nIn the experiment whose results are presented below, fluorescence quenching on equal amounts of three mutated forms of an enzyme (mutant forms 1,2, and 3) was measured after addition of various concentrations of $\\mathrm{KI}(0-0.6 \\mathrm{M})$. Excitation and emission wavelengths used were specific for tryptophan. Quenching data were analysed in terms of the Stern-Volmer constant, $\\mathrm{K}_{\\mathrm{Sv}}$, which can be calculated from the ratio of the unquenched and the quenched fluorescence intensities, $F_{0} / F$, using the relationship\n\n$$\nF_{0} / F=1+K_{\\mathrm{SV}}[Q]\n$$\n\n$[Q]=$ the molar concentration of the quencher.\n\n[figure1]\nA: Among the three mutated proteins, protein 1 is expected to have the lowest enzymatic activity.\nB: Iodide ions have higher accessibility to the tryptophan residues of protein 2 as compared to protein 3 .\nC: Protein 3 has the highest KSV.\nD: Protein 1 does not have tryptophan residues.\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nEnzyme activity usually correlates with its conformational flexibility, such that higher flexibility (lower rigidity) is usually accompanied with higher activity. Tryptophan residues which emit fluorescence are most commonly located within the nonpolar interior environment of the proteins. An excellent way to experimentally determine the exposure of tryptophan residues to solution is by measuring the quenching (decrease) of their fluorescence. The effect of mutations on the accessibility of tryptophan residues can be studied by measuring amount of fluorescence quenching by potassium iodide (KI). Iodide ions selectively quench fluorescence emitted by exposed tryptophan residues.\n\nIn the experiment whose results are presented below, fluorescence quenching on equal amounts of three mutated forms of an enzyme (mutant forms 1,2, and 3) was measured after addition of various concentrations of $\\mathrm{KI}(0-0.6 \\mathrm{M})$. Excitation and emission wavelengths used were specific for tryptophan. Quenching data were analysed in terms of the Stern-Volmer constant, $\\mathrm{K}_{\\mathrm{Sv}}$, which can be calculated from the ratio of the unquenched and the quenched fluorescence intensities, $F_{0} / F$, using the relationship\n\n$$\nF_{0} / F=1+K_{\\mathrm{SV}}[Q]\n$$\n\n$[Q]=$ the molar concentration of the quencher.\n\n[figure1]\n\nA: Among the three mutated proteins, protein 1 is expected to have the lowest enzymatic activity.\nB: Iodide ions have higher accessibility to the tryptophan residues of protein 2 as compared to protein 3 .\nC: Protein 3 has the highest KSV.\nD: Protein 1 does not have tryptophan residues.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_3ea20e3eb34d66f4fdc3g-07.jpg?height=668&width=1048&top_left_y=1134&top_left_x=501"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Biology_1301",
        "problem": "The enzyme 'lock and key' model implies that every enzyme has a substrate that it binds to at a specific site and it then catalyses a reaction. If the model below is an enzyme, which of the substrates does it fit?\n\n[figure1]\nA: ![]([figure2])\nB: ![]([figure3])\nC: ![]([figure4])\nD: ![]([figure5])\n",
        "prompt": "You are participating in an international Biology competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe enzyme 'lock and key' model implies that every enzyme has a substrate that it binds to at a specific site and it then catalyses a reaction. If the model below is an enzyme, which of the substrates does it fit?\n\n[figure1]\n\nA: ![]([figure2])\nB: ![]([figure3])\nC: ![]([figure4])\nD: ![]([figure5])\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-06.jpg?height=331&width=534&top_left_y=337&top_left_x=658",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-06.jpg?height=169&width=686&top_left_y=818&top_left_x=134",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-06.jpg?height=163&width=671&top_left_y=821&top_left_x=1115",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-06.jpg?height=169&width=674&top_left_y=1109&top_left_x=148",
            "https://cdn.mathpix.com/cropped/2024_03_14_b9e515217a571029676eg-06.jpg?height=157&width=669&top_left_y=1115&top_left_x=1119"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Biology",
        "language": "EN",
        "modality": "multi-modal"
    }
]