📚 AS OCR Biology: Case Study Practice Drill | AS OCR 生物：案例分析实战演练
In AS OCR Biology, case study questions assess your ability to apply theoretical knowledge to real-world scenarios. This drill provides a series of typical case studies, each with a scenario, question, and guided analysis. Working through these examples will help you develop the skills needed to interpret data, draw conclusions, and link concepts across the specification.
在 AS OCR 生物中，案例分析题考查你将理论知识应用于实际情境的能力。本演练提供了一系列典型案例，每个案例包含场景、问题和引导性分析。通过这些练习，你将培养解读数据、得出结论以及跨章节联系概念的技能。
1. Enzyme Inhibition in Drug Development | 药物研发中的酶抑制
A pharmaceutical company tested a new drug that inhibits an enzyme involved in bacterial metabolism. The enzyme activity was measured at various substrate concentrations under three conditions: no inhibitor, inhibitor X, and inhibitor Y. The kinetic parameters were determined and are summarised below.
一家制药公司测试了一种抑制细菌代谢酶的新药。在无抑制剂、抑制剂 X 和抑制剂 Y 三种条件下，测定了不同底物浓度时的酶活性，其结果参数汇总如下。
| Condition | Vmax (arbitrary units) | Km (μM) |
| No inhibitor | 100 | 20 |
| Inhibitor X | 50 | 20 |
| Inhibitor Y | 100 | 50 |
Identify the type of inhibition shown by X and by Y. Explain your reasoning using the kinetic data.
请分别判断抑制剂 X 和 Y 的抑制类型，并运用动力学数据解释你的理由。
Inhibitor X is a non-competitive inhibitor. It reduces Vmax to 50 without changing the Km. This indicates that X binds to an allosteric site, not the active site. Even when substrate concentration increases, the maximum rate cannot be restored because fewer functional enzyme molecules are available. Inhibitor Y is a competitive inhibitor. Its presence increases Km to 50 while Vmax remains 100. Y competes with the substrate for the active site; at high substrate concentrations, the substrate outcompetes the inhibitor, so Vmax can still be reached.
抑制剂 X 为非竞争性抑制剂。它将 Vmax 降低至 50 而不改变 Km，说明 X 结合在酶的别构位点而非活性位点上。即便提高底物浓度，最大反应速率也无法恢复，因为可用的功能酶分子数量减少了。抑制剂 Y 为竞争性抑制剂，其存在使 Km 增至 50 而 Vmax 保持 100。Y 与底物竞争活性位点；当底物浓度足够高时，底物能竞争胜出，故 Vmax 仍可达到。
2. Membrane Permeability and Temperature | 膜透性与温度
Beetroot cylinders of equal size were rinsed and placed in distilled water at five different temperatures: 0 °C, 20 °C, 40 °C, 60 °C and 80 °C. After 30 minutes, the absorbance of the bathing solution was measured at 540 nm. The results are shown in the table.
将大小相等的甜菜根圆柱体冲洗后，分别放入 0 °C、20 °C、40 °C、60 °C 和 80 °C 的蒸馏水中。30 分钟后，在 540 nm 处测定浸泡液的吸光度。结果见下表。
| Temperature / °C | Absorbance (arbitrary units) |
| 0 | 0.05 |
| 20 | 0.06 |
| 40 | 0.15 |
| 60 | 0.60 |
| 80 | 1.20 |
Explain the trend in absorbance with increasing temperature, referring to the structure of cell membranes.
请解释吸光度随温度升高的变化趋势，并结合细胞膜结构进行说明。
At 0–20 °C, the membrane has low fluidity because phospholipids are closely packed; little pigment leaks out. As temperature rises to 40 °C, the phospholipid bilayer becomes more fluid, increasing permeability. At 60 °C and especially 80 °C, the membrane proteins denature, and the lipid bilayer loses its integrity, creating large gaps. This causes a massive loss of betalain pigment, giving a sharp rise in absorbance.
在 0–20 °C 时，膜流动性低，磷脂紧密排列，色素几乎不泄漏。当温度升至 40 °C 时，磷脂双分子层流动性增加，通透性升高。在 60 °C 尤其是 80 °C 时，膜蛋白变性，脂质双分子层结构被破坏，产生大的空隙，导致甜菜红素大量流失，吸光度急剧上升。
3. Mitotic Index and Cancer Diagnosis | 有丝分裂指数与癌症诊断
Two tissue samples, A and B, were examined under a light microscope. For each sample, 1,000 cells were counted. Sample A had 20 cells in some stage of mitosis; sample B had 150 cells in mitosis.
在光学显微镜下观察两个组织样品 A 和 B。每个样品各计数 1000 个细胞。样品 A 有 20 个细胞处于有丝分裂某个时期，样品 B 有 150 个细胞处于有丝分裂。
Calculate the mitotic index for each sample. Identify which sample is likely from a malignant tumour and justify your answer.
分别计算两份样品的有丝分裂指数，指出哪个样品可能来自恶性肿瘤，并说明理由。
Mitotic index = (number of cells in mitosis / total number of cells) × 100
Mitotic index A = (20 / 1000) × 100 = 2.0%. Mitotic index B = (150 / 1000) × 100 = 15.0%. Sample B shows a much higher mitotic index, indicating rapid, uncontrolled cell division. This is characteristic of cancerous tissue, where the normal regulation of the cell cycle has broken down. Sample A represents healthy tissue with a low rate of mitosis.
有丝分裂指数 A = (20 / 1000) × 100 = 2.0%。有丝分裂指数 B = (150 / 1000) × 100 = 15.0%。样品 B 的有丝分裂指数远高于 A，表明细胞进行着快速、不受控制的分裂，这是癌组织的特征，其细胞周期的正常调控已经失效。样品 A 代表健康组织，分裂率很低。
4. Cardiovascular Risk Factors | 心血管风险因素
A 52-year-old male patient has the following data: systolic blood pressure 150 mmHg, total blood cholesterol 6.5 mmol dm⁻³, smokes 15 cigarettes per day, and has a sedentary lifestyle. Using the risk model provided, his 10-year cardiovascular disease (CVD) risk is calculated as 28%.
一位 52 岁男性患者的数据如下：收缩压 150 mmHg，总血胆固醇 6.5 mmol dm⁻³，每天吸烟 15 支，生活方式久坐。根据所提供的风险模型，其未来十年心血管疾病 (CVD) 风险为 28%。
Explain how each of the factors contributes to the development of CVD, and suggest two lifestyle changes that could reduce his risk.
请解释各项因素如何促进 CVD 的发生，并建议两项可降低其风险的生活方式调整。
High blood pressure damages the endothelial lining of arteries, making them prone to atherosclerosis. Elevated cholesterol, particularly LDL, accumulates in the damaged wall and forms plaque, narrowing the lumen. Smoking introduces chemicals like nicotine that increase blood pressure and carbon monoxide that reduces oxygen transport; it also promotes platelet aggregation. A sedentary lifestyle correlates with obesity and higher blood pressure. To reduce risk, the patient should quit smoking and increase regular aerobic exercise, such as brisk walking, for at least 150 minutes per week.
高血压损伤动脉内皮，使其容易发生粥样硬化。升高的胆固醇，特别是低密度脂蛋白，在受损的血管壁内沉积并形成斑块，使管腔变窄。吸烟引入尼古丁等化学物质使血压升高，一氧化碳削弱氧气运输，还促进血小板聚集。久坐生活方式与肥胖及高血压相关。为降低风险，患者应戒烟并增加规律的有氧运动，例如每周至少 150 分钟的快走。
5. Antibiotic Resistance and Natural Selection | 抗生素耐药性与自然选择
In 2010, a hospital laboratory recorded the zone of inhibition for the antibiotic ciprofloxacin against E. coli as 25 mm. By 2020, the average inhibition zone had decreased to 10 mm. The antibiotic was widely prescribed over that decade.
2010 年，某医院实验室记录抗生素环丙沙星对大肠杆菌的抑菌圈直径为 25 mm。到 2020 年，平均抑菌圈降至 10 mm。该抗生素在这十年间被广泛使用。
Explain how the evolution of antibiotic resistance could account for this trend. Use the principles of natural selection in your explanation.
解释抗生素耐药性的进化如何造成这一趋势，并结合自然选择的原理进行说明。
Within the bacterial population, random mutations produced some individuals with genes conferring reduced susceptibility to ciprofloxacin. When the antibiotic was applied, susceptible bacteria died, but resistant ones survived and reproduced. The resistant allele was passed to offspring. Over generations, the frequency of the resistance allele increased, causing the average zone of inhibition to shrink. This is directional natural selection driven by the antibiotic acting as a selection pressure.
在细菌种群中，随机突变产生了对环丙沙星敏感性降低的个体。当使用抗生素时，敏感的细菌死亡，而耐药细菌存活并繁殖，将耐药等位基因传给后代。经过多个世代，耐药等位基因频率上升，导致平均抑菌圈缩小。这是由抗生素作为选择压力所驱动的定向自然选择。
6. Hormonal Control of Plant Tissue Culture | 植物组织培养的激素控制
A plant scientist cultured tobacco pith explants on media containing different ratios of auxin to cytokinin. The results are as follows:
- High auxin, low cytokinin: only roots developed.
- Low auxin, high cytokinin: only shoots developed.
- Equal moderate concentrations: callus growth, no organ differentiation.
一位植物学家在含有不同生长素与细胞分裂素比例的培养基上培养烟草髓外植体。结果如下：
- 高生长素、低细胞分裂素：只形成根。
- 低生长素、高细胞分裂素：只形成芽。
- 两者浓度中等且相等：形成愈伤组织，无器官分化。
Explain how the relative concentrations of these plant hormones influence organogenesis in tissue culture.
解释这些植物激素的相对浓度如何影响组织培养中的器官发生。
Auxin promotes root initiation, while cytokinin promotes shoot initiation. When the auxin-to-cytokinin ratio is high, genes controlling root development are preferentially expressed, leading to root formation. When the ratio is low, shoot development is favoured. At a balanced ratio, the cells divide to produce an undifferentiated callus mass. This demonstrates that morphogenesis is determined not by absolute hormone amounts but by their relative balance.
生长素促进根的发生，细胞分裂素促进芽的发生。当生长素与细胞分裂素比值较高时，控制根发育的基因优先表达，导致生根；比值较低时则有利于生芽。当比例均衡时，细胞分裂产生未分化的愈伤组织。这表明形态建成取决于激素的相对平衡，而非绝对浓度。
7. Blood Glucose Regulation in Diabetes | 糖尿病患者的血糖调节
After an overnight fast, a healthy person and a person with type 1 diabetes each consumed 75 g of glucose. Blood glucose concentration was measured every 30 minutes for 3 hours. The healthy individual’s glucose peaked at 7.8 mmol dm⁻³ at 1 h and returned to 5.0 mmol dm⁻³ by 2.5 h. The diabetic individual’s glucose reached 14.2 mmol dm⁻³ and remained above 10 mmol dm⁻³ at 3 h.
一夜禁食后，一名健康人与一名 1 型糖尿病患者各摄入 75 g 葡萄糖。每 30 分钟测一次血糖浓度，持续 3 小时。健康人的血糖在 1 小时达到峰值 7.8 mmol dm⁻³，并在 2.5 小时恢复到 5.0 mmol dm⁻³。糖尿病患者的血糖升至 14.2 mmol dm⁻³，且在 3 小时后仍高于 10 mmol dm⁻³。
Explain these results with reference to the actions of insulin and the pathophysiology of type 1 diabetes.
请结合胰岛素的作用和 1 型糖尿病的病理生理学解释这些结果。
In the healthy person, the rise in blood glucose after the meal stimulates pancreatic beta cells to release insulin. Insulin increases the permeability of muscle and liver cells to glucose and activates enzymes for glycogenesis, lowering blood glucose back to normal. In type 1 diabetes, the beta cells are destroyed by an autoimmune attack, so little or no insulin is produced. Without insulin, glucose cannot enter cells efficiently, and the liver continues to break down glycogen and produce glucose, leading to sustained hyperglycaemia.
在健康人体内，餐后血糖升高刺激胰腺 β 细胞分泌胰岛素。胰岛素增加肌肉和肝细胞对葡萄糖的通透性，并激活糖原合成的酶，从而将血糖降至正常水平。在 1 型糖尿病中，β 细胞被自身免疫攻击破坏，胰岛素几乎不分泌。缺乏胰岛素时，葡萄糖无法有效进入细胞，肝脏还持续分解糖原并生成葡萄糖，导致持续性高血糖。
8. Respiratory Quotient and Metabolic Substrates | 呼吸商与代谢底物
A student used a simple respirometer to measure the oxygen consumption and carbon dioxide production of a person at rest over 10 minutes. The results were:
- Volume of O₂ consumed = 250 cm³
- Volume of CO₂ produced = 200 cm³
一名学生使用简易呼吸计测量了一名静息者 10 分钟内的耗氧量和 CO₂ 生成量。结果如下：
- 消耗的 O₂ 体积 = 250 cm³
- 产生的 CO₂ 体积 = 200 cm³
Respiratory Quotient (RQ) = volume of CO₂ produced ÷ volume of O₂ consumed
Calculate the RQ and deduce the likely predominant metabolic substrate. Explain the reasoning.
计算呼吸商，推测主要的代谢底物，并解释推理过程。
RQ = 200 / 250 = 0.8. The RQ for exclusive carbohydrate oxidation is 1.0, for lipid oxidation approximately 0.7, and for protein around 0.8–0.9. An RQ of 0.8 suggests a mixed diet with a significant proportion of protein and some fat, or protein being the predominant fuel at the time of measurement. Since the person was at rest and had not recently eaten, the body likely catabolised a mixture of substrates, with RQ closer to 0.8 indicating less reliance on pure carbohydrates.
RQ = 200 / 250 = 0.8。纯碳水化合物氧化的 RQ 为 1.0，脂类约为 0.7，蛋白质约为 0.8–0.9。RQ 为 0.8 提示混合膳食，含有较高比例的蛋白质和一些脂肪，或者当时蛋白质为主要燃料。由于受试者静息且未近期进食，身体可能分解多种底物，RQ 接近 0.8 表明对纯碳水化合物的依赖较小。
9. Ecological Sampling with Quadrats | 使用样方进行生态取样
To estimate the abundance of daisies in a meadow, a student placed ten 0.5 m × 0.5 m quadrats randomly. The number of daisy plants per quadrat was: 8, 12, 0, 5, 15, 6, 10, 7, 9, 11. The total area of the meadow is 2000 m².
为估算一片草地中雏菊的丰度，一名学生随机放置了 10 个 0.5 m × 0.5 m 的样方。每个样方中的雏菊植株数分别为：8, 12, 0, 5, 15, 6, 10, 7, 9, 11。草地总面积为 2000 m²。
Calculate the mean density of daisies per square metre and estimate the total population size. Explain how random sampling improves the reliability of the data.
计算每平方米雏菊的平均密度，并估算总种群大小。解释随机取样如何提高数据的可靠性。
Mean number per quadrat = (8+12+0+5+15+6+10+7+9+11) / 10 = 8.3. Each quadrat area = 0.5 × 0.5 = 0.25 m². Mean density = 8.3 / 0.25 = 33.2 plants per m². Estimated total population = density × total area = 33.2 × 2000 = 66,400 plants. Random sampling avoids investigator bias and ensures that the sample is representative of the entire meadow, making the estimation more accurate and the data reproducible.
每个样方平均株数 = (8+12+0+5+15+6+10+7+9+11) / 10 = 8.3。每个样方面积 = 0.5 × 0.5 = 0.25 m²。平均密度 = 8.3 / 0.25 = 33.2 株 / m²。估计总种群数 = 密度 × 总面积 = 33.2 × 2000 = 66,400 株。随机取样避免了调查者偏差，确保样本对整片草地具有代表性，从而使估算更精确，数据可重复。
10. Pedigree Analysis of a Genetic Disorder | 遗传病系谱分析
Below is a description of a pedigree. A couple (both unaffected) have three children: two unaffected daughters and one affected son. The affected son marries an unaffected woman, and they have one affected daughter and one unaffected son. The disorder appears in both males and females and can skip generations.
以下是一个系谱的描述。一对表型正常的夫妇育有三个孩子：两个正常女儿和一个患病的儿子。患病的儿子与一名正常女性结婚，生下一个患病的女儿和一个正常的儿子。该疾病在男性和女性中均有出现，且可隔代出现。
Deduce the most likely mode of inheritance for this disorder. Support your conclusion with evidence from the pedigree, and calculate the probability that the first couple’s third pregnancy would produce another affected child.
推断该疾病最可能的遗传方式，用系谱证据支持你的结论，并计算第一对夫妇第三次怀孕生出患病孩子的概率。
The disorder is most likely autosomal recessive. Evidence: unaffected parents have an affected son, meaning both must be heterozygous carriers. The trait appears in both sexes, ruling out Y-linked. It skips generations, which is characteristic of recessive inheritance. The fact that an affected man and an unaffected woman have an affected daughter indicates the mother is a carrier, consistent with autosomal recessive rather than X-linked recessive (if X-linked, an affected father would pass the allele to all his daughters, but their daughter could be affected only if the mother is a carrier; however, the unaffected son suggests it’s not X-linked dominant). Both parents are Aa. The probability of their next child being affected (aa) is 1/4 or 25%.
该疾病最可能为常染色体隐性遗传。证据：表型正常的父母生出了患病的儿子，说明两者均为杂合携带者。疾病出现在两性中，排除了 Y 连锁。可隔代出现，符合隐性遗传特征。患病的男性与正常女性生下患病女儿，说明母亲是携带者，这符合常染色体隐性而非 X 连锁隐性（若是 X 连锁隐性，患病的父亲会将等位基因传给所有女儿，但女儿患病仅当母亲为携带者时可能；且生下的正常儿子进一步支持常染色体隐性）。父母基因型均为 Aa，他们下一个孩子患病 (aa) 的概率为 1/4，即 25%。
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