📚 Year 11 OCR Biology: Case Study Practical Drills | 11年级 OCR 生物：案例分析实战演练
Mastering case study questions is essential for success in the OCR GCSE Biology exam. These questions test your ability to apply knowledge, interpret data, and think critically about real-world biological scenarios. This article provides 10 hands-on case study drills covering key topics from the Year 11 syllabus, each designed to build your confidence and analytical skills.
掌握案例分析题是 OCR GCSE 生物考试成功的关键。这些问题考查你运用知识、解释数据以及批判性思考现实生物情境的能力。本文提供 10 个实践案例演练，涵盖 11 年级大纲的核心主题，每个案例都旨在培养你的信心和分析能力。
1. Case Study 1: Antibiotic Resistance in Bacteria | 案例1：细菌的抗生素耐药性
A hospital monitored the percentage of Klebsiella pneumoniae isolates resistant to three antibiotics (X, Y, Z) from 2019 to 2023. The data reveal a worrying trend for antibiotic X, demanding an evolutionary explanation.
某医院监测了 2019 至 2023 年间肺炎克雷伯菌分离株对三种抗生素（X、Y、Z）的耐药百分比。数据显示抗生素 X 出现令人担忧的趋势，需要从进化角度进行解释。
The table below summarises the resistance data collected annually.
下表总结了每年收集的耐药数据。
| Year | % Resistant to X | % Resistant to Y | % Resistant to Z |
|---|---|---|---|
| 2019 | 5 | 2 | 1 |
| 2020 | 12 | 3 | 2 |
| 2021 | 23 | 3 | 2 |
| 2022 | 35 | 4 | 3 |
| 2023 | 42 | 4 | 3 |
Resistance to antibiotic X increased dramatically from 5% to 42% over five years. This can be explained by natural selection. Random mutations produced some bacteria with alleles that allowed them to survive in the presence of antibiotic X. When the antibiotic was administered, susceptible bacteria died, whereas resistant individuals survived and reproduced. These resistant bacteria passed on the advantageous allele to their offspring. Over many generations, the frequency of the resistance allele rose, causing the population to become largely resistant.
抗生素 X 的耐药性在五年内从 5% 急剧上升到 42%。这可用自然选择来解释。随机突变产生了一些具有等位基因的细菌，这些等位基因使它们能够在抗生素 X 存在下存活。当使用该抗生素时，敏感细菌死亡，而耐药个体存活并繁殖。这些耐药细菌将有利的等位基因传递给后代。经过多代，耐药等位基因的频率上升，导致种群变得大部分耐药。
In contrast, resistance to antibiotics Y and Z remained low because these drugs were used far less frequently, so the selection pressure was weak. This case highlights why hospitals must use antibiotics carefully and rotate treatments to slow the spread of resistance.
相比之下，对 Y 和 Z 抗生素的耐药性保持较低，因为这些药物使用频率低得多，因此选择压力较弱。本案例突显了医院为何必须谨慎使用抗生素并轮换治疗，以减缓耐药性的传播。
2. Case Study 2: Enzyme Activity and pH | 案例2：酶活性与pH
A student investigated the effect of pH on the activity of an enzyme found in the small intestine. The enzyme was incubated with its substrate at 37 °C across a range of pH values, and the initial rate of reaction was measured. Results are shown in the table.
一名学生研究了 pH 对一种存在于小肠的酶活性的影响。酶与底物在 37 °C、不同 pH 值下孵育，并测定了初始反应速率。结果如表所示。
| pH | Relative activity (%) |
|---|---|
| 2 | 10 |
| 4 | 30 |
| 6 | 85 |
| 7 | 100 |
| 8 | 90 |
| 10 | 40 |
| 12 | 5 |
The optimum pH for this enzyme is around 7, where activity reaches 100%. At pH 6 and 8, activity remains high because the active site is still complementary to the substrate. However, at pH values far from the optimum, activity falls sharply. This occurs because excessive H⁺ or OH⁻ ions disrupt the ionic and hydrogen bonds that maintain the enzyme's tertiary structure. The active site loses its precise shape and can no longer bind the substrate effectively - a process called denaturation.
该酶的最适 pH 大约为 7，此时活性达到 100%。在 pH 6 和 8 时，活性仍然较高，因为活性位点仍与底物互补。然而，当 pH 远离最适值时，活性急剧下降。这是因为过多的 H⁺ 或 OH⁻ 离子破坏了维持酶三级结构的离子键和氢键。活性位点失去其精确形状，无法再有效结合底物——这一过程称为变性。
At the extremes of pH 2 and 12, denaturation is largely irreversible, explaining the very low activities observed. In the human body, this intestinal enzyme works best in the neutral environment maintained by bile and pancreatic secretions.
在 pH 2 和 12 的极端条件下，变性在很大程度上是不可逆的，这解释了观察到的非常低的活性。在人体内，这种肠道酶在胆汁和胰腺分泌物维持的中性环境中效果最佳。
3. Case Study 3: Investigating Photosynthesis Using Pondweed | 案例3：利用金鱼藻探究光合作用
A common class practical uses pondweed (Elodea) to measure the rate of photosynthesis. A lamp was placed at different distances from a beaker containing pondweed, and the number of oxygen bubbles produced per minute was counted. The results are recorded below.
一个常见的课堂实验使用金鱼藻（伊乐藻）测量光合作用速率。将一盏灯放在距离装有金鱼藻的烧杯不同距离处，记录每分钟产生的氧气气泡数。结果记录如下。
| Distance from lamp (cm) | Bubbles per minute |
|---|---|
| 10 | 45 |
| 20 | 42 |
| 30 | 35 |
| 40 | 22 |
| 50 | 10 |
As the distance increases, light intensity decreases according to the inverse square law. The bubble count drops, indicating a lower rate of photosynthesis. Between 10 cm and 20 cm, the rate barely changes, suggesting that another factor such as carbon dioxide concentration or temperature is limiting the rate. Beyond 30 cm, light intensity becomes the limiting factor, causing the rate to fall significantly.
随着距离增加，光照强度按照平方反比定律降低。气泡数下降，表明光合作用速率降低。在 10 cm 和 20 cm 之间，速率几乎不变，这表明另一个因素如二氧化碳浓度或温度正在限制速率。超过 30 cm 后，光照强度成为限制因素，导致速率显著下降。
The overall equation for photosynthesis can be written as:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
This investigation demonstrates that oxygen production is a reliable indicator of the light-dependent reactions and that growers must balance multiple factors to optimise plant growth.
光合作用的总方程式可写为：
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
本实验证明氧气产量是光反应可靠的指标，种植者必须平衡多个因素以优化植物生长。
4. Case Study 4: The Carbon Cycle and Deforestation | 案例4：碳循环与森林砍伐
Deforestation of tropical rainforests releases large quantities of carbon dioxide into the atmosphere and disrupts the carbon cycle. The table below presents data on forest cover and atmospheric CO₂ concentration over three decades.
热带雨林的砍伐向大气释放大量二氧化碳，并破坏碳循环。下表呈现了过去三十年森林覆盖面积和大气 CO₂ 浓度的数据。
| Year | Forest area (million ha) | Atmospheric CO₂ (ppm) |
|---|---|---|
| 1990 | 4000 | 350 |
| 2000 | 3800 | 370 |
| 2010 | 3500 | 390 |
| 2020 | 3100 | 415 |
As forest area declined by 900 million hectares, atmospheric CO₂ rose from 350 ppm to 415 ppm. Trees act as carbon sinks, removing CO₂ during photosynthesis. When forests are cut down and burned, stored carbon is oxidised and released directly. Moreover, fewer trees remain to absorb CO₂ produced by respiration and combustion of fossil fuels. The result is a net increase in atmospheric CO₂, which intensifies the greenhouse effect and contributes to global warming.
随着森林面积缩减了 9 亿公顷，大气 CO₂ 从 350 ppm 升至 415 ppm。树木作为碳汇，在光合作用中移除 CO₂。当森林被砍伐和焚烧时，储存的碳被氧化并直接释放。此外，剩余树木减少，吸收由呼吸作用和化石燃料燃烧产生的 CO₂ 的能力降低。其结果是大气 CO₂ 净增加，从而加剧温室效应并导致全球变暖。
Restoring forests and protecting existing ones are vital strategies for rebalancing the carbon cycle and mitigating climate change.
恢复森林并保护现有森林是重新平衡碳循环和缓解气候变化的重要策略。
5. Case Study 5: Diabetes and Insulin Regulation | 案例5：糖尿病与胰岛素调节
A 14-year-old patient complained of thirst, frequent urination, and fatigue. An oral glucose tolerance test was performed: after fasting, the patient drank a glucose solution, and blood glucose concentrations were measured over time. The results, alongside a healthy control, are tabulated.
一名 14 岁患者诉说口渴、尿频和疲劳。进行了口服葡萄糖耐量测试：禁食后，患者饮用葡萄糖溶液，并随时间测量血糖浓度。结果与健康对照一同列表。
| Time (min) | Control glucose (mmol/L) | Patient glucose (mmol/L) |
|---|---|---|
| 0 | 5.0 | 8.0 |
