📚 Year 12 OCR Biology: Case Study Practice | Year 12 OCR 生物：案例分析实战演练
This article presents a series of case studies designed to help Year 12 students apply their knowledge of OCR Biology A to unfamiliar scenarios. By working through these examples, you will practise analysing data, interpreting graphs, and linking biological concepts to experimental evidence. Each case study focuses on a key topic from Modules 2, 3 and 4, mirroring the style of exam questions.
本文提供一系列案例分析，旨在帮助 Year 12 学生将 OCR 生物学 A 的知识应用于陌生情境。通过演练这些示例，你将练习分析数据、解读图表，并将生物学概念与实验证据联系起来。每个案例都聚焦于模块 2、3 和 4 的核心主题，模拟考题风格。
1. Enzyme Action and Temperature: The Trypsin–Casein Reaction | 酶作用与温度：胰蛋白酶-酪蛋白反应
Scenario: A student investigated the effect of temperature on the rate of hydrolysis of casein by the enzyme trypsin. Trypsin breaks down casein, a protein in milk, causing the cloudy suspension to clear. The time taken for the suspension to become colourless was recorded at 20, 30, 40, 50 and 60 °C. The results are shown in Table 1.
情景：一名学生研究了温度对胰蛋白酶水解酪蛋白速率的影响。胰蛋白酶将牛奶中的蛋白质酪蛋白分解，使浑浊悬浮液变澄清。记录了在 20、30、40、50 和 60 °C 下悬浮液变为无色所需的时间，结果如表 1 所示。
| Temperature / °C | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|
| Time to clear / s | 180 | 90 | 45 | 60 | no clearing |
Q1: Calculate the rate of reaction at 30 °C and explain the trend observed between 20 and 40 °C.
问题 1：计算 30 °C 下的反应速率，并解释观察到的 20 至 40 °C 间的趋势。
Working: Rate = 1 / time. At 30 °C, rate = 1 ÷ 90 = 0.0111 s⁻¹. From 20 to 40 °C, the rate increases because the enzyme and substrate molecules gain more kinetic energy. This leads to more frequent successful collisions, so more enzyme–substrate complexes form per unit time. The optimum temperature is near 40 °C for this trypsin.
解答：速率 = 1/时间。30 °C 时，速率 = 1 ÷ 90 = 0.0111 s⁻¹。从 20 到 40 °C，速率上升，因为酶和底物分子获得更多动能。这导致更频繁的有效碰撞，因此单位时间内形成更多的酶-底物复合物。该胰蛋白酶的最适温度在 40 °C 附近。
Q2: At 60 °C no clearing occurred. Explain why, using your knowledge of enzyme structure.
问题 2：60 °C 时未发生澄清现象。请运用酶结构的知识解释原因。
Explanation: At 60 °C the high temperature caused the enzyme to denature. The increased kinetic energy breaks hydrogen bonds and other weak interactions maintaining the tertiary structure of trypsin. The active site changes shape, so it is no longer complementary to the casein substrate. Consequently, no enzyme–substrate complexes can form and the reaction stops.
解释：60 °C 的高温使酶变性。增加的动能破坏了维持胰蛋白酶三级结构的氢键及其他弱相互作用。活性位点形状改变，因此不再与酪蛋白底物互补。结果无法形成酶-底物复合物，反应停止。
2. Membrane Permeability: Beetroot Cell Pigment Leakage | 膜通透性：甜菜根细胞色素渗漏
Scenario: Discs of beetroot were washed and placed in water baths at 30, 40, 50, 60 and 70 °C for 10 minutes. The absorbance of the surrounding water was measured with a colorimeter using a blue-green filter. Higher absorbance indicates more betacyanin pigment leakage.
情景：将甜菜根圆片洗净后，分别置于 30、40、50、60 和 70 °C 的水浴中 10 分钟。用蓝绿色滤光片的比色计测量周围水的吸光度。吸光度越高表示渗漏的甜菜红素越多。
| Temperature / °C | 30 | 40 | 50 | 60 | 70 |
|---|---|---|---|---|---|
| Absorbance / au | 0.05 | 0.10 | 0.25 | 0.80 | 1.20 |
Q: Explain the shape of the curve relating absorbance to temperature, making specific reference to membrane structure.
问题：请解释吸光度随温度变化曲线的形状，并具体提及膜结构。
Explanation: Between 30 and 50 °C, the increase is gradual. The phospholipid bilayer gains fluidity as kinetic energy rises, creating small transient gaps through which betacyanin can leak. Above 50 °C, the sharp rise in absorbance indicates extensive membrane damage. At these temperatures, the hydrogen bonds in membrane proteins begin to break, causing carrier and channel proteins to denature. This disrupts the selective permeability of the tonoplast and plasma membrane, leading to uncontrolled pigment loss. The phospholipids may also move so violently that the bilayer disintegrates.
解释：30 至 50 °C 之间，吸光度增加缓慢。随着动能增加，磷脂双分子层流动性增强，产生小的瞬时孔隙，甜菜红素可以从中渗漏。50 °C 以上，吸光度急剧上升表明膜受到广泛损伤。在此温度下，膜蛋白中的氢键开始断裂，导致载体蛋白和通道蛋白变性。这破坏了液泡膜和细胞膜的选择透过性，导致色素不受控制地流失。磷脂分子也可能运动过于剧烈使双分子层解体。
3. Osmosis in Plant Tissue: Potato Strips in Sucrose Solutions | 植物组织渗透：土豆条在不同浓度蔗糖溶液中
Scenario: Potato strips of equal dimensions were weighed and placed in sucrose solutions of 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm⁻³. After 30 minutes, the strips were reweighed and the percentage change in mass calculated. The results were: +15%, +6%, -2%, -9%, -14%, -19%.
情景：将相同尺寸的土豆条称重后分别放入 0.0、0.2、0.4、0.6、0.8 和 1.0 mol dm⁻³ 的蔗糖溶液中。30 分钟后重新称重并计算质量变化百分比。结果为：+15%, +6%, -2%, -9%, -14%, -19%。
Q1: Plot these data and determine the water potential of potato tissue. Explain the reasoning.
问题 1：绘制数据图并确定土豆组织的水势。解释推理过程。
Approach: Plot percentage change in mass against sucrose concentration. Draw a line of best fit. The point where the line crosses zero mass change represents the solution concentration that is isotonic to the potato cells. From the graph, this occurs at approximately 0.28 mol dm⁻³. Using a calibration curve, this sucrose concentration corresponds to a water potential of around -750 kPa. At this point, the net movement of water is zero, so the water potential inside the cells equals that of the external solution.
方法：绘制质量变化百分比与蔗糖浓度的关系图，画出最佳拟合线。该线与零质量变化相交的点代表与土豆细胞等渗的溶液浓度。从图中可读出约为 0.28 mol dm⁻³。根据校准曲线，此蔗糖浓度对应的水势约为 -750 kPa。此时水的净移动为零，因此细胞内的水势等于外部溶液的水势。
Q2: Explain why at 0.8 mol dm⁻³ the mass decreased, using the term incipient plasmolysis.
问题 2：用初始质壁分离的概念解释为何在 0.8 mol dm⁻³ 时质量减少。
Explanation: At 0.8 mol dm⁻³ sucrose, the solution has a more negative water potential than the potato cell vacuole. Water moves out of the cell by osmosis, causing the protoplast to shrink and pull away from the cell wall. This is incipient plasmolysis. The loss of turgor pressure reduces the mass of the tissue. Eventually, the protoplast becomes fully plasmolysed, and the mass continues to decrease.
解释：在 0.8 mol dm⁻³ 蔗糖溶液中，溶液的水势比土豆细胞液泡的更负。水通过渗透作用从细胞中流出，导致原生质体收缩并与细胞壁分离，即为初始质壁分离。膨压的丧失降低了组织质量。最终原生质体完全质壁分离，质量继续下降。
4. Cholesterol and Cardiovascular Risk: Interpreting LDL Data | 胆固醇与心血管风险：解读低密度脂蛋白数据
Scenario: A cohort study followed 2500 adults. Blood LDL cholesterol was measured and participants were monitored for 10 years for coronary heart disease. The data table shows relative risk of CHD for increasing LDL levels, with LDL < 2.6 mmol dm⁻³ as reference group.
情景：一项队列研究追踪了 2500 名成年人。测量了其血液低密度脂蛋白胆固醇，并监测 10 年内冠心病 (CHD) 的发生情况。数据表显示了随 LDL 水平升高冠心病相对风险的变化，以 LDL < 2.6 mmol dm⁻³ 作为参照组。
| LDL level / mmol dm⁻³ | <2.6 | 2.6–3.3 | 3.4–4.1 | 4.2–4.9 | ≥5.0 |
|---|---|---|---|---|---|
| Relative risk | 1.0 | 1.5 | 2.3 | 3.8 | 5.6 |
Q: Describe the link between LDL and CHD risk, and explain the role of LDL in the development of atherosclerosis.
问题：描述 LDL 与冠心病风险之间的联系，并解释 LDL 在动脉粥样硬化发展中的作用。
Answer: The relative risk of CHD increases markedly as LDL levels rise above 3.4 mmol dm⁻³, with a more than fivefold risk in the highest group. LDL carries cholesterol in the blood. When LDL levels are chronically high, cholesterol is deposited in damaged endothelial linings of arteries. This triggers an inflammatory response where macrophages engulf the cholesterol, forming foam cells and fatty streaks. Over time, smooth muscle cells proliferate and a fibrous cap develops, creating an atheroma. The plaque narrows the artery lumen, raising blood pressure and increasing the likelihood of thrombus formation.
回答：随着 LDL 水平升至 3.4 mmol dm⁻³ 以上，冠心病相对风险显著增加，最高组风险超过五倍。LDL 在血液中运输胆固醇。当 LDL 水平长期偏高时，胆固醇沉积在受损动脉内皮衬里中。这引发炎症反应，巨噬细胞吞噬胆固醇形成泡沫细胞和脂肪条纹。随时间推移，平滑肌细胞增殖并形成纤维帽，产生粥样斑块。斑块使动脉管腔变窄，升高血压并增加血栓形成的可能性。
5. Transpiration and Water Transport: Using a Potometer | 蒸腾与水分运输：使用蒸腾计
Scenario: A potometer was set up using a leafy shoot and the rate of bubble movement was recorded under four conditions: still air, moving air (fan), humid air (plastic bag), and 30 °C bright light. Results: still air – 2.0 cm per min; fan – 5.2 cm per min; humid – 1.2 cm per min; bright light 30 °C – 7.8 cm per min.
情景：用带叶枝条设置蒸腾计，记录在四种条件下气泡移动的速率：静止空气、流动空气（风扇）、潮湿空气（塑料袋）和 30 °C 强光。结果：静止空气 – 每分钟 2.0 cm；风扇 – 5.2 cm；潮湿 – 1.2 cm；强光 30 °C – 7.8 cm。
Q: Explain how each variable affected the transpiration rate, linking to the cohesion-tension theory.
问题：结合内聚力-张力理论，解释每个变量如何影响蒸腾速率。
Answer: Moving air removes the boundary layer of saturated water vapour near the stomata, maintaining a steep water vapour concentration gradient, so transpiration accelerates. The fan increased the rate over still air. High humidity reduces the gradient, so water vapour diffuses out more slowly. Bright light triggers stomatal opening for photosynthesis, allowing more water vapour to exit. The 30 °C temperature also increases kinetic energy of water molecules, speeding evaporation. According to the cohesion-tension theory, the evaporation of water from mesophyll cell walls generates a tension (negative pressure) that pulls the water column up the xylem. The cohesive forces between water molecules and adhesion to xylem walls maintain the continuous column.
回答：流动空气带走了气孔附近的水蒸气饱和边界层，维持了陡峭的水蒸气压差，因此蒸腾加快。风扇比静止空气速率更高。高湿度降低了梯度，水蒸气向外扩散减慢。强光促使气孔张开进行光合作用，允许更多水蒸气逸出。30 °C 的温度也增加了水分子动能，加速蒸发。根据内聚力-张力理论，水从叶肉细胞壁蒸发产生张力（负压），拉动木质部中的水柱向上运动。水分子间的内聚力以及与木质部管壁的附着力维持了连续水柱。
6. Gas Exchange in Fish: Countercurrent Flow Efficiency | 鱼的换气：逆流交换效率
Scenario: A researcher measured oxygen concentration in water and in the blood along a fish gill. Data show: at point where water enters gill, water O₂ = 100%, blood O₂ = 10%; at midpoint, water O₂ = 60%, blood O₂ = 40%; at exit, water O₂ = 30%, blood O₂ = 80%.
情景：研究者沿鱼鳃测量了水中和血液中的氧浓度。数据显示：在进水端，水 O₂ = 100%，血液 O₂ = 10%；中点处水 O₂ = 60%，血液 O₂ = 40%；出水端水 O₂ = 30%，血液 O₂ = 80%。
Q: Explain how these data demonstrate the countercurrent exchange principle and state the advantage over parallel flow.
问题：解释这些数据如何体现逆流交换原理，并说明与并流相比的优势。
Answer: Throughout the lamella contact, the water always has a higher oxygen concentration than the blood flowing in the opposite direction. This maintains a diffusion gradient for oxygen along the entire length of the gill lamella. Even at the point where blood exits, its oxygen concentration (80%) is still lower than the incoming water (100%), so the gradient persists. In a parallel flow system, the gradient would diminish to zero before full saturation, limiting oxygen uptake to about 50%. Countercurrent flow thus allows fish to extract over 80% of dissolved oxygen, a crucial adaptation for efficient gas exchange in water.
回答：在整个薄片接触过程中，水的氧气浓度始终高于反向流动的血液。这维持了沿鳃薄片全长的氧气扩散梯度。即使在血液离开时，其氧浓度 (80%) 仍低于刚进入的水 (100%)，因此梯度持续存在。在并流系统中，梯度会在完全饱和前降至零，将氧气吸收限制在约 50%。逆流交换因此使鱼类能够提取超过 80% 的溶解氧，是水中高效气体交换的关键适应。
7. Immune Response: Haemagglutination Inhibition in Influenza Vaccination | 免疫反应：流感疫苗的血凝抑制试验
Scenario: During an influenza outbreak, researchers tested patient sera for antibodies against haemagglutinin (HA). A higher haemagglutination inhibition (HI) titre indicates more antibodies. Two groups were compared: recently vaccinated and unvaccinated. Mean HI titres were 1:160 and 1:20 respectively.
情景：在一次流感暴发期间，研究人员检测了患者血清中抗血凝素 (HA) 的抗体。血凝抑制 (HI) 效价越高表明抗体越多。比较了近期接种疫苗与未接种疫苗两组人群。平均 HI 效价分别为 1:160 和 1:20。
Q: Explain why vaccination leads to a higher HI titre and how this prevents influenza infection.
问题：解释为什么疫苗接种导致更高的 HI 效价，以及这如何预防流感感染。
Answer: The influenza vaccine contains haemagglutinin antigens that stimulate a primary immune response. Specific B‑cells proliferate and differentiate into plasma cells that secrete anti‑HA antibodies. Memory B‑cells are also produced. Upon exposure to the actual virus, these memory cells rapidly divide to generate large quantities of antibodies. The antibodies bind to haemagglutinin on the viral surface, preventing the virus from attaching to host cell receptors and agglutinating red blood cells. The HI titre measures the highest dilution at which antibodies can still inhibit haemagglutination; a higher titre reflects a stronger humoral response.
回答：流感疫苗含有血凝素抗原，刺激初级免疫应答。特异性 B 细胞增殖并分化为浆细胞，分泌抗血凝素抗体。记忆 B 细胞也会产生。当暴露于真实病毒时，这些记忆细胞迅速分裂产生大量抗体。抗体与病毒表面的血凝素结合，阻止病毒附着于宿主细胞受体并凝集红细胞。HI 效价测量的是抗体仍能抑制血凝反应的最高稀释度；效价越高反映了更强的体液免疫应答。
8. DNA Replication: Analysing Semi‑Conservative Replication Evidence | DNA 复制：分析半保留复制的证据
Scenario: Meselson and Stahl grew E. coli for many generations in medium containing ¹⁵N (heavy). They then transferred the bacteria to ¹⁴N (light) medium and extracted DNA after zero, one and two generations. The DNA was centrifuged. After 0 generations: one band at heavy position. After 1 generation: one band at hybrid position. After 2 generations: two bands – one at hybrid, one at light.
情景：梅塞尔森和斯塔尔将大肠杆菌在含 ¹⁵N（重）的培养基中培养多代，然后转移到 ¹⁴N（轻）培养基中，并在零代、一代和两代后提取 DNA 进行离心。0 代后：一条重链带。1 代后：一条杂合链带。2 代后：两条带——一条杂合链带，一条轻链带。
Q: Explain how these results support semi‑conservative replication and refute the conservative model.
问题：解释这些结果如何支持半保留复制并驳斥全保留模型。
Answer: In the conservative model, the original double helix would remain intact and a completely new copy made from light nitrogen would be produced. After one generation, both heavy and light bands would be visible. This was not observed; only a hybrid band was seen, showing that each DNA molecule contained one ¹⁵N strand and one ¹⁴N strand. After two generations, the appearance of both a hybrid and a light band matches the semi‑conservative prediction: half of the molecules are hybrid and half are fully light. Dispersive replication would have produced a continuous spread of intermediate bands, which was not seen.
回答：在全保留模型中，原始双螺旋保持完整，并产生一条由轻氮构成的全新拷贝。经过一代后，应同时可见重链和轻链带。实验中只观察到杂合链带，表明每个 DNA 分子含有一条 ¹⁵N 链和一条 ¹⁴N 链。两代后，同时出现杂合与轻链带，与半保留预测相符：一半分子杂合，一半全轻。分散复制会形成一系列连续的中间带，而未被观察到，由此可以排除。
9. Mitotic Index: Onion Root Tip Data | 有丝分裂指数：洋葱根尖数据
Scenario: A student prepared a stained squash of an onion root tip and observed cells under high power. In one field of view, she counted 13 cells in interphase
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