📚 Case Study Practice Exercises for Year 10 AQA Biology | Year 10 AQA 生物：案例分析实战演练
This article provides a series of case study exercises designed to develop your data analysis and scientific thinking skills for the AQA GCSE Biology course. Each case links to a required practical or key concept and includes data or scenarios, followed by guided questions and model answers to help you practise interpreting evidence and drawing conclusions. Strong exam performance depends on your ability to apply knowledge to unfamiliar contexts, and working through these examples will build your confidence in handling graphs, tables, calculations and explain-style questions.
本文提供一系列案例分析练习，旨在培养你在AQA GCSE生物课程中的数据分析与科学思维能力。每个案例都关联一项必做实验或重要概念，包含数据或情景，随后有引导性问题和参考答案，帮助你练习解释证据、得出结论。出色的考试成绩有赖于你将知识应用于陌生情境的能力，完成这些习题将增强你处理图表、表格、计算和解释型问题的信心。
1. Enzyme Activity and Temperature | 酶活性与温度
A student investigated how temperature affects amylase activity. Starch solution and amylase were mixed at different temperatures, and the time taken for the starch to be fully digested (a negative iodine test) was recorded. The results are shown below.
一名学生研究了温度如何影响淀粉酶活性。在不同温度下将淀粉溶液与淀粉酶混合，记录淀粉完全消化的时间（碘液测试呈阴性）。结果如下表所示。
| Temperature / °C | Time for starch to disappear / s |
|---|---|
| 10 | 125 |
| 20 | 65 |
| 30 | 32 |
| 40 | 22 |
| 50 | 45 |
| 60 | 150 |
Q1. Describe the trend in the time taken from 10 °C to 40 °C.
问题1. 描述从10 °C到40 °C所需时间的变化趋势。
Q2. Explain why the time increases significantly at 60 °C.
问题2. 解释为什么60 °C时所需时间大幅增加。
A1. As temperature rises from 10 °C to 40 °C, the time decreases from 125 s to 22 s, meaning the rate of reaction increases. This is because enzyme and substrate molecules gain kinetic energy, move faster and collide more frequently. More collisions have the activation energy required for the reaction, so the starch is broken down more quickly.
答案1. 随着温度从10 °C升至40 °C，时间从125秒缩短至22秒，即反应速率加快。这是因为酶和底物分子获得动能，运动加快，碰撞更频繁。更多碰撞具有反应所需的活化能，因此淀粉分解更快。
A2. At 60 °C the time jumps to 150 s because the high temperature causes the enzyme's active site to change shape irreversibly. The enzyme has been denatured, so the starch substrate can no longer bind and the reaction almost stops. Some activity may remain briefly at slightly lower temperatures, but above the optimum the rate rapidly falls.
答案2. 在60 °C时时间猛增至150秒，因为高温不可逆地改变了酶活性部位的形状。酶已变性，淀粉底物无法再结合，反应近乎停止。在稍低温度下可能仍有短暂活性，但超过最适温度后速率迅速下降。
2. Enzyme Activity and pH | 酶活性与pH
Catalase, extracted from potato tissue, was tested at different pH values. The volume of oxygen gas produced per minute was measured. The data are given in the table.
将从土豆组织中提取的过氧化氢酶在不同pH值下测试，测量每分钟产生的氧气体积。数据见表。
| pH | Oxygen produced / cm³ min⁻¹ |
|---|---|
| 4.0 | 10 |
| 5.0 | 22 |
| 6.0 | 30 |
| 7.0 | 32 |
| 8.0 | 27 |
| 9.0 | 8 |
Q1. Estimate the optimum pH for catalase and justify your answer.
问题1. 估计过氧化氢酶的最适pH，并给出理由。
Q2. Explain the decrease in oxygen production at pH 4 and pH 9.
问题2. 解释pH 4和pH 9时氧气产量下降的原因。
A1. The optimum pH is around 7.0, because this value gives the highest rate of oxygen production (32 cm³ min⁻¹). At this pH the shape of the enzyme's active site is most complementary to the substrate, allowing the fastest formation of enzyme-substrate complexes.
答案1. 最适pH约为7.0，因为该值下氧气产率最高（32 cm³ min⁻¹）。在此pH下，酶活性部位的形状与底物最匹配，能最快形成酶-底物复合物。
A2. At extreme pH values, hydrogen and ionic bonds that maintain the enzyme's tertiary structure are disrupted. The active site changes shape and can no longer bind the substrate effectively, so the enzyme is denatured. At pH 4 and pH 9 the conditions are too far from the optimum and catalase activity drops sharply.
答案2. 在极端pH值下，维持酶三级结构的氢键和离子键被破坏。活性部位形状改变，无法有效结合底物，因此酶变性。pH 4和pH 9条件下距离最适值太远，过氧化氢酶活性急剧下降。
3. Photosynthesis and Light Intensity | 光合作用与光照强度
A student placed a lamp at different distances from a beaker of pondweed and counted the number of oxygen bubbles produced per minute. The room temperature and the concentration of sodium hydrogencarbonate solution were kept constant.
一名学生将灯放在距离水草烧杯不同距离处，记录每分钟产生的氧气气泡数。室温与碳酸氢钠溶液浓度保持不变。
| Distance of lamp / cm | Bubbles per minute |
|---|---|
| 10 | 48 |
| 20 | 28 |
| 30 | 15 |
| 40 | 9 |
Q1. Describe the relationship between lamp distance and rate of photosynthesis.
问题1. 描述灯距与光合作用速率的关系。
Q2. Explain why the bubble count falls as the lamp is moved further away, referring to the limiting factor.
问题2. 解释为何拉远灯距后气泡数减少，请提及限制因素。
Q3. Suggest why the student used sodium hydrogencarbonate solution.
问题3. 说明学生为何使用碳酸氢钠溶液。
A1. As the lamp distance increases, the number of bubbles per minute decreases. For example, at 10 cm the rate is 48 bubbles min⁻¹, falling to only 9 bubbles min⁻¹ at 40 cm. This inverse relationship shows that photosynthesis slows when light intensity drops.
答案1. 随着灯距增加，每分钟气泡数减少。例如10 cm处速率为48个气泡/分钟，40 cm处降至仅9个气泡/分钟。这种反比关系表明，当光强度降低时光合作用减慢。
A2. Moving the lamp further away reduces the light intensity reaching the pondweed. Light intensity is inversely proportional to the square of the distance, so even small increases in distance cause a large drop. When light intensity is low, it becomes the limiting factor for photosynthesis, so the rate decreases despite other factors being optimal.
答案2. 将灯移远减少了到达水草的光强度。光强度与距离的平方成反比，因此即使距离小幅增加也会导致光强大幅下降。当光强度较低时，它成为光合作用的限制因素，因此即使其他条件最佳，速率也会下降。
A3. Sodium hydrogencarbonate solution provides a source of dissolved carbon dioxide for the pondweed. Without added CO₂, photosynthesis might be limited by the natural concentration of CO₂ in the water, which would prevent observing the effect of light alone.
答案3. 碳酸氢钠溶液为水草提供溶解的二氧化碳来源。若不添加CO₂，光合作用可能受水中自然CO₂浓度限制，从而无法单独观察光的影响。
4. Osmosis in Potato Cylinders | 土豆条的渗透作用
Potato cylinders were weighed, placed in different concentrations of sucrose solution for 24 hours and then re-weighed. The percentage change in mass was calculated.
将土豆条称重，放入不同浓度的蔗糖溶液中浸泡24小时后再次称重，计算质量变化百分比。
| Sucrose concentration / mol dm⁻³ | % Change in mass |
|---|
