📚 Case Study Practice in Cambridge IGCSE Biology | 剑桥IGCSE生物案例分析实战演练
Case study analysis is one of the most powerful ways to master IGCSE Biology. Instead of simply recalling facts, you learn to apply concepts to unfamiliar scenarios – just like real scientists do. The questions that follow each scenario will test your ability to interpret data, identify patterns, and explain biological mechanisms using precise terminology. This article walks you through ten carefully selected case studies spanning the entire Cambridge IGCSE syllabus. Each case is presented with sample data, followed by guided analysis and model answers, so you can see exactly how to structure a high-scoring response.
案例分析是掌握 IGCSE 生物学最有效的方法之一。你不再只是死记硬背知识点，而是像真正的科学家一样，把概念应用到陌生的情境中。每个案例后面的问题会检验你解读数据、识别规律以及用准确术语解释生物学机制的能力。本文精心选取了十个贯穿剑桥 IGCSE 大纲的案例，每个案例都配有样本数据，随后是引导性分析和标准答案示范，让你清晰了解高分回答的组织方式。
1. Introduction to Case Study Analysis | 案例分析简介
In IGCSE Biology, case studies often present experimental results, patient symptoms, ecological surveys, or pedigree charts. To tackle them effectively, you must first identify the relevant topic – is it about enzymes, plant transport, immunity, or genetics? Then, carefully note the variables: the independent variable (what you change), the dependent variable (what you measure), and any controlled variables. Always back up your explanations with core principles, such as 'enzymes are specific because their active site has a complementary shape to the substrate.'
在 IGCSE 生物考试中，案例分析通常会呈现实验结果、患者症状、生态调查或家系图。要高效应对这些问题，你必须先锁定相关主题——是关于酶、植物运输、免疫还是遗传？然后，仔细辨别变量：自变量（你改变的量）、因变量（你测量的量）以及控制变量。你的解释一定要有核心原理作支撑，比如“酶具有专一性是因其活性位点与底物形状互补”。
Another key skill is data interpretation. Look for trends, anomalies, and optimum points. If a graph levels off, consider limiting factors. If a result is significantly outside the range, suggest a reason – perhaps an equipment error or a misread measurement. Finally, many case studies require you to apply knowledge to a new context, such as explaining why a person with a blocked bile duct has fatty stools. Connect the symptom to the process: bile emulsifies fats, so without it, fat digestion is impaired.
另一项关键技能是解读数据。要寻找趋势、异常值和最优点。如果曲线趋于平缓，就要想到限制因素。如果某个数据明显超出范围，就要给出原因——可能是仪器误差或读数错误。最后，许多案例分析要求你把知识迁移到新情境中，比如解释为什么胆管阻塞患者会排出脂肪便。要把症状与过程联系起来：胆汁能乳化脂肪，因此缺少胆汁，脂肪的消化就会受阻。
2. Case Study 1: Enzyme Activity and Temperature | 案例1：酶活性与温度
A student investigated the effect of temperature on the activity of catalase, an enzyme found in potato tissue. Small discs of potato were immersed in hydrogen peroxide solution at different temperatures. The volume of oxygen gas produced per minute was recorded. The results are shown in the table.
一名学生研究了温度对马铃薯组织中过氧化氢酶活性的影响。将马铃薯小圆片浸入不同温度的过氧化氢溶液中，记录每分钟产生的氧气体积。数据见下表。
| Temperature (°C) | Rate of oxygen production (cm³/min) |
|---|---|
| 10 | 2.0 |
| 20 | 5.5 |
| 30 | 10.5 |
| 40 | 12.0 |
| 50 | 7.5 |
| 60 | 0.5 |
1. Describe the trend shown in the data. 2. Explain why the rate increases from 10°C to 40°C. 3. Explain why the rate drops steeply above 50°C. 4. Predict the result at 65°C and justify your answer.
1. 描述数据呈现的趋势。2. 解释为什么速率从 10°C 升高到 40°C 时会加快。3. 解释为什么 50°C 以上速率急剧下降。4. 预测 65°C 时的结果并说明理由。
The rate of oxygen production increases as temperature rises, peaking at 40°C, then declines rapidly. Between 10°C and 40°C, the enzyme and substrate molecules have more kinetic energy, so they collide more frequently and with greater energy, forming more enzyme-substrate complexes. The reaction follows: 2H₂O₂ → 2H₂O + O₂.
氧气产生速率随温度升高而加快，在 40°C 达到峰值，随后迅速下降。在 10°C 至 40°C 之间，酶和底物分子的动能增加，它们碰撞更频繁、更剧烈，形成更多的酶-底物复合物。反应方程式为：2H₂O₂ → 2H₂O + O₂。
2H₂O₂ → 2H₂O + O₂
Above 50°C, the enzyme begins to denature. The high temperature breaks the hydrogen bonds holding the tertiary structure of the enzyme together, so the active site changes shape and can no longer bind the substrate. At 65°C, the enzyme would be fully denatured, so the rate would be zero or near zero.
超过 50°C，酶开始变性。高温破坏了维持酶三级结构的氢键，活性位点形状发生改变，无法再与底物结合。在 65°C，酶已经完全变性，因此速率会降至零或接近零。
3. Case Study 2: Limiting Factors in Photosynthesis | 案例2：光合作用的限制因素
A farmer grew tomato plants in a greenhouse and measured the rate of photosynthesis under different light intensities and carbon dioxide concentrations. The oxygen output (arbitrary units) is recorded below.
一位农民在温室中种植番茄，测量了不同光照强度和二氧化碳浓度下的光合作用速率。氧气产量（任意单位）记录如下。
| Light intensity (lux) | O₂ output at 0.04% CO₂ | O₂ output at 0.12% CO₂ |
|---|---|---|
| 0 | 0 | 0 |
| 1000 | 20 | 30 |
| 2000 | 40 | 60 |
| 3000 | 55 | 85 |
| 4000 | 60 | 100 |
| 5000 | 60 | 100 |
Identify the limiting factor at light intensities of 1000 lux and 4000 lux for the 0.04% CO₂ condition. Explain why increasing CO₂ concentration raised the oxygen output at 2000 lux. Suggest why the rate did not increase beyond 4000 lux even at the higher CO₂ level.
在 0.04% CO₂ 条件下，找出光照强度为 1000 lux 和 4000 lux 时的限制因素。解释在 2000 lux 条件下，提高 CO₂ 浓度为什么会增加氧气产量。并说明为什么在 4000 lux 以上，即使 CO₂ 浓度较高，光合速率也不再增加。
At 1000 lux and 0.04% CO₂, light intensity is the limiting factor because the rate rises as light increases. At 4000 lux, the rate plateaus at 60 units, so CO₂ concentration has become the limiting factor – the plant has enough light but not enough CO₂ to proceed faster. Increasing CO₂ to 0.12% at 2000 lux provides more raw material for the Calvin cycle, so more glucose and oxygen can be produced. Beyond 4000 lux, even with 0.12% CO₂, the rate plateaus at 100 units, which suggests another factor is limiting, such as temperature or the availability of chlorophyll.
在 0.04% CO₂、1000 lux 条件下，光照强度是限制因素，因为速率随光强增加而上升。在 4000 lux 条件下，速率稳定在 60 单位，因此 CO₂ 浓度成为限制因素——植物有足够的光，但 CO₂ 不足，无法进行更快的反应。在 2000 lux 时把 CO₂ 提高到 0.12%，为卡尔文循环提供了更多原料，因而可以产生更多的葡萄糖和氧气。而当超过 4000 lux 时，即使 CO₂ 浓度为 0.12%，速率仍稳定在 100 单位，这说明另一个因素（如温度或叶绿素数量）成为了限制因素。
4. Case Study 3: Digestion and Absorption | 案例3：消化与吸收
A patient complained of fatigue and weight loss. A stool sample revealed undigested starch grains and fats. Blood tests showed low glucose levels. The doctor suspected a blockage of the pancreatic duct. Explain how this blockage leads to the observed symptoms.
一名患者主诉疲劳和体重下降。粪便样本中发现了未经消化的淀粉颗粒和脂肪。血液检测显示血糖水平偏低。医生怀疑其胰管阻塞。请解释胰管阻塞如何导致所观察到的症状。
The pancreatic duct carries pancreatic juice containing pancreatic amylase, lipase, and trypsin. If blocked, these enzymes cannot reach the duodenum. Starch remains undigested because pancreatic amylase cannot break it into maltose and glucose. Fats are not emulsified or digested efficiently since lipase is absent and bile may also be trapped. Consequently, little glucose and fatty acids are absorbed in the ileum. This explains the low blood glucose (fatigue) and the presence of undigested starch and fats in the stool, leading to weight loss.
胰管输送含有胰淀粉酶、脂肪酶和胰蛋白酶的胰液。如果发生阻塞，这些酶就无法到达十二指肠。由于缺少胰淀粉酶，淀粉不能被分解为麦芽糖和葡萄糖。因为缺少脂肪酶，且胆汁也可能淤积，脂肪无法被充分乳化与消化。因此，在回肠中被吸收的葡萄糖和脂肪酸都很少。这就解释了为什么患者血糖偏低（感到疲劳），以及粪便中出现未消化的淀粉和脂肪，最终导致体重下降。
Additionally, the absorption of glucose in the ileum relies on active transport and facilitated diffusion. With little glucose produced, the concentration gradient is too small for efficient uptake. The villi have a rich blood supply, but without sufficient products of digestion, absorption is severely limited.
此外，回肠对葡萄糖的吸收依赖于主动运输和协助扩散。如果葡萄糖生成量极少，就无法形成足够的浓度梯度来进行高效吸收。尽管小肠绒毛血管丰富，但消化产物不足时，吸收会受到严重限制。
5. Case Study 4: Blood Circulation and Heart Rate | 案例4：血液循环与心率
A student recorded her heart rate before, during, and after a five-minute run. The data: resting: 72 bpm; just after exercise: 140 bpm; 2 min after exercise: 100 bpm; 5 min after exercise: 78 bpm. Explain these changes in terms of the need for oxygen and removal of waste products. Use the terms cardiac output, stroke volume, and anaerobic respiration.
一名学生记录了自己在五分钟跑步前、中、后的心率。数据如下：静息时 72 次/分；刚运动完 140 次/分；运动后 2 分钟 100 次/分；运动后 5 分钟 78 次/分。从需氧量和废物清除的角度解释这些变化。请使用心输出量、每搏输出量和无氧呼吸等术语。
ore carbon dioxide. The heart responds by increasing both stroke volume (volume pumped per beat) and heart rate, thereby raising cardiac output (stroke volume × heart rate). Heart rate climbs to 140 bpm to deliver oxygen and remove CO₂ quickly. After exercise, oxygen debt remains because anaerobic respiration produced lactic acid. The fast heart rate continues to supply oxygen for lactic acid breakdown in the liver. Gradually, as the oxygen debt is repaid, heart rate returns to near resting levels.
运动时，肌肉收缩更频繁，需要更多 ATP。有氧呼吸加剧，细胞需要更多氧气，并产生更多二氧化碳。心脏通过增加每搏输出量（每次搏动泵出的血液量）和心率来提升心输出量（每搏输出量 × 心率）。心率升至 140 次/分，以便快速输送氧气、运走 CO₂。运动结束后，由于无氧呼吸产生了乳酸，仍有氧债存在。心率保持较快，就是为了提供氧气让肝脏分解乳酸。随着氧债逐渐还清，心率便回落至接近静息水平。
6. Case Study 5: Breathing and Gas Exchange | 案例5：呼吸与气体交换
An investigation compared the lung function of non-smokers and long-term smokers. The table shows the average volume of air forcefully exhaled in one second (FEV₁).
一项调查对比了非吸烟者和长期吸烟者的肺功能。下表显示了用力呼气 1 秒量 (FEV₁) 的平均值。
| Group | FEV₁ (litres) |
|---|---|
| Non-smokers | 3.2 |
| Smokers (10 years) | 2.4 |
| Smokers (20 years) | 1.8 |
Explain how cigarette smoke leads to reduced FEV₁. Refer to cilia, mucus, alveoli, and emphysema.
解释香烟烟雾如何导致 FEV₁ 下降。请提及纤毛、黏液、肺泡和肺气肿。
Tar in cigarette smoke paralyzes and destroys the cilia lining the trachea and bronchi. Without cilia sweeping mucus upward, mucus accumulates, narrowing the airways and causing smoker's cough. This increases airflow resistance, so less air can be forcefully exhaled in one second. Over time, the smoke also damages the walls between alveoli, breaking them down and reducing the surface area for gas exchange – a condition called emphysema. The loss of elastic tissue makes it harder to expel air, further decreasing FEV₁. The data clearly show that the longer the smoking history, the greater the loss of lung function.
香烟烟雾中的焦油会麻痹并破坏气管和支气管内壁的纤毛。没有纤毛将黏液向上推送，黏液就会积聚，使气道变窄，并引发吸烟者咳嗽。这会增加气流阻力，导致 1 秒内能被用力呼出的气体量减少。久而久之，烟雾还会破坏肺泡壁，使肺泡破裂，减少气体交换的表面积——这就是肺气肿。弹性组织的丧失令呼气更加困难，进一步降低了 FEV₁。数据清楚地表明，吸烟史越长，肺功能丧失越严重。
7. Case Study 6: Plant Transport and Transpiration | 案例6：植物运输与蒸腾作用
taken for the bubble to move a fixed distance was recorded. (Shorter time = higher rate).
一名学生使用气泡蒸腾计测量了一条带叶枝条在不同条件下的蒸腾速率。记录气泡移动固定距离所用的时间（时间越短，速率越高）。
| Condition | Time for bubble to move (s) |
|---|---|
| Still air, room humidity | 45 |
| Fan blowing air | 20 |
| Bright light | 25 |
| High humidity (bagged leaf) | 80 |
Explain the effect of each condition on the transpiration rate, referring to the mechanisms of water movement in the xylem and stomatal opening.
解释每种条件对蒸腾速率的影响，同时要提到水分在水质部中的运输机制以及气孔的开闭。
The fan increases air movement, which sweeps away water vapour from around the stomata, maintaining a steep concentration gradient of water vapour between the leaf and the air. This increases the rate of evaporation, so transpiration is faster (20 s). Bright light causes stomata to open wider for carbon dioxide uptake, which also allows more water vapour to escape; hence the rate increases. High humidity (bagged leaf) traps water vapour, reducing the concentration gradient, so transpiration slows down dramatically (80 s). Water is pulled up the xylem in a continuous stream under tension – the transpiration stream – driven by evaporation at the leaves.
风扇加快空气流动，带走气孔周围的水蒸气，使叶片与空气之间维持较大的水蒸气浓度梯度。这加快了蒸发速率，因此蒸腾作用更快（20 秒）。明亮的光照使气孔张开更大以吸收二氧化碳，这同时也让更多水蒸气散失，因此速率提高。高湿度（套袋的叶片）会困住水蒸气，降低浓度梯度，于是蒸腾速度显著减慢（80 秒）。水分在张力作用下沿着木质部连续向上运输——这就是蒸腾流，其动力来自叶片的蒸发作用。
8. Case Study 7: Ecology and Food Chains | 案例7：生态学与食物链
A pond ecosystem contains algae, Daphnia (water fleas), small fish, and kingfishers. Scientists estimated the biomass at each trophic level: algae: 1000 g/m²; Daphnia: 150 g/m²; small fish: 20 g/m²; kingfisher: 2 g/m². Only about 10% of the energy is transferred from one trophic level to the next.
一个池塘生态系统包含藻类、水蚤、小鱼和翠鸟。科学家估算了每个营养级的生物量：藻类 1000 g/m²，水蚤 150 g/m²，小鱼 20 g/m²，翠鸟 2 g/m²。能量从上一营养级传递到下一营养级的效率大约只有 10%。
Calculate the approximate energy transfer efficiency between algae and Daphnia, and suggest two reasons why so much energy is lost. Also, predict what would happen to the Daphnia population if the small fish were removed, and comment on the stability of the ecosystem.
计算藻类与水蚤之间的能量传递效率，并给出能量大量流失的两个原因。同时预测如果小鱼被移除，水蚤的数量会发生什么变化，并评述生态系统的稳定性。
Energy transfer efficiency = (150/1000) × 100% = 15% ion, released as heat, or lost in egested faeces. If small fish are removed, the Daphnia would experience less predation, so their population would increase. This could lead to overgrazing of algae, eventually causing the Daphnia population to crash as food runs out. A four-level food chain is often less stable, and removal of one species can cause dramatic fluctuations.
能量传递效率 = (150/1000) × 100% = 15%。能量损失的原因包括：并非所有藻类都被吃掉，一部分未被取食而死亡；此外，被摄入的能量中有很大一部分用于呼吸作用，以热能形式散失，或在排遗中损失。如果移走小鱼，水蚤被捕食的压力就会减小，其数量将上升。这可能导致藻类被过度啃食，最终当食物耗尽时，水蚤的数量又会暴跌。一条包含四个营养级的食物链通常不太稳定，移除其中某一物种就可能引起剧烈的波动。
9. Case Study 8: Genetic Disorders and Pedigree Analysis | 案例8：遗传病与家系图分析
Cystic fibrosis (CF) is a recessive genetic disorder. The pedigree below shows a family where some individuals are affected. (Not drawn: unaffected parents of affected child must both be carriers). A couple, both carriers (Ff), plan to have a child.
囊性纤维化 (CF) 是一种隐性遗传病。下面的家系图中显示一个家族中有部分成员患病。（未画出：患病孩子的父母双方必定都是携带者。）一对均为携带者 (Ff) 的夫妇计划要一个孩子。
eir child will be a carrier but not have the disease. Explain what a carrier is and why the disease only manifests when two recessive alleles are present.
画出 Ff × Ff 杂交的庞纳特方格，并指出他们的孩子是携带者而不患病的概率。解释什么是携带者，以及为什么只有在两个隐性等位基因同时存在时疾病才会表现出来。
rier), ff (affected). Probability of being a carrier without the disease = 2/4 = 1/2 or 50% ominant allele produces enough functional CFTR protein. The disease only appears when an individual inherits two copies of the recessive allele (ff), so no functional protein is made, leading to thick mucus in lungs and pancreas.
庞纳特方格的结果为：FF（正常），Ff（携带者），Ff（携带者），ff（患者）。孩子是携带者且不患病的概率 = 2/4 = 1/2，即 50%。携带者是指同时拥有一个显性正常等位基因和一个隐性缺陷等位基因 (Ff) 的个体。他们不表现出症状，因为正常的显性等位基因足以产生足够的功能性 CFTR 蛋白。只有当个体从父母双方各继承一个隐性等位基因 (ff) 时，才会因无法合成正常的蛋白质而发病，导致
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