📚 Year 11 Cambridge IGCSE Biology: Case Study Practice | 剑桥11年级生物：案例分析实战演练
Case studies in Cambridge IGCSE Biology challenge you to interpret data, design experiments and explain real-world phenomena. They require more than just memorising facts — you must apply core concepts to unfamiliar scenarios and build clear, evidence-based answers. This article presents ten worked case studies that mirror exam-style questions, each demonstrating how to break down a scenario and structure a high-scoring response. Use these drills to build confidence and refine your analytical skills.
在剑桥 IGCSE 生物中，案例分析考验你解读数据、设计实验并解释真实世界现象的能力。它们不仅需要记忆事实，更要求你将核心概念应用于陌生情景，构建清晰、基于证据的答案。本文呈现十个模仿考试题型的实战案例，每个案例展示如何拆解情景并组织高分回答。借助这些练习增强自信、提升你的分析技巧。
1. Why Case Studies Matter | 为什么案例分析重要
Examiners use case studies to assess your understanding of scientific methods, data handling and application of biological principles. A strong answer always refers to specific evidence from the scenario and uses correct scientific terminology. Approach every case by first identifying the topic area, then reading the data or description carefully, noting key variables and finally linking observations to the relevant theory.
考官通过案例分析评估你对科学方法、数据处理和生物学原理应用的理解。高分答案总是引用情景中的具体证据并使用正确的科学术语。解答每个案例时，先识别主题领域，再仔细阅读数据或描述，标记关键变量，最后将观察结果与相关理论联系起来。
2. Osmosis in Potato Strips | 土豆条的渗透作用
A student placed potato strips of equal length (5.0 cm) into beakers containing sucrose solutions of different concentrations: 0.0, 0.2, 0.4 and 0.6 mol dm⁻³. After 30 minutes, they removed the strips, blotted them dry and measured the final length. The changes in length were: 0.0 mol dm⁻³: +2.5 mm; 0.2 mol dm⁻³: +1.0 mm; 0.4 mol dm⁻³: -0.5 mm; 0.6 mol dm⁻³: -3.0 mm.
一名学生将等长(5.0 cm)的土豆条放入不同浓度的蔗糖溶液中：0.0、0.2、0.4 和 0.6 mol dm⁻³。30分钟后取出土豆条，吸干水分并测量最终长度。长度变化为：0.0 mol dm⁻³: +2.5 mm；0.2 mol dm⁻³: +1.0 mm；0.4 mol dm⁻³: -0.5 mm；0.6 mol dm⁻³: -3.0 mm。
Explain why the potato strip in 0.0 mol dm⁻³ increased in length while the strip in 0.6 mol dm⁻³ decreased.
解释为什么土豆条在0.0 mol dm⁻³溶液中长度增加，而在0.6 mol dm⁻³溶液中长度减少。
In distilled water (0.0 mol dm⁻³), the water potential outside the cells is higher than inside. Water molecules move into the cells by osmosis down a water potential gradient. The cells swell and become turgid, pushing against each other and increasing the overall length of the strip. In 0.6 mol dm⁻³ sucrose, the external solution has a lower water potential than the cell contents. Water leaves the cells by osmosis; the cytoplasm shrinks and the cell membrane pulls away from the wall (plasmolysis). This loss of turgor causes the strip to decrease in length.
在蒸馏水(0.0 mol dm⁻³)中，细胞外的水势高于细胞内。水分子通过渗透作用顺水势梯度进入细胞。细胞吸水膨胀，变得硬挺，互相挤压，从而使土豆条整体长度增加。在0.6 mol dm⁻³蔗糖溶液中，外部溶液的水势低于细胞内部。水分通过渗透离开细胞；细胞质皱缩，细胞膜与细胞壁分离(质壁分离)。膨压丧失导致土豆条长度减少。
3. Enzyme Activity and Temperature | 酶活性与温度
A student investigated the effect of temperature on catalase activity using potato cubes and hydrogen peroxide. They recorded the volume of oxygen produced in 30 seconds at different temperatures: 10 °C: 3 cm³; 20 °C: 7 cm³; 30 °C: 14 cm³; 40 °C: 18 cm³; 50 °C: 10 cm³; 60 °C: 2 cm³.
一名学生用土豆块和过氧化氢研究温度对过氧化氢酶活性的影响。他们记录了不同温度下30秒内产生的氧气体积：10 °C: 3 cm³；20 °C: 7 cm³；30 °C: 14 cm³；40 °C: 18 cm³；50 °C: 10 cm³；60 °C: 2 cm³。
Describe and explain the trend shown in these results.
描述并解释这些结果所显示的趋势。
From 10 °C to 40 °C, the volume of oxygen produced increases, showing that the rate of reaction rises. This happens because higher temperatures give substrate and enzyme molecules more kinetic energy, so they move faster and collide more frequently. More successful enzyme-substrate complexes form per second. The optimum temperature appears to be around 40 °C, where the rate is highest. Above 40 °C, the oxygen volume drops sharply: at 50 °C the rate is much lower, and at 60 °C very little oxygen is produced. This decline occurs because the high temperature disrupts the hydrogen bonds and other forces maintaining the enzyme’s tertiary structure. The active site changes shape, so the substrate no longer fits — the enzyme is denatured.
从10 °C到40 °C，氧气产量逐渐增加，表明反应速率上升。这是因为较高的温度给予底物和酶分子更多动能，它们移动更快，碰撞更频繁。每秒钟有更多成功的酶-底物复合物形成。最适温度约在40 °C，此时速率最高。超过40 °C后，氧气量急剧下降：50 °C时速率大大降低，60 °C几乎不产生氧气。这种下降是因为高温破坏了维持酶三级结构的氢键和其他作用力。活性部位形状改变，底物不再适配——酶已变性。
4. Heart Rate and Exercise | 心率与运动
A student measured their resting heart rate (70 beats per minute) and then ran on the spot for five minutes. Immediately after exercise, their heart rate was 135 bpm. The student also measured a higher breathing rate.
一名学生测量了静息心率(70次/分)，然后原地跑步五分钟。运动后即刻心率为135次/分，同时呼吸频率也升高。
Explain why the heart rate increases during vigorous exercise.
解释剧烈运动时心率为何增加。
During exercise, muscle cells contract more frequently and with greater force. This increases their demand for oxygen and glucose for aerobic respiration, and they produce more carbon dioxide. The rise in CO₂ concentration is detected by chemoreceptors in the carotid arteries and aorta. These receptors send impulses to the medulla oblongata in the brain. The medulla responds by increasing the frequency of impulses sent along the sympathetic nerve to the sinoatrial node (SAN), the heart’s pacemaker. As a result, the SAN fires more often and heart rate rises. Adrenaline released from the adrenal glands also stimulates the SAN, further accelerating heart rate. The faster heartbeat delivers more oxygenated blood to the muscles and removes carbon dioxide more rapidly.
运动时，肌肉细胞更频繁且更有力地收缩。这增加了它们对有氧呼吸所需氧气和葡萄糖的需求，同时产生更多二氧化碳。颈动脉和主动脉中的化学感受器探测到CO₂浓度升高。这些感受器向脑干延髓发送神经冲动。延髓通过增加沿交感神经传至窦房结(SAN，心脏起搏器)的冲动频率作出响应。因此SAN更频繁放电，心率上升。肾上腺释放的肾上腺素也会刺激SAN，进一步加快心率。加速的心跳向肌肉输送更多含氧血液，并更快地清除二氧化碳。
5. Anaerobic Respiration in Yeast | 酵母的无氧呼吸
A student mixed yeast, sugar solution and warm water in a conical flask. A delivery tube connected the flask to a test tube of clear limewater, and a balloon was fitted tightly over the flask mouth. After 20 minutes, the limewater turned milky and the balloon had inflated.
一名学生将酵母、糖溶液和温水在锥形瓶中混合。用导管将烧瓶与装有澄清石灰水的试管连接，并在瓶口紧紧套上一个气球。20分钟后，石灰水变浑浊，气球膨胀。
Explain these observations fully.
完整解释这些观察结果。
Yeast is a single-celled fungus that can respire aerobically when oxygen is present, but in a sealed flask oxygen soon becomes limited. The yeast switches to anaerobic respiration (fermentation). In this process, glucose is broken down without oxygen to produce ethanol, carbon dioxide and a small amount of energy. The balanced word equation is: Glucose → Ethanol + Carbon dioxide (+ energy). The chemical equation can be summarised as: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. The carbon dioxide gas produced travels through the delivery tube into the limewater, reacting with calcium hydroxide to form insoluble calcium carbonate, which turns the solution milky. The same CO₂ gas accumulates inside the balloon, causing it to inflate. No oxygen is taken in from the outside, confirming anaerobic conditions.
酵母是一种单细胞真菌，有氧时可进行有氧呼吸，但在密封瓶中氧气很快耗尽。酵母转而进行无氧呼吸(发酵)。在此过程中，葡萄糖在没有氧气的情况下降解，产生乙醇、二氧化碳和少量能量。平衡文字方程为：葡萄糖 → 乙醇 + 二氧化碳(+ 能量)。化学方程式可总结为：C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂。产生的二氧化碳气体经导管进入石灰水，与氢氧化钙反应生成不溶性碳酸钙，使溶液变浑浊。同样的CO₂气体在气球内积聚，使气球膨胀。外部没有氧气进入，证实为无氧条件。
6. Genetics – A Monohybrid Cross | 遗传学 – 单因子杂交
In rabbits, the allele for brown fur (B) is dominant to the allele for white fur (b). A breeder crosses a heterozygous brown rabbit (Bb) with a homozygous white rabbit (bb).
在兔子中，棕色皮毛的等位基因(B)对白色皮毛的等位基因(b)为显性。饲养者将一只杂合棕色兔(Bb)与一只纯合白色兔(bb)杂交。
Predict the expected phenotype ratio of the offspring. Use a genetic diagram in your answer.
预测后代的预期表现型比例。请用遗传图解作答。
Parental genotypes: Bb × bb. Gametes: the heterozygous brown parent produces gametes containing either B or b; the white parent produces only gametes containing b. When we combine these randomly, the possible offspring genotypes are Bb and bb. A Punnett square shows:
亲本基因型：Bb × bb。配子：杂合棕色
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