📚 Year 10 CAIE Biology: Case Study Practice | 案例分析实战演练
Case studies are a vital part of learning biology because they show how theoretical knowledge is applied to real-world situations. In the CAIE Year 10 syllabus, you often encounter data-based questions, experimental scenarios, and health-related problems that require you to analyse, interpret, and draw conclusions. This revision guide walks you through ten practical case studies, each linking to a core topic. By working through these examples, you will sharpen your skills in identifying variables, explaining trends, and using biological principles to solve problems.
案例分析是学习生物学的重要环节，因为它展示了理论知识如何应用于实际情境。在CAIE十年级考纲中，你经常会遇到基于数据的题目、实验情景和与健康相关的问题，需要你进行分析、解读并得出结论。这份复习指南将带你演练十个与核心主题相关的实践案例。通过逐一攻破这些案例，你将提高识别变量、解释变化趋势以及运用生物学原理解题的能力。
1. Introduction to Case Study Skills | 案例分析技能入门
Before diving into specific scenarios, it is helpful to review the general approach to any case study. Always begin by reading the background information carefully and identifying the independent variable (what you change), the dependent variable (what you measure), and the control variables (what you keep the same). Look for trends or patterns in data tables or graphs. When explaining results, link back to key concepts such as enzyme activity, diffusion, or photosynthesis. Use scientific vocabulary accurately and remember to quote data where instructed.
在深入具体案例之前，先回顾处理任何案例分析的一般方法会很有帮助。首先要仔细阅读背景信息，确定自变量（你改变的变量）、因变量（你测量的变量）和控制变量（你保持不变的条件）。然后寻找数据表格或图表中的趋势或规律。在解释结果时，要联系酶活性、扩散或光合作用等关键概念。准确使用科学术语，并记得在要求时引用数据。
2. Case Study 1: Enzyme Activity and Temperature | 案例一：酶活性与温度
Researchers investigated the effect of temperature on the rate of reaction of catalase, an enzyme found in liver cells. They placed equal masses of liver in test tubes with 10 cm³ of hydrogen peroxide solution at different temperatures. The volume of oxygen gas produced in 60 seconds was recorded. The results are shown in the table below.
研究人员探究了温度对过氧化氢酶（一种存在于肝细胞中的酶）反应速率的影响。他们将等质量的肝脏放入不同温度的试管中，每个试管含有10 cm³过氧化氢溶液。记录了60秒内产生的氧气体积。结果如下表所示。
| Temperature / °C | Volume of O₂ produced / cm³ |
|---|---|
| 5 | 4 |
| 15 | 12 |
| 25 | 24 |
| 35 | 32 |
| 45 | 28 |
| 55 | 8 |
The volume of oxygen produced increased as the temperature rose from 5 °C to 35 °C. This is because the enzyme and substrate molecules gained kinetic energy, leading to more frequent successful collisions and a higher rate of reaction. The optimum temperature appeared to be around 35 °C.
随着温度从5 °C上升到35 °C，产生的氧气体积增加。这是因为酶和底物分子获得了动能，导致成功碰撞的频率增加，反应速率提高。最适温度大约在35 °C左右。
Above 35 °C, the rate of reaction decreased sharply. This is due to the heat breaking the hydrogen bonds and other forces that maintain the active site’s specific shape. The enzyme’s active site denatures. The substrate can no longer fit into the active site, so the enzyme–substrate complex cannot form, and the reaction slows down. This denaturation is permanent.
温度超过35 °C后，反应速率急剧下降。这是因为热量破坏了维持活性位点特定形状的氢键和其他力的作用。酶的活性位点发生变性。底物不再能进入活性位点，因此酶–底物复合物无法形成，反应减缓。这种变性是永久性的。
3. Case Study 2: Digestive Enzyme Efficiency | 案例二：消化酶效率
A student wanted to model the digestion of protein in the stomach. She placed cubes of boiled egg white (protein) in test tubes containing pepsin, hydrochloric acid, and water at pH 2.0. Another set of test tubes contained pepsin and water at pH 7.0 but no acid. The time taken for the egg white to disappear was measured at different temperatures.
一位学生想模拟蛋白质在胃中的消化过程。她将煮熟的蛋白块放入含有胃蛋白酶、盐酸和水（pH 2.0）的试管中。另一组试管含有胃蛋白酶和水（pH 7.0）但不含酸。在不同温度下测量蛋白块消失所需的时间。
The results showed that at pH 2.0, the egg white was digested much faster, especially at body temperature (37 °C). At pH 7.0, very little digestion occurred. This confirms that pepsin works optimally in acidic conditions similar to the stomach. Without hydrochloric acid, the enzyme’s active site is not in the correct conformation to bind to the protein.
结果显示，在pH 2.0时蛋白消化速度明显更快，特别是在体温（37 °C）下。在pH 7.0时几乎没有发生消化。这证实了胃蛋白酶在类似胃的酸性条件下发挥最佳活性。没有盐酸，酶的活性位点无法保持与蛋白质结合的正确构象。
This case highlights why stomach ulcers can be made worse by a reduction in stomach acid, because the enzyme pepsin becomes less effective, leading to poor protein digestion.
这个案例说明了为什么胃酸减少会使胃溃疡恶化，因为胃蛋白酶效率降低，导致蛋白质消化不良。
4. Case Study 3: Osmosis and Plant Cells | 案例三：渗透作用与植物细胞
A class experiment studied the effect of sucrose concentration on potato strip mass. Potato strips of identical length and diameter were weighed, then placed in solutions of 0.0, 0.2, 0.4, 0.6, 0.8, and 1.0 mol/dm³ sucrose for 30 minutes. After blotting dry, the strips were weighed again and the percentage change in mass calculated. A negative percentage means the potato lost mass; a positive percentage means it gained mass.
班里进行了一项实验，研究蔗糖浓度对土豆条质量的影响。将长度和直径相同的土豆条称重，然后分别放入0.0、0.2、0.4、0.6、0.8和1.0 mol/dm³的蔗糖溶液中浸泡30分钟。吸干水分后再次称重，并计算质量变化百分比。负百分比表示土豆质量减少；正百分比表示质量增加。
In the 0.0 mol/dm³ solution (distilled water), the potato strips gained mass because the water potential inside the potato cells was lower (more negative) than the surrounding solution. Water moved in by osmosis, causing the cells to become turgid. In very concentrated solutions, the potato lost mass because water moved out of the cells into the solution where the water potential was more negative. The cells became plasmolysed as the cell membrane pulled away from the cell wall.
在0.0 mol/dm³溶液（蒸馏水）中，土豆条质量增加，因为土豆细胞内部的水势低于周围溶液。水通过渗透作用进入，使细胞变得坚挺。在极浓的溶液中，土豆质量减少，因为水从细胞移出至水势更低的溶液。随着细胞膜与细胞壁分离，细胞发生质壁分离。
The concentration at which no net change in mass occurs can be taken as the water potential of the potato cells. Understanding this principle is essential for explaining how plants absorb water from the soil and why over-fertilising can cause wilting.
质量无净变化的浓度点可视为土豆细胞的水势。理解这一原理对于解释植物如何从土壤中吸收水分以及过度施肥为何导致萎蔫至关重要。
5. Case Study 4: Photosynthesis Rate and Light Intensity | 案例四：光合作用速率与光照强度
A student used a piece of pondweed to investigate the rate of photosynthesis. The plant was placed in a beaker of water, and a lamp was positioned at distances of 10, 20, 30, 40, and 50 cm from the beaker. The number of bubbles released per minute was counted. Sodium hydrogencarbonate powder was added to the water to ensure a constant supply of CO₂.
一位学生使用水草来研究光合作用速率。将植物放入烧杯水中，并将灯放置在距烧杯10、20、30、40和50 cm处，计数每分钟释放的气泡数量。水中加入碳酸氢钠粉末以确保稳定的CO₂供应。
As the lamp moved closer, the number of bubbles increased, indicating a higher rate of photosynthesis. Light is needed to provide energy for the light-dependent reactions. While light intensity increases, photosynthesis rate rises until another factor, such as CO₂ concentration or temperature, becomes limiting. At a very close distance, if bubble count plateaued, it would indicate a different limiting factor.
随着灯靠近，气泡数量增加，表明光合作用速率提高。光为光反应提供能量。随着光强增加，光合速率上升，直到另一个因素，如CO₂浓度或温度，成为限制因素。如果灯距很近时气泡数不再增加，则表明存在其他限制因素。
This experiment also illustrates a control measure: sodium hydrogencarbonate keeps carbon dioxide concentration in excess, so it is not limiting. The results can be used to plot a graph and determine the point of light saturation.
此实验还体现了一个控制措施：碳酸氢钠使二氧化碳浓度保持在过量状态，因此它不是限制因素。可以依据结果绘制图表并找出光饱和点。
6. Case Study 5: Starch Test and Leaf Experiment | 案例五：淀粉实验与叶片检测
A plant was destarched by leaving it in the dark for 48 hours. A leaf was then partly covered with aluminium foil, leaving some parts exposed. The plant was placed in sunlight for 6 hours. The leaf was then removed, boiled in water, then boiled in ethanol to decolourise it, and finally rinsed and spread onto a white tile. Iodine solution was added.
将一株植物在黑暗中放置48小时以去除淀粉。然后用铝箔覆盖一片叶子的一部分，留出其余部分。将植物放在阳光下6小时。之后摘下叶子，在水中煮沸，然后在乙醇中煮沸脱色，最后漂洗并铺在白色瓷砖上。滴加碘液。
The exposed parts of the leaf turned blue-black, showing that starch was present. The covered parts remained brown, indicating no starch. This demonstrates that light is necessary for photosynthesis. The starch is a product of photosynthesis and can be tested with iodine solution. The boiling in ethanol removes chlorophyll so colour changes are visible.
叶子被光照的部分变成蓝黑色，表明存在淀粉。被遮盖的部分保持棕色，说明无淀粉产生。这证明了光是光合作用所必需的。淀粉是光合作用的产物，可用碘液检测。在乙醇中煮沸是为了脱去叶绿素，使颜色变化清晰可见。
This case study is a classic practical that tests understanding of experimental design and control of variables. The destarching step ensures that any starch detected was produced during the experiment.
这个案例是一项经典实验，考查对实验设计和变量控制的理解。去除淀粉的步骤确保检测到的任何淀粉都是在实验过程中产生的。
7. Case Study 6: Lung Volume and Exercise | 案例六：肺活量与运动
A spirometer was used to compare the lung volumes of an athlete and a non-athlete at rest and after a five-minute run. The following data were collected.
使用肺活量计比较了一名运动员和一名非运动员在休息和跑步5分钟后的肺容积。收集了以下数据。
| Subject | Tidal volume at rest (dm³) | Tidal volume after exercise (dm³) | Breathing rate at rest (breaths/min) | Breathing rate after exercise (breaths/min) |
|---|---|---|---|---|
| Athlete | 0.6 | 2.1 | 12 | 22 |
| Non-athlete | 0.5 | 1.2 | 14 | 32 |
After exercise, both tidal volume and breathing rate increased. This is because working muscles require more oxygen for aerobic respiration to release energy, and they produce more carbon dioxide that needs to be removed. The athlete showed a much larger tidal volume after exercise and a smaller increase in breathing rate, suggesting more efficient lungs and better gas exchange.
运动后，潮气量和呼吸频率都增加了。这是因为工作的肌肉需要更多氧气进行有氧呼吸以释放能量，并产生更多需要排出的二氧化碳。运动员运动后的潮气量大得多，而呼吸频率增加较少，表明肺部效率更高，气体交换更佳。
The differences highlight the effects of training: cardiac and respiratory muscles become stronger, and the alveoli surface area may be better utilised. This kind of data analysis appears frequently in exams.
这些差异突出了训练的效果：心脏和呼吸肌变得更强壮，肺泡表面积也能更好地利用。此类数据分析在考试中经常出现。
8. Case Study 7: Heart Rate and Physical Activity | 案例七：心率与体力活动
In another investigation, two students measured their resting heart rate and then performed step-ups for 2 minutes. Their heart rates were recorded every minute during recovery. Student A had a resting heart rate of 65 bpm and took 4 minutes to return to resting levels. Student B had a resting heart rate of 78 bpm and took 7 minutes to recover.
在另一项研究中，两名学生测量了安静心率，然后完成了2分钟踏阶运动。每分钟记录恢复期间的心率。学生A安静心率为65 bpm，4分钟后恢复到安静水平。学生B安静心率为78 bpm，7分钟后才恢复。
A lower resting heart rate and faster recovery time are indicators of better cardiovascular fitness. The heart is a muscle; with regular exercise, the stroke volume increases, so the heart does not need to beat as frequently to pump the same amount of blood. This is an application of the topic of circulatory system adaptation.
较低的安静心率和较短的恢复时间是心血管更健康的标志。心脏是一块肌肉；规律运动能增加每搏输出量，因此心脏无需频繁跳动就能泵出等量血液。这是循环系统适应性的应用。
During exercise, the heart rate increases to deliver more oxygen and glucose to muscles and to remove lactate and carbon dioxide. Control of heart rate is influenced by adrenaline and the nervous system. Case studies like this often require you to suggest reasons for individual variation, such as age, gender, and fitness level.
运动时心率加快，以向肌肉输送更多氧气和葡萄糖，并清除乳酸和二氧化碳。心率受肾上腺素和神经系统控制。此类案例常要求你推断个体差异的原因，如年龄、性别和体能水平。
9. Case Study 8: Diffusion in the Lungs | 案例八：肺部扩散作用
Gaseous exchange in the alveoli relies on diffusion. Consider a patient suffering from emphysema, a condition where the walls of the alveoli break down, reducing the surface area for gas exchange. Their arterial oxygen saturation is measured at 90%, compared with a normal 97%.
肺泡内的气体交换依赖于扩散作用。一位肺气肿患者，肺泡壁受损，气体交换表面积减少。测得动脉血氧饱和度为90%，而正常为97%。
The reduced surface area slows down the rate of oxygen diffusion into the blood, as described by Fick’s law. The smaller the surface area, the lower the rate of diffusion. Additionally, the damaged alveoli may become enlarged, increasing the diffusion distance, which further reduces the efficiency. A lower oxygen concentration in the blood means less oxygen is delivered to respiring tissues, causing fatigue.
根据菲克定律，表面积减小会降低氧气扩散入血的速率。表面积越小，扩散速率越低。此外，受损的肺泡可能扩大，增加了扩散距离，进一步降低效率。血液中较低的氧浓度意味着输送到呼吸组织的氧气减少，导致疲劳。
This case study demonstrates how the structure of the lungs is adapted for efficient diffusion: the millions of alveoli create a huge surface area, thin walls (one cell thick), and a rich blood supply to maintain a steep concentration gradient. A disruption in any of these features leads to clinical symptoms.
本案例展示了肺部结构如何适应高效扩散：数百万肺泡提供了巨大的表面积，薄薄的一层细胞壁以及丰富的血液供应维持了陡峭的浓度梯度。任何一项特征出现问题都会引发临床症状。
10. Case Study 9: Food Tests and Nutrient Identification | 案例九：食物检测与营养物质鉴定
A lab technician discovered an unlabelled white powder. To identify it, she performed a series of food tests. She added iodine solution, and it turned blue-black, indicating the presence of starch. Then she heated a solution of the powder with Benedict’s solution: no colour change, indicating no reducing sugar. The biuret test gave a purple colour, showing protein. When she mixed the powder with ethanol and poured it into water, a cloudy white emulsion formed, confirming the presence of lipids.
一位实验室技术员发现了一包未标记的白色粉末。为了鉴定其成分，她进行了一系列食物检测。加入碘液后变成蓝黑色，表明存在淀粉。然后她将粉末溶液与本尼迪克特试剂加热：无颜色变化，说明没有还原糖。双缩脲检测呈紫色，显示有蛋白质。当她将粉末与乙醇混合并倒入水中时，形成乳白色浑浊，证实含有脂质。
These results mean the powder was a mixture of starch, protein, and fats. Knowing the chemical composition helps determine its use and energy content. The tests are based on specific chemical reactions: iodine traps inside the coiled starch molecule; copper(II) sulfate in Benedict’s solution is reduced by reducing sugars to form a brick-red precipitate; biuret reagent reacts with peptide bonds; and ethanol extracts lipids into an emulsion.
这些结果表明该粉末是淀粉、蛋白质和脂肪的混合物。了解化学成分有助于确定其用途和能量含量。这些测试基于特定的化学反应：碘分子嵌入螺旋形淀粉分子圈内；本尼迪克特溶液中的硫酸铜被还原糖还原，形成砖红色沉淀；双缩脲试剂与肽键反应；乙醇将脂质萃取成乳浊液。
Case studies like this evaluate your ability to interpret qualitative observations and link them to biochemical knowledge. Remember to always describe both the initial colour and the final colour change in tests.
此类案例评估你解读定性观察并将其与生物化学知识联系起来的能力。记住在描述测试时，要同时说明初始颜色和最终的颜色变化。
11. Case Study 10: Control of Blood Glucose | 案例十：血糖调节
After a meal containing carbohydrates, a person’s blood glucose concentration rises. The pancreas detects the rise and secretes insulin. Insulin stimulates the liver and muscle cells to take up glucose and convert it to glycogen for storage. In people with type 1 diabetes, the pancreas does not produce enough insulin, so blood glucose remains high after eating. A glucose tolerance test was carried out on a diabetic patient and a healthy control. Blood samples were taken every 30 minutes after drinking a glucose solution.
进食含碳水化合物的食物后，人的血糖浓度会升高。胰腺检测到升高后分泌胰岛素。胰岛素刺激肝脏和肌肉细胞摄取葡萄糖，并将其转化为糖原储存起来。对于1型糖尿病患者，胰腺无法产生足够的胰岛素，因此进食后血糖持续偏高。对一名糖尿病患者和一名健康对照者进行了葡萄糖耐量测试。在饮用葡萄糖溶液后，每30分钟采集一次血样。
The healthy person’s blood glucose peaked at 30 minutes and then returned to baseline by 120 minutes. The diabetic’s glucose level peaked higher and stayed elevated, barely dropping after 2 hours. This indicates a lack of insulin activity. Without adequate insulin, glucose cannot be effectively removed from the blood and stored as glycogen; the liver also continues to release glucose via gluconeogenesis.
健康人的血糖在30分钟达到峰值，随后在120分钟内回到基线。糖尿病患者的血糖峰值更高，且持续偏高，2小时后几乎没有下降。这表明缺乏胰岛素活性。没有足够的胰岛素，葡萄糖无法有效从血液中移除并储存为糖原；肝脏还持续通过糖异生释放葡萄糖。
Homeostasis is the maintenance of a constant internal environment. The negative feedback loop involving insulin and glucagon keeps blood glucose within a narrow range. This case study links to topics of enzymes (insulin is a protein hormone), diffusion (glucose transport), and disease. It is typical of the application questions found in paper 2 or 4.
稳态是指维持稳定的内环境。涉及胰岛素和胰高血糖素的负反馈回路将血糖维持在狭窄的范围内。这个案例联系了酶（胰岛素是一种蛋白质激素）、扩散（葡萄糖转运）和疾病等知识点。这类应用题型在试卷二或试卷四中很常见。
12. Conclusion: Applying Case Study Skills | 结语：应用案例分析技巧
Through these ten case studies, you have practised interpreting data, explaining biological phenomena, and linking theory to practical scenarios. Remember that the key to success in CAIE biology is not just recalling facts, but applying them to unfamiliar contexts. When you read a case study, always ask: ‘What biological concept is being tested? What is the relationship between the variables? How can I justify my answer with evidence and scientific reasoning?’
通过这十个案例分析，你已经练习了如何解读数据、解释生物学现象，并将理论与实际情境联系起来。要记住，在CAIE生物考试中取得成功的关键不仅仅是记忆事实，而是将它们应用于不熟悉的背景中。当你阅读一个案例时，始终要问自己：“这个题目在考查哪个生物学概念？变量之间有何关系？我如何用证据和科学推理来合理解释我的答案？”
Practise with past paper questions and design your own mini-experiments to strengthen these skills. With a solid understanding of core principles and a systematic approach, you will be able to tackle any case study that comes your way.
通过练习历年真题中的案例分析，并且设计你自己的小实验来强化这些技能。有了对核心原理的扎实理解和系统性的解题方法，你将能够应对考试中可能出现的任何案例分析题目。
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