📚 Year 9 CAIE Biology: Case Study Practice | Year 9 CAIE 生物：案例分析实战演练
Case study questions in Year 9 CAIE Biology test not only your knowledge of facts but also your ability to apply concepts to unfamiliar situations, analyse data, and draw evidence-based conclusions. These exercises will guide you through common scenarios, highlight exam command words, and show you how to structure high-scoring answers.
Year 9 CAIE 生物案例题不仅考查你对知识点的记忆，更注重你在陌生情境中应用概念、分析数据并得出基于证据的结论的能力。以下实战演练将带你走近常见情境，剖析考题指令词，并展示如何构建高分答案。
1. Understanding Command Words in Case Studies | 理解案例题指令词
Before tackling any case study, you must identify the command word. Words like 'describe' ask for factual recall, 'explain' requires reasons or mechanisms, and 'suggest' invites you to apply logic to a new situation. A common mistake is to give a description when explanation is needed.
在处理任何案例之前，你必须先识别指令词。'describe'要求陈述事实，'explain'需要给出原因或机制，而'suggest'则邀请你将逻辑应用到新情境中。常见错误是在需要解释时只做了描述。
In an exam, underline the command word. For example: 'Explain why the heart rate increases during exercise.' Here you must link increased muscle activity to higher carbon dioxide levels and the brain's response, not just state that heart rate goes up.
考试时记得画出指令词。比如：'Explain why the heart rate increases during exercise.' 你就要将肌肉活动增加与血液中二氧化碳浓度升高以及脑部反应联系起来，而不是仅仅说心率上升。
Similarly, 'calculate' expects a numerical answer with units, 'compare' needs both similarities and differences, and 'evaluate' asks for advantages and disadvantages plus a concluding judgement. Practise matching your answer structure to the command word every time.
同样，'calculate'要求给出带单位的数值答案，'compare'需要异同点，而'evaluate'则要求分析优缺点并做出总结性评判。始终练习让答案结构与指令词相匹配。
2. Case Study 1: Enzyme Activity and Temperature | 案例一：酶活性与温度
A student investigated the effect of temperature on the activity of the enzyme catalase using hydrogen peroxide. The volume of oxygen produced per minute was recorded at temperatures from 10°C to 60°C. The data showed an optimum at 37°C, with a sharp decline in activity above 45°C.
一名学生用过氧化氢研究温度对过氧化氢酶活性的影响。记录了从10°C到60°C每分钟产生的氧气量。数据显示最适温度为37°C，超过45°C后活性急剧下降。
Explain why catalase activity increases as temperature rises from 10°C to 37°C. At low temperatures, both enzyme and substrate molecules have low kinetic energy, so collisions are infrequent. As temperature rises, kinetic energy increases, leading to more frequent successful collisions and more enzyme-substrate complexes forming per second.
解释为何过氧化氢酶活性在10°C升至37°C时会增加。低温时酶分子和底物分子动能都较低，碰撞频率低。温度升高使动能增加，带来更频繁的有效碰撞，每秒形成更多的酶-底物复合物。
Describe the structural changes that occur above 45°C. High temperatures cause the enzyme's polypeptide chains to vibrate violently, breaking the hydrogen bonds and ionic interactions that maintain the specific shape of the active site. The active site becomes denatured and can no longer bind the substrate, so the reaction rate drops dramatically.
描述45°C以上发生的结构变化。高温使酶的多肽链剧烈振动，维持活性位点特定形态的氢键和离子相互作用被破坏。活性位点变性，无法再与底物结合，因此反应速率急剧下降。
Suggest two control variables for this investigation and explain their importance. The pH must be kept constant using a buffer because changes in pH can also denature the enzyme. The concentration of hydrogen peroxide and the amount of catalase must be kept the same, or the rate of oxygen production would be affected independently of temperature.
请为这项研究提出两个控制变量并说明其重要性。必须用缓冲液保持pH恒定，因为pH变化同样会使酶变性。过氧化氢浓度和过氧化氢酶的量也必须一致，否则产氧速率会在温度之外受到独立影响。
3. Case Study 2: Photosynthesis and Light Intensity | 案例二：光合作用与光照强度
An aquatic plant was placed in water containing sodium hydrogencarbonate to supply carbon dioxide. The number of bubbles of oxygen produced per minute was recorded as a lamp was moved to different distances from the plant. The bubble count increased as the lamp was brought closer, but plateaued at very short distances.
将水生植物放在含有碳酸氢钠（提供二氧化碳）的水中。随着灯与植物的距离改变，记录每分钟产生的氧气气泡数。灯越近，气泡数越多，但在非常近的距离处达到平台。
Write the balanced symbol equation for photosynthesis.
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
写出光合作用的平衡符号方程。
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
Explain the plateau in oxygen production at high light intensity. Light intensity is no longer the limiting factor; another factor such as carbon dioxide concentration or temperature has become limiting. All chloroplasts are working at maximum capacity, so increasing light further cannot raise the rate.
解释高光照强度下氧气产量出现平台的原因。此时光照强度已不再是限制因素，另一个因素如二氧化碳浓度或温度变成了限制因素。所有叶绿体都在满负荷工作，继续增加光照已无法提升速率。
Suggest how you could demonstrate that carbon dioxide is the new limiting factor. By adding more sodium hydrogencarbonate to increase CO₂ concentration, you would expect the bubble count to rise again if CO₂ had been limiting. If temperature is the limiting factor, gently warming the water should increase the rate.
请提出如何证明二氧化碳是新的限制因素。加入更多碳酸氢钠以提高CO₂浓度，如果之前CO₂是限制因素，气泡数应会再次上升。若温度是限制因素，稍稍加热水应能提高速率。
4. Case Study 3: Diffusion and Surface Area to Volume Ratio | 案例三：扩散与表面积体积比
Agar cubes of different sizes (1 cm, 2 cm, 3 cm) containing phenolphthalein and sodium hydroxide were placed in dilute hydrochloric acid. The time taken for the pink colour to completely disappear was measured. The 1 cm cube cleared much faster than the 3 cm cube.
将含有酚酞和氢氧化钠的不同大小琼脂块（1 cm、2 cm、3 cm）放入稀盐酸中，测量粉红色完全消失所需的时间。1 cm立方块比3 cm立方块脱色快得多。
Explain why the smallest cube decolourised fastest. Small cubes have a higher surface area to volume ratio. This means a larger proportion of the cube's total volume is in contact with the acid, so diffusion of hydrogen ions to all parts of the cube occurs more quickly.
解释为什么最小的琼脂块脱色最快。小立方块具有更高的表面积与体积比。这意味着立方块总体积中有更大比例与酸接触，因此氢离子扩散到块内各处更迅速。
A student says that larger organisms need specialised transport systems. Use this investigation to justify the statement. As an organism grows, its surface area to volume ratio decreases. Diffusion alone becomes too slow to supply all cells with oxygen and nutrients and to remove wastes. Therefore, large multicellular organisms have evolved circulatory systems to speed up internal transport.
有学生说大型生物需要专门的运输系统。请用本探究来证明该观点。随着生物体变大，其表面积与体积比下降。单靠扩散太慢，无法为所有细胞供应氧气和养分并带走废物。因此，大型多细胞生物进化出循环系统以加速内部运输。
5. Case Study 4: Digestion and Absorption | 案例四：消化与吸收
A patient had part of their stomach removed. After surgery, they experienced fatigue and pale skin. Blood tests showed low levels of vitamin B12, which is normally absorbed by the small intestine after binding with a substance produced in the stomach lining.
一名患者切除了部分胃。术后出现疲劳和皮肤苍白。血检显示维生素B12水平偏低，而B12通常需与胃壁产生的一种物质结合后才能在小肠被吸收。
Describe the role of the stomach in protein digestion. The stomach wall secretes hydrochloric acid, which kills bacteria and creates an acidic optimum pH for pepsin. Chief cells release pepsinogen, which is converted to active pepsin by the acid. Pepsin breaks proteins into shorter chains of amino acids, beginning the chemical digestion of proteins.
描述胃在蛋白质消化中的作用。胃壁分泌盐酸，可杀灭细菌并为胃蛋白酶创造酸性最适pH。主细胞分泌胃蛋白酶原，胃酸将其活化为有活性的胃蛋白酶。胃蛋白酶将蛋白质分解为较短的氨基酸链，从而启动蛋白质的化学消化。
Explain how the small intestine is adapted for efficient absorption of nutrients. The inner lining is folded into villi and microvilli, which massively increase the surface area. Each villus contains a dense network of blood capillaries and a lacteal. The epithelium is only one cell thick, providing a short diffusion distance. These features allow rapid absorption of digested food molecules into the blood and lymph.
解释小肠如何适应高效吸收营养物质。内壁折叠成绒毛和微绒毛，极大地增加了表面积。每条绒毛内含致密的毛细血管网和一条乳糜管。上皮仅有一个细胞的厚度，缩短了扩散距离。这些结构使消化后的小分子能快速被吸收进入血液和淋巴。
6. Case Study 5: Blood Circulation and Exercise | 案例五：血液循环与运动
An athlete's heart rate, stroke volume, and cardiac output were measured at rest and during vigorous running. At rest, heart rate was 70 bpm and stroke volume 70 ml per beat. During exercise, heart rate rose to 150 bpm and stroke volume increased to 120 ml per beat.
测量了一名运动员在安静状态下和剧烈跑步时的心率、每搏输出量和心输出量。安静时心率为70次/分，每搏输出量70毫升/博。运动时心率升至150次/分，每博输出量增至120毫升。
: 70 × 70 = 4900 ml/min (4.9 L/min). During exercise: 150 × 120 = 18000 ml/min (18 L/min).
计算安静时和运动时的心输出量。使用公式：心输出量 = 心率 × 每搏输出量。安静时：70 × 70 = 4900 ml/min (4.9 L/min)。运动时：150 × 120 = 18000 ml/min (18 L/min)。
Explain why cardiac output must increase during exercise. Working muscles respire more rapidly, needing more oxygen for aerobic respiration and generating more carbon dioxide. The increased cardiac output delivers oxygen faster and removes carbon dioxide more efficiently, helping to maintain a steep concentration gradient for gas exchange at the muscles.
解释运动时心输出量必须增加的原因。运动的肌细胞呼吸更快，需氧量增加并产生更多二氧化碳。增加的心输出量能更快地输送氧气并更有效地带走二氧化碳，有助于在肌肉处维持利于气体交换的陡浓度梯度。
7. Case Study 6: Transpiration in Plants | 案例六：植物的蒸腾作用
A potometer was set up to measure water uptake by a leafy shoot under different environmental conditions. The air temperature was increased from 20°C to 30°C; the rate of water uptake rose sharply. When a plastic bag was tied around the leaves, water uptake slowed significantly.
设置了蒸腾计来测量不同环境下带叶枝条的吸水量。气温从20°C升至30°C时，吸水速率急剧增加。当用塑料袋套住叶子时，吸水速率明显减慢。
Explain the effect of increasing temperature on transpiration rate. Higher temperatures increase the kinetic energy of water molecules, so they evaporate from the mesophyll cell surfaces more quickly. The water vapour concentration inside the leaf rises, and the steeper diffusion gradient to the drier outside air speeds up water loss through the stomata.
解释升温对蒸腾速率的影响。较高温度增加了水分子的动能，使水分从叶肉细胞表面蒸发更快。叶内水蒸气浓度升高，与较干燥外界空气之间的扩散梯度变大，加快了通过气孔散失水分的速度。
Describe how a plastic bag around the leaves reduces water uptake. The bag traps water vapour, raising humidity around the leaf. This reduces the water vapour concentration gradient between the leaf air spaces and the outside, so less water vapour diffuses out of the stomata. Transpiration slows, and therefore water uptake decreases.
描述塑料袋套着叶子为何减少吸水量。塑料袋截留了水蒸气，提高了叶片周围的湿度。这减小了叶内气隙与外界之间的水蒸气浓度梯度，因此经气孔扩散出去的水蒸气变少。蒸腾减弱，从而吸水量下降。
8. Case Study 7: Ecological Relationships and Food Webs | 案例七：生态关系与食物网
In a pond ecosystem, the following organisms were recorded: pondweed, algae, Daphnia, water fleas, small fish, dragonfly nymphs, and a single species of water bird. A food web was constructed showing that small fish eat both Daphnia and dragonfly nymphs.
在一个池塘生态系统中，记录了以下生物：水草、藻类、水蚤、剑水蚤、小鱼、蜻蜓若虫和一种水鸟。构建的食物网显示小鱼既吃水蚤也吃蜻蜓若虫。
Identify a producer, a primary consumer, and a secondary consumer from this web. Producers include pondweed and algae (they photosynthesise). Primary consumers include Daphnia and water fleas (herbivores feeding on algae). Small fish act as secondary consumers when they eat Daphnia and as tertiary consumers when they eat dragonfly nymphs that have consumed other animals.
从该网中找出一个生产者、一个初级消费者和一个次级消费者。生产者包括水草和藻类（进行光合作用）。初级消费者有水蚤和剑水蚤（以藻类为食的植食者）。小鱼吃水蚤时是次级消费者，吃蜻蜓若虫（若虫已吃其他动物）时则处于更高级。
Predict what might happen to the Daphnia population if the dragonfly nymphs were removed. Dragonfly nymphs prey on Daphnia, so removing them would reduce predation pressure. The Daphnia population would likely increase initially. However, increased Daphnia numbers could reduce the algae population, and competition for food might eventually slow Daphnia growth or cause a subsequent decline.
预测如果移除蜻蜓若虫，水蚤种群会发生什么变化。蜻蜓若虫捕食水蚤，移除它们会降低捕食压力。水蚤数量初期可能增加。然而，水蚤增多会使藻类减少，对食物的竞争最终可能减缓水蚤的增长或导致其后数量回落。
9. Case Study 8: Inheritance and Blood Groups | 案例八：遗传与血型
The ABO blood group system in humans is controlled by three alleles: IA, IB, and IO. IA and IB are codominant, and both are dominant over IO. A mother with blood group A (genotype IAIO) and a father with blood group B (genotype IBIO) are expecting a child.
人类ABO血型系统由三个等位基因控制：IA、IB和IO。IA 和 IB 为共显性，两者都对 IO 为显性。一位A型血的母亲（基因型 IAIO）和一位B型血的父亲（基因型 IBIO）正期待一个孩子。
and IO. Gametes from father: IB and IO. Offspring genotypes: IAIB (group AB), IBIO (group B), IAIO (group A), IOIO (group O). The phenotype ratio is 1 AB : 1 A : 1 B : 1 O, each with a 25% probability.
构建遗传图解展示孩子的可能基因型和表现型。亲代基因型：IAIO × IBIO。母亲产生的配子：IA 和 IO。父亲产生的配子：IB 和 IO。子代基因型：IAIB (AB型), IBIO (B型), IAIO (A型), IOIO (O型)。表现型比例为 1 AB : 1 A : 1 B : 1 O，各占25%概率。
Explain why blood group AB is described as an example of codominance. The IA allele codes for A antigen, and IB codes for B antigen. In an IAIB individual, both antigens are present on the surface of red blood cells, meaning both alleles are expressed equally in the phenotype; neither masks the other.
解释为何AB血型被称为共显性的例子。IA等位基因编码A抗原，IB编码B抗原。IAIB个体中，红细胞表面同时存在两种抗原，意味着两个等位基因在表现型中同等表达，互不遮盖。
10. Case Study 9: Respiration and Yeast Fermentation | 案例九：呼吸作用与酵母发酵
Yeast cells were added to a glucose solution in a closed flask with a gas syringe attached. After one hour, the plunger had moved out, and the gas collected turned limewater milky. The yeast suspension was then boiled and re-tested; no gas was produced.
将酵母细胞加入盛有葡萄糖溶液的密闭烧瓶中，连接气体注射器。一小时后活塞外移，收集到的气体使石灰水变浑浊。将酵母悬浮液煮沸后再次测试，没有气体产生。
rbon dioxide (+ energy). This is anaerobic respiration in yeast, also called fermentation.
说出所产生的气体并写出该过程的文字方程式。该气体是二氧化碳。文字方程：葡萄糖 → 乙醇 + 二氧化碳（+能量）。这是酵母的无氧呼吸，又称发酵。
Explain why no gas was produced after boiling. Boiling denatures the enzymes involved in respiration, including those in the glycolysis pathway. Without functional enzymes, yeast cells cannot break down glucose, so no carbon dioxide is released. Living, active yeast is essential for fermentation.
解释为什么煮沸后没有气体产生。煮沸使参与呼吸作用的酶变性，包括糖酵解途径中的酶。没有功能性酶，酵母细胞无法分解葡萄糖，因而不会释放二氧化碳。活性的活酵母是发酵所必需的。
Compare the maximum ATP yield per glucose molecule between aerobic respiration and fermentation in yeast. Aerobic respiration can yield up to about 32 molecules of ATP per glucose, whereas fermentation yields only 2 ATP per glucose. The majority of energy remains locked in the ethanol, which is why alcoholic beverages still contain chemical potential energy.
比较酵母有氧呼吸与发酵每分子葡萄糖的最大ATP产量。有氧呼吸每分子葡萄糖最多产生约32个ATP，而发酵每葡萄糖仅产生2个ATP。大部分能量仍锁在乙醇中，这也是酒精饮料仍含有化学潜能的原因。
11. Summary of Case Study Answering Techniques | 案例分析作答技巧总结
Always read the scenario carefully and highlight key data such as numbers, temperatures, or species names. Match your answer length to the mark allocation; a three-mark question usually requires three distinct points. Use biological terminology precisely, e.g. 'denatured' rather than 'killed' for enzymes.
务必仔细阅读情境，高亮关键数据，如数字、温度或物种名称。根据配分决定答案长度；3分题通常需要三个不同的点。准确使用生物学术语，例如对酶要用'变性'而非'杀死'。
When data shows a trend, quote figures from the table or graph in your answer to support your explanation. For calculation questions, always show your working and include the correct units. In 'suggest' questions, there may not be a single correct answer, but your idea must be scientifically plausible and based on known principles.
当数据显示某种趋势时，在答案中引用表格或图表中的数字来佐证你的解释。计算题要始终展示步骤并带上正确单位。在'suggest'题中，可能并没有唯一正解，但你的想法必须科学合理并基于已知原理。
Finally, if a question asks you to evaluate, don't just list advantages. You must also mention disadvantages or limitations and end with a clear, justified conclusion. Practising these strategies with the cases above will build confidence and improve your exam performance.
最后，如果题目要求'evaluate'，不要只列优点，还应提及缺点或局限性，并以明确、有理有据的结论收尾。用上述案例反复练习这些策略，你将增强信心并提高考试成绩。
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