Year 11 CAIE Biology: Case Study Practice Drills | CAIE 生物：案例分析实战演练

📚 Year 11 CAIE Biology: Case Study Practice Drills | CAIE 生物：案例分析实战演练

In the CAIE IGCSE Biology examination, case study questions assess your ability to apply knowledge to unfamiliar real‑world scenarios. These questions appear extensively in Paper 4 (Theory Extended) and Paper 6 (Alternative to Practical), where you may be asked to interpret experimental data, analyse graphs, explain trends or suggest improvements. Mastering a systematic approach to case studies will help you unlock high marks by demonstrating both biological understanding and scientific reasoning.

在 CAIE IGCSE 生物考试中，案例分析题考察你将知识应用于陌生真实情境的能力。这类题目大量出现在卷四（理论拓展）和卷六（实验替代），要求你解读实验数据、分析图表、解释趋势或提出改进建议。掌握一套系统性的案例分析策略，能够帮助你展示生物学理解和科学推理，从而获得高分。

1. Understanding the Question Context | 理解题目背景

Every case study begins with a short description of an investigation, a health issue or an ecological observation. Before you start writing, underline the key biological terms in the text and pay close attention to the command words such as ‘describe’, ‘explain’, ‘calculate’ or ‘suggest’. Knowing what the examiner wants helps you to tailor your answer precisely rather than just recalling facts.

每个案例分析都以一段实验调查、健康问题或生态观察的简短描述开头。动笔之前，务必划出文中关键的生物学术语，并特别注意指令词，例如 “describe” (描述)、“explain” (解释)、“calculate” (计算) 或 “suggest” (建议)。清楚考官的意图能够帮助你精准作答，而不是仅仅回忆知识点。

For example, if the question says ‘Describe the trend shown in the graph’, you only need to state what the graph shows (increase, decrease, plateau) without giving reasons. If it asks ‘Explain the results’, you must refer to biological processes such as enzyme denaturation or diffusion.

例如，若题目要求“描述图中所示的趋势”，你只需说出图表示的内容（上升、下降、平台期），无需说明原因。若要求“解释结果”，则必须提及酶变性、扩散等生物学过程。

2. Identifying Variables and Control | 识别变量与对照

Many case studies are based on experiments. Here you should identify the independent variable (the factor you change), the dependent variable (what you measure) and the controlled variables (factors kept constant). A well‑written answer often includes at least two controlled variables, such as volume of solution, temperature, pH or light intensity.

许多案例分析以实验为基础。此时你应识别自变量（你改变的因素）、因变量（你测量的结果）和控制变量（保持不变的因素）。一份优质的答案通常至少包含两个控制变量，例如溶液体积、温度、pH 或光照强度。

Being able to state how a variable was controlled is equally important. For instance, ‘Temperature was controlled by using a thermostatically controlled water bath at 25°C’ demonstrates precision and earns marks in Paper 6.

能够说明变量如何被控制同样重要。例如，“通过使用恒温水浴锅将温度控制在25°C”展示了实验的精确性，能在卷六中赢得分数。

3. Case 1: Enzyme Activity and Temperature | 案例一：酶活性与温度

A student investigated the effect of temperature on amylase activity. Five test tubes containing equal volumes of starch solution and amylase were incubated at different temperatures. The time for the starch to be completely broken down was measured using iodine solution, which turns blue‑black in the presence of starch.

一名学生研究了温度对淀粉酶活性的影响。将装有等量淀粉溶液和淀粉酶的五支试管分别置于不同温度下保温。通过碘液测试测定淀粉完全分解所需的时间（碘液遇淀粉变为蓝黑色）。

The data show that as temperature rises from 10°C to 40°C, the reaction time decreases, meaning the enzyme works faster. At its optimum (close to 40°C), amylase is most efficient. However, at 50°C the time jumps, revealing a sharp loss of activity. In terms of enzyme action, heat increases the kinetic energy of molecules, leading to more frequent successful collisions until the enzyme’s active site is denatured and the substrate can no longer bind.

数据显示，温度从10°C上升到40°C时，反应时间缩短，说明酶的工作速度加快。在其最适温度（约40°C），淀粉酶效率最高。然而在50°C时反应时间激增，表明活性急剧丧失。从酶的作用机制来看，热量增加了分子的动能，导致更频繁的有效碰撞，直到酶的活性部位变性，底物无法再与之结合。

When answering such case studies, always quote data, e.g. ‘The time dropped from 125 s to 12 s between 10°C and 40°C, then rose to 210 s at 50°C’. If asked to plot a graph, choose a suitable scale, label both axes with units, and draw a smooth curve – not straight lines between points.

在回答此类案例分析时，务必引用数据，例如“时间从10°C的125 s下降到40°C的12 s，然后在50°C升至210 s”。如果要求绘图，请选择合适的刻度，为两条轴标上单位，并绘制平滑曲线——而非点与点之间的直线。

4. Case 2: Osmosis and Potato Strips | 案例二：渗透作用与马铃薯条

An experiment was carried out to investigate osmosis in potato tissue. Potato strips of equal length and mass were placed in sucrose solutions of different concentrations for 30 minutes. The percentage change in mass was calculated.

进行了一项实验以研究马铃薯组织中的渗透作用。将长度和质量相同的马铃薯条分别放入不同浓度的蔗糖溶液中，静置30分钟。计算了质量变化的百分比。

The positive mass change in low sucrose concentrations indicates water entered the potato cells by osmosis because the surrounding solution had a higher water potential than the cell sap. As the external sucrose concentration increases, the water potential decreases, and water leaves the cells, causing negative mass change. The point where the graph of percentage change crosses zero corresponds to the water potential of the potato cells – here approximately 0.35 mol dm⁻³.

低浓度蔗糖溶液中质量增加，表明水通过渗透作用进入马铃薯细胞，因为周围溶液的水势高于细胞液的水势。随着外部蔗糖浓度升高，水势降低，水离开细胞，导致质量减少。质量变化百分比为零的点对应马铃薯细胞的水势——此处约为0.35 mol dm⁻³。

Common exam questions ask you to estimate the solute concentration inside potato cells, explain why blotting the strips before weighing is important, or suggest why repeats are needed. Remember to state that surface water must be removed to measure the true mass of the tissue, not adhering water.

常见的考题包括：估计马铃薯细胞内的溶质浓度、解释称重前吸干薯条表面水分为何重要，或建议为何需要重复实验。记住要说明必须除去表面水分，才能测量的是组织的真实质量，而非附着的水。

广告 / Sponsored

牛津AQA国际GCSE生物?OxfordAQA International GCSE Biology (9201): Onl?进口原版

¥399.0

📱 去京东购买

京东精选

5. Case 3: Photosynthesis and Light Intensity | 案例三：光合作用与光照强度

A student used a submerged aquatic plant (Elodea) to measure the rate of photosynthesis. A lamp was placed at different distances from the plant, and the number of oxygen bubbles released per minute was counted. The results are tabulated below.

一位学生利用沉水水生植物（伊乐藻）测量光合作用速率。将灯置于离植物不同的距离处，计数每分钟释放的氧气泡数量。结果如下表所示。

The results show that as the distance increases (light intensity decreases), the rate of photosynthesis drops. This is because light provides the energy for the light‑dependent stage, where water is split and ATP is produced. At low light intensity, these products become limiting, so fewer bubbles are produced.

结果表明，随着距离增加（光照强度减弱），光合作用速率下降。这是因为光为光反应阶段提供能量，在该阶段水被分解并产生 ATP。在低光照强度下，这些产物成为限制因素，因此产生的气泡更少。

If the question asks for a control, a common answer is to add a fixed amount of sodium hydrogen carbonate to ensure CO₂ is not limiting, and to maintain a constant temperature with a water bath. Also note that counting bubbles can be inaccurate; measuring volume of gas collected in a syringe is more reliable.

如果题目要求设立对照，常见的答案是加入定量的碳酸氢钠以确保二氧化碳不是限制因素，并使用水浴保持恒温。还需注意，计数气泡可能不准确；用注射器收集并测量气体体积更可靠。

6. Case 4: Pedigree Chart and Genetic Disease | 案例四：家系图与遗传病

A pedigree chart is used to trace the inheritance of a genetic disorder. In the case study below, square symbols (□) represent unaffected males, circles (○) represent unaffected females, and filled symbols (■, ●) show individuals with the condition. The first generation consists of a normal father (□) and a normal mother (○), who have three children: a normal son, a normal daughter, and an affected son (■). The affected son marries an unaffected woman (○) and they have an affected daughter (●) and an unaffected son (□).

家系图用于追踪遗传病的遗传方式。在以下案例中，方形符号（□）代表未患病男性，圆形符号（○）代表未患病女性，实心符号（■, ●）表示患病个体。第一代由正常父亲（□）和正常母亲（○）组成，他们育有三个子女：一个正常儿子、一个正常女儿和一个患病儿子（■）。该患病儿子与一名未患病女性（○）结婚，他们育有一个患病女儿（●）和一个正常儿子（□）。

From the pattern, we can deduce that the disorder is recessive because two unaffected parents produced an affected child. It cannot be X‑linked recessive, because an affected father (■) would pass his X‑linked allele to all daughters, yet here the affected man and an unaffected woman produced an unaffected son – which would be impossible if it were X‑linked recessive since the father’s affected X would go to all daughters and not to sons. The evidence suggests it is autosomal recessive.

从这一模式可以推断，该病症为隐性遗传，因为两个未患病的父母生出了患病的孩子。不可能是 X 连锁隐性遗传，因为如果父亲患病（■），他定会将带病的 X 等位基因传给所有女儿，但此处患病男子与正常女子生下一个正常儿子——若为 X 连锁隐性，这不可能，因为父亲的致病 X 染色体只会传给女儿而不会传给儿子。证据表明这是常染色体隐性遗传。

ut genotypes: the parents in generation I must both be Aa. Then show Punnett squares or probability calculations if required.

作答时，分配等位基因：设 A = 正常，a = 致病等位基因。推导出基因型：第一代的父母必定都是 Aa。如果题目要求，再画出庞纳特方格或进行概率计算。

7. Case 5: Natural Selection – Peppered Moth | 案例五：自然选择——桦尺蛾

The peppered moth (Biston betularia) exists in two forms: light‑coloured and dark‑coloured. Before the industrial revolution, tree trunks were covered with pale lichens, and light moths were well camouflaged. With industrial pollution, soot darkened the tree trunks and killed lichens, favouring the dark form. The table shows the percentage of each form caught by birds over several decades.

桦尺蛾（Biston betularia）具有浅色和深色两种类型。工业革命之前，树干被浅色地衣覆盖，浅色蛾伪装效果良好。随着工业污染，煤烟使树干变黑、地衣死亡，深色类型获得优势。下表显示了数十年间被鸟类捕食的各类型百分比。

Year	Light moths (%)	Dark moths (%)
1850	98	2
1900	55	45
1950	20	80
2000	广告 / Sponsored 现货 Cambridge Biology for the IB Diploma coursebook 剑桥国际文凭 I… 生物 备考教辅 · 正版现货 · 与本文内容相关 查看详情 / View on Taobao参考价 ¥546.0 更多咨询请联系16621398022（同微信） 备考资讯每日更新，关注不错过 考纲变化 · 科目分析 · 学习方法 在微信里长按二维码识别关注 或截图后用「扫一扫 → 相册」选择本图 公众号：tutorhao