📚 Year 10 CIE Biology: Case Study Practice | 十年级CIE生物：案例分析实战演练
Case studies are a vital part of CIE IGCSE Biology. They require you to apply theoretical knowledge to real or simulated experimental data, interpret graphs, spot trends, and evaluate methods. This article presents 12 typical case studies covering key Year 10 topics, showing you how to approach data-based questions step by step. Work through these examples to sharpen your analysis skills and boost your exam confidence.
案例研究是 CIE IGCSE 生物考试的重要组成部分。你需要将理论知识应用于真实或模拟的实验数据中，解释图表、识别趋势并评价实验方法。本文精选了 12 个典型的案例研究，覆盖十年级核心主题，一步步教你如何应对数据分析题。通过这些实例进行练习，能够提高你的分析能力，增强考试信心。
1. Enzyme Activity and pH | 酶活性与 pH
A student extracted catalase from potato discs and placed them in hydrogen peroxide solutions buffered at different pH levels. The volume of oxygen produced in 60 seconds was recorded.
一名学生从马铃薯圆片中提取了过氧化氢酶，并将其放入不同 pH 缓冲液配制的过氧化氢溶液中，记录了 60 秒内产生的氧气体积。
The following table shows the results:
下表显示了实验结果：
| pH | 2 | 4 | 6 | 7 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|---|
| Oxygen (cm³) | 3 | 8 | 15 | 19 | 14 | 5 | 1 |
First, identify the optimum pH (around 7) and describe the shape of the graph: activity rises steadily to a peak, then falls sharply as pH moves away from the optimum. Use the idea of enzyme denaturation at extreme pH values, where the active site shape changes and substrate no longer fits.
首先找出最适 pH（约为 7）并描述图形形状：活性稳步上升至峰值，之后随着 pH 偏离最适值而急剧下降。用极端 pH 下酶变性的观点来解释，即活性位点形状改变，底物不再匹配。
For full marks, always link shape change to active site distortion and loss of complementary binding. Remember, simply stating 'the enzyme denatures' is not enough; explain that the bonds maintaining the tertiary structure break.
想拿满分，一定要将形状变化与活性位点扭曲、互补结合丧失联系起来。记住，仅仅说“酶变性”是不够的，要解释维持三级结构的键被破坏。
2. Photosynthesis and Light Intensity | 光合作用与光照强度
An aquatic plant (Elodea) was placed in a beaker of water, and the number of bubbles released per minute was counted at different distances from a lamp. The table below gives the results.
将水生植物（伊乐藻）放入盛水的烧杯中，在不同距离的灯光下计数每分钟释放的气泡数。结果如下表。
| Distance (cm) | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| Bubbles/min | 42 | 20 | 10 | 5 | 2 |
You may be asked to calculate the rate of photosynthesis as 1/distance or to state the limiting factor. At close distances, light intensity is high; the rate may plateau if another factor (e.g. CO₂ concentration) becomes limiting.
你可能需要计算光合作用速率（如 1/距离）或指出限制因子。距离近时，光照强，如果另一个因子（如二氧化碳浓度）成为限制，速率可能会趋于平缓。
Describe the relationship: as light intensity increases, the rate of photosynthesis increases up to a point. Always mention the light-dependent reactions: more light provides more energy to split water and produce ATP and NADPH.
描述关系：随着光照强度增加，光合作用速率提高，直至达到某一点。一定要提及光反应：更多光能提供更多能量来分解水，产生 ATP 和 NADPH。
3. Osmosis in Potato Strips | 土豆条的渗透作用
A student cut potato strips of equal mass and placed them in sucrose solutions from 0.0 mol/dm³ to 1.0 mol/dm³. After 30 minutes, the strips were blotted and reweighed. The percentage change in mass was calculated.
一名学生切取等质量的土豆条，分别放入浓度从 0.0 mol/dm³ 到 1.0 mol/dm³ 的蔗糖溶液中。30 分钟后吸干表面水分再称重，计算质量变化百分比。
Typical data:
0.0 M: +18%
0.2 M: +9%
0.4 M: -2%
0.6 M: -11%
0.8 M: -20%
1.0 M: -28%
典型数据：
0.0 M: +18%
0.2 M: +9%
0.4 M: -2%
0.6 M: -11%
0.8 M: -20%
1.0 M: -28%
Plot a graph and find the concentration where the line crosses the zero-change axis. This is the approximate water potential of the potato cells. Explain mass gain by water entering by osmosis down the water potential gradient.
绘制图表，找出变化为零时对应的浓度，这大约是土豆细胞的水势。解释质量增加是因为水沿水势梯度通过渗透进入细胞。
Evaluate the method: results could be improved by using more repeats, controlling temperature, and fully drying the strips to the same degree each time.
评价实验方法：可通过增加重复次数、控制温度、每次用相同方式吸干土豆条来改进结果。
4. Surface Area to Volume Ratio and Diffusion | 表面积与体积比和扩散
Agar cubes containing phenolphthalein (pink in alkali) were cut into sides of 1 cm, 2 cm and 3 cm. They were placed in dilute hydrochloric acid, and the time taken for the pink colour to disappear was measured.
将含酚酞（碱中呈粉红色）的琼脂块切成边长 1 cm、2 cm 和 3 cm 的立方体，放入稀盐酸中，记录粉红色消失所需的时间。
e = 6:1; 2 cm = 3:1; 3 cm = 2:1. The smaller cube turned colourless fastest because diffusion distance is shorter.
随着立方体增大，表面积与体积比减小，因此酸扩散到中心所需时间更长。计算 SA:V 比：1 cm 立方体 = 6:1；2 cm = 3:1；3 cm = 2:1。较小的立方体最先褪色，因为扩散距离更短。
Relate this to organisms: small organisms can rely on diffusion alone, whereas larger organisms need transport systems to supply all cells efficiently.
将此与生物体联系起来：小生物可以仅靠扩散，而较大生物则需要运输系统以有效供应所有细胞。
5. Food Chains and Energy Transfer | 食物链与能量传递
snake. Biomass data (g/m²) were collected: grass 800, grasshopper 80, frog 10, snake 1.2.
草原生态系统中的一条食物链：草 → 蚱蜢 → 青蛙 → 蛇。收集到的生物量数据（g/m²）为：草 800，蚱蜢 80，青蛙 10，蛇 1.2。
of energy transferred between trophic levels. For grass to grasshopper: (80/800)×100 = 10%. This aligns with the typical 10% rule. Explain energy losses through respiration, uneaten parts, egestion and excretion.
计算营养级之间的能量传递百分比。草到蚱蜢：(80/800)×100 = 10%，符合常见的 10% 规律。解释能量损失：呼吸作用、未食用部分、排遗和排泄。
Draw a pyramid of biomass: it will be upright. Explain why pyramids of numbers may sometimes be inverted (e.g. one oak tree supports many insects).
绘制生物量金字塔：应为正锥形。解释数量金字塔有时为何倒置（如一棵橡树供养许多昆虫）。
6. Bacterial Growth Curve | 细菌生长曲线
A population of E. coli was cultured in nutrient broth at 37°C. Viable cell counts were taken every hour for 24 hours. The typical growth curve shows lag, log (exponential), stationary and death phases.
将大肠杆菌种群培养在 37°C 的营养肉汤中，每小时进行活菌计数，共 24 小时。典型的生长曲线显示延滞期、对数（指数）期、稳定期和衰亡期。
In the lag phase, cells are adjusting to the environment, synthesising enzymes, but not dividing rapidly. In the log phase, nutrients are plentiful and division is rapid – the population doubles at regular intervals.
在延滞期，细胞适应环境，合成酶，但分裂不活跃。对数期营养物质充足，分裂迅速，种群数量每隔固定时间翻倍。
Stationary phase occurs when nutrient depletion and waste accumulation slow growth; the death rate equals the birth rate. Finally, death phase sees a decline as toxins build up and nutrients run out.
稳定期出现在营养耗尽、废物积累导致生长减缓时，死亡速率等于新生速率。最后衰亡期由于毒素累积和营养枯竭，细胞数量下降。
om the log phase gradient. Use the formula: g = t/n, where n is the number of generations.
你可能需要根据对数期的斜率估算代时。使用公式：g = t/n，其中 n 为代数。
7. Antibiotic Resistance and Zone of Inhibition | 抗生素耐药性与抑菌圈
. After incubation, the diameters of the clear zones were: A = 24 mm, B = 10 mm, C = 0 mm.
将三种抗生素纸片（A、B、C）置于布满细菌的琼脂平板上。培养后，透明抑菌圈的直径分别为：A = 24 mm，B = 10 mm，C = 0 mm。
Interpretation: antibiotic A is the most effective against this bacterium; C shows complete resistance. When comparing, always measure the zone diameter and refer to the type of antibiotic.
解读：抗生素 A 对该细菌最有效；C 显示完全耐药。比较时，始终测量抑菌圈直径，并说明抗生素种类。
Discuss how resistance arises through mutation and natural selection. Overuse of antibiotics selects for resistant strains, making infections harder to treat.
讨论耐药性如何通过突变和自然选择产生。滥用抗生素会筛选出耐药菌株，使感染更难治疗。
8. Transpiration and Wind Speed | 蒸腾作用与风速
A potometer was used to measure the rate of water uptake by a cut shoot under different wind speeds (created by a fan). Results: still air: 0.5 cm³/min; low wind: 1.2 cm³/min; high wind: 2.0 cm³/min.
使用蒸腾计测量不同风速（由风扇产生）下切枝的吸水速率。结果：静止空气：0.5 cm³/min；低风速：1.2 cm³/min；高风速：2.0 cm³/min.
Explain that wind removes the layer of humid air around the stomata, increasing the water vapour concentration gradient between leaf and air, so transpiration speeds up.
解释：风将气孔周围潮湿的空气层吹走，增大了叶片与空气间的水蒸气浓度梯度，因此蒸腾加快。
Identify other factors that affect transpiration: temperature, humidity, light intensity. Emphasise that a potometer measures water uptake, not transpiration directly, because some water is used in photosynthesis or stored.
找出影响蒸腾作用的其它因素：温度、湿度、光照强度。强调蒸腾计测量的是吸水速率，而非直接测量蒸腾，因为部分水用于光合作用或被储存。
9. Digestion: Amylase and Temperature | 消化：淀粉酶与温度
A student mixed amylase solution with starch suspension at various temperatures and tested for starch using iodine solution every 30 seconds. The time taken for the iodine to remain yellow/brown was recorded.
一名学生在不同温度下将淀粉酶溶液与淀粉悬液混合，每隔 30 秒用碘液检测淀粉，记录碘液保持黄棕色的时间。
Data: at 20°C, 8 minutes; at 30°C, 4 minutes; at 40°C, 2 minutes; at 50°C, 3 minutes; at 60°C, no colour change (starch still present).
数据：20°C 需 8 分钟；30°C 需 4 分钟；40°C 需 2 分钟；50°C 需 3 分钟；60°C 时碘液不变色（淀粉仍存在）。
The optimum temperature is near 40°C. Above 50°C the enzyme denatures, so the active site cannot bind starch. Mention that kinetic energy increases with temperature up to the optimum, leading to more frequent collisions.
最适温度接近 40°C。超过 50°C 酶变性，活性位点无法结合淀粉。补充说明，在最适温度以下，动能随温度升高而增加，导致碰撞更频繁。
10. Respiration and Respirometer | 呼吸作用与呼吸计
A simple respirometer was set up with germinating peas in one tube and glass beads (equal volume) in a control tube. Coloured liquid moved towards the peas tube because oxygen was consumed and carbon dioxide absorbed by soda lime.
设置一个简单呼吸计，一个管放有萌发的豌豆，对照管放等体积的玻璃珠。有色液体向豌豆管移动，因为氧气被消耗，产生的二氧化碳被钠石灰吸收。
The distance moved by the liquid over 10 minutes was recorded at different temperatures: 10°C: 2 mm; 20°C: 8 mm; 30°C: 18 mm. Calculate the rate of oxygen uptake at each temperature.
记录不同温度下 10 分钟内液体移动距离：10°C: 2 mm；20°C: 8 mm；30°C: 18 mm。计算各温度下的氧气消耗速率。
Explain that warmer temperatures increase enzyme activity in respiration (more kinetic energy), so oxygen is used faster. Be careful to state that CO₂ is removed so the volume change is due only to oxygen uptake.
解释：较高温度增强呼吸作用中酶的活性（动能增大），因此氧气消耗更快。要注意说明二氧化碳被移除，因此体积变化仅由氧气消耗导致。
11. Sampling Techniques: Quadrats and Transects | 取样技术：样方与样带
Students investigated the distribution of plantain (Plantago major) on a school field. They placed 1 m² quadrats at 10 m intervals along a transect from a trampled area to an undisturbed area.
学生们调查了学校草地上车前草的分布。他们沿一条样带，从踩踏区到无人干扰区，每间隔 10 米放置一个 1 平方米的样方。
Results: disturbed zone: 2 plants/quadrat; transition: 8; undisturbed: 22. Abundance increases away from trampling. Describe how to make the sampling valid: randomise quadrats at each distance or take multiple transects.
结果：干扰区：每样方 2 株；过渡区：8 株；未干扰区：22 株。丰度随着远离踩踏而增加。描述如何使取样有效：在每个距离处随机放置样方或取多个样带。
Calculate the mean density and estimate the total population if the field area is known. Discuss the difference between percentage cover and frequency as measures of abundance.
计算平均密度，若已知草地面积可估算总种群数量。讨论覆盖百分比和频度作为丰度衡量指标的区别。
12. Heart Rate and Exercise | 心率与运动
A student recorded their resting heart rate (65 bpm), then completed 4 minutes of step-ups and measured heart rate every minute during recovery. Recovery data: 0 min post-exercise: 142 bpm; 1 min: 110; 2 min: 85; 3 min: 72; 4 min: 66.
一名学生记录静息心率（65 次/分），然后完成了 4 分钟踏步运动，并在恢复期每分钟测量心率。恢复数据：运动后即刻：142；1 分钟：110；2 分钟：85；3 分钟：72；4 分钟：66。
Plot a recovery curve. The heart rate falls quickly at first, then gradually returns to resting. Explain the need for extra oxygen after exercise to oxidise lactic acid produced during anaerobic respiration – this is the oxygen debt.
绘制恢复曲线。心率起初快速下降，随后缓慢恢复至静息水平。解释运动后需额外氧气来氧化无氧呼吸产生的乳酸——这就是氧债。
Link to control by the medulla oblongata and the hormone adrenaline. A shorter recovery time indicates better cardiovascular fitness.
联系到延髓的控制作用和肾上腺素。恢复时间越短，表明心血管适能越好。
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