Case Study Practice in Biology | 生物学案例分析实战演练

📚 Case Study Practice in Biology | 生物学案例分析实战演练

Case studies are a fantastic way to apply your KS3 biology knowledge to real or simulated situations. In this article, you will work through eight practical scenarios that involve reading data tables, spotting patterns, and drawing scientific conclusions. Each case is built around common CAIE-style investigations, helping you to sharpen your analytical skills and become confident in answering ‘suggest and explain’ type questions.

案例研究是将你的KS3生物学知识应用到真实或模拟情境中的绝佳方式。在本文中，你将完成八个实践场景，涉及阅读数据表、发现规律并得出科学结论。每个案例都围绕CAIE风格常见的探究活动设计，帮助你磨砺分析技能，在回答“提出并解释”类问题时更加自信。

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1. Enzyme Activity and Temperature | 酶活性与温度

An experiment was set up to investigate how temperature affects the activity of amylase. A starch solution was mixed with amylase at different temperatures, and the time taken for the starch to be fully broken down (until the iodine test no longer turned blue-black) was recorded. The results are shown below.

一项实验考察温度如何影响淀粉酶的活性。在不同温度下将淀粉溶液与淀粉酶混合，记录淀粉完全分解（直到碘液测试不再变为蓝黑色）所需的时间。结果如下所示。

Temperature (°C)	Time for starch breakdown (seconds)
10	210
20	130
30	55
37	30
45	48
60	195

From the data, the shortest breakdown time occurs at 37°C, which is close to human body temperature. This is the optimum temperature for amylase because the enzyme and substrate molecules have the most successful collisions, and the active site fits the substrate perfectly. As the temperature rises above 37°C, the time increases again because the enzyme starts to denature – its active site changes shape and can no longer bind to starch effectively. At 60°C the time is nearly as long as at 10°C, showing that most enzyme molecules are denatured.

从数据中看出，分解时间最短的是37°C，接近人体体温。这是淀粉酶的最适温度，因为酶分子与底物分子的有效碰撞最多，活性位点与底物完美契合。当温度升至37°C以上时，所需时间再次增加，这是因为酶开始变性——其活性位点形状改变，无法有效结合淀粉。在60°C时，所需时间几乎与10°C时一样长，说明大部分酶分子已经变性。

We can calculate the rate of reaction as 1/time (s⁻¹) for a simple comparison. At 37°C, rate = 1/30 ≈ 0.033 s⁻¹; at 10°C, rate = 1/210 ≈ 0.0048 s⁻¹. This shows the reaction is nearly seven times faster at the optimum temperature.

我们可以用反应速率 = 1/时间 (s⁻¹) 来进行简单比较。在37°C时，速率 = 1/30 ≈ 0.033 s⁻¹；在10°C时，速率 = 1/210 ≈ 0.0048 s⁻¹。这表明在最适温度下反应几乎快了七倍。

Rate = 1 / time (seconds)

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2. Photosynthesis and Light Intensity | 光合作用与光照强度

A student investigated the effect of light intensity on the rate of photosynthesis in aquatic plants. Pondweed was placed in a beaker of water, and a lamp was moved to different distances to vary the light intensity reaching the plant. The number of oxygen bubbles released per minute was counted. The results are summarised in the table.

一位学生探究了光照强度对水生植物光合作用速率的影响。将水草放入盛水的烧杯中，移动光源以改变到达植物的光照强度，并记录每分钟释放的氧气气泡数。结果总结如下表。

Distance of lamp from plant (cm)	Bubbles per minute
100	4
60	12
30	25
10	45
5	48

As the lamp is brought closer, light intensity increases, and the rate of photosynthesis rises sharply at first and then levels off. The decline in the rate of increase beyond 10 cm suggests that another factor, such as carbon dioxide concentration or temperature, is limiting the process. At the highest light intensity, the plant cannot photosynthesise any faster because the limiting factor is in short supply.

随着光源移近，光照强度增大，光合作用速率先急剧上升然后趋于平缓。在10 cm之后速率增幅减小，表明另一个因素（如二氧化碳浓度或温度）正成为限制因素。在最高光照强度下，植物无法再进行更快的光合作用，因为限制因素供给不足。

To compare rates more precisely, one could plot a graph of bubbles per minute against light intensity (or 1/distance²). The plateau indicates that the chloroplasts are saturated with light, and any extra light energy cannot be used unless the availability of CO₂ or water is increased.

为了更精确地比较速率，可以绘制每分钟气泡数对光照强度（或1/距离²）的图。曲线趋于平坦表明叶绿体对光的吸收已经饱和，除非增加二氧化碳或水分的供应，否则额外的光能无法被利用。

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3. Heart Rate and Exercise | 心率与运动

To investigate the effect of exercise on heart rate, a student measured her pulse while at rest, immediately after jogging for 2 minutes, after sprinting for 1 minute, and then again after 5 minutes of rest. The data are shown below.

为探究运动对心率的影响，一名学生测量了自己安静时、慢跑2分钟后立即、冲刺1分钟后立即、以及休息5分钟后的脉搏。数据如下所示。

Activity level	Heart rate (beats per minute)
Resting	72
After jogging	115
After sprinting	152
5 min after sprinting	88

Heart rate increases during exercise because muscles require more oxygen and glucose for aerobic respiration, and they produce more carbon dioxide that must be removed. The heart pumps faster to deliver these substances through the blood. Sprinting caused a greater increase than jogging because more muscle cells were working harder, demanding even more oxygen. After 5 minutes of rest, the heart rate dropped but had not yet returned to resting level, showing that the body was still repaying the oxygen debt built up during the intense burst.

运动时心率增加，因为肌肉需要更多的氧气和葡萄糖进行有氧呼吸，并产生更多需要清除的二氧化碳。心脏加快跳动，通过血液输送这些物质。冲刺比慢跑引起更大的心率提升，因为更多的肌肉细胞在更剧烈地工作，需要更多的氧气。休息5分钟后心率下降，但尚未恢复至安静水平，说明身体仍在偿还剧烈运动期间积累的氧债。

The recovery rate can be described as the difference between peak heart rate and the heart rate after 5 min, divided by 5. Here: (152 – 88) / 5 = 12.8 beats per minute decrease per minute. A fitter person would typically recover faster.

恢复速率可以用峰值心率与5分钟后心率之差除以5来描述。此处：(152 – 88) / 5 = 12.8 次/分/分钟的心率下降。体能更好的人通常恢复得更快。

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4. Osmosis in Potato Strips | 土豆条的渗透作用

Five identical potato strips were weighed, then placed into sugar solutions of different concentrations. After 30 minutes they were blotted dry and reweighed. The percentage change in mass was calculated. Results:

五根相同的土豆条称重后放入不同浓度的蔗糖溶液中。30分钟后吸干表面水分再次称重，计算质量变化百分比。结果如下：

Sugar concentration (mol/dm³)	Percentage change in mass (%)
0.0	+10.5
0.2	+4.2
0.4	–0.1
0.6	–5.8
0.8	–11.3

A positive change means the potato gained mass because water moved into the cells by osmosis from a region of higher water potential (the dilute solution) to a lower water potential inside the cells. A negative change means water moved out of the cells into the concentrated sugar solution. The point where the mass does not change (around 0.4 mol/dm³) represents the concentration of the potato cell sap – the water potentials are equal, so no net water movement occurs.

正变化表示土豆质量增加，因为水通过渗透作用从较高水势的稀溶液进入水势较低的细胞内。负变化说明水从细胞中流出进入浓糖溶液。质量几乎不变的点（约0.4 mol/dm³）代表土豆细胞液的浓度——此时水势相等，没有净水分移动。

From the table, we can estimate the isotonic point by interpolation: the line between +4.2% and –0.1% crosses zero near 0.38 mol/dm³. This is a classic experiment to determine the water potential of plant tissue.

从表中我们可以通过插值估算等渗点：+4.2%与–0.1%之间的连线在约0.38 mol/dm³处穿过零。这是测定植物组织水势的经典实验。

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5. Food Webs and Population Changes | 食物网与种群变化

Consider a simple food web in a grassland ecosystem: Grass → Rabbit → Fox. Additionally, a small insect (grasshopper) also feeds on grass, and the fox occasionally eats the grasshopper. A disease suddenly kills 80% of the rabbit population. Predict and explain the likely effects on the other populations over the following year.

考虑一个草原生态系统的简单食物网：草→兔→狐狸。此外，一种小昆虫（蚱蜢）也以草为食，狐狸偶尔会吃蚱蜢。一场疾病突然导致80%的兔子种群死亡。预测并解释接下来一年中其他种群可能发生的变化。

With far fewer rabbits, the fox population loses its main food source. Fox numbers are likely to decrease due to starvation or migration. Meanwhile, the grass population will increase because fewer rabbits are grazing. The grasshopper population may also increase initially because there is more grass available and less competition from rabbits. However, if fox numbers drop significantly, the predation pressure on grasshoppers might relax, allowing their numbers to rise even further. Eventually, the increased grasshopper numbers might reduce the grass, and the fox population might switch to eating more grasshoppers, partially regulating their numbers. The ecosystem would gradually stabilise at new population levels.

由于兔子数量骤减，狐狸失去了主要食物来源，狐狸数量可能因饥饿或迁移而下降。同时，由于啃食草的兔子减少，草的数量会增加。蚱蜢数量最初也可能增加，因为有更多草可用，且与兔子的竞争减弱。然而，如果狐狸数量显著下降，对蚱蜢的捕食压力会减轻，蚱蜢数量可能进一步上升。最终，增多的蚱蜢可能会减少草的数量，而狐狸可能转而更多地捕食蚱蜢，部分地控制其数量。生态系统将在新的种群水平上逐渐稳定下来。

This case illustrates how a change in one population sends ripple effects through the food web. It also shows the importance of alternative food sources in buffering the impact—foxes did not face total extinction because they could switch prey.

本案例说明了单一物种种群的变化如何在食物网中引发连锁反应。也体现了替代食物来源在缓冲冲击中的重要性——狐狸没有完全灭绝，因为它们能够转换猎物。

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6. Fermentation and Carbon Dioxide Production | 发酵与二氧化碳产生

Yeast cells were added to four different sugar solutions (glucose, fructose, sucrose, and lactose) in test tubes, each topped with a balloon to capture carbon dioxide. After one hour at 30°C, the circumference of each balloon was measured as an indicator of CO₂ produced. The results:

在四支试管中分别加入含有不同糖类（葡萄糖、果糖、蔗糖和乳糖）的溶液，并加入酵母细胞，每支试管用气球封口以捕集二氧化碳。在30°C下放置一小时后，测量每个气球的周长作为产生CO₂的指标。结果：

Sugar type	Balloon circumference (cm)
Glucose	28
Fructose	26
Sucrose	20
Lactose	6

Yeast ferments sugars anaerobically to produce ethanol and carbon dioxide. Glucose and fructose, being monosaccharides, are used directly by yeast for respiration, leading to large balloon circumferences. Sucrose is a disaccharide and must first be broken down by yeast enzymes, so the rate of CO₂ production is slightly lower. Lactose is poorly utilised by brewer’s yeast because the yeast lacks enough of the enzyme lactase to split lactose into its constituent sugars; hence very little CO₂ is produced.

酵母无氧发酵糖类产生乙醇和二氧化碳。葡萄糖和果糖是单糖，可直接被酵母利用进行呼吸，因此气球周长较大。蔗糖是二糖，必须首先被酵母酶分解，因此CO₂产生速率稍低。酿酒酵母不能很好地利用乳糖，因为缺乏足够的乳糖酶将其分解为组成部分，所以产生的CO₂极少。

If you calculate the fermentation rate as circumference per hour, glucose gives 28 cm/h, while lactose gives only 6 cm/h. This experiment helps students understand enzyme specificity and the role of sugars as respiratory substrates.

若将发酵速率计为每小时周长，葡萄糖为28 cm/h，而乳糖仅为6 cm/h。本实验有助于学生理解酶的专一性以及糖类作为呼吸底物的作用。

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7. Transpiration and Environmental Factors | 蒸腾作用与环境因素

A potometer was used to measure the rate of water uptake by a leafy shoot under different environmental conditions. The movement of an air bubble was recorded over 10 minutes. Four conditions were tested: still air, windy (fan), humid (plastic bag covering the shoot), and high-light intensity. Results:

使用气泡蒸腾计测量了不同环境条件下带叶枝条的吸水速率。记录气泡在10分钟内的移动距离。测试了四种条件：静止空气、有风（风扇）、湿润（塑料袋罩住枝条）和强光。结果：

Condition	Distance moved by bubble in 10 min (mm)
Still air, room light	18
Windy (fan)	34
High humidity (bag)	7
High light intensity	29

Wind increases transpiration because it removes the layer of moist air around the stomata, maintaining a steeper concentration gradient of water vapour. High light intensity opens stomata wider for photosynthesis, which also increases water loss. High humidity reduces transpiration because the air is already saturated with water, so the diffusion gradient is shallow.

风加快蒸腾作用，因为它吹走了气孔周围湿润的空气层，保持了较陡的水蒸气浓度梯度。强光使气孔为光合作用而开得更大，这也增加了水分流失。高湿度降低了蒸腾作用，因为空气已经充满水汽，扩散梯度较小。

The transpiration rate can be expressed in mm/min: windy, 3.4 mm/min; high light, 2.9 mm/min; still air, 1.8 mm/min; humid, 0.7 mm/min. In the bag, water vapour accumulated, slowing further evaporation from the mesophyll cells.

蒸腾速率可用mm/min表示：有风，3.4 mm/min；强光，2.9 mm/min；静止空气，1.8 mm/min；湿润，0.7 mm/min。塑料袋内水汽积聚，减缓了叶肉细胞水分的进一步蒸发。

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8. Diffusion and Temperature | 扩散与温度

A crystal of potassium permanganate was placed in a beaker of cold water (5°C) and in an identical beaker of warm water (40°C). The time taken for the purple colour to spread evenly throughout each beaker was recorded. The experiment was repeated with larger crystals and with stirring.

将一颗高锰酸钾晶体分别放入盛有冷水(5°C)和温水(40°C)的相同烧杯中，记录紫色均匀扩散至整个烧杯所需的时间。并重复实验，分别使用较大的晶体和进行搅拌。

Condition	Time for even colour (seconds)
Cold water (5°C), small crystal, unstirred	480
Warm water (40°C), small crystal, unstirred	95
Warm water (40°C), large crystal, unstirred	240
Warm water (40°C), small crystal, stirred	30

Diffusion was much faster in warm water because the water molecules and potassium permanganate particles have more kinetic energy, moving more rapidly and colliding more frequently. The larger crystal took longer to dissolve and diffuse simply because there was more solid to disperse. Stirring accelerated the process dramatically by distributing the particles through convection, eliminating the reliance on diffusion alone.

在温水中扩散快得多，因为水分子和高锰酸钾颗粒具有更大的动能，运动更快，碰撞更频繁。较大的晶体溶解和扩散耗时更长，仅仅因为需要分散的固体更多。搅拌通过流体对流使颗粒分布，显著加快了过程，消除了单纯依赖扩散的限制。

This simple investigation reinforces the concept that diffusion rate depends on temperature, concentration gradient, and particle size. In living organisms, diffusion of oxygen into blood and carbon dioxide out of cells obeys these same principles.

这个简单的探究强化了以下概念：扩散速率取决于温度、浓度梯度和颗粒大小。在生物体内，氧气进入血液和二氧化碳离开细胞的扩散都遵循同样的原理。

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By working through these case studies, you have practised interpreting data, identifying patterns, linking outcomes to core biological concepts, and performing basic calculations. Always read the question carefully, pay attention to units, and support your answers with evidence from the data provided. This approach will prepare you well for KS3 CAIE biology assessments and beyond.

通过完成这些案例研究，你已经练习了解读数据、识别规律、将结果与核心生物学概念联系起来并进行基本计算。始终仔细阅读问题、注意单位，并用所提供的数据来支持你的答案。这种方法将为你的KS3 CAIE生物学评估及后续学习打下扎实的基础。

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