📚 KS3 Cambridge Biology: Case Study Practice | KS3剑桥生物：案例分析实战演练
Case studies are a vital part of learning biology. They allow you to apply theoretical knowledge to real-world scenarios, develop analytical skills, and prepare for examinations. This article presents ten engaging case studies that cover key topics in the KS3 Cambridge biology syllabus. Follow each step to practise your scientific thinking.
案例研究是学习生物的重要部分。它们让你将理论知识应用于真实情境，培养分析能力，并为考试做好准备。本文提供了十个贴合 KS3 剑桥生物大纲的趣味案例。跟随每一步来练习你的科学思维。
1. Digestion of a Cheese Sandwich | 奶酪三明治的消化
A student eats a cheese sandwich made of bread, cheese and lettuce. The digestive system must break down carbohydrates, proteins and fats into small, soluble molecules that can be absorbed into the blood.
一名学生吃了一个由面包、奶酪和生菜制成的奶酪三明治。消化系统必须将碳水化合物、蛋白质和脂肪分解成小的可溶性分子，以便被吸收进入血液。
To analyse digestion, first identify the main nutrients. Bread provides starch, a carbohydrate. Cheese contains protein and fat. Lettuce has fibre but very little digestible nutrient, though it aids peristalsis.
要分析消化过程，首先要确定主要营养成分。面包提供淀粉，一种碳水化合物。奶酪含有蛋白质和脂肪。生菜虽含有纤维但几乎无可消化营养，不过它有助于肠蠕动。
Next, match each nutrient with the enzyme that breaks it down. Salivary and pancreatic amylase digest starch into maltose and then glucose. Protease (pepsin in the stomach and trypsin from the pancreas) breaks protein into amino acids. Lipase digests fats into fatty acids and glycerol, helped by bile that emulsifies fat.
接下来，为每种营养成分匹配分解它的酶。唾液和胰淀粉酶将淀粉消化为麦芽糖，再变为葡萄糖。蛋白酶（胃蛋白酶和胰蛋白酶）将蛋白质分解为氨基酸。脂肪酶将脂肪分解为脂肪酸和甘油，胆汁则帮助乳化脂肪。
Finally, link each product to absorption. Glucose and amino acids pass into the blood in the small intestine. Fatty acids and glycerol are absorbed into lacteals and later enter the blood. This case shows how different food groups require specific enzymes and produce distinct end-products.
最后，将每种产物与吸收联系起来。葡萄糖和氨基酸在小肠进入血液。脂肪酸和甘油被吸收进乳糜管，随后进入血液。这个案例说明不同食物组需要特定的酶，并产生不同的最终产物。
2. Light Intensity and Photosynthesis | 光照强度与光合作用
An experiment uses an aquatic plant such as Elodea. A lamp is placed at different distances from the plant, and the number of oxygen bubbles produced per minute is counted.
一项实验使用水草如伊乐藻。将一盏灯放在距离水草不同的位置，数出每分钟产生的氧气气泡数。
| Distance (cm) | Bubbles per minute |
|---|---|
| 10 | 42 |
| 20 | 35 |
| 30 | 26 |
| 40 | 15 |
| 50 | 6 |
What trend do you see? As the distance from the light source increases, the number of bubbles decreases. This is because less light reaches the plant, so the rate of photosynthesis falls.
你看到什么趋势？随着与光源距离增加，气泡数量减少。这是因为到达水草的光变少，所以光合作用速率下降。
The independent variable is the distance of the lamp; the dependent variable is the rate of photosynthesis, measured by bubble count. Controlled variables include temperature, carbon dioxide concentration and the type of plant.
自变量是灯的距离；因变量是光合作用速率，用气泡数衡量。控制变量包括温度、二氧化碳浓度和水草种类。
At very high light intensity, the rate may level off – another factor such as CO₂ or temperature becomes limiting. If we were to add sodium hydrogencarbonate to provide extra CO₂, we could see whether carbon dioxide was the limiting factor.
在很高的光照强度下，速率可能趋于平稳——另一个因素如二氧化碳或温度成为限制因子。如果加入碳酸氢钠以提供额外二氧化碳，就能检验二氧化碳是否为限制因子。
3. Energy Transfer in a Grassland Food Chain | 草原食物链的能量传递
Consider the food chain: grass → grasshopper → frog → snake → hawk. The arrows represent the direction of energy flow.
考虑这个食物链：草 → 蚱蜢 → 青蛙 → 蛇 → 鹰。箭头代表能量流动的方向。
Only about 10% of the energy in one trophic level is passed to the next. The rest is lost through movement, heat, respiration and uneaten parts.
每一营养级中的能量只有大约10%传递给下一级。其余部分经运动、热量、呼吸和未食用部分而散失。
| Trophic level | Energy available (kJ) |
|---|---|
| Grass (producer) | 10 000 |
| Grasshopper (primary consumer) | 1 000 |
| Frog (secondary consumer) | 100 |
| Snake (tertiary consumer) | 10 |
| Hawk (quaternary consumer) | 1 |
If the grass starts with 10 000 kJ of energy, only about 1 000 kJ reaches the grasshopper, 100 kJ reaches the frog, and so on. The hawk receives only 1 kJ. This explains why food chains rarely exceed five trophic levels – there is simply not enough energy to support another level.
如果草最初有10 000 kJ 能量，只有约1 000 kJ 到达蚱蜢，100 kJ 到达青蛙，依此类推。鹰只得到1 kJ。这解释了为什么食物链很少超过五个营养级——根本没有足够能量支持再高一级。
When constructing a pyramid of numbers or biomass, the producer level is almost always the largest. However, a pyramid of numbers can be inverted – for instance, one oak tree may support many insects.
在构建数量或生物量金字塔时，生产者层级几乎总是最大的。然而，数量金字塔有可能呈现倒置——例如一棵橡树可以支撑许多昆虫。
4. Bacterial Growth and Temperature | 细菌生长与温度
A class investigates how temperature affects bacterial growth. Nutrient agar plates are inoculated with the same volume of bacterial culture and incubated at different temperatures for 48 hours. The diameter of the clear zone around a disinfectant disc is measured, but here we simply record colony size.
一个班级研究温度如何影响细菌生长。他们在营养琼脂平板上接种等量细菌培养液，并在不同温度下培养48小时。测量消毒剂纸片周围的透明圈直径，但本例中我们只记录菌落大小。
| Temperature (°C) | Mean colony diameter (mm) |
|---|---|
| 5 | 2 |
| 15 | 6 |
| 25 | 14 |
| 37 | 24 |
| 50 | 8 |
| 65 | 0 |
Analysis shows that bacterial growth increases as temperature rises towards 37°C, the optimal temperature for many human pathogens. Beyond 37°C, growth slows and at 65°C the bacteria are killed because enzymes denature and the cell structure is damaged.
分析表明，当温度升至接近37°C时细菌生长加快，这是许多人类病原体的最适温度。超过37°C后生长变慢，在65°C时细菌被杀死，因为酶变性且细胞结构被破坏。
At 5°C growth is very slow but still possible; this is why refrigeration preserves food – it slows down microbial reproduction but does not necessarily kill all bacteria.
在5°C下生长十分缓慢但仍有可能；这就是冷藏能保存食物的原因——它减慢了微生物繁殖，但不一定能杀死所有细菌。
This case highlights the importance of controlling temperature in food storage and laboratory incubations. In school labs, plates are never incubated at 37°C to avoid growing human pathogens.
这个案例凸显了在食品储存和实验室培养中控制温度的重要性。在学校实验室中，培养皿绝不会在37°C下培养，以避免滋长人类病原体。
5. Response to Exercise | 运动中的身体反应
A student monitors his breathing rate and heart rate at rest, during light jogging and after sprinting. The data are recorded in a table.
一名学生监测自己在休息、慢跑和短跑冲刺后的呼吸频率和心率。数据记录在表格中。
| Activity | Breathing rate (breaths/min) | Heart rate (bpm) |
|---|---|---|
| Rest | 14 | 70 |
| Light jog | 24 | 105 |
| Sprinting | 38 | 160 |
During light jogging, muscles work harder and require more oxygen for aerobic respiration. The breathing rate increases to bring in more oxygen, and heart rate rises to pump oxygenated blood faster to the muscles. Carbon dioxide produced by respiration is removed more quickly.
在慢跑时，肌肉更努力工作，需要通过有氧呼吸获得更多氧气。呼吸频率加快以吸入更多氧，心率升高以更快地将含氧血泵至肌肉。呼吸产生的二氧化碳也被更快地清除。
After sprinting, the demand for oxygen is so great that the body cannot supply enough to the muscles. Muscles switch to anaerobic respiration, producing lactic acid. This causes muscle fatigue and cramps. The increased breathing rate after exercise helps repay the oxygen debt by converting lactic acid back to glucose in the liver.
短跑冲刺后，需氧量大到身体无法为肌肉提供充足氧气。肌肉转向无氧呼吸，产生乳酸。这导致肌肉疲劳和抽筋。运动后呼吸频率加快有助于偿还氧债，在肝脏将乳酸重新转化为葡萄糖。
6. Genetics of Pea Plant Height | 豌豆株高的遗传
Gregor Mendel crossed pure-breeding tall pea plants with pure-breeding short pea plants. All the F₁ offspring were tall. When F₁ plants were self-pollinated, the F₂ generation showed about 75% tall plants and 25% short plants.
格雷戈尔·孟德尔将纯种高茎豌豆与纯种矮茎豌豆杂交。所有F₁子代都是高茎。当F₁植株自花授粉后，F₂代出现了约75%的高茎植株和25%的矮茎植株。
Let the allele for tall be T (dominant) and the allele for short be t (recessive). The parental genotypes are TT and tt. The F₁ genotype is Tt, and all are tall because the dominant allele masks the recessive one.
假设高茎等位基因为 T（显性），矮茎等位基因为 t（隐性）。亲本基因型为 TT 和 tt。F₁ 基因型为 Tt，全部为高茎，因为显性等位基因掩盖了隐性等位基因。
Using a Punnett square for F₁ self-cross:
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
The expected offspring genotypes are TT, Tt, Tt and tt. This gives a phenotype ratio of 3 tall : 1 short. This classic 3:1 ratio demonstrates Mendel’s law of segregation.
预期的子代基因型为 TT、Tt、Tt 和 tt。这呈现表型比为3高茎 : 1矮茎。经典的3:1比例证明了孟德尔的分离定律。
If you are given actual results that differ slightly from the expected ratio, you can explain that fertilisation is a random process and that small sample sizes may cause deviation. Repeating the experiment many times brings results closer to the predicted ratio.
如果实际结果与预期比例略有不同，可以解释为受精是随机过程，且小样本量可能导致偏差。多次重复实验会使结果更接近预测比例。
7. Estimating a Plant Population Using Quadrats | 用样方估计植物种群数量
Ecologists want to estimate the number of dandelions in a school field measuring 50 m × 30 m. Ten 1 m² quadrat samples are taken using random coordinates.
生态学家想估算一片50米×30米的校园草地中蒲公英的数量。他们利用随机坐标取了十个1平方米的样方。
| Quadrat | Dandelions counted |
|---|---|
| 1 | 5 |
| 2 | 7 |
| 3 | 4 |
| 4 | 6 |
| 5 | 8 |
| 6 | 5 |
| 7 | 9 |
| 8 | 6 |
| 9 | 4 |
| 10 | 7 |
First calculate the mean number of dandelions per quadrat: (5+7+4+6+8+5+9+6+4+7) / 10 = 6.1. The total area is 50 × 30 = 1500 m². Estimated total dandelions = mean per m² × total area = 6.1 × 1500 = 9150.
首先计算每个样方的平均蒲公英数：(5+7+4+6+8+5+9+6+4+7) / 10 = 6.1。总面积是50 × 30 = 1500平方米。估算的蒲公英总数 = 每平方米平均值 × 总面积 = 6.1 × 1500 = 9150株。
The method relies on truly random sampling. If all quadrats were placed in sunny patches, the estimate might be too high. Random coordinates generated by a computer or a random number table reduce bias. Several replicates improve reliability.
该方法依赖真正的随机取样。如果所有样方都放在阳光充足的区域，估算值可能偏高。使用电脑或随机数表生成随机坐标可以减少偏差。多次重复能提高可靠性。
8. Modelling the Spread of a Cold Virus | 模拟感冒病毒的传播
In a class of 30 students, one person comes to school with a cold. Each day, every infected person passes the virus to two new people. Those who catch the cold stay infectious for two days and then recover and become immune.
在一个有30名学生的班级中，有一人带着感冒来到学校。每天每个感染者将病毒传给两个新人。感染者持续两天具有传染性，之后康复并获得免疫。
We can model the cumulative number of cases in a table:
| Day | New cases today | Total infected so far |
|---|---|---|
| 0 | 1 | 1 |
| 1 | 2 | 3 |
| 2 | 4 | 7 |
| 3 | 8 | 15 |
| 4 | 16 | 31 |
On day 4 the total exceeds 30. This shows how quickly a disease can spread in a confined population. In reality, spread slows down as fewer susceptible people remain, and public health measures such as hand washing and isolating sick students reduce transmission.
到第4天总感染人数超过30。这说明疾病在封闭人群中传播速度有多快。现实中，随着易感人群减少，传播会慢下来；诸如洗手和隔离生病学生等公共卫生措施可减少传播。
This exercise helps explain the shape of an epidemic curve. The initial exponential rise occurs when every case generates more than one new case. The effective reproduction number (R) is greater than 1.
这个练习有助于解释流行曲线的形状。当每个病例产生超过一个新病例时，会出现初始的指数增长。有效再生数(R)大于1。
9. Osmosis in Potato Strips | 马铃薯条的渗透作用
Identical potato strips are weighed and placed into five beakers containing different sucrose solutions: 0.0 M, 0.2 M, 0.4 M, 0.6 M, 0.8 M. After 30 minutes, the strips are blotted dry and reweighed. The percentage change in mass is calculated.
质量相同的马铃薯条被称重后放入五个含有不同蔗糖溶液的烧杯中：0.0 M、0.2 M、0.4 M、0.6 M、0.8 M。30分钟后，吸干表面水分并再次称重。计算质量变化百分比。
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