Year 10 Edexcel Biology: Case Study Practice | Year 10 Edexcel 生物：案例分析实战演练

📚 Year 10 Edexcel Biology: Case Study Practice | Year 10 Edexcel 生物：案例分析实战演练

Case study questions are an important part of Edexcel GCSE Biology. They test your ability to apply knowledge to unfamiliar scenarios, analyse data, and draw conclusions. This article provides eight practice cases with step-by-step guidance to help you build confidence.

案例分析题是Edexcel GCSE生物考试的重要组成部分。它们考察你将知识应用于陌生情境、分析数据并得出结论的能力。本文提供八个练习案例，并附有逐步指导，帮助你建立自信。

1. Investigating Enzyme Activity | 探究酶活性

A student investigated the effect of temperature on the activity of catalase. Catalase breaks down hydrogen peroxide (H₂O₂) into water (H₂O) and oxygen (O₂). She measured the volume of oxygen produced in 30 seconds at different temperatures.

一名学生探究了温度对过氧化氢酶活性的影响。过氧化氢酶将过氧化氢 (H₂O₂) 分解为水 (H₂O) 和氧气 (O₂)。她测量了在不同温度下30秒内产生的氧气体积。

Temperature (°C)	Volume of O₂ (cm³) in 30 s
10	5
20	15
30	28
35	32
40	20
50	2

Question: Identify the optimum temperature for this enzyme. Explain why enzyme activity decreases at temperatures above the optimum.

问题：确定该酶的最适温度。解释为何在高于最适温度时酶活性下降。

Guidance: Look for the temperature at which the highest volume of oxygen was produced. Recall that high temperatures can denature the enzyme by changing the shape of its active site.

指导：寻找产生氧气体积最大的温度。回忆高温会使酶变性，改变其活性位点的形状。

Answer: The optimum temperature is 35 °C. Above this, the rate decreases because the enzyme is denatured. The active site no longer fits the substrate, so fewer enzyme-substrate complexes form.

答案：最适温度为35 ℃。高于此温度，速率下降，因为酶变性。活性位点不再适合底物，因此形成更少的酶-底物复合物。

2. Osmosis in Plant Tissue | 植物组织中的渗透作用

A student placed potato cylinders of equal mass into different concentrations of sucrose solution. After 24 hours, she recorded the change in mass. Her results are shown below.

一名学生将等质量的土豆条放入不同浓度的蔗糖溶液中。24小时后，她记录了质量变化。结果如下所示。

Sucrose concentration (mol/dm³)	Percentage change in mass (%)
0.0	+12
0.2	+5
0.4	−1
0.6	−8
0.8	−15

Question: Estimate the concentration of sucrose solution that has the same water potential as the potato cells. Explain how osmosis causes the mass changes.

问题：估算与土豆细胞水势相等的蔗糖溶液浓度。解释渗透作用如何导致质量变化。

Guidance: The point where mass change is zero indicates no net water movement. If the solution is more dilute than the cell contents, water enters and mass increases; if more concentrated, water leaves and mass decreases.

指导：质量变化为零的点表明没有净水流动。如果溶液比细胞内容物稀释，水进入，质量增加；如果更浓，水流失，质量减少。

Answer: The isotonic concentration is approximately 0.35 mol/dm³ (between 0.2 and 0.4 where mass change crosses zero). In 0.0 and 0.2 mol/dm³, water entered by osmosis from a high to a low water potential, so mass increased. In higher concentrations, water left the cells, so mass decreased.

答案：等渗浓度约为0.35 mol/dm³（介于0.2和0.4之间，质量变化过零点）。在0.0和0.2 mol/dm³中，水通过渗透作用从高水势流向低水势，因此质量增加。在更高浓度中，水离开细胞，所以质量减少。

3. Food Tests Mystery | 食物测试谜题

A nutritionist tested an unknown food sample. She obtained these results:

一位营养学家测试了一种未知食物样本。她得到了如下结果：

Biuret test: solution turned purple

双缩脲测试：溶液变为紫色
Iodine test: stayed orange-brown

碘液测试：保持橙棕色
Benedict’s test after heating: brick-red precipitate

加热后本尼迪克特测试：砖红色沉淀
Ethanol emulsion test: cloudy white layer formed

乙醇乳化测试：形成云雾状白色层

Question: Identify the nutrients present in the sample. Explain which nutrient is absent.

问题：确定样本中存在的营养物质。解释哪种营养物质不存在。

Guidance: Recall what each test detects: Biuret for protein, iodine for starch, Benedict’s for reducing sugars (glucose, etc.), and ethanol emulsion for lipids (fats/oils).

指导：回忆每个测试检测什么：双缩脲检测蛋白质，碘液检测淀粉，本尼迪克特检测还原糖（葡萄糖等），乙醇乳化检测脂类（脂肪/油）。

Answer: The sample contains protein (purple Biuret), reducing sugar (brick-red Benedict’s), and lipids (cloudy emulsion). Starch is absent because the iodine test remained orange-brown, not blue-black.

答案：样本含有蛋白质（双缩脲呈紫色）、还原糖（本尼迪克特砖红色）和脂类（云雾状乳液）。淀粉不存在，因为碘液测试保持橙棕色，而非蓝黑色。

4. Heart Rate and Exercise | 心率与运动

A student measured her resting heart rate as 68 bpm. She then ran on the spot for 5 minutes and recorded her heart rate every minute during recovery.

一名学生测量了她的静息心率为68次/分。然后她原地跑步5分钟，并在恢复期间每分钟记录一次心率。

Time after exercise (min)	Heart rate (bpm)
0	152
1	130
2	108
3	90
4	78
5	70

Question: Calculate the percentage increase in heart rate from resting to immediately after exercise. Explain why the heart rate remains elevated during recovery.

问题：计算从静息到运动后即刻心率的百分比增加。解释为何恢复期间心率仍较高。

Guidance: Percentage increase = (increase ÷ original) × 100. During exercise, muscles respire more, producing extra carbon dioxide and requiring more oxygen. After exercise, oxygen debt must be repaid, so the heart continues to pump faster.

指导：百分比增加 = (增加量 ÷ 原值) × 100。运动期间，肌肉呼吸增强，产生更多二氧化碳并需要更多氧气。运动后需偿还氧债，因此心脏继续快速泵血。

Answer: Increase = (152 − 68) / 68 × 100 = 123.5%. Heart rate stays high to deliver oxygen for aerobic respiration and to remove lactic acid built up during anaerobic respiration. It gradually falls as the oxygen debt is repaid.

答案：增加 = (152 − 68) / 68 × 100 = 123.5%。心率保持高位以输送氧气用于有氧呼吸，并清除无氧呼吸期间积累的乳酸。随着氧债被偿还，心率逐渐下降。

5. Genetics: Breeder’s Problem | 遗传学：育种者问题

A plant breeder crosses two pea plants. One parent is heterozygous tall (Tt) and the other is homozygous dwarf (tt). The allele for tall (T) is dominant over dwarf (t).

一位植物育种者将两株豌豆杂交。一个亲本是杂合高茎 (Tt)，另一个是纯合矮茎 (tt)。高茎等位基因 (T) 对矮茎 (t) 为显性。

Question: Draw a Punnett square to show the possible genotypes of the offspring. What is the expected ratio of tall to dwarf plants?

问题：绘制邦尼特方阵显示后代可能的基因型。预期高茎与矮茎的比例是多少？

Guidance: Place the gametes of one parent along the top and the other along the side. Combine alleles to fill the grid. Then identify phenotypes.

指导：将一个亲本的配子放在顶部，另一个放在侧面。组合等位基因填充网格。然后确定表现型。

Answer: Gametes from Tt: T and t; from tt: t and t. Punnett square:

答案：来自 Tt 的配子：T 和 t；来自 tt 的配子：t 和 t。邦尼特方阵：

	t	t
T	Tt (tall)	Tt (tall)
t	tt (dwarf)	tt (dwarf)

Genotype ratio: 2 Tt : 2 tt. Phenotype ratio: 1 tall : 1 dwarf (or 50% tall, 50% dwarf).

基因型比例：2 Tt : 2 tt。表现型比例：1 高 : 1 矮（或 50% 高，50% 矮）。

6. Ecosystem Sampling | 生态系统取样

Students used a 0.5 m × 0.5 m quadrat to estimate the population of clover plants in a field. They threw the quadrat randomly and repeated ten times. The counts of clover plants per quadrat were: 8, 12, 9, 0, 15, 7, 11, 10, 6, 13. The field is 100 m × 40 m.

学生使用0.5 m × 0.5 m的样方估算田地里三叶草的数量。他们随机投掷样方，重复十次。每个样方三叶草植株数量为：8, 12, 9, 0, 15, 7, 11, 10, 6, 13。田地面积为100 m × 40 m。

Question: Calculate the mean number of clovers per quadrat. Then estimate the total population of clovers in the field.

问题：计算每个样方三叶草的平均数量。然后估算田地里三叶草的总种群数量。

Guidance: Mean = sum of counts ÷ number of quadrats. Area of quadrat = 0.25 m². To estimate total, multiply mean by (total field area ÷ quadrat area).

指导：平均值 = 计数总和 ÷ 样方数量。样方面积 = 0.25 m²。估算总量时，用平均值乘以（田地总面积 ÷ 样方面积）。

Answer: Sum = 8+12+9+0+15+7+11+10+6+13 = 91. Mean = 91 ÷ 10 = 9.1 clovers per quadrat. Field area = 4000 m². Number of quadrats in field = 4000 ÷ 0.25 = 16000. Estimated total = 9.1 × 16000 = 145,600 clover plants.

答案：总和 = 8+12+9+0+15+7+11+10+6+13 = 91。平均值 = 91 ÷ 10 = 9.1 株/样方。田地面积 = 4000 m²。样方数 = 4000 ÷ 0.25 = 16000。估算总量 = 9.1 × 16000 = 145,600 株三叶草。

7. Photosynthesis Investigation | 光合作用探究

A student placed pondweed in a beaker of water and counted the number of oxygen bubbles released per minute at different light intensities. Light intensity was varied by moving a lamp. She kept the same distance for each reading and added sodium hydrogencarbonate to provide CO₂.

一名学生将水草放入烧杯水中，在不同光照强度下计数每分钟释放的氧气气泡数。通过移动台灯改变光照强度。她每次保持同一距离读数，并加入碳酸氢钠提供 CO₂。

Results:

结果：

Distance of lamp (cm)	Bubbles per minute
10	45
20	33
30	21
40	12
50	3

Question: Describe the relationship between light intensity and the rate of photosynthesis. Explain why the rate changes as the lamp is moved further away.

问题：描述光照强度与光合作用速率之间的关系。解释为什么随着台灯移远，速率发生变化。

Guidance: As distance increases, light intensity decreases (inverse square law). Photosynthesis requires light energy to split water and produce ATP. Less light means slower light-dependent reactions.

指导：随着距离增加，光照强度降低（平方反比定律）。光合作用需要光能分解水并产生 ATP。光越少，光依赖反应越慢。

Answer: As light intensity decreases, the rate of photosynthesis falls (indicated by fewer bubbles). This is because less light energy is available to drive the light-dependent reactions, reducing the production of ATP and reduced NADP. Less CO₂ is fixed, so oxygen production drops.

答案：随着光照强度降低，光合作用速率下降（表现为气泡减少）。这是因为可利用的光能减少，驱动光依赖反应的能力降低，减少了 ATP 和还原型辅酶Ⅱ的生成。固定的 CO₂ 减少，故氧气产量下降。

8. Disease Transmission | 疾病传播

An outbreak of influenza occurred in a school. Public health officials recommended vaccination, hand washing, and staying home when ill. Data showed that classes where more students were vaccinated had fewer secondary cases.

一所学校爆发了流感。公共卫生官员建议接种疫苗、洗手以及生病时居家。数据显示，接种学生较多的班级继发病例较少。

Question: Explain how vaccination helps control the spread of influenza. Suggest two other ways the school could reduce transmission, using biological knowledge.

问题：解释接种疫苗如何帮助控制流感传播。利用生物学知识，提出另外两种学校可以降低传播的方法。

Guidance: Vaccination stimulates the immune system to produce memory cells, so if the pathogen enters later, a rapid response prevents illness and reduces spread. Other ways: good hygiene kills pathogens on surfaces; covering coughs reduces airborne droplets; isolation reduces contact.

指导：疫苗刺激免疫系统产生记忆细胞，如果之后病原体侵入，快速反应可防止生病并减少传播。其他方法：良好卫生可杀灭表面病原体；咳嗽遮掩减少飞沫传播；隔离减少接触。

Answer: Vaccination causes the body to produce antibodies and memory lymphocytes specific to influenza. If the vaccinated person encounters the virus, their secondary immune response is faster and stronger, often preventing infection and reducing the chance of passing it on. The school could also enforce regular hand washing with soap (destroying viral envelopes) and ensure proper ventilation to disperse droplets. Ill students should stay home to break the chain of transmission.

答案：疫苗接种使身体产生针对流感的抗体和记忆淋巴细胞。如果接种者遇到病毒，他们的二次免疫反应更快更强，通常能防止感染并降低传播机会。学校还可以强制使用肥皂洗手（破坏病毒包膜），并确保适当通风以驱散飞沫。生病学生应居家，以切断传播链。

Published by TutorHao | Biology Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Year 10 Edexcel Biology: Case Study Practice | Year 10 Edexcel 生物：案例分析实战演练