📚 Case Study Practice Drills for Year 10 AQA Biology | AQA 10年级生物案例分析实战演练
Welcome to this collection of case study practice drills designed specifically for Year 10 AQA Biology. In these real-world scenarios, you will apply your knowledge of cells, organisation, infection and response, bioenergetics, homeostasis, inheritance, and ecology. Each case study challenges you to think like a scientist, analyse data, and draw evidence-based conclusions.
欢迎参与这套专为AQA 10年级生物课程设计的案例分析实战演练。通过这些真实世界的情境，你将运用细胞、组织、感染与反应、生物能学、稳态、遗传和生态学等知识。每个案例都要求你像科学家一样思考，分析数据并得出基于证据的结论。
1. Identifying a Digestive System Disorder | 消化系统疾病的识别
A 45-year-old patient reports severe pain in the upper abdomen, especially after eating fatty meals. An ultrasound reveals small, hard deposits in the gallbladder. The doctor suspects that bile release is obstructed.
一位45岁患者报告上腹部剧烈疼痛，尤其在进食油腻食物后。超声检查显示胆囊内存在小而坚硬的沉积物。医生怀疑胆汁排放受阻。
Which substance in bile is instrumental in breaking down large fat globules into smaller droplets, and what physical process is this?
胆汁中哪种物质有助于将大脂肪球分解成小油滴，这属于什么物理过程？
Bile contains bile salts, which emulsify fats. This emulsification increases the surface area of fats for lipase enzymes to act upon more efficiently. It is a physical, not chemical, digestion step.
胆汁含有胆盐，能够乳化脂肪。这种乳化作用增大了脂肪的表面积，使脂肪酶能更高效地作用。这是一个物理消化步骤而非化学消化。
Suggest why symptoms worsen after a fatty meal. Your answer should refer to the role of bile in digestion.
请解释为何症状在食用油腻食物后加重。你的答案应提到胆汁在消化中的作用。
After a fatty meal, the gallbladder normally contracts to release bile. If the duct is blocked by gallstones, bile cannot reach the small intestine. Fat remains undigested, causing discomfort, bloating, and pain.
进食油腻食物后，胆囊通常会收缩并释放胆汁。如果胆管被结石堵塞，胆汁无法到达小肠。脂肪停留在未消化状态，引起不适、腹胀和疼痛。
2. Investigating the Effect of Temperature on Enzyme Activity | 探究温度对酶活性的影响
Students conducted an experiment using the enzyme amylase, which breaks down starch. They mixed amylase with starch solution at different temperatures and recorded the time taken for starch to be completely digested, using iodine solution to test for the presence of starch.
学生们使用分解淀粉的淀粉酶进行实验。他们在不同温度下将淀粉酶与淀粉溶液混合，并记录淀粉完全消化所需的时间，用碘液测试淀粉是否存在。
| Temperature (°C) | Time for starch to disappear (seconds) |
|---|---|
| 10 | 180 |
| 20 | 90 |
| 30 | 45 |
| 40 | 20 |
| 50 | 35 |
| 60 | No digestion |
Explain why the reaction time decreases from 10°C to 40°C, using the concept of kinetic energy and enzyme-substrate collisions.
用动能和酶-底物碰撞的概念解释为什么反应时间从10°C到40°C逐渐缩短。
As temperature rises, the kinetic energy of enzyme and substrate molecules increases. They move faster and collide more often, leading to a higher rate of successful enzyme-substrate complex formation and faster starch breakdown.
随着温度升高，酶和底物分子的动能增加。它们移动更快，碰撞更频繁，导致酶-底物复合物成功形成的速率提高，淀粉分解加快。
What has happened at 60°C to prevent starch digestion entirely? Describe the structural change in the enzyme.
在60°C时淀粉完全不被消化，发生了什么？描述酶的结构变化。
At 60°C, the heat energy breaks hydrogen bonds and other interactions that maintain the enzyme’s tertiary structure. The active site loses its specific shape; the enzyme becomes denatured and can no longer bind starch, so no reaction occurs.
在60°C时，热能破坏了维持酶三级结构的氢键和其他作用力。活性位点失去其特定形状；酶变性，不能再结合淀粉，因此无反应发生。
3. Interpreting a Family Pedigree for Cystic Fibrosis | 解读囊性纤维化的家族系谱
Cystic fibrosis (CF) is a recessive genetic disorder caused by a mutation in the CFTR gene. The following diagram shows a family pedigree. Squares represent males, circles represent females. Shaded symbols indicate individuals with cystic fibrosis. Unshaded symbols are unaffected. Individual 3 and 4 are parents of children 7, 8, and 9.
囊性纤维化(CF)是一种由CFTR基因突变引起的隐性遗传病。下图展示一个家族系谱。方块代表男性，圆圈代表女性。阴影符号表示患有囊性纤维化的个体。无阴影者为未患病。个体3和4是子女7、8、9的父母。
ed) and 4 (female, unaffected). 3 marries 4. Generation II children: 5 (female, unaffected), 6 (male, CF), 7 (female, unaffected), 8 (male, CF), 9 (female, unaffected). Assume that individual 1 is homozygous dominant and individual 2 is heterozygous.
系谱：第I代：1（男性，未患病）与2（女性，未患病）结婚 → 子女：3（男性，未患病）和4（女性，未患病）。3与4结婚。第II代子女：5（女，未患病），6（男，CF），7（女，未患病），8（男，CF），9（女，未患病）。假设个体1是显性纯合子，个体2是杂合子。
Deduce the genotypes of individuals 3 and 4. Use F for the normal allele and f for the cystic fibrosis allele.
推断个体3和4的基因型。用F表示正常等位基因，f表示囊性纤维化等位基因。
Because cystic fibrosis is recessive, affected individuals 6 and 8 must have genotype ff. Both parents (3 and 4) are unaffected but have produced affected children, so each parent must be a carrier with genotype Ff.
由于囊性纤维化是隐性遗传，患病的个体6和8 必定是基因型ff。父母（3和4）均未患病但生出了患病子女，因此双方必定是携带者，基因型均为Ff。
Calculate the probability that the next child born to individuals 3 and 4 will have cystic fibrosis. Show your reasoning using a Punnett square.
计算个体3和4的下一个孩子患囊性纤维化的概率。用旁氏方格展示推理过程。
A cross between two Ff parents produces a genotypic ratio of 1 FF : 2 Ff : 1 ff. Only the ff offspring are affected. Therefore, the probability is 1/4 or 25%.
两个Ff亲本杂交产生的基因型比例为1 FF : 2 Ff : 1 ff。只有ff后代患病。因此概率为1/4或25%。
4. Evaluating the Impact of Vaccination on Whooping Cough | 评估百日咳疫苗接种的影响
Whooping cough (pertussis) is a highly contagious bacterial infection. The graph below shows the number of reported whooping cough cases in a region from 1970 to 1990. In 1974, a national immunisation programme using the DTP vaccine was introduced. Before 1974, annual cases averaged 80,000. By 1978, cases dropped to 2,000. However, in 1982, adverse media reports led to a decline in vaccine uptake, and cases rose again to 15,000 by 1986.
百日咳是一种高传染性细菌感染。下图显示某地区1970年至1990年间报告的百日咳病例数。1974年，全国推行使用百白破疫苗的免疫计划。1974年前，年均病例为80,000例。到1978年，病例降至2,000例。然而，1982年负面媒体报道导致疫苗接种率下降，病例到1986年再次上升至15,000例。
Explain how vaccination provided herd immunity that reduced cases so dramatically by 1978.
解释疫苗接种如何提供群体免疫，使病例到1978年大幅减少。
The DTP vaccine contains weakened or inactivated forms of the pathogen, stimulating the immune system to produce memory lymphocytes. When a high proportion of the population is vaccinated, the pathogen cannot spread easily because most individuals are immune. This protects even those who are not vaccinated, a phenomenon called herd immunity.
百白破疫苗含有减毒或灭活的病原体，刺激免疫系统产生记忆淋巴细胞。当人口中大部分接种疫苗后，病原体难以传播，因为多数人有免疫力。这甚至能保护未接种者，此现象称为群体免疫。
Analyse why a drop in vaccination uptake led to a resurgence of whooping cough cases after 1982. Link your answer to the concept of immunisation thresholds.
分析为何疫苗接种率下降导致1982年后百日咳病例回升。将你的答案与免疫阈值概念联系起来。
Herd immunity requires a critical percentage of the population to be immune (the threshold, often above 90% for highly contagious diseases). When vaccine uptake fell, the proportion of susceptible individuals increased, breaking the herd immunity barrier. The pathogen circulated more widely, causing more infections.
群体免疫需要人口中达到一个关键免疫比例（阈值，对于高传染性疾病通常需90%以上）。当疫苗接种率下降，易感个体比例上升，打破了群体免疫屏障。病原体传播更广，导致感染增加。
5. Understanding the Evolution of Antibiotic Resistance | 理解抗生素耐药性的进化
Staphylococcus aureus is a common bacterium that can cause skin infections. Methicillin-resistant Staphylococcus aureus (MRSA) has become a serious problem in hospitals. The initial population of bacteria included a few cells that possessed a resistance gene due to a random mutation. When treated with methicillin, most bacteria died, but resistant ones survived and reproduced.
金黄色葡萄球菌是一种常见的可引发皮肤感染的细菌。耐甲氧西林金黄色葡萄球菌(MRSA)已成为医院中的严重问题。初始细菌群体中包含少数因随机突变而拥有耐药基因的细胞。使用甲氧西林处理后，大多数细菌死亡，但耐药菌存活并繁殖。
Describe how natural selection has led to the rise of MRSA. Include the terms variation, selection pressure, survival advantage, and allele frequency.
描述自然选择如何导致MRSA的兴起。需包含变异、选择压力、生存优势和等位基因频率等术语。
Within the bacterial population, genetic variation existed: some bacteria carried a resistance allele. The antibiotic methicillin acted as a strong selection pressure. Sensitive bacteria were killed, while resistant ones had a survival advantage. They survived and reproduced, passing the resistance allele to offspring. Over generations, the frequency of the resistance allele increased, leading to a resistant population.
在细菌群体内存在遗传变异：一些细菌携带耐药等位基因。抗生素甲氧西林作为强大的选择压力。敏感菌被杀死，而耐药菌具有生存优势。它们存活并繁殖，将耐药等位基因传递给后代。经过多代，耐药等位基因频率升高，形成耐药群体。
Suggest why hospitals now use combinations of different antibiotics to treat serious infections, rather than a single antibiotic.
提出为何医院现在使用多种抗生素联合治疗严重感染，而不是单一抗生素。
A bacterium with a mutation conferring resistance to one antibiotic is unlikely to simultaneously carry resistance to multiple unrelated antibiotics. Using a combination reduces the chance that any bacterium will survive and reproduce. This slows the evolution of multi-drug resistance.
对一种抗生素产生耐药突变的细菌，不太可能同时携带对多种不相关抗生素的耐药性。联合用药降低了任何细菌存活并繁殖的机会，从而减缓多重耐药性的进化。
6. Evaluating Risk Factors for Coronary Heart Disease | 评估冠心病的风险因素
A health study followed 10,000 adults aged 40-60 for 15 years, recording lifestyle factors and incidence of coronary heart disease (CHD). Key data: smokers had a CHD rate of 12.5%, non-smokers 5.2%. Those with a diet high in saturated fat had a CHD rate of 14.0% compared to 6.1% for those with low saturated fat intake. Individuals with high blood pressure (above 140/90 mmHg) showed a CHD rate of 18.3%, while those with normal blood pressure showed 4.8%. Participants who were physically inactive had a CHD rate of 11.0% vs. 5.5% for active individuals.
一项健康研究对10,000名40-60岁成年人进行15年随访，记录生活方式因素和冠心病(CHD)发病率。关键数据：吸烟者CHD发病率为12.5%，非吸烟者5.2%。饱和脂肪摄入高的人群CHD率为14.0%，低饱和脂肪摄入者为6.1%。高血压患者（高于140/90 mmHg）CHD率为18.3%，血压正常者为4.8%。不锻炼者CHD率11.0%，经常锻炼者5.5%。
Identify the one factor from the data that shows the strongest correlation with increased CHD risk, and explain the physiological reason.
从数据中找出与CHD风险增加相关性最强的一个因素，并解释其生理原因。
High blood pressure shows the strongest correlation (18.3% vs 4.8%). High blood pressure damages the smooth endothelial lining of arteries, increasing the likelihood of atheroma formation. Fatty deposits build up, narrowing the lumen and increasing the risk of blood clots that can block coronary arteries.
高血压表现出最强相关性（18.3%对4.8%）。高血压会损伤动脉光滑的内皮衬里，增加动脉粥样斑块形成的可能性。脂肪沉积积聚，使管腔变窄，增加可阻塞冠状动脉的血栓风险。
Explain how a diet high in saturated fat can lead to atherosclerosis and eventually a heart attack.
解释高饱和脂肪饮食如何导致动脉粥样硬化并最终引发心脏病发作。
Saturated fats raise the level of low-density lipoprotein (LDL) cholesterol in the blood. Excess LDL cholesterol accumulates in arterial walls, triggering an inflammatory response and forming fatty streaks, which develop into plaques. These plaques narrow arteries and can rupture, causing blood clots. If a clot blocks a coronary artery, the heart muscle is deprived of oxygen, leading to a myocardial infarction (heart attack).
饱和脂肪会升高血液中低密度脂蛋白胆固醇水平。过多的LDL胆固醇积聚在动脉壁，引发炎症反应并形成脂肪条纹，进而发展为斑块。斑块使动脉变窄并可能破裂，形成血栓。如果血栓阻塞冠状动脉，心肌缺氧，导致心肌梗死（心脏病发作）。
7. Investigating the Limiting Factors of Photosynthesis | 探究光合作用的限制因素
A student placed Canadian pondweed (Elodea) in a beaker of water and measured the rate of photosynthesis by counting oxygen bubbles released per minute. The experiment was performed at different light intensities, carbon dioxide concentrations, and temperatures. At 20°C, 0.04% CO₂, and medium light intensity, the rate was 25 bubbles/min. When light intensity was increased further, the rate plateaued at 40 bubbles/min. Increasing CO₂ to 0.08% raised the plateau to 65 bubbles/min, but raising temperature to 30°C caused the rate to drop to 30 bubbles/min despite high light and CO₂.
一名学生将加拿大伊乐藻放入盛水的烧杯中，通过计数每分钟释放的氧气泡数测量光合作用速率。实验在不同光照强度、二氧化碳浓度和温度下进行。在20°C、0.04% CO₂和中等光照强度下，速率为25气泡/分钟。进一步增加光强时，速率在40气泡/分钟处趋于平稳。将CO₂浓度升至0.08%，平台值提高到65气泡/分钟，但将温度升至30°C后，尽管高光照和高CO₂，速率却降至30气泡/分钟。
Use the data to identify the limiting factor at 25 bubbles/min before the light increase, and explain your choice.
利用数据指出在光强增加前速率25气泡/分钟时的限制因素，并解释你的选择。
When light intensity was increased, the rate also increased from 25 to 40 bubbles/min. This indicates that light intensity was the limiting factor initially, because increasing it resulted in a higher rate of photosynthesis. If CO₂ or temperature were limiting, increasing light would not have had this effect.
当光照强度增加时，速率也从25上升到40气泡/分钟。这表明最初光照强度是限制因素，因为增加光强导致了光合速率提高。如果CO₂或温度是限制因素，增加光照就不会有此效果。
Explain why raising the temperature to 30°C caused the rate to decrease, even though light and CO₂ were abundant. Refer to enzyme action in your answer.
解释为什么将温度升至30°C会导致速率下降，尽管光照和CO₂充足。回答中需提及酶的作用。
Photosynthesis involves enzymes, such as rubisco. While moderate warming can increase enzyme activity, temperatures above the optimum denature these enzymes. At 30°C, some enzymes may have begun to denature, losing their specific active site shape, so the overall rate of the light-independent reactions fell, reducing oxygen production.
光合作用涉及多种酶，如rubisco。虽然适度升温可提高酶活性，但超过最适温度后会使酶变性。在30°C时，部分酶可能开始变性，失去特定的活性位点形状，因此暗反应总速率下降，氧气产量减少。
8. Interpreting Osmosis Data from Potato Tuber Strips | 解读马铃薯块茎条的渗透数据
Students investigated osmosis in potato strips placed in different sucrose concentrations. They measured the initial and final lengths after 30 minutes and calculated the percentage change. Results are shown below.
学生研究了放置在不同蔗糖浓度溶液中的马铃薯条的渗透作用。他们测量了初始长度和30分钟后的长度，并计算了变化百分比。结果如下所示。
| Sucrose concentration (mol/dm³) | % change in length |
|---|---|
| 0.0 | +12.5 |
| 0.2 | +6.0 |
| 0.4 | -2.0 |
| 0.6 | -8.5 |
| 0.8 | -14.0 |
| 1.0 | -18.5 |
Explain why the potato strips increased in length when placed in 0.0 mol/dm³ sucrose solution. Use the term water potential.
解释为什么马铃薯条在0.0 mol/dm³蔗糖溶液中长度增加。使用水势这一术语。
Distilled water (0.0 mol/dm³ sucrose) has a higher water potential than the potato cells’ cytoplasm and vacuole sap. Water moves by osmosis from the region of higher water potential (the solution) into the cells, which have a lower water potential due to dissolved solutes. The influx of water causes cells to become turgid, increasing strip length.
蒸馏水（0.0 mol/dm³蔗糖）的水势高于马铃薯细胞质和液泡液。水通过渗透作用从水势较高处（溶液）进入水势较低的细胞（因含溶解溶质）。水流入使细胞变得饱满，导致薯条长度增加。
Estimate the sucrose concentration that would be isotonic to the potato tissue, and describe how you reached this conclusion.
估计与马铃薯组织等渗的蔗糖浓度，并描述你是如何得出此结论的。
The isotonic point occurs where there is 0% change in length. Between 0.2 mol/dm³ (+6.0%) and 0.4 mol/dm³ (-2.0%), the line crosses zero. By interpolation, the isotonic concentration is approximately 0.32 mol/dm³. At this concentration, the water potential of the external solution equals that of the potato cells, so no net movement of water occurs.
等渗点出现在长度变化为0%时。在0.2 mol/dm³ (+6.0%)与0.4 mol/dm³ (-2.0%)之间，曲线穿过零点。通过内插法，等渗浓度约为0.32 mol/dm³。在此浓度下，外部溶液的水势与马铃薯细胞相等，因此没有水分的净移动。
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