AS AQA Biology: Case Study Practice Drills | AS AQA 生物：案例分析实战演练

📚 AS AQA Biology: Case Study Practice Drills | AS AQA 生物：案例分析实战演练

In AS AQA Biology, case study questions require you to apply theoretical knowledge to novel scenarios. These drills will sharpen your analytical skills by guiding you through real-world examples from across the specification. Each case includes a scenario, key data, and step-by-step reasoning to help you master the art of constructing well-evidenced answers.

在 AS AQA 生物考试中，案例分析题要求你将理论知识应用于全新的情境。以下的实战演练将通过覆盖考纲的各类真实案例，帮助你提升分析能力。每个案例包含情景描述、核心数据以及逐步推理过程，让你掌握构建有理有据答案的技巧。

1. Enzyme Inhibition: Methotrexate and Cancer Therapy | 酶抑制：甲氨蝶呤与癌症治疗

Chemotherapy drug methotrexate inhibits dihydrofolate reductase (DHFR), an enzyme that converts dihydrofolate to tetrahydrofolate, a cofactor essential for nucleotide synthesis. By blocking this step, methotrexate slows DNA replication in rapidly dividing cancer cells.

化疗药物甲氨蝶呤能抑制二氢叶酸还原酶 (DHFR)，该酶将二氢叶酸转化为四氢叶酸，而四氢叶酸是核苷酸合成所必需的辅因子。通过阻断此步骤，甲氨蝶呤减缓了快速分裂癌细胞的 DNA 复制。

This inhibition is competitive: methotrexate has a structure very similar to the substrate dihydrofolate and competes for the enzyme’s active site. The degree of inhibition can be reduced by increasing the concentration of the natural substrate.

这种抑制属于竞争性抑制：甲氨蝶呤的结构与底物二氢叶酸非常相似，会竞争酶的活性部位。通过提高天然底物的浓度，可以减弱抑制程度。

[Dihydrofolate] (μmol dm⁻³)	Rate without inhibitor (μmol min⁻¹)	Rate with methotrexate (μmol min⁻¹)
0.2	12	4
0.5	24	12
1.0	36	24
2.0	44	36

From the data, the maximum velocity (Vmax) eventually reaches a similar value in both conditions (~48 μmol min⁻¹), while the apparent Km increases dramatically in the presence of methotrexate. This pattern is characteristic of competitive inhibition.

从数据可以看出，最大反应速度 (Vmax) 在两种条件下最终都接近相似值（约 48 μmol min⁻¹），而甲氨蝶呤存在时表观 Km 大幅升高。这种模式是竞争性抑制的典型特征。

When answering exam questions on this case, always highlight the structural similarity between inhibitor and substrate, the competition for the active site, and the diagnostic kinetic changes—Vmax unchanged, Km increased. Link these to the clinical effect: reduced dNTP supply for DNA synthesis.

在回答与此案例相关的考题时，务必强调抑制剂与底物的结构相似性、对活性位点的竞争以及动力学特征变化——Vmax 不变、Km 升高。将这些变化与临床效果联系起来：DNA 合成所需 dNTP 的供应减少。

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2. Membrane Transport: Cystic Fibrosis and CFTR Protein | 膜运输：囊性纤维化与 CFTR 蛋白

Cystic fibrosis (CF) arises from a mutation in the gene encoding the cystic fibrosis transmembrane conductance regulator (CFTR), a chloride ion channel. The most common mutation, ΔF508, disrupts protein folding, so the channel fails to reach the cell surface membrane.

囊性纤维化源于编码 CFTR（囊性纤维化跨膜传导调节因子）氯离子通道的基因发生突变。最常见的突变 ΔF508 会破坏蛋白质折叠，导致通道无法到达细胞表面膜。

In healthy airways, CFTR allows Cl⁻ to exit epithelial cells into the mucus, drawing Na⁺ and water out by osmotic effect. This keeps the mucus fluid and mobile. When CFTR is absent, Cl⁻ transport is blocked, and the mucus becomes thick and sticky.

在健康的气道中，CFTR 让 Cl⁻ 从上皮细胞进入黏液层，通过渗透效应将 Na⁺ 和水一同带出。这使黏液保持液态且易于移动。若缺乏 CFTR，Cl⁻ 运输受阻，黏液就变得浓稠粘滞。

This thickened mucus obstructs pancreatic ducts, lung airways, and other passages, causing recurrent infections and digestive problems. The condition beautifully illustrates how a single channel protein defect can have systemic consequences through disrupted transmembrane gradients.

这种浓稠黏液会堵塞胰管、肺气道及其他管道，引发反复感染和消化问题。该疾病完美地演示了一个单一通道蛋白的缺陷如何通过破坏跨膜梯度而导致全身性后果。

When analysing CF case studies, remember to trace the sequence: genotype → altered protein structure → defective ion movement → osmotic imbalance → pathological symptoms. Use terms like ‘facilitated diffusion’, ‘channel protein’, ‘water potential gradient’, and ‘golgi trafficking’ to show precise understanding.

在分析囊性纤维化案例时，要记住追溯因果链：基因型 → 蛋白质结构改变 → 离子运动异常 → 渗透不平衡 → 病症。使用诸如「易化扩散」「通道蛋白」「水势梯度」「高尔基体转运」等术语，以体现精准的理解。

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3. Immunology: Influenza Vaccination and Memory Cells | 免疫学：流感疫苗接种与记忆细胞

Each year, the influenza vaccine is reformulated to match circulating strains. The vaccine contains inactivated viral particles that carry haemagglutinin and neuraminidase antigens but cannot cause disease. Injection stimulates the primary immune response without illness.

每年，流感疫苗都会根据流行毒株重新配制。疫苗含有灭活病毒颗粒，携带血凝素和神经氨酸酶抗原，却无法致病。注射后可在不引起疾病的情况下激发初次免疫应答。

B‑lymphocytes with complementary receptors undergo clonal selection and expansion, producing plasma cells that secrete specific antibodies. Simultaneously, memory B‑cells and memory T‑cells are generated and persist for years, enabling a faster, stronger secondary response upon real infection.

具有互补受体的 B 淋巴细胞经历克隆选择与扩增，产生浆细胞分泌特异性抗体。同时，记忆 B 细胞和记忆 T 细胞形成并存活多年，使得真实感染时能产生更快、更强的二次应答。

Some individuals still catch influenza after vaccination, often because the surface antigens have changed (antigenic drift). Memory cells may not recognise the new shape, and the immune system responds as if to a primary pathogen. This explains why annual revaccination is needed.

有些人接种后仍会感染流感，常因表面抗原发生改变（抗原漂移）。记忆细胞可能无法识别新形状，免疫系统像初次遭遇病原体一样作出应答。这就解释了为什么需要每年重新接种。

In a case study question, you might be given data on antibody titres before and after vaccination or infection. Highlight the lag phase, the rise in specific IgG, and the rapid anamnestic response. Link these to the concepts of immunological memory and herd immunity.

在案例分析题中，可能会提供接种或感染前后的抗体滴度数据。要指出迟滞期、特异性 IgG 的升高以及快速回忆应答。将这些现象与免疫记忆和群体免疫的概念联系起来。

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4. Lung Function: Analysing a Spirometer Trace in Asthma | 肺功能：分析哮喘患者的肺活量图

A 22‑year‑old with suspected asthma underwent spirometry, and the trace showed a prolonged forced expiratory volume in the first second (FEV₁) relative to forced vital capacity (FVC), giving an FEV₁/FVC ratio of 65%, well below the normal >75%. After inhaling a bronchodilator, the ratio improved to 78%.

一名 22 岁疑似哮喘患者进行了肺活量测定，描记图显示第一秒用力呼气量 (FEV₁) 占用力肺活量 (FVC) 的比例仅为 65%，远低于正常值（>75%）。在吸入支气管扩张剂后，该比值改善至 78%。

The reduced ratio indicates obstructive airways disease; bronchoconstriction and thickened mucus narrow the airways, increasing resistance to airflow. The post‑bronchodilator improvement confirms that the obstruction was at least partially reversible, a hallmark of asthma.

比值降低表明存在阻塞性气道疾病；支气管收缩和黏液增厚使气道变窄，气流阻力增大。使用支气管扩张剂后数值改善，证实阻塞至少部分可逆，这是哮喘的特征。

In answering questions on such traces, always label tidal volume, inspiratory and expiratory reserve volumes, and vital capacity. Explain that during an asthma attack, smooth muscle in the bronchioles contracts, reducing lumen diameter and making expiration harder, especially the later phase.

回答此类曲线题时，要始终标明潮气量、补吸气量、补呼气量以及肺活量。解释哮喘发作时，细支气管平滑肌收缩，管腔直径减小，呼气变得更加困难，尤其在呼气后期。

Relate the physiology to the spirometer data: a low FEV₁ indicates difficulty exhaling quickly; the FVC may remain relatively normal, but the ratio drops. Discuss triggers like allergens causing histamine release and how inhaled corticosteroids reduce inflammation long‑term.

将生理学与肺活量数据相联系：低 FEV₁ 表明快速呼气困难；FVC 可能保持相对正常，但比值下降。讨论过敏原等诱因如何引发组胺释放，以及吸入性皮质类固醇如何长期减轻炎症。

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5. Digestion: Lactose Intolerance and Enzyme Replacement | 消化：乳糖不耐受与酶替代疗法

Lactose, the disaccharide in milk, must be hydrolysed by lactase into glucose and galactose before absorption. Some adults lose lactase expression after weaning, leading to lactose intolerance; undigested lactose is fermented by gut bacteria, causing bloating, cramps and diarrhoea.

乳糖是乳汁中的二糖，必须被乳糖酶水解为葡萄糖和半乳糖后才能被吸收。部分成年人在断奶后乳糖酶表达下降，导致乳糖不耐受；未经消化的乳糖被肠道细菌发酵，引起腹胀、痉挛和腹泻。

Lactase enzyme supplements, taken orally with dairy meals, can supply the missing enzyme activity. The pills contain fungal or yeast‑derived lactase that works in the small intestine’s alkaline pH, mirroring the natural human enzyme’s action.

随乳制品口服的乳糖酶补充剂可提供缺失的酶活性。这些药片含有源自真菌或酵母的乳糖酶，能在小肠碱性 pH 环境中发挥作用，模拟天然人乳糖酶的功能。

In a case study, patients often report improved symptoms when using such supplements. You could be asked to explain why the supplement does not cure the condition permanently: it provides functional enzyme only during the few hours after ingestion, as the protein is digested further along the alimentary canal.

在案例分析中，患者常报告使用此类补充剂后症状改善。你可能被要求解释为什么补充剂无法根治此症：它仅在摄入后几小时内提供功能性酶，因为该蛋白质随后会在消化道中被进一步消化分解。

Link this to the lock‑and‑key model of enzyme action, the specificity of lactase for the β‑1,4 glycosidic bond, and the importance of pH and temperature for enzyme activity. Consider also the products of lactose hydrolysis—glucose and galactose—and how they are actively transported into epithelial cells by sodium‑dependent cotransporters.

将这联系到酶的锁钥模型、乳糖酶对 β-1,4 糖苷键的特异性，以及 pH 和温度对酶活性的重要性。同时还要考虑乳糖水解产物葡萄糖和半乳糖，如何通过钠依赖性协同转运蛋白被主动转运进入上皮细胞。

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6. Cardiovascular Disease: Atherosclerosis and Cholesterol | 心血管疾病：动脉粥样硬化与胆固醇

Atherosclerosis begins with endothelial damage, often triggered by high blood pressure or toxins from cigarette smoke. Low‑density lipoproteins (LDLs) infiltrate the damaged endothelium and accumulate in the arterial wall, where they become oxidised.

动脉粥样硬化始于内皮损伤，常由高血压或香烟烟雾中的毒素引发。低密度脂蛋白 (LDL) 渗入受损的内皮并在动脉壁内积聚，进而被氧化。

Monocytes migrate into the area, differentiate into macrophages, and engulf oxidised LDLs via phagocytosis, becoming foam cells. These foam cells accumulate along with calcium deposits and fibrous tissue, forming an atheroma plaque that narrows the lumen and stiffens the artery.

单核细胞迁移至该区域，分化为巨噬细胞，通过吞噬作用吞噬氧化的 LDL，形成泡沫细胞。这些泡沫细胞与钙沉积以及纤维组织一同积聚，形成动脉粥样硬化斑块，使管腔变窄、动脉壁变硬。

If the plaque ruptures, clotting factors are exposed, and a thrombus (blood clot) forms. In a coronary artery, this can block blood flow to heart muscle, causing a myocardial infarction. A case study may present a patient’s blood lipid profile: high total cholesterol, elevated LDL, and perhaps low HDL.

若斑块破裂，凝血因子暴露，就会形成血栓（血凝块）。在冠状动脉中，这会阻断流向心肌的血流，引发心肌梗死。案例分析可能提供患者的血脂谱：总胆固醇高、LDL 升高，或许 HDL 偏低。

When answering, explain why HDL is protective (it transports cholesterol back to the liver for excretion, a process called reverse cholesterol transport) and why LDL is harmful (it delivers cholesterol to tissues, including arteries). Discuss lifestyle and drug interventions like statins that reduce LDL production in the liver.

回答时，要解释为何 HDL 具有保护作用（它将胆固醇运回肝脏排出，这一过程称为胆固醇逆向转运），以及为何 LDL 有害（它将胆固醇输送到包括动脉在内的组织）。讨论生活方式干预和诸如他汀类药物的治疗，这类药物可减少肝脏中 LDL 的生成。

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7. DNA Mutation: Sickle Cell Anaemia | DNA 突变：镰状细胞贫血

Sickle cell anaemia results from a single base substitution in the gene for the β‑globin chain of haemoglobin. The mutation changes the codon GAG to GTG, resulting in the replacement of glutamic acid (a hydrophilic amino acid) with valine (a hydrophobic one) at position 6.

镰状细胞贫血是血红蛋白 β-珠蛋白链基因中一个碱基替换的结果。突变使密码子 GAG 变为 GTG，导致第 6 位的谷氨酸（亲水性氨基酸）被缬氨酸（疏水性氨基酸）取代。

This subtle change alters the haemoglobin molecule’s shape, especially when deoxygenated. Molecules of HbS (sickle haemoglobin) aggregate into long fibres, distorting red blood cells into a stiff sickle shape that can block capillaries and is prematurely destroyed, causing anaemia.

这一微小的变化改变了血红蛋白分子的形状，尤其是在脱氧状态下。HbS（镰状血红蛋白）分子聚集成细长纤维，使红细胞扭曲成僵硬的镰刀状，从而堵塞毛细血管，并因过早破坏而导致贫血。

In case studies, you might be given DNA sequences to compare between normal and sickle cell alleles, or data showing malaria prevalence correlating with the sickle cell trait. Heterozygous individuals (HbA/HbS) have some protective advantage against severe malaria because the altered red cell environment hinders parasite survival.

案例分析中，可能会给出正常与镰状细胞等位基因的 DNA 序列供比较，或者提供疟疾流行率与镰状细胞性状相关的数据。杂合子个体 (HbA/HbS) 对重症疟疾具有一定保护优势，因为改变的红细胞环境不利于疟原虫存活。

Explain the mutation at both the DNA and protein levels, using terms like ‘missense mutation’, ‘codon’, ‘primary structure’, ‘tertiary structure’, and ‘deductive reasoning’. Emphasise how a single nucleotide change can dramatically alter the phenotype through disrupted folding and solubility.

从 DNA 和蛋白质两个层面解释突变，使用「错义突变」「密码子」「一级结构」「三级结构」和「演绎推理」等术语。强调单个核苷酸的变化如何通过破坏折叠和溶解性而显著改变表型。

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8. Genetics: ABO Blood Group Inheritance | 遗传学：ABO 血型的遗传

The ABO blood group system is controlled by a single gene with three alleles: Iᴬ, Iᴮ and i. Iᴬ and Iᴮ are codominant, both expressed in the phenotype when present together (type AB), while i is recessive. A case scenario often involves parents and child blood groups to deduce genotypes.

ABO 血型系统由一个有三种等位基因的单一基因控制：Iᴬ、Iᴮ 和 i。Iᴬ 与 Iᴮ 为共显性，同时存在时都会在表型中表达（AB 型），而 i 为隐性。案例情景通常涉及父母与子女的血型，据此推断基因型。

For example, if a mother is blood type O (genotype ii) and the father is type AB (genotype IᴬIᴮ), their offspring can only be type A (Iᴬi) or type B (Iᴮi). No child can be type O or AB. Punnett squares easily confirm this, showing equal probabilities of each outcome.

例如，若母亲是 O 型血（基因型 ii），父亲是 AB 型血（基因型 IᴬIᴮ），其后代只能是 A 型 (Iᴬi) 或 B 型 (Iᴮi)。不可能出现 O 型或 AB 型的孩子。用庞尼特方格可轻松验证，显示每种结果的概率相等。

In exam questions, you may be presented with a disputed paternity case or a blood transfusion compatibility table. Always recall that type O is the universal donor (no A or B antigens on red cells) and type AB is the universal recipient (no anti‑A or anti‑B antibodies in plasma).

在考题中，可能会出现亲子鉴定争议或输血相容性表格。要始终记住 O 型是万能供血者（红细胞表面无 A 或 B 抗原），AB 型是万能受血者（血浆中无抗 A 或抗 B 抗体）。

When writing answers, show clear genetic diagrams, label alleles correctly (using superscripts), and connect genotype to phenotype through the expression of glycosyltransferase enzymes that add specific sugar residues to the H antigen on red blood cells.

书写答案时，需展示清晰的遗传图解，正确标注等位基因（使用上标），并将基因型通过糖基转移酶的表达与表型联系起来，该酶会将特定糖残基添加到红细胞表面的 H 抗原上。

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9. Evolution: Antibiotic Resistance in Bacteria | 进化：细菌的抗生素耐药性

A clinical case might track MRSA (methicillin‑resistant Staphylococcus aureus) incidence in a hospital. The use of methicillin selects for bacteria carrying a resistance gene (mecA) that encodes an altered penicillin‑binding protein with low drug affinity.

临床案例可能追踪某医院 MRSA（耐甲氧西林金黄色葡萄球菌）的发生情况。使用甲氧西林会筛选出携带耐药基因 (mecA) 的细菌，该基因编码一种对药物亲和力低的改变了的青霉素结合蛋白。

Before antibiotic exposure, resistant variants may exist at very low frequency due to spontaneous mutation. Once methicillin is used, susceptible bacteria die, but resistant ones survive and reproduce, passing the advantageous allele to their offspring. Over generations, the population shifts to predominantly resistant cells.

在接触抗生素之前，耐药变体可能因自发突变而以极低频率存在。一旦使用甲氧西林，敏感菌被杀死，耐药菌却能存活并繁殖，将有利等位基因传递给后代。经若干代后，种群转变为以耐药菌为主。

This is natural selection in action: genetic variation (resistance alleles), selection pressure (antibiotic), differential survival, and heritability. The case study may provide a graph showing the increase in the percentage of resistant isolates over time, correlated with antibiotic usage data.

这就是自然选择的实践体现：遗传变异（耐药等位基因）、选择压力（抗生素）、差异性生存以及可遗传性。案例分析可能提供图表，显示随时间推移，耐药菌株比例的增加与抗生素用量数据相关。

When confronting such data, explicitly label the graph trends, identify the point where resistance begins to surge, and discuss how horizontal gene transfer (conjugation, transduction) can spread resistance genes between different bacteria, accelerating the crisis.

面对此类数据时，要明确标注图表趋势，指出耐药性开始激增的时间点，并讨论水平基因转移（接合、转导）如何在不同的细菌之间传播耐药基因，从而加剧危机。

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10. Ecology: Estimating Population Size Using Mark‑Release‑Recapture | 生态学：用标记重捕法估算种群大小

A conservation team studying a population of woodlice in a woodland floor used pitfall traps. In the first capture, 120 woodlice were caught, marked with a dot of non‑toxic paint, and released. The next day, 100 woodlice were captured, of which 15 were marked.

一个保护研究小组利用陷阱法调查林地地表鼠妇的种群。第一次捕获了 120 只鼠妇，用无毒颜料点上标记后释放。次日捕获 100 只，其中 15 只带有标记。

The Lincoln index estimates population size N: N = (C₁ × C₂)/M, where C₁ is number first captured and marked, C₂ is total in second capture, and M is the number marked in the second capture. Plugging in values: N = (120 × 100)/15 = 800 woodlice.

林肯指数用于估算种群大小 N：N = (C₁ × C₂)/M，其中 C₁ 为首次捕获并标记的数量，C₂ 为第二次捕获总数，M 为第二次捕获中的标记个体数。代入数值：N = (120 × 100)/15 = 800 只鼠妇。

The method rests on several assumptions: the marks do not affect survival or behaviour, the marked individuals mix randomly with the population, no immigration or emigration occurs, and the population is closed. In a case study, you may be asked to evaluate why the estimate might be inaccurate.

该方法基于若干假设：标记不影响个体的生存或行为，标记个体与种群随机混合，没有迁入或迁出，且种群处于封闭状态。案例分析中，可能要求你评估此估算可能不准确的原因。

For instance, if the paint made woodlice more visible to predators, fewer marked individuals would be recaptured, making M smaller and N overestimated. Alternatively, reproduction between samples could increase unmarked numbers, also inflating the estimate. Always propose improvements, such as using a more subtle marking technique or taking samples over multiple days.

例如，若标记涂料使鼠妇更容易被捕食者发现，重捕的标记个体就会减少，M 变小，导致 N 被高

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