📚 Case Study Practice Drills for Year 12 CAIE Biology | Year 12 CAIE 生物案例分析实战演练
In CAIE Year 12 Biology, case studies offer a powerful way to bridge theory and application, sharpening the analytical skills needed for high-scoring exam answers. This article presents ten practice drills spanning the AS syllabus, each built around a realistic scenario, guided data interpretation, and detailed explanation of the underlying principles. Use these examples to strengthen your understanding of enzymes, membranes, disease, genetics and more.
在 CAIE 12 年级生物课程中，案例分析是连接理论与应用的绝佳途径，能有效磨练高分答题所需的分析能力。本文围绕 AS 阶段大纲设计了十个实战练习，每个练习都包含真实情境、数据解读指导以及对核心原理的详细阐释。通过这些案例，进一步巩固你对酶、细胞膜、疾病、遗传等主题的理解。
1. Enzyme Activity and pH | 酶活性与 pH
A student incubated the enzyme pepsin with egg-white protein at different pH values and measured the time taken for the solution to clear. The shortest clearing time occurred at pH 2.0. At pH 7.0, no clearing was observed even after 30 minutes.
一名学生在不同 pH 条件下将胃蛋白酶与蛋清蛋白共同孵育，并测量溶液变澄清所需的时间。在 pH 2.0 时澄清时间最短；在 pH 7.0 时，即使经过 30 分钟也未观察到任何澄清。
Pepsin is a protease adapted to the highly acidic environment of the stomach. Its tertiary structure, shaped by ionic and hydrogen bonds, is stable at pH 2.0. At neutral pH, the excess of OH⁻ ions disrupts these bonds, causing the enzyme to denature irreversibly and lose its catalytic function.
胃蛋白酶是一种适应胃内强酸环境的蛋白酶。其三级结构依靠离子键和氢键维持，在 pH 2.0 时稳定。在中性 pH 下，过量的 OH⁻ 离子会破坏这些键，导致酶不可逆变性并丧失催化功能。
The active site of pepsin contains aspartate residues that must be protonated (Asp–COOH) to bind substrate. Only at low pH are these residues in the correct protonation state, illustrating the link between pH, active-site chemistry and reaction rate.
胃蛋白酶的活性位点含有天冬氨酸残基，这些残基必须处于质子化状态（Asp–COOH）才能结合底物。只有在低 pH 条件下，这些残基才能维持正确的质子化形态，这体现了 pH、活性位点化学环境与反应速率之间的联系。
| pH | Clearing time (s) |
|---|---|
| 1.5 | 85 |
| 2.0 | 42 |
| 3.0 | 70 |
| 5.0 | 210 |
| 7.0 | No clearing |
The data confirm a sharp optimum near pH 2.0, consistent with pepsin's physiological role. In exam-style analysis, you would link the loss of activity at higher pH to denaturation and explain why denaturation is permanent because the primary structure remains intact but the specific 3D conformation is lost.
数据证实了一个接近 pH 2.0 的尖锐最适值，这与胃蛋白酶的生理功能相符。在考试分析中，你需要将较高 pH 下活性的丧失与变性联系起来，并解释为何变性是永久性的——因为一级结构虽然完整，但特定的三维构象已经丧失。
2. Membrane Transport and Cystic Fibrosis | 膜运输与囊性纤维化
Cystic fibrosis (CF) is caused by a mutation in the CFTR gene, leading to a faulty chloride ion channel. A lung epithelial cell from a CF patient shows abnormally thick mucus, persistent infections, and a high sweat Cl⁻ concentration.
囊性纤维化（CF）由 CFTR 基因突变引起，导致氯离子通道功能缺陷。来自 CF 患者的肺上皮细胞表现为异常黏稠的黏液、反复感染以及汗液中高浓度的 Cl⁻。
In normal airway epithelia, CFTR channels allow Cl⁻ to leave the cell into the lumen, drawing Na⁺ and water out by osmosis, which keeps the surface liquid hydrated. In CF, Cl⁻ export is reduced, so less water follows; the mucus layer becomes dehydrated and viscous, impairing ciliary clearance.
在正常气道上皮中，CFTR 通道允许 Cl⁻ 离开细胞进入管腔，并通过渗透作用带动 Na⁺ 和水向外移动，使表面液体保持湿润。在 CF 中，Cl⁻ 外流减少，跟随的水量也减少；黏液层脱水变稠，阻碍了纤毛清除功能。
In the sweat gland, the role of CFTR is reversed: it normally reabsorbs Cl⁻ from the duct. A defective channel prevents this reabsorption, so Cl⁻ and accompanying Na⁺ remain in the sweat, producing the classic salty sweat used in the diagnostic sweat test.
在汗腺中，CFTR 的作用相反：它通常从导管中重吸收 Cl⁻。通道缺陷阻止了这种重吸收，因此 Cl⁻ 和与之相伴的 Na⁺ 残留在汗液中，形成了用于诊断的经典“咸汗”现象。
The ΔF508 mutation removes a phenylalanine at position 508, causing the protein to misfold in the rough ER. Quality-control mechanisms tag it for proteasomal degradation, so very little CFTR reaches the apical membrane. This illustrates how a single base-pair deletion can alter the primary structure, preventing correct tertiary folding and final functional placement.
ΔF508 突变删除了第 508 位的苯丙氨酸，导致蛋白质在粗面内质网中错误折叠。质量控制系统将其标记并送往蛋白酶体降解，因此极少有 CFTR 能抵达顶膜。这体现了单个碱基对的缺失如何改变一级结构，进而阻止正确的三级折叠和最终的功能定位。
3. Mitosis, Cancer and Chemotherapy | 有丝分裂、癌症与化疗
A biopsy from a breast tumour shows cells with a high mitotic index, irregular nuclear shapes, and broken chromosomes. The oncologist prescribes a drug that stabilises microtubules, preventing spindle fibre disassembly.
一份乳腺肿瘤活检显示细胞具有高有丝分裂指数、不规则的细胞核形态和断裂的染色体。肿瘤科医生开具了一种能稳定微管、阻止纺锤体纤维解聚的药物。
Cancer cells divide uncontrollably because mutations have disabled cell-cycle checkpoints, particularly the G₁/S and G₂/M transitions. A high mitotic index indicates rapid proliferation, which is a hallmark of malignancy.
癌细胞因突变导致细胞周期检查点（尤其是 G₁/S 和 G₂/M 转换点）失效而失控分裂。高有丝分裂指数表明增殖迅速，这是恶性转化的标志之一。
The prescribed drug, a taxane, locks tubulin dimers in the polymerised state. During metaphase, chromosomes align at the equator, but the spindle cannot shorten in anaphase because the kinetochore microtubules fail to depolymerise. Cells arrest in mitosis and eventually undergo apoptosis.
所开具的药物（一种紫杉烷类）将微管蛋白二聚体锁定在聚合状态。在中期的细胞里，染色体排列在赤道板上，但到了后期纺锤体无法缩短，因为着丝粒微管不能解聚。细胞停滞在有丝分裂阶段并最终走向凋亡。
Chemotherapy exploits the high division rate of cancer cells, but rapidly renewing normal tissues (e.g., bone marrow, hair follicles) are also affected, causing side effects such as immunosuppression and hair loss. This case study connects targeted drug action with the events of mitosis and the concept of selective toxicity.
化疗利用了癌细胞分裂速度快的特性，但分裂同样活跃的正常组织（如骨髓、毛囊）也会受到影响，从而产生免疫抑制和脱发等副作用。本案例将靶向药物作用与有丝分裂事件以及选择毒性的概念联系起来。
4. Infectious Disease Dynamics: R₀ Calculation | 传染病动力学：R₀ 计算
A novel respiratory virus is spreading in a city with no prior immunity. Epidemiologists estimate that each infected person infects 3 others on average, the infectious period lasts 10 days, and contacts are 12 per day. The transmission probability per contact is unknown.
一种新型呼吸道病毒在无先前免疫的城市中传播。流行病学家估计，每名感染者平均传染 3 人，传染期持续 10 天，每天接触人数为 12。每次接触的传播概率未知。
R₀ = β × κ × D
the duration of infectiousness (D). Rearranging, β = R₀ / (κ × D) = 3 / (12 × 10) = 0.025.
R₀（基本再生数）等于每日接触人数 (κ)、每次接触传播概率 (β) 和传染期持续时间 (D) 三者的乘积。整理公式可得 β = R₀ / (κ × D) = 3 / (12 × 10) = 0.025。
5 × 6 × 10 = 1.5. An R₀ above 1 means the disease will keep spreading; below 1, it dies out. This simple model underpins strategies like social distancing and isolation.
尽管 β 数值很小，但高接触率维持了疫情的传播。若公共卫生措施将接触率减半为每天 6 次，则 R₀ = 0.025 × 6 × 10 = 1.5。R₀ 大于 1 意味着疾病会继续扩散；小于 1 则逐渐消失。这一简单模型支撑着保持社交距离、隔离等防控策略。
In the AS syllabus, you may be asked to interpret such mathematical models and discuss their limitations: they assume homogeneous mixing, constant infectiousness, and ignore changes in behaviour. Yet they remain vital for predicting the initial trajectory of an outbreak and guiding policy.
在 AS 大纲中，你可能会被要求解读这类数学模型并讨论其局限：它们假定人群均匀混合、传染性恒定不变，且忽略了行为的改变。然而，这些模型对于预测疫情初期走势和指导政策仍至关重要。
5. Investigating Transpiration with a Potometer | 用蒸腾计研究蒸腾作用
A class set up a bubble potometer using a leafy shoot of laurel in the following conditions: (A) still air, 25°C, moderate light; (B) moving air from a fan, 25°C, moderate light; (C) still air, 35°C, moderate light. The distance moved by the air bubble in 10 minutes was recorded.
一个班级使用月桂带叶枝条搭建气泡蒸腾计，并在以下条件下记录： (A) 静止空气、25°C、中等光照； (B) 风扇制造流动空气、25°C、中等光照； (C) 静止空气、35°C、中等光照。记录 10 分钟内气泡移动的距离。
- Condition A: 12 mm
- Condition B: 21 mm
- Condition C: 18 mm
The bubble movement estimates water uptake, which closely reflects the transpiration rate when the plant is well hydrated. Condition B produced the highest rate because moving air removes the boundary layer of water vapour near the stomata, maintaining a steep water-vapour concentration gradient between the leaf air spaces and the external atmosphere.
气泡移动距离用于估算水分吸收量，在植物水分充足时能很好地反映蒸腾速率。条件 B 的速率最高，因为流动的空气带走了气孔附近的水蒸气边界层，使叶片内部空气与外界大气之间维持了较陡的水蒸气浓度梯度。
Condition C (elevated temperature) raised transpiration relative to A because warmer air can hold more water vapour, increasing the saturation deficit and accelerating evaporation from mesophyll cell walls. Sunken stomata, a waxy cuticle, and ABA-induced stomatal closure would all reduce bubble movement and are effective xerophytic adaptations.
条件 C（升高温度）使蒸腾速率高于条件 A，因为温暖空气能容纳更多水蒸气，增大了饱和差，加快了叶肉细胞壁表面的蒸发。凹陷的气孔、蜡质角质层以及脱落酸诱导的气孔关闭都会减少气泡移动，是有效的旱生植物适应特征。
A potometer does not measure transpiration directly but water uptake; a small fraction of water is used in photosynthesis and to maintain cell turgidity. Despite this, it provides a reliable comparative measure of transpiration rate under different environmental factors.
蒸腾计并非直接测量蒸腾作用，而是测定水分吸收量；一小部分水分用于光合作用和维持细胞膨压。尽管如此，它仍为比较不同环境因素下的蒸腾速率提供了可靠的衡量方法。
6. Haemoglobin and the Oxygen Dissociation Curve | 血红蛋白与氧解离曲线
Blood samples were taken from an elite athlete training at high altitude and from a sedentary person at sea level. Their oxygen dissociation curves were plotted. The athlete's curve was shifted to the right, with a pO₂ at 50% saturation of 35 mmHg versus the sea-level subject's 27 mmHg.
从高海拔训练的精英运动员和在海平面久坐之人身上采集了血样，并绘制了氧解离曲线。运动员的曲线右移，50% 饱和时的 pO₂ 为 35 mmHg，而海平面受试者为 27 mmHg。
The rightward shift (Bohr effect) reduces haemoglobin's affinity for oxygen, so more O₂ is unloaded in respiring tissues. At high altitude, the partial pressure of oxygen is lower, making it harder to load oxygen in the lungs. The body compensates by increasing 2,3-bisphosphoglycerate (2,3-BPG) concentration in red blood cells, which stabilises the T-state of haemoglobin and promotes oxygen release to tissues.
曲线右移（波尔效应）降低了血红蛋白对氧的亲和力，因此在呼吸组织中能释放更多的 O₂。在高海拔地区，氧分压较低，增大了肺部载氧的难度。机体通过增加红细胞内 2,3-二磷酸甘油酸（2,3-BPG）的浓度进行代偿，后者稳定了血红蛋白的 T 态，促进向组织释氧。
During exercise, respiring muscles produce heat, CO₂ and lactic acid, all of which further shift the curve to the right. This case illustrates how the cooperative binding of oxygen to haemoglobin's four haeme groups, combined with allosteric effectors, fine-tunes oxygen delivery to match metabolic demand.
运动时，呼吸肌产生热量、CO₂ 和乳酸，这些因素都使曲线进一步右移。本案例表明，氧气与血红蛋白四个血红素基团的协同结合，加上别构效应物的作用，精确调节了氧的输送以匹配代谢需求。
Foetal haemoglobin (HbF) has a higher affinity for O₂ than adult haemoglobin (HbA), resulting in a left-shifted curve. This adaptation allows the foetus to extract oxygen from the mother's blood across the placenta – a comparison often tested in exam questions.
胎儿血红蛋白（HbF）与 O₂ 的亲和力高于成人血红蛋白（HbA），表现为曲线左移。这种适应使胎儿能够跨越胎盘从母血中获取氧气，这是考题中常见的对比点。
7. Monohybrid Inheritance and Pedigree Analysis | 单基因遗传与系谱分析
A couple, Mark and Lisa, have a child with cystic fibrosis, yet neither parent shows symptoms. Lisa's brother also has CF, but Mark has no family history. Draw a pedigree and determine the probability that their next child will be a carrier but not affected.
一对夫妇马克和丽莎的孩子患有囊性纤维化，但父母均无症状。丽莎的兄弟也患有 CF，而马克无家族史。请绘制系谱并确定他们下一个孩子是携带者但不受影响的概率。
CF is an autosomal recessive disorder. Since Lisa is unaffected but has a child with CF and an affected brother, Lisa must be a heterozygous carrier (Aa). Mark has no family history, so the most probable genotype is homozygous normal (AA), but their affected child proves Mark must also be a carrier (Aa) because an affected child requires two recessive alleles.
CF 是一种常染色体隐性遗传病。丽莎未患病却生育了 CF 患儿且有一个患病的兄弟，因此丽莎必然是杂合子携带者（Aa）。马克无家族史，其最可能基因型为纯合正常（AA），但他们患病的子女证明马克也必定是携带者（Aa），因为患病需要两个隐性等位基因。
Gametes from parents: A, a (Lisa) × A, a (Mark)
(Aa) are 2 out of 3. So the chance that the next child is a carrier but unaffected is 2/3 × 3/4 = 1/2, or 50%.
用庞纳特方格得出基因型比例为 1 AA : 2 Aa : 1 aa。不受影响的概率为 3/4，而不受影响的孩子中携带者（Aa）占 2/3。因此下一个孩子是携带者但不受影响的概率为 2/3 × 3/4 = 1/2，即 50%。
es are unaffected carriers, giving a direct 2/4 = 1/2 probability. The distinction between 'carrier' and 'affected' is a common source of marks and must be explicitly linked to the recessive pattern.
也可直接列出子代：AA（正常、非携带者）、Aa（携带者）、aA（携带者）、aa（患病）。四个中有两个是未患病的携带者，直接得出 2/4 = 1/2 的概率。“携带者”与“患病”之间的区别是常见的得分点，必须明确联系隐性遗传模式作答。
8. Immune Response: Primary and Secondary | 免疫反应：初次与再次应答
A child receives two doses of the MMR vaccine: the first at 12 months and the second at 4 years. Antibody titres were measured over time. After the first dose, IgM appeared at day 5 and peaked at day 10, while IgG appeared at day 7, peaking around day 21. After the second dose, IgG levels rose sharply within 2 days, reaching a peak five times higher than the primary peak.
一名儿童接种了两剂 MMR 疫苗：第一剂在 12 个月时，第二剂在 4 岁时。随时间测量抗体滴度。第一剂后，IgM 于第 5 天出现，第 10 天达峰；IgG 于第 7 天出现，约第 21 天达峰。第二剂后，IgG 水平在 2 天内急剧上升，峰值是初次应答峰值的五倍。
The first exposure (whether natural or vaccine) activates naive B cells that differentiate into plasma cells and memory B cells. IgM is produced first because the heavy-chain gene rearrangement allows rapid secretion of pentameric IgM. Class switching to IgG requires T-helper cell cytokines and takes days, hence the lag in IgG production.
初次暴露（无论是自然感染还是疫苗）激活初始 B 细胞，分化为浆细胞和记忆 B 细胞。IgM 最先产生，因为重链基因重排允许快速分泌五聚体 IgM。转换为 IgG 需要辅助性 T 细胞分泌的细胞因子，该过程需时数日，因此 IgG 的产生存在滞后。
Upon re-exposure, memory B cells that bear high-affinity, class-switched receptors are rapidly activated. They proliferate and differentiate within hours, producing a much faster, stronger and predominantly IgG response. This explains the observed rapid surge of IgG after the booster.
再次暴露时，携带高亲和力、已类别转换受体的记忆 B 细胞被迅速激活。它们在数小时内增殖和分化，产生更快、更强且以 IgG 为主的应答。这解释了加强针后观察到的 IgG 急剧飙升。
| Feature | Primary response | Secondary response |
|---|---|---|
| Lag phase | 5–10 days | 1–3 days |
| Antibody class | IgM then IgG | Mainly IgG |
| Peak antibody level | Lower | 5–10 times higher |
| Affinity | Lower | Higher (affinity maturation) |
The MMR booster exploits immunological memory to ensure long-term protection, overriding any individual variation in primary response and closing the immunity gap. This principle is examined frequently, often requiring you to interpret graphs and explain the cellular basis of the difference.
MMR 加强针利用免疫记忆确保长期保护，克服了个体间
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