📚 Case Study Practice for Year 13 WJEC Biology | Year 13 WJEC 生物案例分析实战演练
Case study questions are a cornerstone of the Year 13 WJEC Biology examination. They require you to apply knowledge from across the entire specification to unfamiliar scenarios, testing not just recall but the ability to analyse data, evaluate evidence, and construct coherent scientific arguments. This article provides a structured approach to tackling such questions, followed by worked examples that mirror the style and depth you will encounter in your exams. By practising these case studies, you will sharpen your analytical skills and build confidence for the final assessment.
案例分析题是 Year 13 WJEC 生物考试的核心题型。它们要求你将整个考纲中的知识应用于陌生情境，考查的不仅是记忆，更是分析数据、评估证据和构建连贯科学论证的能力。本文提供一套结构化解题方法，并配以贴近考试风格和深度的实战案例。通过练习这些案例，你将磨炼分析技能，为最终考试建立信心。
1. What Are Case Study Questions? | 什么是案例分析题？
Case study questions present a passage or a set of data (graphs, tables, diagrams) describing a biological scenario. They are designed to assess AO2 (application of knowledge) and AO3 (analysis, evaluation, and synthesis) skills. Typically, you will need to identify patterns, explain underlying mechanisms, make predictions, or suggest experimental improvements. These questions often combine topics; for example, a single case may link photosynthesis, respiration, and the effect of environmental changes on ecosystems.
案例分析题给出描述某个生物场景的短文或一组数据（图表、表格、示意图），旨在考查 AO2（知识应用）和 AO3（分析、评价与综合）能力。你通常需要识别模式、解释深层机制、作出预测或提出实验改进措施。这类题目常常综合多个主题，比如一个案例可能同时涉及光合作用、呼吸作用以及环境变化对生态系统的影响。
In WJEC Unit 3 and Unit 4 papers, such questions carry significant marks and demand extended prose. Mark schemes reward the use of precise biological terminology and a logical flow of ideas. The key is to treat the case as a puzzle: all the clues are in the stimulus material, and your existing knowledge is the tool to decode them.
在 WJEC 单元 3 和单元 4 试卷中，此类题目分值高，要求扩展性回答。评分标准奖励使用精确的生物术语和清晰的逻辑链。关键在于将案例视作谜题：所有线索都在题干材料中，你的已有知识便是解码的工具。
2. Effective Strategies for Tackling Case Studies | 应对案例分析题的有效策略
Begin by reading the entire case study and all sub-questions carefully. Highlight command words such as ‘explain’, ‘suggest’, ‘evaluate’, and ‘calculate’. Then annotate the data: circle anomalies, slope changes on graphs, or maximum values. Next, mentally link the scenario to the relevant topic area – is this about nerve impulses, population growth, or immune responses? Before writing, plan how your answer will move from observation to biological interpretation.
首先通读整个案例和所有小题，圈出指令词如“解释”、“建议”、“评价”和“计算”。然后在图表上做标记：圈出异常点、曲线斜率的变化或最大值。接着在脑中把场景与相关主题领域联系起来——这涉及神经冲动、种群增长还是免疫应答？动笔前规划如何从观察过渡到生物学解释。
Data analysis questions often require you to describe trends and then give a reason. For instance, if a graph shows oxygen production levelling off at high light intensity, state that the rate becomes constant, then explains it using the concept of limiting factors, such as carbon dioxide concentration or temperature. Use the formula ‘describe + because’ to ensure you address both aspects. Always include units and quote figures from the stimulus to support your reasoning.
数据分析题经常要求你先描述趋势，再给出原因。例如，若曲线显示高光强下氧气产量趋于平稳，应先指出速率不再增加，再用限制因子（如 CO₂ 浓度或温度）的概念加以解释。运用“描述 + 因为”的模式确保两点都覆盖。务必写明单位，并引用题干中的数据来支撑你的论证。
For evaluative questions, weigh up evidence for and against a conclusion. If a vaccine trial shows a drop in disease incidence but a small rise in mild side effects, you must acknowledge both the efficacy and the safety concern, using the data to quantify the effect. A balanced conclusion that references specific numbers will score highly.
对于评价类问题，要权衡支持与反对某一结论的证据。若疫苗试验显示发病率下降但轻微副作用略有上升，你必须同时承认有效性和安全性隐忧，并用数据量化效应。引用具体数字的平衡结论会得到高分。
Finally, check your answers against the question’s mark allocation. A three-mark ‘explain’ question expects three distinct scientific points. Avoid writing a long paragraph that only makes one point repeatedly.
最后，对照分值检查答案。一道 3 分的“解释”题需要 3 个不同的科学要点。避免写一段冗长文字却只重复一个观点。
3. Case Study 1: Limiting Factors in Photosynthesis | 案例一：光合作用的限制因子
Scenario: A student investigated the effect of light intensity on the rate of photosynthesis in Elodea. The number of bubbles released per minute was recorded under different light intensities at a constant CO₂ concentration (0.04 %) and temperature (25 °C). The data are shown in the table below.
场景：一位学生探究了光照强度对黑藻光合速率的影响。在恒定 CO₂ 浓度（0.04%）和温度（25 °C）下，记录不同光强下每分钟释放的气泡数。数据如下表。
| Light intensity / µmol m⁻² s⁻¹ | Bubbles per minute |
|---|---|
| 0 | 0 |
| 200 | 14 |
| 400 | 27 |
| 600 | 38 |
| 800 | 42 |
| 1000 | 43 |
| 1200 | 43 |
Describe and explain the relationship shown by the data. (4 marks)
请描述并解释数据所示的关系。（4 分）
Model answer: At light intensities from 0 to 600 µmol m⁻² s⁻¹, the rate of photosynthesis increases rapidly, as shown by the rise in bubble count from 0 to 38 min⁻¹. Between 600 and 1000 µmol m⁻² s⁻¹, the rate continues to increase but more slowly, reaching 43 bubbles min⁻¹. Beyond 1000 µmol m⁻² s⁻¹, the rate remains constant at 43 bubbles min⁻¹. Initially, light intensity is the limiting factor because it directly drives the light-dependent reactions, producing ATP and NADPH. As light intensity rises further, another factor – likely CO₂ concentration (only 0.04 %) – becomes limiting, so the rate plateaus despite more light being available.
参考答案：在 0 至 600 µmol m⁻² s⁻¹ 光强范围内，光合速率快速上升，气泡数从 0 增至 38 min⁻¹。600 至 1000 µmol m⁻² s⁻¹ 之间，速率继续上升但变缓，达到 43 min⁻¹。超过 1000 µmol m⁻² s⁻¹ 后，速率稳定在 43 min⁻¹。起初光强是限制因子，因为它直接驱动光反应，产生 ATP 和 NADPH。随着光强进一步提高，另一个因子——很可能是 CO₂ 浓度（仅 0.04%）——变为限制因子，因此尽管光强增加，速率不再上升。
Tip: Notice how the answer systematically ‘describes’ the trend in three phases and then ‘explains’ each phase using the concept of limiting factors. Always refer to the data points to substantiate your description.
点拨：注意答案如何系统地将趋势“描述”为三个阶段，并用限制因子概念“解释”每个阶段。描述时务必引用数据点。
4. Case Study 2: Action Potentials and Nerve Impulses | 案例二：动作电位与神经冲动
Scenario: Figure 1 shows the changes in membrane potential of a squid giant axon during an action potential. At time 0 ms, the membrane potential is −70 mV. A stimulus is applied at 1 ms, causing the potential to rise sharply to +35 mV by 2.5 ms, then fall below −70 mV to −80 mV at 4 ms, and finally return to −70 mV by 5 ms.
场景：图 1 显示枪乌贼巨轴突动作电位期间膜电位的变化。0 ms 时，膜电位为 −70 mV。1 ms 时施加刺激，电位急剧上升，在 2.5 ms 时达到 +35 mV，然后下降至 −80 mV（4 ms），最终在 5 ms 时恢复至 −70 mV。
Explain the changes in membrane potential between 1 ms and 4 ms. (5 marks)
解释 1 ms 到 4 ms 间膜电位的变化。（5 分）
Model answer: The stimulus causes voltage-gated sodium ion channels to open. Sodium ions (Na⁺) diffuse rapidly into the axon down their electrochemical gradient, causing depolarisation. This inward movement of positive charge drives the membrane potential from −70 mV up to +35 mV. At the peak, sodium channels inactivate and voltage-gated potassium channels open. Potassium ions (K⁺) diffuse out of the axon, repolarising the membrane. The efflux of K⁺ is so large that the membrane potential briefly becomes more negative than the resting potential (hyperpolarisation, −80 mV), before the potassium channels close and the sodium-potassium pump restores resting ionic distribution.
参考答案：刺激使电压门控钠离子通道开放。钠离子（Na⁺）顺着电化学梯度迅速内流，引起去极化。正电荷的内移使膜电位从 −70 mV 升至 +35 mV。在峰值处，钠通道失活，电压门控钾通道开放。钾离子（K⁺）外流出轴突，使膜复极化。K⁺ 的外流量极大，导致膜电位短暂变得比静息电位更负（超极化，−80 mV），随后钾通道关闭，钠钾泵恢复静息离子分布。
You could also be asked to calculate the absolute refractory period from the graph or to explain why the action potential is all-or-nothing. Always link the opening and closing of specific ion channels to the shape of the curve.
你可能还需根据曲线计算绝对不应期，或解释动作电位为何是全或无的。始终将特定离子通道的开闭与曲线形状联系起来。
5. Case Study 3: Immunity and Vaccination Response | 案例三：免疫与疫苗接种反应
Scenario: A child received two doses of the MMR vaccine at 12 months and 4 years. Blood antibody levels against measles were measured over 8 years. At 12 months (first dose), antibody level was 0.2 arbitrary units (AU). It rose to 6.0 AU by 18 months, then declined to 2.0 AU at 4 years. After the booster at 4 years, antibody level jumped to 15.0 AU within 2 months and remained above 8.0 AU until age 8.
场景：一名儿童在 12 个月和 4 岁时接种了两剂 MMR 疫苗。测量了 8 年间麻疹抗体血液水平。12 个月（首剂）时抗体为 0.2 任意单位 (AU)，18 个月时升至 6.0 AU，4 岁时降至 2.0 AU。4 岁加强针后，抗体水平在 2 个月内跃升至 15.0 AU，并维持在 8.0 AU 以上直至 8 岁。
Explain why the secondary response is faster and produces a higher antibody concentration than the primary response. (4 marks)
解释为何二次应答比初次应答更快、产生更高抗体浓度。（4 分）
Model answer: During the primary response, naive B lymphocytes specific to measles antigens are activated. They proliferate and differentiate into plasma cells that secrete antibodies, and into memory B cells. The primary response is relatively slow because few specific lymphocytes are present initially. After the primary response, the population of long-lived memory B cells remains in the body. Upon re-exposure (the booster), these memory cells quickly recognise the antigen, proliferate, and differentiate into plasma cells much more rapidly. This produces a larger clone of antibody-secreting cells, leading to a higher and more sustained antibody concentration.
参考答案：初次应答中，对麻疹抗原特异的初始 B 淋巴细胞被激活。它们增殖分化为分泌抗体的浆细胞和记忆 B 细胞。初次应答较慢，因为最初体内特异性淋巴细胞很少。初次应答后，长寿记忆 B 细胞群体留存在体内。再次暴露（加强针）时，这些记忆细胞快速识别抗原，更迅速地增殖分化为浆细胞，产生更大的抗体分泌克隆，从而出现更高且更持久的抗体浓度。
Examiners expect you to use the terms ‘memory B cells’, ‘clone’, and ‘differentiation’ explicitly. The shape of the antibody-level curves reflects the dynamics of clonal selection.
考官希望你明确使用“记忆 B 细胞”、“克隆”和“分化”等术语。抗体水平曲线的形状反映了克隆选择动力学。
6. Case Study 4: Population Dynamics and Predator-Prey Relationships | 案例四：种群动态与捕食者-猎物关系
Scenario: A lake was stocked with pike (predator) and perch (prey). Population sizes were estimated annually. The perch population grew from 200 to 1200 over 3 years, then declined to 300 in the next 2 years while pike numbers rose from 50 to 350. Subsequently, perch numbers rose again to 1100 as pike fell back to 100.
场景：某湖泊引入白斑狗鱼（捕食者）与河鲈（猎物）。每年估算种群数量。河鲈在 3 年内从 200 增至 1200，随后 2 年降至 300，此时狗鱼从 50 升至 350。随后河鲈数量又升至 1100，狗鱼则回落至 100。
Using the Lotka–Volterra model, explain the cyclic fluctuations in the two populations. (5 marks)
运用洛特卡-沃尔泰拉模型解释两个种群的周期性波动。（5 分）
Model answer: When the perch population is high, there is abundant food for pike, so pike survival and reproduction increase, causing the pike population to rise. The increased predation pressure reduces the perch population. As perch numbers fall, pike experience food shortage, leading to increased mortality and reduced reproduction, so the pike population declines. With fewer predators, the perch population can recover, and the cycle repeats. The predator peak lags behind the prey peak because it takes time for pike numbers to respond to increased food availability. This generates the out-of-phase, wave-like dynamics characteristic of predator-prey interactions described by the Lotka–Volterra equations.
参考答案：当河鲈数量很高时，狗鱼食物充足，存活率与繁殖率提高，狗鱼种群上升。增强的捕食压力使河鲈数量下降。随着河鲈减少，狗鱼面临食物短缺，死亡率升高、繁殖降低，狗鱼种群下降。捕食者减少后河鲈得以恢复，周而复始。捕食者峰值滞后于猎物峰值，因为狗鱼数量需要时间响应食物增加。由此形成捕食者-猎物相互作用特有的异步波动，如洛特卡-沃尔泰拉方程所描述。
Mathematically, the rate of change in prey population N is dN/dt = rN – aNP, where r is per capita growth rate, a is attack rate, P is predator number. The predator equation is dP/dt = faNP – qP, where f is conversion efficiency and q is mortality. This underpins the graphical cycles.
数学上，猎物数量变化率 dN/dt = rN – aNP，r 为内禀增长率，a 为攻击率，P 为捕食者数量。捕食者方程为 dP/dt = faNP – qP，f 为转化效率，q 为死亡率。这支撑了周期性图形。
7. Case Study 5: Gene Expression – The lac Operon | 案例五：基因表达——乳糖操纵子
Scenario: E. coli is grown in a medium containing glucose but no lactose. The enzymes β-galactosidase and lactose permease are not produced. When the bacteria are transferred to a medium with lactose as the sole carbon source, the enzymes appear within minutes. If lactose is removed, enzyme production stops.
场景：大肠杆菌培养在含葡萄糖而无乳糖的培养基中，不产生 β-半乳糖苷酶与乳糖透性酶。当细菌转入以乳糖为唯一碳源的培养基后，几分钟内酶即出现。若除去乳糖，酶合成停止。
Explain the role of the lac operon in this switch. (6 marks)
解释 lac 操纵子在此转换中的作用。（6 分）
Model answer: The lac operon consists of a promoter, operator, and three structural genes (lacZ, lacY, lacA). In the absence of lactose, a repressor protein binds to the operator, blocking RNA polymerase from transcribing the structural genes. When lactose is present, it is converted to allolactose, which acts as an inducer. Allolactose binds to the repressor, changing its shape so it can no longer bind the operator. RNA polymerase can then transcribe the genes, producing mRNA for β-galactosidase and permease. Additionally, when glucose is absent, cAMP levels rise; cAMP binds to CAP, and the cAMP–CAP complex binds to the promoter, enhancing RNA polymerase binding and maximising transcription. When lactose is removed, the repressor becomes active again and binds the operator, switching off the operon.
参考答案：lac 操纵子含启动子、操纵基因和三个结构基因（lacZ, lacY, lacA）。无乳糖时，阻遏蛋白与操纵基因结合，阻止 RNA 聚合酶转录结构基因。有乳糖时，乳糖转变为异乳糖（诱导物）。异乳糖与阻遏蛋白结合，改变其形状，使其无法与操纵基因结合。随后 RNA 聚合酶转录基因，产生 β-半乳糖苷酶和透性酶的 mRNA。此外，无葡萄糖时 cAMP 水平上升；cAMP 与 CAP 结合，cAMP-CAP 复合物结合于启动子，增强 RNA 聚合酶结合，使转录最大化。除去乳糖后，阻遏蛋白重新活化并结合操纵基因，关闭操纵子。
This example illustrates both negative control (repressor) and positive control (CAP) of gene expression, a core concept in WJEC Unit 4.
本例展示了基因表达的负调控（阻遏蛋白）与正调控（CAP），是 WJEC 单元 4 的核心概念。
8. Case Study 6: Respiration and Metabolic Rate | 案例六：呼吸作用与代谢率
Scenario: A respirometer was used to measure O₂ consumption and CO₂ production of germinating seeds using different respiratory substrates. With glucose, the ratio CO₂ produced / O₂ consumed (RQ) was 1.0. With a lipid-rich seed extract, RQ was 0.7. With a protein-rich extract, RQ was 0.9.
场景：用呼吸计测定不同呼吸底物下萌发种子的耗 O₂ 量和 CO₂ 产量。使用葡萄糖时，呼吸商 (RQ = 产生 CO₂ / 消耗 O₂) 为 1.0。使用富含脂质的种子提取物时，RQ 为 0.7。使用富含蛋白质的提取物时，RQ 为 0.9。
Explain why the RQ values differ between the substrates and what this indicates about the metabolic pathways involved. (4 marks)
解释为何底物间 RQ 值不同，并说明这反映了所涉及的代谢途径。（4 分）
Model answer: Glucose is a carbohydrate with a general formula (CH₂O)ₙ. Its complete oxidation requires equal numbers of O₂ molecules and CO₂ molecules produced, giving RQ = 1.0. Lipids have a much lower proportion of oxygen atoms, so more O₂ is needed to oxidise fatty acids, resulting in an RQ less than 1 (often 0.7). Proteins have a variable RQ around 0.9 because their amino acids contain more oxygen than lipids but also require some O₂ for deamination and urea production. The RQ value thus allows us to infer the predominant substrate being respired. In this experiment, the shift from 1.0 to 0.7 when lipid extract was used confirms a switch from carbohydrate to lipid metabolism.
参考答案：葡萄糖是碳水化合物，通式为 (CH₂O)ₙ。其完全氧化消耗的 O₂ 分子数与产生的 CO₂ 分子数相等，RQ = 1.0。脂质氧原子比例低得多，氧化脂肪酸需更多 O₂，导致 RQ 小于 1（常为 0.7）。蛋白质约 0.9，因其氨基酸含氧量高于脂质，但脱氨和尿素生成仍需额外 O₂。因此 RQ 值可推断主要呼吸底物。本实验中，使用脂质提取物时 RQ 从 1.0 降至 0.7，证实了代谢从碳水化合物转向脂质。
In WJEC, you may also need to calculate RQ from given volumes and link it to anaerobic respiration (RQ > 1). Practice such calculations to become fluent with the units.
在 WJEC 考试中，你可能还需要根据给定体积计算 RQ，并将其与无氧呼吸（RQ > 1）联系起来。多加练习，熟练单位和计算。
9. Case Study 7: Ecological Succession | 案例七：生态演替
Scenario: A field was abandoned after farming ceased. Vegetation was surveyed over 50 years. Year 0: bare soil with annual weeds. Year 5: grasses and perennial herbs dominate. Year 20
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