📚 Case Study Practice for WJEC AS Biology | WJEC AS 生物学案例分析实战演练
Case study questions form a significant part of the WJEC AS Biology examination. They require you to apply knowledge from Unit 1 (Basic Biochemistry and Cell Organisation) and Unit 2 (Biodiversity and Physiology of Body Systems) to unfamiliar scenarios. This guide offers a structured approach to tackling these questions through real-world examples, data interpretation, and evaluation of experimental design. By practising with the case studies provided, you will improve your ability to extract relevant information, analyse numerical data, and construct clear, logical answers that meet the Assessment Objectives.
案例分析题在 WJEC AS 生物考试中占有重要地位。这类题目要求你将单元一（基础生物化学与细胞组织）和单元二（生物多样性与身体系统生理学）的知识应用到陌生情境中。本文通过真实案例、数据解读与实验设计评价，为你提供一套系统攻克此类题的训练方法。通过演练这些案例，你将提升提取关键信息、分析数值数据以及构建条理清晰答案的能力，从而更好地达成考核目标。
1. What Are Case Study Questions in WJEC? | 什么是 WJEC 考试中的案例分析题？
Case study questions present a short passage describing an experiment, a biological phenomenon, or a set of observations. They often include tables, graphs, or diagrams. You are expected to analyse the data, identify trends, explain the underlying biology, and evaluate the methodology. Marks are awarded for precise use of scientific terminology and linking data to biological concepts.
案例分析题会提供一小段描述实验、生物学现象或一组观察结果的文字，通常还配有表格、曲线图或示意图。你需要在分析数据的基础上，识别变化趋势、解释背后的生物学原理并评价研究方法。评分点在于能否准确使用科学术语，并用数据支撑生物学概念。
In WJEC AS papers, these questions test Assessment Objective 2 (application of knowledge) and Assessment Objective 3 (evaluation of information). They may draw on topics such as enzyme activity, membrane transport, genetic inheritance, succession, or the cardiac cycle. The data are often given in unfamiliar units or with control groups that you must interpret.
在 WJEC AS 试卷中，这类题同时考查 AO2（知识应用）和 AO3（信息评价）。题干可以来自酶活性、膜运输、遗传传递、演替或心动周期等主题。给出的数据可能使用陌生单位或设置有对照组，需要你去解读。
2. Key Skills for Tackling Case Studies | 应对案例分析的关键技能
Start by reading the introductory text carefully. Underline the independent variable (the factor being changed), the dependent variable (what is measured), and any controlled variables mentioned. This helps you understand the design of the investigation before you look at the results.
首先要仔细阅读引导文字。划出自变量（被改变的因素）、因变量（被测量的量）和所有提到的控制变量。这能让你在查看结果之前先理解实验设计。
Next, scan any data table or graph to identify major patterns: is there an increase, a decrease, an optimum point, or a plateau? Look for anomalies — data points that do not fit the overall trend. Many questions ask you to suggest reasons for such anomalies, using your biological knowledge about variability or limitations of technique.
接着快速浏览数据表或图表，找出主要规律：是上升、下降、出现最优点还是趋于平稳？同时注意异常值——不符合整体趋势的数据点。很多题目会要求你用关于生物变异或技术局限的知识，解释异常值出现的可能原因。
Finally, when answering, use the “PEEL” structure: Point, Evidence (with data quoting), Explanation (biological mechanism), and Link back to the question. For example, “The rate of reaction increased between 10°C and 40°C (from 2.3 to 6.7 arbitrary units) because higher temperature provides more kinetic energy to enzyme and substrate molecules, leading to more frequent successful collisions.”
最后作答时可使用 “PEEL” 结构：提出观点、给出证据（引用数据）、解释（生物学机理）、回扣题目。例如：“反应速率在 10℃ 至 40℃ 之间上升（从 2.3 升至 6.7 任意单位），因为温度升高为酶和底物分子提供了更多动能，导致成功碰撞频率增加。”
3. Interpreting Data Tables and Graphs | 解读数据表格与图表
WJEC case studies frequently use line graphs, bar charts, and scattergrams. When you see a line graph, check whether the independent variable is continuous. If it is, describe the trend as “linear increase”, “exponential rise”, or “bell-shaped curve”. Always quote numbers from the axes to support your description.
WJEC 案例常常出现折线图、条形图和散点图。看到折线图时，先判断自变量是否连续。如果是连续的，可以描述趋势为“线性增长”“指数上升”或“钟形曲线”。描述时必须引用坐标轴上的数值作为证据。
For tables, calculate percentage changes or ratios if the question requires comparison. For instance, if a table shows the number of bacterial colonies at different antibiotic concentrations, you might compute the percentage inhibition: (colonies without antibiotic – colonies with antibiotic) / colonies without antibiotic × 100%. Show your working, as it may be awarded separate marks.
对于表格，若题目要求比较，可计算百分比变化或比值。比如，一张表给出不同抗生素浓度下的细菌菌落数，你可以计算抑制百分率：（无抗生素菌落数 − 含抗生素菌落数）/ 无抗生素菌落数 × 100%。写出计算过程，因为步骤可单独得分。
Pay attention to error bars or standard deviation values if provided. Overlapping error bars suggest that any observed difference may not be statistically significant. In WJEC AS, you are not required to carry out statistical tests, but you should be able to comment on the reliability of differences from the spread of data.
如果给出误差线或标准差值，要特别关注。若误差线相互重叠，说明观察到的差异可能不具有统计学显著性。在 WJEC AS 阶段无需进行统计检验，但应能根据数据的分散程度评价差异的可靠性。
4. Case Study 1: Enzyme Activity and Temperature | 案例一：酶活性与温度
An investigation was carried out into the effect of temperature on the activity of amylase. The enzyme was mixed with starch solution at six different temperatures, and the time taken for the complete breakdown of starch was recorded using iodine tests. The results are shown below.
一项实验研究了温度对淀粉酶活性的影响。将酶与淀粉溶液混合，分别在六个温度条件下反应，通过碘液测试记录淀粉完全分解所需的时间。结果如下表所示。
| Temperature / °C | Time for starch breakdown / s | Rate of reaction / s⁻¹ (×10⁻³) |
|---|---|---|
| 10 | 245 | 4.1 |
| 20 | 98 | 10.2 |
| 30 | 42 | 23.8 |
| 40 | 25 | 40.0 |
| 50 | 33 | 30.3 |
| 60 | 180 | 5.6 |
The calculated rate (1/time) is plotted against temperature. Describe and explain the trend shown. The rate increases from 10°C to 40°C, with the highest rate at 40°C (40.0 × 10⁻³ s⁻¹). This is because as temperature rises, kinetic energy of molecules increases, leading to more frequent collisions and more enzyme-substrate complexes formed. Beyond 40°C, the rate drops sharply; at 60°C it is only 5.6 × 10⁻³ s⁻¹, suggesting that the enzyme has been denatured. High temperature breaks hydrogen and ionic bonds in the tertiary structure, altering the active site’s shape.
将计算得到的反应速率（1/时间）对温度作图。请描述并解释变化趋势。速率从 10℃ 到 40℃ 上升，在 40℃ 时达到最高（40.0 × 10⁻³ s⁻¹）。这是因为温度升高使分子动能增加，碰撞更频繁，更多的酶-底物复合物得以形成。超过 40℃ 后速率急剧下降；60℃ 时仅为 5.6 × 10⁻³ s⁻¹，表明酶已变性。高温破坏了三级结构中的氢键和离子键，改变了活性位点的形状。
The optimum temperature appears to be 40°C, but in a standard classroom experiment with human amylase, the expected optimum is around 37°C. In a WJEC mark scheme, you might be asked to account for this discrepancy. Possible reasons include the use of a thermostatically controlled water bath that overshot the set temperature, or that a plant amylase with higher optimum was used. Acknowledging such limitations demonstrates evaluative skill.
最适温度来看是 40℃，但在标准课堂实验中，人体淀粉酶的最适温度约为 37℃。在 WJEC 评分方案中，可能会让你解释这一差异。可能原因包括恒温水浴超调，或使用了具有更高最适温度的植物淀粉酶。承认这些局限性能够体现你的评价能力。
5. Identifying Variables and Controls | 识别变量与对照
Every well-designed investigation has a clear independent variable (IV), dependent variable (DV), and at least one control group. In the amylase example above, the IV is temperature, the DV is time for starch breakdown (or calculated rate), and the controlled variables could include pH, enzyme concentration, and starch concentration. A control group might involve a tube with boiled enzyme to confirm that breakdown is enzyme-mediated.
每个设计严谨的实验都应有明确的自变量（IV）、因变量（DV）和至少一个对照组。在上述淀粉酶实验中，自变量是温度，因变量是淀粉分解时间（或计算得到的速率），控制变量包括 pH、酶浓度和淀粉浓度。对照可以设置一组含有煮沸酶液的试管，以证实分解反应确实由酶催化。
In WJEC case studies, you may be presented with an experiment that lacks certain controls. You should be able to point out, for instance, that no measure was taken to keep pH constant, which could confound the effect of temperature if the enzyme’s pH optimum is narrow. Similarly, if the experiment was not replicated, the reliability of the data is questionable. Always link your criticism to the specific context.
在 WJEC 案例中，题目可能会呈现一个缺少某些对照的实验。你应该能够指出，比如，未采取措施保持 pH 恒定，若酶的最适 pH 范围很窄，则会干扰温度效应的判断。同样，如果实验没有设置重复，数据的可靠性就值得怀疑。批评时务必与具体情境相结合。
6. Case Study 2: Inheritance and Pedigree Charts | 案例二：遗传与家系图
A pedigree chart is displayed showing the inheritance of a rare condition in a family. Affected individuals are shaded. The trait skips generations and appears in both males and females. From this pattern, you can deduce that the allele is recessive and not sex-linked (since an affected female has an unaffected father). The genotype of the affected individuals can be assigned as homozygous recessive (aa), while unaffected parents must be heterozygous (Aa) if they have an affected child.
题干给出某家族中一种罕见病症的家系图，患病个体以阴影表示。该性状隔代出现，且在男性和女性中均有分布。由此可推断该等位基因为隐性，且不位于性染色体上（因为一名患病女性的父亲未患病）。可推定患病个体基因型为隐性纯合子（aa），而拥有患病子女的正常父母必定是杂合子（Aa）。
A typical subsequent question asks for the probability that an unaffected sibling is a carrier. Using a Punnett square: Aa × Aa yields a genotypic ratio of 1 AA : 2 Aa : 1 aa. Since the sibling is unaffected, the aa possibility is eliminated, leaving a 2/3 chance of being a carrier. Clearly state this reasoning and show your Punnett square in the answer space.
常见追问是计算某一正常兄弟姐妹是携带者的概率。使用庞纳特方格：Aa × Aa 的基因型比为 1 AA : 2 Aa : 1 aa。因为该个体表现正常，可排除 aa 的可能性，故携带者概率为 2/3。作答时应清晰阐述推理过程并画出方格。
You might also be asked to explain why the condition cannot be dominant. If the condition were dominant, at least one parent of each affected individual would have to be affected, which is not the case here as unaffected parents produced affected children. This form of reasoning is frequently tested in WJEC Unit 1.
题目还可能要求解释为何该病症不可能是显性遗传。若为显性，每个患病个体至少有一名亲本患病，但本题中正常父母生出了患病子女，故不成立。这种推理方式是 WJEC 单元一的常见考点。
7. Statistical Analysis: Calculating Means and Standard Deviation | 统计分析：计算平均值与标准差
Case studies sometimes provide raw repeated measurements. You will be expected to calculate a mean (x̄) and, in some cases, a standard deviation (s) to assess dispersion. The formula for standard deviation may be given, but you should be able to substitute values correctly.
s = √[ Σ(x – x̄)² / (n – 1) ]
案例有时会给出原始的重复测量数据。你需要计算平均值（x̄），有时还要计算标准差（s）以评价离散程度。标准差公式通常会给出，但你要能正确代入数值。
s = √[ Σ(x – x̄)² / (n – 1) ]
For example, the following readings of heart rate (bpm) were taken for a student at rest: 62, 65, 64, 63, 67. The mean is 64.2 bpm. Calculate the standard deviation: first, find the deviations (−2.2, +0.8, −0.2, −1.2, +2.8), square them, sum them (4.84+0.64+0.04+1.44+7.84=14.8). Divide by n−1 (4) to give 3.7, then take the square root: s ≈ 1.92 bpm. This small standard deviation indicates that the readings were tightly clustered around the mean, showing good precision.
例如，某学生静息心率（bpm）测量值为：62, 65, 64, 63, 67。平均值为 64.2 bpm。计算标准差：先求出偏差（−2.2, +0.8, −0.2, −1.2, +2.8），平方后求和（4.84+0.64+0.04+1.44+7.84=14.8），除以 n−1（4）得 3.7，再开平方根：s ≈ 1.92 bpm。标准差较小，说明数据集中在均值附近，精密度较好。
In a WJEC context, you might compare two sets of data. If mean blood pressure before exercise is 120 ± 5 mmHg and after is 140 ± 12 mmHg, the overlapping ranges (120±5 gives 115–125; 140±12 gives 128–152) do not overlap; therefore, the difference is likely real. However, if the error bars overlapped, you would state that the difference may not be significant and more replicates are needed.
在 WJEC 答题中，你可能需要比较两组数据。如果运动前平均血压为 120 ± 5 mmHg，运动后为 140 ± 12 mmHg，两个范围（120±5 给出 115–125；140±12 给出 128–152）不重叠，说明差异很可能真实存在。如果误差条相互覆盖，则应指出差异可能不显著，需增加重复。
8. Case Study 3: Energy Flow in an Ecosystem | 案例三：生态系统能量流动
A case study presents the energy content in a grassland food chain: producers (grass) contain 20 000 kJ m⁻² yr⁻¹; primary consumers (rabbits) contain 2 000 kJ m⁻² yr⁻¹; secondary consumers (foxes) contain 200 kJ m⁻² yr⁻¹. The efficiency of energy transfer can be calculated as (energy at trophic level / energy at previous level) × 100%. From grass to rabbits: (2000 / 20 000) × 100% = 10%. From rabbits to foxes: (200 / 2000) × 100% = 10%.
案例给出一条草原食物链的能量数据：生产者（草）含 20 000 kJ m⁻² yr⁻¹；初级消费者（兔）含 2 000 kJ m⁻² yr⁻¹；次级消费者（狐）含 200 kJ m⁻² yr⁻¹。能量传递效率 = （某一营养级的能量 / 上一级能量）× 100%。从草到兔：（2000 / 20 000）× 100% = 10%；从兔到狐：（200 / 2000）× 100% = 10%。
Explain why so much energy is lost: energy is lost as heat from respiration, as undigested material in faeces, and through excretion of urea. In WJEC mark schemes, simply stating “respiration” is insufficient; you must specify that it is heat energy released during respiration, which cannot be reused by the next trophic level. The low efficiency limits the length of food chains, typically to four or five levels.
解释为何能量损耗如此巨大：能量通过呼吸作用以热的形式散失、以粪便中未消化物质的形式排出，以及通过尿素排泄丢失。在 WJEC 评分方案中，只写“呼吸作用”是不够的，必须明确指出是呼吸过程释放的热能，不能被下一营养级再利用。低传递效率限制了食物链的长度，通常只有四到五个营养级。
You may also be asked to calculate the percentage of energy originally captured by producers that remains in the secondary consumers. This is (200 / 20 000) × 100% = 1%. Using such examples helps you practise the arithmetic that is an integral part of AS Biology cases.
还可能要求计算生产者最初固定的能量中，有多少保留到次级消费者。即（200 / 20 000）× 100% = 1%。通过这类例子可以训练算术能力，这是 AS 生物案例中不可分割的一部分。
9. Evaluating Experimental Design and Reliability | 评价实验设计与可靠性
Evaluation questions ask, “How could the investigation be improved?” Your response must go beyond generic statements like “use more repeats”. Be specific: “Repeat the experiment at each temperature at least three times and calculate the mean to reduce the effect of random errors” or “Use a colorimeter to measure starch concentration instead of relying on subjective iodine colour judgement, which increases accuracy and objectivity.”
评价类问题会问：“可以如何改进这个探究？”你的回答不能只笼统地说“多做重复”，而要具体：“每个温度至少重复实验三次并计算平均值，以减小随机误差的影响”或“用比色计测量淀粉浓度，而不是依靠主观的碘液颜色判断，从而提高准确性和客观性”。
Another aspect of reliability is sample size. In a study on lung capacity, if only one male and one female are tested, the results cannot be generalised. Suggest: “Increase the number of participants to at least 10 per gender and control for variables such as age, height, and fitness level.” This shows awareness of the factors that contribute to biological variability.
可靠性的另一个方面是样本量。在一项肺活量研究中，如果只测量了一男一女，其结果无法推广。可以建议：“每个性别至少增加至 10 名参与者，并控制年龄、身高和体能等变量。”这表现出你对导致生物变异性因素的认识。
When an anomaly is present in the data, do not ignore it. Acknowledge it and suggest a biological reason, such as “This plant may have been in the shade, reducing its photosynthetic rate”, or a methodological reason, “The syringe may have leaked, causing a false reading.” Always recommend repeating the measurement to verify it.
数据中出现异常值时，不能忽略。要承认它的存在，并提出生物学原因，例如“这株植物可能处于阴蔽处，导致光合速率下降”，或方法学原因，“注射器可能漏气，造成假读数”。最后总要建议重测以进行验证。
10. Case Study 4: Immunity and Antibody Response | 案例四：免疫与抗体应答
A patient’s blood was tested for antibody concentration after two exposures to the same antigen, six weeks apart. The graph shows a small primary response peaking at day 10 with concentration 15 arbitrary units, followed by a large secondary response peaking at day 5 after the second exposure with 120 arbitrary units. The concentration then declined more slowly in the secondary response.
某患者两次接触同一抗原（间隔六周）后，检测其血液中的抗体浓度。图线显示，初次应答在接触后第 10 天达到峰值（15 任意单位），之后大幅下降；二次接触后第 5 天即出现强烈的二次应答，峰值为 120 任意单位，且下降速度较慢。
Explain the faster, stronger secondary response: after the primary infection, memory B cells and memory T cells are produced and remain in the circulation. Upon re-exposure, these memory cells rapidly divide and differentiate into plasma cells that produce large quantities of specific antibodies. This is the basis of vaccination. In your answer, emphasise the role of clonal selection and the fact that fewer pathogen cells survive long enough to cause disease.
解释为何二次应答更快、更强：初次感染后产生了记忆 B 细胞和记忆 T 细胞，它们停留在循环系统中。再次接触相同抗原时，这些记忆细胞迅速分裂、分化为浆细胞，产生大量特异性抗体。这正是疫苗接种的原理。作答时应强调克隆选择的作用，并指出病原体尚未来得及大量增殖就已受到抑制，因此疾病减轻。
You could be given data on a person who has a compromised immune system, showing a much reduced secondary response. This allows you to link to white blood cell counts or the effects of HIV/AIDS, directly connecting the case study with Unit 2 content on disease and immunity.
题目可能给出免疫功能受损者的数据，其二抗反应严重削弱。借此可联系白细胞计数或 HIV/AIDS 的影响，将案例与单元二关于疾病与免疫的内容直接挂钩。
11. Common Pitfalls and How to Avoid Them | 常见错误及避免方法
Many students lose marks by not quoting data when asked to “describe” or “compare”. If a question says “Compare the effect of temperature on enzyme X and enzyme Y”, you must state which is higher at a specific temperature and provide numerical differences. Using phrases like “Enzyme X had a greater activity” without values is insufficient.
很多学生失分是因为在“描述”或“比较”时没有引用数据。如果题目要求“比较温度对酶 X 和酶 Y 的影响”，你必须说明在某个具体温度下谁的活性更高，并给出数值差异。只写“酶 X 活性更高”而没有数值是不够的。
Another pitfall is confusing correlation with causation. The data might show that as carbon dioxide concentration rises, plant growth increases, but you must explain via photosynthesis: CO₂ is a substrate for the Calvin cycle, where it is fixed by RuBisCO into GP. Simply saying “more CO₂ causes more growth” is descriptive, not explanatory.
另一个陷阱是混淆相关与因果。数据可能显示随着二氧化碳浓度上升，植物生长量增加，但你需要通过光合作用来解释：CO₂ 是卡尔文循环的底物，被 RuBisCO 固定成 GP。只说“CO₂ 多引起生长多”只是描述，而不是解释。
Time management is also critical. In longer case studies, plan your answer before writing. Allocate marks proportionally: a 5-mark question deserves more detail and more data points than a 2-mark one. Practise under timed conditions to build the pace required for the WJEC AS paper.
时间管理同样关键。遇到较长的案例分析题，应先规划再动笔。按分数比例分配时间：5 分的题要比 2 分的题写出更多细节、引用更多数据。建议在限时条件下进行练习，以适应 WJEC AS 试卷的节奏。
12. Practice Questions and Model Answers | 模拟习题与参考答案
Question 1: A student investigated the effect of pH on catalase activity, measuring the volume of oxygen produced in 2 minutes. The results: pH 4 – 3 cm³, pH 5 – 8 cm³, pH 6 – 14 cm³, pH 7 – 19 cm³, pH 8 – 12 cm³, pH 9 – 3 cm³. Describe and explain these results. (4 marks)
题目 1：某学生研究了 pH 对过氧化氢酶活性的影响，记录 2 分钟内产生的氧气体积。结果：pH 4 – 3 cm³，pH 5 – 8 cm³，pH 6 – 14 cm³，pH 7 – 19 cm³，pH 8 – 12 cm³，pH 9 – 3 cm³。描述并解释这些结果。（4 分）
Model Answer: The volume of oxygen produced increases from pH 4 (3 cm³) to a maximum at pH 7 (19 cm³), then decreases to 3 cm³ at pH 9. The highest activity occurs at pH 7, which is close to the optimum pH for catalase. At pH values above or below the optimum, the enzyme’s active site is distorted because hydrogen and ionic bonds maintaining the tertiary structure are disrupted, reducing formation of enzyme-substrate complexes.
参考答案：氧气产量从 pH 4（3 cm³）增加到 pH 7 的最大值（19 cm³），随后降至 pH 9 的 3 cm³。最大活性出现在 pH 7，接近过氧化氢酶的最适 pH。在高于或低于最适 pH 时，维持
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