📚 Case Study Practice Drill for Year 10 WJEC Biology | WJEC生物案例分析实战演练
Case study questions in WJEC GCSE Biology challenge you to apply your knowledge to real‑world scenarios, interpret data, and write structured explanations. In this article, you will work through four complete case studies covering osmosis, enzyme action, biomagnification, and disease transmission. Each section builds the practical skills assessed in Year 10 – from calculating percentage changes and plotting graphs to evaluating experimental design and justifying scientific conclusions.
WJEC GCSE生物考试中的案例分析题要求你将知识应用到实际情景中，解读数据并写出有条理的解释。本文将通过四个完整的案例——渗透作用、酶作用、生物放大和疾病传播——带你实战演练。每一节都会训练Year 10考试所考查的实用技能：从计算百分比变化、绘制图表，到评价实验设计并论证科学结论。
1. What Makes a WJEC Case Study Question? | WJEC案例分析题的特点
WJEC case study questions typically provide a scenario, a data table or graph, and a series of short‑answer and extended‑response parts. They expect you to recall biological facts, perform calculations, describe trends, explain mechanisms using correct terminology, and suggest improvements to investigations. Marks are awarded for clear, logical reasoning and for quoting evidence from the data.
WJEC案例分析题通常会给出一个情景、一张数据表或图表，并设置一系列简答与扩展性问题。题目要求你回忆生物学事实、进行计算、描述趋势、用正确的术语解释机制，并提出改进实验的方法。清晰、逻辑严密的推理以及引用数据中的证据都会获得相应分数。
2. Understanding Command Words | 理解指令词
‘Describe’ asks you to state what the data shows without explaining why. ‘Explain’ requires scientific reasoning – you must link cause and effect using concepts such as water potential or enzyme‑substrate complexes. ‘Calculate’ means showing your working and giving units. ‘Evaluate’ expects you to weigh up strengths and weaknesses of an experiment or a conclusion. ‘Suggest’ invites you to apply your knowledge to a new situation.
‘Describe’ 要求你说出数据呈现的趋势，不需要解释原因。‘Explain’ 则要求给出科学推理——你必须运用水势或酶‑底物复合物等概念把因果关系联系起来。‘Calculate’ 意味着要展示计算步骤并写出单位。‘Evaluate’ 需要你权衡实验或结论的优点与缺点。‘Suggest’ 是邀请你把知识迁移到新情境中。
3. Case Study 1: Osmosis in Potato Strips – Working with Data | 案例1：土豆条的渗透作用——数据处理
A student investigated how sucrose solution concentration affects potato strip mass. Five solutions were prepared: 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm⁻³. Potato strips were blotted dry, weighed, placed in each solution for 30 minutes, then reweighed. The raw data are shown below.
一名学生探究了蔗糖溶液浓度对土豆条质量的影响。配制了0.0、0.2、0.4、0.6、0.8和1.0 mol dm⁻³五种溶液。将土豆条吸干、称重后放入每种溶液中浸泡30分钟，然后再次称重。原始数据如下表所示。
| Concentration (mol dm⁻³) | Initial mass (g) | Final mass (g) |
|---|---|---|
| 0.0 | 5.2 | 5.8 |
| 0.2 | 5.0 | 5.3 |
| 0.4 | 5.1 | 5.0 |
| 0.6 | 5.3 | 4.8 |
| 0.8 | 5.1 | 4.3 |
| 1.0 | 5.0 | 4.0 |
Before analysing the data, you must calculate the percentage change in mass for each concentration. This allows a fair comparison because the initial masses were not identical.
在分析数据之前，你必须先计算每个浓度下的质量变化百分比。这样能进行公平比较，因为土豆条的初始质量并不完全相同。
% change in mass = (final mass − initial mass) / initial mass × 100%
For example, at 0.0 mol dm⁻³: (5.8 − 5.2) / 5.2 × 100% = +11.5%. At 0.6 mol dm⁻³: (4.8 − 5.3) / 5.3 × 100% = −9.4%. Always attach a positive sign for an increase and a negative sign for a decrease.
例如，0.0 mol dm⁻³ 浓度下：(5.8 − 5.2) / 5.2 × 100% = +11.5%。0.6 mol dm⁻³ 时：(4.8 − 5.3) / 5.3 × 100% = −9.4%。增加时记得标上正号，减少时标上负号。
4. Case Study 1: Plotting a Graph and Describing the Trend | 案例1：绘制图表并描述趋势
You would plot concentration on the x‑axis (independent variable) and percentage change in mass on the y‑axis (dependent variable). Use a sharp pencil, draw axes with linear scales, label them with units, and plot points as small crosses. Join the points with a smooth curve of best fit. The graph should show a downward trend, crossing the x‑axis where the change is 0% – this is the point of no net water movement, allowing you to estimate the water potential of the potato tissue.
你将浓度（自变量）标在 x 轴，质量变化百分比（因变量）标在 y 轴。用削尖的铅笔画出坐标轴、标上单位，将数据点用细小的十字标出，并用平滑的曲线连接各点。图形应呈现下降趋势，并在变化为 0% 处穿过 x 轴——该点代表没有净水移动的时刻，借此可以估算土豆组织的水势。
Describe the trend: As sucrose concentration increases from 0.0 to 1.0 mol dm⁻³, the percentage change in mass decreases, from a positive value to an increasingly negative value. The potato strips gain mass in distilled water but lose mass in concentrated sugar solutions.
描述趋势：当蔗糖浓度从 0.0 mol dm⁻³ 增加到 1.0 mol dm⁻³ 时，质量变化百分比下降，从正值变为越来越大的负值。土豆条在蒸馏水中质量增加，而在浓糖溶液中质量减少。
5. Case Study 1: Explaining the Osmosis Results | 案例1：用科学原理解释结果
Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane. In 0.0 mol dm⁻³ solution, the water potential outside the potato cells is higher than inside; water enters the cells by osmosis, making them turgid and increasing the mass of the strips. In 1.0 mol dm⁻³ sucrose, the external solution has a very low water potential, so water leaves the cells, making them flaccid and causing a decrease in mass. When the percentage change is zero, the water potentials inside and outside are equal – this is the isotonic point.
渗透作用是水分子通过选择性半透膜从水势较高的区域向水势较低的区域净移动。在 0.0 mol dm⁻³ 溶液中，土豆细胞外的水势高于细胞内；水通过渗透进入细胞，细胞变得饱满，土豆条质量增加。在 1.0 mol dm⁻³ 蔗糖溶液中，外界溶液的水势非常低，因此水离开细胞，细胞变得萎软，质量下降。当百分比变化为零时，膜内外水势相等，这就是等渗点。
6. Case Study 1: Evaluating the Osmosis Experiment | 案例1：评价渗透实验
A valid evaluation identifies sources of error and suggests realistic improvements. The student should have blotted the strips for a consistent time and force to remove surface water; variation in blotting could lead to random errors. Repeating the experiment three times at each concentration and calculating a mean would increase reliability. The range of concentrations could be extended beyond 1.0 mol dm⁻³ to confirm the isotonic point more accurately. A water bath would control temperature, because higher temperatures increase the rate of osmosis and could confound the results.
有效的评价需要识别误差来源并提出切实可行的改进措施。学生每次擦干土豆条的时长和力度应该一致，不一致擦干会引入随机误差。在每个浓度下重复实验三次并计算平均值可提高可靠性。可以把浓度范围扩展到 1.0 mol dm⁻³ 以上，以便更精确地确定等渗点。使用恒温水浴控制温度，因为温度升高会加快渗透速率，从而干扰结果。
7. Case Study 2: Enzyme Activity – Rate Calculations | 案例2：酶活性——速率计算
A group of students investigated the effect of temperature on the activity of catalase, an enzyme that breaks down hydrogen peroxide (H₂O₂) into water and oxygen. They placed potato discs containing catalase into hydrogen peroxide solution, recorded the volume of oxygen produced in 60 seconds, and repeated the experiment at six temperatures. Their results are given below.
一组学生探究了温度对过氧化氢酶活性的影响。过氧化氢酶能将过氧化氢（H₂O₂）分解成水和氧气。他们将含有过氧化氢酶的土豆圆片放入过氧化氢溶液中，记录 60 秒内产生的氧气体积，并在六种温度下重复了实验。结果如下。
| Temperature (°C) | Oxygen produced in 60 s (cm³) | Rate of reaction (cm³ s⁻¹) |
|---|---|---|
| 10 | 12 | 0.20 |
| 20 | 24 | 0.40 |
| 30 | 36 | 0.60 |
| 40 | 30 | 0.50 |
| 50 | 10 | 0.17 |
| 60 | 3 | 0.05 |
de the units in your answer. Plotting a graph of rate against temperature reveals an optimum temperature around 30–40 °C.
反应速率的计算方法是产生的氧气总体积除以所用时间。比如 30 °C 时：36 cm³ ÷ 60 s = 0.60 cm³ s⁻¹。回答时务必写上单位。以温度对反应速率作图可以显示最适温度大约在 30–40 °C。
8. Case Study 2: Explaining the Effect of Temperature | 案例2：解释温度对酶的影响
As temperature increases, enzyme and substrate molecules gain kinetic energy and move faster, so collisions occur more frequently and more enzyme‑substrate complexes form per second. This increases the rate of reaction up to the optimum temperature. Beyond the optimum, the enzyme’s active site loses its specific shape because the weak bonds holding its tertiary structure break. The substrate no longer fits the active site – the enzyme is denatured – and the rate falls sharply. At 60 °C almost all catalase is denatured, hence the very low oxygen production.
温度升高时，酶和底物分子的动能增加，移动更快，因此碰撞更频繁，每秒形成的酶‑底物复合物更多，反应速率随之上升，直至最适温度。超过最适温度后，维持酶三级结构的弱键断裂，活性位点失去了特定形状。底物不再与活性位点契合——酶已变性——反应速率急剧下降。60 °C 时几乎所有的过氧化氢酶都已变性，所以氧气产量极低。
9. Case Study 3: Biomagnification in a Food Chain | 案例3：食物链中的生物放大作用
DDT is a pesticide that remains in the environment for a long time. The table below shows the concentration of DDT found in the tissues of organisms in a freshwater food chain.
DDT 是一种能在环境中长期残留的杀虫剂。下表显示了淡水食物链中各种生物组织里 DDT 的浓度。
| Trophic level | Organism | DDT concentration (mg kg⁻¹) |
|---|---|---|
| Producer | Algae | 0.04 |
| Primary consumer | Zooplankton | 0.5 |
| Secondary consumer | Small fish | 2.0 |
| Tertiary consumer | Large fish | 5.0 |
| Quaternary consumer | Heron | 25.0 |
Biomagnification occurs because DDT is absorbed by algae, stored in fatty tissues, and not broken down. When primary consumers eat large quantities of algae, the toxin accumulates in their bodies. At each successive trophic level, the concentration increases, reaching potentially lethal levels in top predators such as herons. This explains why the heron has a DDT concentration over 600 times greater than the algae.
生物放大效应发生的原因是 DDT 被藻类吸收，储存在脂肪组织中且不易分解。当初级消费者大量摄食藻类时，毒素便在其体内积累。每上升一个营养级，浓度就进一步增加，到了鹭这类顶级捕食者体内就可能达到致死水平。这解释了为什么鹭体内的 DDT 浓度比藻类高出 600 多倍。
10. Case Study 4: Disease Transmission and Herd Immunity | 案例4：疾病传播与群体免疫
Measles is a highly contagious viral disease. Health authorities collected data on the number of measles cases in a town before and after a vaccination programme. In 2015, with only 60% of children immunised, there were 400 cases. By 2018, after raising coverage to 92%, cases dropped to 25. The following graph shows this trend.
麻疹是一种传染性极强的病毒性疾病。卫生部门收集了一个小镇在开展疫苗接种计划前后的麻疹病例数据。2015 年，仅有 60% 的儿童接种了疫苗，麻疹病例数为 400 例。到 2018 年，接种率提高到 92% 后，病例数降至 25 例。下图反映了这一趋势。
(A simple line graph would be presented with year on the x‑axis and number of cases on the y‑axis; two lines – vaccination coverage and case numbers – declining in opposite directions.)
（这里应有一张简单的折线图：x 轴为年份，y 轴为病例数；两条折线——接种率和病例数——呈反向变动。）
Herd immunity protects unvaccinated individuals because when a high proportion of the population is immune, the pathogen cannot spread effectively. This reduces the basic reproduction number (R₀) below 1, so an outbreak cannot sustain itself. Evaluate the data: the 60% coverage in 2015 was below the 95% threshold usually required to interrupt measles transmission, which explains the high case numbers. After reaching 92%, cases fell dramatically, confirming that vaccination is an effective control measure.
群体免疫保护了未接种者，因为当人群中免疫个体的比例很高时，病原体就无法有效传播。这使得基本传染数（R₀）降至 1 以下，疫情便难以持续。评价数据：2015 年 60% 的接种率低于阻断麻疹传播通常需要的 95% 阈值，这解释了病例数为何居高不下。接种率达到 92% 后，病例数急剧下降，证实了接种疫苗是一项有效的控制措施。
11. Top Tips for WJEC Case Study Questions | WJEC案例分析题的高分技巧
Read the scenario carefully and highlight key data. When asked to calculate, always show your working and include the correct units. In graph questions, use a sharp pencil, label axes with quantity and unit, and choose an appropriate scale. For ‘explain’ questions, structure your answer by stating the biological process, linking it to the specific data, and using relevant scientific vocabulary like ‘water potential’, ‘active site’, or ‘biomagnification’. When evaluating, give at least one strength and one weakness, and suggest a specific improvement. Finally, mention whether the data supports or refutes the hypothesis presented in the stem.
仔细阅读情景描述，标出关键数据。计算题务必展示步骤并写出正确单位。作图题用削尖的铅笔，坐标轴标上量和单位，选择合适的刻度。回答‘解释’类题目时，按“陈述生物过程 → 联系具体数据 → 使用‘水势’‘活性位点’‘生物放大’等科学词汇”的结构组织答案。评价实验时，至少给出一个优点和一个缺点，并提出具体的改进方法。最后，指出数据是支持还是否定了题干给出的假设。
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