📚 Year 11 CCEA Biology: Case Study Practice | CCEA 生物：案例分析实战演练
Case study questions are a key component of CCEA GCSE Biology exams. They test your ability to apply scientific knowledge to real-world scenarios, interpret data, and draw valid conclusions. This article provides a series of practical exercises to sharpen your case study skills.
案例研究题是CCEA GCSE生物考试的重要组成部分。它们考察你将科学知识应用于现实情境、解读数据并得出有效结论的能力。本文提供一系列实战练习，帮助你提升案例研究技巧。
1. Why Case Studies Matter | 为什么案例分析如此重要
CCEA examiners design case studies to assess higher-order thinking skills. Rather than simply recalling facts, you must analyse unfamiliar data, link concepts from different topics, and evaluate scientific methods. These questions often carry high marks, so mastering them can significantly boost your grade.
CCEA考官设计案例分析题是为了评估高阶思维能力。你不仅需要回忆事实，还必须分析陌生的数据、联系不同主题的概念并评价科学方法。这类题目通常分值较高，因此掌握它们能显著提高你的成绩。
2. Step-by-Step Approach to Case Studies | 案例分析的分步方法
A structured method turns a daunting case study into manageable steps. First, read the scenario twice and highlight key biological terms. Second, identify the command word (e.g. ‘calculate’, ‘explain’, ‘evaluate’) to understand what the examiner wants. Third, extract and organise the data, looking for patterns. Fourth, apply relevant biological principles. Finally, write your answer using scientific terminology and always link back to the data provided.
一个系统化的方法可以将棘手的案例分析变得易于掌控。首先，仔细阅读情境两遍，标记关键生物学术语。其次，识别指令词（如“计算”、“解释”、“评价”），理解考官的意图。第三，提取并整理数据，寻找模式。第四，应用相关的生物学原理。最后，用科学术语书写答案，并始终与所提供的数据相联系。
3. Case Study 1: Osmosis in Potato Strips (Background) | 案例一：土豆条渗透作用（背景）
A student investigated the effect of sucrose concentration on the mass of potato strips. Six strips of potato, each with an initial mass of 2.00 g, were placed in solutions of different sucrose concentration (0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm⁻³). After 30 minutes, the strips were blotted dry and reweighed. The student calculated the percentage change in mass for each strip.
一名学生研究了蔗糖浓度对土豆条质量的影响。六条土豆条，每条初始质量为2.00 g，分别放入不同蔗糖浓度（0.0, 0.2, 0.4, 0.6, 0.8 和 1.0 mol dm⁻³）的溶液中。30分钟后，吸干土豆条表面水分并重新称重。该学生计算了每条土豆条的质量变化百分比。
4. Case Study 1: Data and Calculations | 案例一：数据与计算
Table 1 shows the results. To find the percentage change in mass, use the equation:
Percentage change = ((Final mass – Initial mass) / Initial mass) × 100%
表1显示了结果。要计算质量变化百分比，请使用以下公式：
质量变化百分比 = ((最终质量 – 初始质量) / 初始质量) × 100%
| Sucrose concentration (mol dm⁻³) | Initial mass (g) | Final mass (g) | Mass change (%) |
|---|---|---|---|
| 0.0 | 2.00 | 2.50 | +25.0 |
| 0.2 | 2.00 | 2.30 | +15.0 |
| 0.4 | 2.00 | 2.10 | +5.0 |
| 0.6 | 2.00 | 1.90 | –5.0 |
| 0.8 | 2.00 | 1.70 | –15.0 |
| 1.0 | 2.00 | 1.50 | –25.0 |
Negative values indicate a loss in mass, while positive values indicate a gain. The student must now plot a graph of percentage change against sucrose concentration and draw a line of best fit.
负值表示质量减少，正值表示质量增加。学生现在必须绘制质量变化百分比对蔗糖浓度的图形，并画出最佳拟合线。
5. Case Study 1: Analysis and Conclusion | 案例一：分析与结论
From the graph, the point where the line crosses the x-axis (0% change) gives the concentration that is isotonic to the potato cell cytoplasm. In this case, the line crosses at about 0.5 mol dm⁻³. Below this concentration, water enters the cells by osmosis because the external solution is hypotonic, causing the strips to gain mass. Above 0.5 mol dm⁻³, the solution is hypertonic, so water leaves the cells and the strips lose mass. This analysis clearly links data to the concept of osmosis and water potential.
从图中，最佳拟合线与x轴（0%质量变化）相交的点给出了与土豆细胞质等渗的浓度。在这个案例中，该线约在0.5 mol dm⁻³处相交。低于此浓度时，由于外部溶液为低渗，水通过渗透作用进入细胞，使土豆条质量增加。高于0.5 mol dm⁻³时，溶液为高渗，水离开细胞，土豆条质量减少。该分析将数据与渗透作用和水势的概念清晰地联系起来。
6. Case Study 2: Enzyme Action on Starch (Background) | 案例二：淀粉的酶作用（背景）
A group of students investigated the effect of pH on the activity of the enzyme amylase. They prepared test tubes containing starch solution and buffer solutions of different pH values (4, 5, 6, 7, 8 and 9). Amylase was added to each tube, and every 10 seconds a sample was taken and tested with iodine solution. The time taken for the iodine to stop turning blue-black was recorded. The shorter the time, the faster the rate of reaction.
一组学生研究了pH对淀粉酶活性的影响。他们准备了含有淀粉溶液和不同pH缓冲液（4, 5, 6, 7, 8 和 9）的试管。向每支试管加入淀粉酶，每10秒取样并用碘液测试。记录碘液不再变为蓝黑色的时间。时间越短，反应速率越快。
7. Case Study 2: Interpreting Results | 案例二：解读结果
Table 2 shows the recorded times. The students calculated a relative rate using the formula: Rate = 1000 / time (s).
表2显示了记录的时间。学生们使用公式：速率 = 1000 / 时间 (s) 计算了相对速率。
| pH | Time for iodine to remain orange-brown (s) | Relative rate (1000/time) |
|---|---|---|
| 4 | 120 | 8.3 |
| 5 | 60 | 16.7 |
| 6 | 30 | 33.3 |
| 7 | 25 | 40.0 |
| 8 | 40 | 25.0 |
| 9 | 90 | 11.1 |
The highest relative rate is at pH 7, indicating that this is the optimum pH for this amylase. At pH 4 and pH 9, the rate is much lower, because the extreme pH disrupts the hydrogen and ionic bonds that maintain the enzyme’s active site shape. This denatures the enzyme, so the substrate no longer fits.
最高相对速率出现在pH 7，表明这是该淀粉酶的最适pH。在pH 4和pH 9，速率要低得多，因为极端的pH破坏了维持酶活性位点形状的氢键和离子键。这使酶变性，因此底物不再适合。
8. Case Study 2: Evaluating the Method | 案例二：评估方法
The method using iodine samples is reliable but has limitations. First, the endpoint is subjective – it relies on human judgement of colour. Second, sampling every 10 seconds may miss the exact moment when all starch is digested. To improve, students could use a colorimeter to measure light absorbance continuously. Furthermore, the experiment should be repeated at least three times at each pH to calculate a mean rate and improve accuracy. Discussing these points in your answer shows evaluation skills.
使用碘液取样的方法是可靠的，但也存在局限性。首先，终点具有主观性——依赖于人眼对颜色的判断。其次，每10秒取样一次可能会错过所有淀粉被消化的精确时刻。为了改进，学生可以使用比色计连续测量光吸收。此外，应在每个pH下至少重复三次实验，以计算平均速率并提高准确性。在答案中讨论这些要点能展示你的评价技能。
9. Case Study 3: Investigating Biodiversity (Field Data) | 案例三：调查生物多样性（野外数据）
Biologists surveyed freshwater invertebrates at three sites along a stream: Site A (upstream near source), Site B (adjacent to a cattle farm), and Site C (500 m downstream from the farm). They used kick-sampling and counted the numbers of certain indicator species. The data are shown in Table 3.
生物学家沿着一条溪流的三个地点调查了淡水无脊椎动物：地点A（上游靠近源头），地点B（紧邻一个养牛场），地点C（养牛场下游500米处）。他们使用踢样法，对某些指示物种进行了计数。数据见表3。
| Indicator species | Sensitivity | Site A count | Site B count | Site C count |
|---|---|---|---|---|
| Stonefly nymph | Very sensitive | 10 | 0 | 2 |
| Mayfly nymph | Sensitive | 12 | 1 | 4 |
| Caddisfly larva | Sensitive | 8 | 0 | 3 |
| Freshwater shrimp | Moderate | 5 | 5 | 10 |
| Water louse | Tolerant | 2 | 20 | 8 |
| Bloodworm (midge larva) | Very tolerant | Published by TutorHao | Year 11 Biology Revision Series | aleveler.com
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