📚 Year 11 AQA Science: Case Study Practice Drills | Year 11 AQA 科学：案例分析实战演练
Case study and data analysis questions are a key part of AQA GCSE Science exams, whether you are taking Combined Science: Trilogy or the separate Biology, Chemistry, and Physics courses. These questions test your ability to interpret graphs, tables, and unfamiliar scientific scenarios, and to apply your knowledge to real-world contexts. Mastering case study drills will help you think like a scientist, spot patterns, evaluate evidence, and use mathematical skills confidently. This article provides a series of worked case studies covering Biology, Chemistry, and Physics, guiding you through the thought process for each type of question you might face in Paper 2 and beyond.
案例研究和数据分析题是 AQA GCSE 科学考试的重要组成部分，无论你参加的是综合科学：三部曲还是单独的物理、化学、生物考试。这些题目测试你解读图表、表格和陌生科学情境的能力，以及将知识应用于实际背景的能力。掌握案例分析实战演练将帮助你像科学家一样思考，发现模式，评估证据，并自信地运用数学技能。本文提供一系列涵盖生物、化学和物理的案例分析，带你走通过每种问题可能遇到的思考过程，为 Paper 2 及更远的考试做好准备。
1. Understanding AQA Case Study Demands | 理解 AQA 案例分析的要求
AQA case study questions typically start with a short passage describing an experiment, a technological application, or a set of results. You are then asked a series of questions that may involve plotting data, calculating means, identifying anomalies, drawing conclusions, and evaluating the method. The key is to read the stem carefully, link it to the required practicals or specification points, and show your working clearly. Marks are allocated for correct units and significant figures, so always double-check.
AQA 案例分析题通常先给出一段描述实验、技术应用或一组结果的短文。随后提出一系列问题，可能涉及绘制数据图、计算平均值、识别异常值、得出结论和评价方法。关键在于仔细阅读题干，将其与规定的实践或大纲要点联系起来，并清晰地展示解题过程。分数会分配给正确的单位及有效数字，因此务必需反复检查。
For example, a Biology case might give you a table of enzyme activity at different temperatures and ask you to calculate the rate of reaction. A Chemistry case could show you a diagram of an electrolysis cell and ask you to explain which ions move where. In Physics, you might be given a graph of current against potential difference and asked to determine if a component is ohmic. Always underline command words such as 'describe', 'explain', 'calculate', and 'evaluate'.
例如，一道生物案例可能给出不同温度下酶活性的表格，要求你计算反应速率。化学案例可能展示一个电解池的示意图，要求你解释离子的移动方向。物理案例中，你可能得到电流-电压的图表，要求判断元件是否为欧姆导体。始终在指令词如下划线："描述"、"解释"、"计算"、"评价"。
2. Biology Case: Enzyme Activity and pH | 生物案例：酶活性与 pH
An investigation looked at how the rate of breakdown of starch by amylase changes with pH. A student mixed amylase and starch solution in a test tube, then measured the time taken for the iodine test to stop showing a blue-black colour. The results are shown in the table below.
一项研究调查了淀粉酶分解淀粉的速率如何随 pH 值变化。一名学生将淀粉酶和淀粉溶液混合在试管中，然后测量碘液测试不再显示蓝黑色所需的时间。结果如下表所示。
| pH | 2 | 4 | 6 | 7 | 8 | 10 |
|---|---|---|---|---|---|---|
| Time (s) | 120 | 45 | 20 | 25 | 60 | 200 |
Rate of reaction can be calculated as 1 / time. Notice that at pH 6 the reaction is fastest because the time is shortest. The data suggest the optimum pH for amylase is around 6-7. Extremes of pH (2 and 10) denature the enzyme by altering the shape of its active site, so the substrate can no longer bind. When evaluating this method, you could mention that using a colorimeter instead of iodine test would give more precise timing, and that the pH should be checked with a pH meter, not just indicator paper.
反应速率可以用 1/时间来计算。注意到 pH 6 时反应最快，因为时间最短。数据表明淀粉酶的最适 pH 在 6-7 左右。极端的 pH 值（2 和 10）通过改变酶活性位点的形状使之变性，底物无法再结合。在评价该方法时，你可以提到使用比色计代替碘液测试会提供更精确的时间记录，并且 pH 值应该用 pH 计检查，而不是只用指示剂试纸。
3. Biology Case: Rate of Photosynthesis | 生物案例：光合作用速率
A student used pondweed to investigate the effect of light intensity on the rate of photosynthesis. The number of oxygen bubbles produced per minute was recorded at different distances of a lamp from the beaker. Here are the processed results:
一名学生用伊乐藻研究光照强度对光合作用速率的影响。记录了灯与烧杯不同距离下每分钟产生的氧气气泡数。处理后的结果如下：
| Distance (cm) | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| Bubbles per min (mean) | 45 | 32 | 20 | 14 | 9 |
ting rate against light intensity shows a curve that levels off at high intensity, indicating the rate becomes limited by carbon dioxide concentration or temperature. In your analysis, always mention that as distance increases, light intensity decreases, slowing the light-dependent reactions. Anomalies could be caused by temperature fluctuations or the pondweed not being fully immersed. To improve, a thermostatic water bath should be used to keep temperature constant.
要用任意单位表示光照强度，使用 1/距离²。对于 10 cm，光照强度 = 1/10² = 0.01。将速率对照光照强度绘图，显示出在高强度下曲线趋于平缓，表明速率受到二氧化碳浓度或温度的限制。在分析中，一定要提到随着距离增加，光照强度降低，从而减慢了光依赖反应的速率。异常值可能是由于温度波动或伊乐藻没有完全浸没水中造成的。改进方法是使用恒温水浴以保持温度恒定。
4. Biology Case: Antibiotic Resistance | 生物案例：抗生素耐药性
Scientists monitored the percentage of Staphylococcus aureus bacteria resistant to the antibiotic methicillin over ten years. The data show a steady rise from 2% in 2014 to 22% in 2024. Explain the trend using Darwin's theory of natural selection. Resistant bacteria survive the antibiotic and reproduce, passing on the resistance gene. The non-resistant ones die. This case often includes a graph or table; you might be asked to calculate a percentage increase.
科学家监测了十年间对甲氧西林耐药的金黄色葡萄球菌百分比。数据显示从 2014 年的 2% 稳步上升到 2024 年的 22%。用达尔文的自然选择学说解释这一趋势。耐药细菌能在抗生素中存活并繁殖，将耐药基因传递下去，非耐药细菌则死亡。这个案例通常包含图表或表格；你可能会被要求计算百分比增加。
Percentage increase = [(22 - 2) / 2] × 100% = 1000% when necessary, and maintain strict hygiene. Also, explain why animal use of antibiotics can promote resistance in human pathogens. This is a classic evaluation question linking data to a scientific explanation.
百分比增加 = [(22 - 2) / 2] × 100% = 1000% 的增长。一个常见的后续问题是问如何减少抗生素耐药性的传播：完成整个抗生素疗程，仅在必要时使用，并保持严格的卫生。还需解释为何动物使用抗生素会促进人类病原体的耐药性。这是一个将数据与科学解释联系起来的经典评价问题。
5. Chemistry Case: Rate of Reaction and Concentration | 化学案例：反应速率与浓度
A student investigated how the concentration of hydrochloric acid affects the rate of reaction with magnesium ribbon. They measured the volume of hydrogen gas produced every 10 seconds. The results for two concentrations are given below.
一名学生研究了盐酸浓度如何影响与镁带的反应速率。他们每 10 秒测量一次产生的氢气体积。两种浓度的结果如下表。
| Time (s) | 0 | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|---|
| Vol. H₂ (cm³) 0.5 mol/dm³ | 0 | 12 | 20 | 26 | 30 | 30 |
| Vol. H₂ (cm³) 1.0 mol/dm³ | 0 | 24 | 35 | 40 | 40 | 40 |
³, rate = 40 cm³ / 30 s = 1.33 cm³/s. The higher concentration has more particles per unit volume, so collision frequency is greater, leading to a faster rate. The reaction stops when the magnesium is used up; you can tell because the gas volume plateaus. In a case study, you might be asked why the same mass of magnesium was used (to make it a fair test) or to draw a tangent at time zero to find the initial rate.
计算 0.5 mol/dm³ 酸在前 30 秒的平均反应速率：速率 = 气体体积 / 时间 = 26 cm³ / 30 s = 0.87 cm³/s。对于 1.0 mol/dm³，速率 = 40 cm³ / 30 s = 1.33 cm³/s。较高浓度的酸单位体积内有更多粒子，因此碰撞频率更大，导致速率更快。当镁用尽时反应停止；你可从气体体积不再变化判断出来。在案例研究中，你可能会被问到为什么使用相同质量的镁（为了公平实验）或在时间零点处画切线以求出初始速率。
6. Chemistry Case: Electrolysis to Purify Copper | 化学案例：电解提纯铜
of 2 A for 30 minutes. Explain the reactions at each electrode. At the anode: Cu(s) → Cu²⁺(aq) + 2e⁻, the impure copper dissolves. Impurities like silver and gold fall as sludge. At the cathode: Cu²⁺(aq) + 2e⁻ → Cu(s), pure copper plates onto the cathode.
铜可以通过电解法进行提纯，使用不纯的铜作阳极、纯铜作阴极，放入硫酸铜溶液中。一名学生通入 2 A 电流，持续 30 分钟。解释每个电极上发生的反应。在阳极：Cu(s) → Cu²⁺(aq) + 2e⁻，不纯的铜溶解。银、金等杂质以阳极泥形式掉落。在阴极：Cu²⁺(aq) + 2e⁻ → Cu(s)，纯铜沉积在阴极上。
00 C. You might be given the relevant half-equation and the value of the Faraday constant, but at GCSE level you would only need to describe that more charge means more copper deposited. A common case study asks: why does the anode lose mass and the cathode gain mass? Or why do the blue colour of the electrolyte remain unchanged? e cathode as fast as they are removed from the anode. This case links to reactivity series – silver is less reactive and does not dissolve.
沉积铜的质量可以与电荷量联系起来。电荷量 Q = I × t = 2 A × (30 × 60 s) = 3600 C。你可能会得到相关的半反应式和法拉第常数的数值，但在 GCSE 水平，你只需描述：通过的电荷越多，沉积的铜就越多。一个常见的案例分析问题会问：为什么阳极质量减少而阴极质量增加？或者为什么电解质的蓝色保持不变？那是因为 Cu²⁺ 离子浓度保持恒定，因为从阴极去除的铜离子速率与阳极补充的一样快。这个案例与金属活动性顺序相关——银的活动性较弱，不会溶解。
7. Chemistry Case: Energy Changes and Bond Energies | 化学案例：能量变化与键能
kJ/mol, O-H 463 kJ/mol, calculate the overall energy change. Break: 4 × (C-H) + 2 × (O=O) = 4(412) + 2(496) = 1648 + 992 = 2640 kJ. Form: 2 × (C=O) + 4 × (O-H) = 2(743) + 4(463) = 1486 + 1852 = 3338 kJ. Energy change = energy in – energy out = 2640 - 3338 = -698 kJ/mol, so the reaction is exothermic.
利用键能可以分析甲烷的燃烧反应 CH₄ + 2O₂ → CO₂ + 2H₂O。已知键能：C-H 412 kJ/mol，O=O 496 kJ/mol，C=O 743 kJ/mol，O-H 463 kJ/mol，计算总能量变化。断键吸热：4 × (C-H) + 2 × (O=O) = 4(412) + 2(496) = 1648 + 992 = 2640 kJ。成键放热：2 × (C=O) + 4 × (O-H) = 2(743) + 4(463) = 1486 + 1852 = 3338 kJ。能量变化 = 吸热 - 放热 = 2640 - 3338 = -698 kJ/mol，因此反应是放热的。
A case study question may provide bond energies in a table and ask you to determine if a different fuel, such as ethanol, releases more or less energy per mole. Always present your working step by step, use the correct formulas, and ensure you multiply by the number of bonds. At GCSE, you are not required to know the formula for mean bond energy calculation in detail, but for Combined Science Higher tier, being able to perform this calculation is essential. Also, be aware that bond energy calculations assume all bonds are in the gaseous state, so compare to experimental values which may differ due to state changes.
一道案例分析题可能会提供键能表格，要求你判断另一种燃料（如乙醇）释放的能量是更多还是更少。务必逐步展示计算过程，使用正确的公式，并确保乘以键的数目。在 GCSE 普通阶段，你不需掌握平均键能计算的详细公式，但对于综合科学高级难度，能够进行这项计算是至关重要的。还要注意，键能计算假设所有化学键都处于气态，因此与实验值相比可能因状态变化而有所不同。
8. Physics Case: Ohm’s Law and Resistance | 物理案例：欧姆定律与电阻
ent investigates a fixed resistor and a filament lamp. The current-potential difference (I– a straight line through the origin, showing that current is directly proportional to p.d. – is ohmic. Resistance R = V/I, so at any point R is constant. For the filament lamp, the I– heats up, because metal ions vibrate more, causing more collisions with flowing electrons.
一名学生研究了一个固定电阻器和一个灯丝灯泡。收集了电流-电位差（I–V）数据。对固定电阻器，电流（y 轴）对照电位差（x 轴）的图是一条过原点的直线，表明电流与电位差成正比——它是欧姆导体。电阻 R = V/I，因此在任何一点 R 都恒定。对灯丝灯泡，I–V 图弯曲，在较高电位差下趋于平缓。这表明随着灯丝升温，电阻增大，因为金属离子振动加剧，与流动电子碰撞增多。
I = 1.5 A → R = 4 Ω. For the lamp, you must read the current at that voltage from the curve, then use R = V/I. Because the graph is not a straight line, the resistance has increased. A common comparison is with a diode: a diode has a very high resistance in reverse bias and low resistance in forward bias; its I– . Being able to sketch and interpret these three components is a typical case study drill.
案例分析题常会给出图表，要求你确定在特定电压下的电阻。对电阻器，选取任意一点，例如 V = 6 V，I = 1.5 A → R = 4 Ω。对灯丝灯泡，你必须从曲线上读取该电压下的电流，然后使用 R = V/I。由于图线不是直线，电阻已经增大。一个常见的比较是与二极管：二极管在反向偏置时电阻极高，正向偏置时电阻较低；其 I–V 图在约 0.7 V 后急剧上升。能够绘制并解释这三种元件是典型的案例分析演练。
9. Physics Case: Specific Heat Capacity and Energy | 物理案例：比热容与能量
The temperature rises from 20 °C to 65 °C. Calculate the specific heat capacity c of the alloy. Energy supplied E = P × t = 40 W × (10 × 60 s) = 24 000 J. Using E = m × c × Δθ, where Δθ = 45 °C, we get c = E / (m × Δθ) = 24 000 / (0.5 × 45) = 24 000 / 22.5 = 1067 J/kg°C. The engineer would compare this with aluminium (c ≈ 900 J/kg°C) to see if it heats up quickly enough. A case study might ask why saucepans should have a low specific heat capacity (so they heat up quickly), whereas saucepan handles should have a high specific heat capacity and low thermal conductivity to stay cool.
一位工程师正在测试一种用于炖锅的新型金属合金。一个 0.5 kg 的合金块用功率为 40 W 的电加热器加热 10 分钟。温度从 20 °C 上升到 65 °C。计算该合金的比热容 c。提供的能量 E = P × t = 40 W × (10 × 60 s) = 24 000 J。利用 E = m × c × Δθ，其中 Δθ = 45 °C，得到 c = E / (m × Δθ) = 24 000 / (0.5 × 45) = 24 000 / 22.5 = 1067 J/kg°C。工程师会将其与铝（c ≈ 900 J/kg°C）比较，看其是否足够快地加热。案例分析可能会问为什么锅身应当具有低的比热容（这样能快速升温），而锅柄则应具有高的比热容和低的导热系数以保持凉爽。
Always identify heat losses as a source of error. The measured temperature rise may be lower than theoretical because energy is lost to the surroundings. Using insulation, a lid, and stirring the alloy will improve accuracy. This case tests your ability to rearrange the equation and convert units correctly – both are common marks in AQA Science.
始终要识别热量损失作为误差来源。测得的升温幅度可能低于理论值，因为能量会散失到周围环境中。使用隔热层、盖子和搅拌合金将提高准确性。这个案例测试你重新排列公式和正确转换单位的能力——两者在 AQA 科学考试中都是常见的得分点。
10. Physics Case: Radioactive Decay and Half-Life | 物理案例：放射性衰变与半衰期
ope. Work stepwise: 800 → 400 (1 half-life) → 200 (2 half-lives) → 100 (3 half-lives). So 3 half-lives equal 18 minutes, hence one half-life is 6 minutes. You could also use the graph provided in a case study: at half the initial count rate, read the time from the curve. Remember that half-life is the time taken for the activity (or number of radioactive nuclei) to halve.
某放射性同位素样品初始活度为 800 Bq。18 分钟后，活度下降到 100 Bq。求该同位素的半衰期。逐步推算：800 → 400（1 个半衰期）→ 200（2 个半衰期）→ 100（3 个半衰期）。因此 3 个半衰期等于 18 分钟，所以一个半衰期为 6 分钟。你也可以使用案例研究中提供的图表：在初始计数率一半处，从曲线上读取时间。要记住半衰期是活度（或放射性原子核数目）减半所需的时间。
A common case study links half-life to real-world uses: isotopes with short half-lives are used as medical tracers because they decay quickly and minimise radiation dose; isotopes with long half-lives are useful for carbon dating or monitoring industrial thickness. You might be asked to evaluate the risk: for example, why would an isotope with a half-life of a few hours be more suitable for medical imaging than one with a half-life of several years? Answer: it provides a useful signal but decays away quickly, reducing long-term exposure. Calculating net decline after a given number of half-lives (e.g. what fraction remains after 4 half-lives: 1/16) is another drill.
一个常见的案例分析将半衰期与现实用途联系起来：半衰期短的同位素被用作医学示踪剂，因为它们衰变快，辐射剂量最小；半衰期长的同位素可用于碳定年法或工业厚度监测。你可能会被要求评价风险：例如，为什么半衰期几小时的放射性同位素比半衰期数年的更适合医学成像？答案：它能提供有用信号但衰变很快，减少长期暴露。计算在给定半衰期数目后的剩余比例（例如 4 个半衰期后剩余多少：1/16）是另一项演练。
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