📚 Case Study Practical Exercises for AQA GCSE Science | AQA GCSE 科学案例分析实战演练
Case study questions in AQA GCSE Science are designed to test your ability to apply scientific knowledge to unfamiliar scenarios. Instead of just recalling facts, you will need to interpret data, analyse experiments, and draw evidence-based conclusions. This revision guide provides practical exercises across Biology, Chemistry and Physics to help you build the skills required for these high-mark questions. Follow each step, read the paired English and Chinese explanations carefully, and practise until you can confidently approach any case study that appears in your Year 10 assessments or final exams.
AQA GCSE 科学中的案例分析题旨在考查你将科学知识应用于陌生情境的能力。你不仅需要回忆事实，更要解读数据、分析实验并得出基于证据的结论。这份备考指南提供了覆盖生物、化学和物理的实战演练，帮助你培养解答高分值题目所需的技能。请跟着每一步走，仔细阅读配对的英文和中文解释，反复练习，直到你能自信地应对 Year 10 测验或大考中出现的任何案例分析。
1. Understanding Case Study Questions | 理解案例分析题
In a typical AQA case study, you will be given a short passage, a table of results, or a diagram showing an experimental setup. Your task is to identify the independent and dependent variables, recognise any control variables, and then use your subject knowledge to explain the trends or anomalies. Always read the question prompt twice before writing an answer.
在典型的 AQA 案例分析中，你会看到一篇短文、一张结果表或一张显示实验装置的图。你的任务是找出自变量和因变量，识别控制变量，然后运用学科知识解释趋势或异常值。写答案之前务必把题目读两遍。
Start by highlighting command words such as 'describe', 'explain', 'evaluate' and 'suggest'. 'Describe' means state what the data shows without giving reasons; 'explain' requires scientific reasoning; 'evaluate' asks you to weigh up evidence and possibly discuss limitations. Keeping these distinctions clear will prevent you from losing marks.
首先标出指令词，如 describe、explain、evaluate 和 suggest。Describe 要求陈述数据展示的趋势但不给原因；explain 要求给出科学推理；evaluate 要求权衡证据，可能还要讨论局限性。分清这些差异能避免丢分。
Wherever possible, back your points with data from the question. For example, if a graph shows a plateau at 40 °C, quote the value and link it to denaturation of an enzyme. Using figures from the case study shows the examiner you are engaging directly with the evidence.
只要有可能，就用题目中的数据支持你的观点。比如，如果图表显示在 40 °C 时趋于平缓，就引用这个温度值并将其与酶的变性联系起来。使用案例中的数据能向阅卷人表明你直接运用了所给证据。
2. Biology Case Study: Enzyme Activity and pH | 生物案例：酶活性与 pH
An experiment was carried out to investigate how pH affects the activity of amylase. A starch solution was mixed with an amylase solution at different pH buffers, and the time taken for the starch to disappear (tested with iodine) was recorded. The results are shown below.
有一项实验研究了 pH 对淀粉酶活性的影响。将淀粉溶液与不同 pH 缓冲液中的淀粉酶溶液混合，记录下淀粉消失所需的时间（用碘液检测）。结果如下表所示。
| pH | Time for starch to disappear / s |
|---|---|
| 2 | 360 |
| 4 | 180 |
| 6 | 45 |
| 7 | 30 |
| 8 | 65 |
| 10 | 340 |
Explain why the reaction is fastest at pH 7 and slowest at pH 2 and pH 10. The enzyme amylase has an optimum pH near 7. At pH 7, the active site has a specific shape that fits the starch substrate perfectly, forming many enzyme‑substrate complexes per second. At extremes of pH, the hydrogen bonds in the enzyme's tertiary structure break; the active site changes shape irreversibly, so the substrate can no longer bind. Enzyme denaturation occurs, which dramatically reduces reaction rate.
解释为什么反应在 pH 7 时最快，而在 pH 2 和 pH 10 时最慢。淀粉酶的最适 pH 接近 7。在 pH 7 时，活性位点具有特定的形状，恰好与淀粉底物契合，每秒能形成大量酶‑底物复合物。在极端的 pH 条件下，酶三级结构中的氢键断裂；活性位点不可逆地改变形状，底物因而无法结合。酶发生变性，反应速率急剧下降。
Use data to support your answer: at pH 2 the time is 360 s, at pH 7 it is only 30 s. The time is inversely proportional to the rate, so a shorter time means a higher rate. This clear difference confirms the dependence on pH.
用数据支持你的答案：pH 2 时用时 360 秒，pH 7 时仅 30 秒。时间与速率成反比，因此更短的时间意味着更高的速率。这一明显差异证实了 pH 对酶活性的影响。
3. Biology Case Study: Osmosis in Plant Cells | 生物案例：植物细胞的渗透作用
A student placed identical potato cylinders in sucrose solutions of different concentrations. After 24 hours, the percentage change in mass was calculated. The results are: 0.0 mol/dm³ (distilled water): +12.5%; 0.2 mol/dm³: +5.0%; 0.4 mol/dm³: -3.5%; 0.6 mol/dm³: -10.0%; 0.8 mol/dm³: -14.5%.
一名学生将相同的土豆条放入不同浓度的蔗糖溶液中。24 小时后计算质量变化百分比。结果如下：0.0 mol/dm³（蒸馏水）：+12.5%；0.2 mol/dm³：+5.0%；0.4 mol/dm³：−3.5%；0.6 mol/dm³：−10.0%；0.8 mol/dm³：−14.5%。
Explain why the mass increased in low concentrations and decreased in high concentrations. Water moves by osmosis from a region of high water potential to a region of low water potential across a partially permeable membrane. In distilled water, the water potential outside the cells is higher than inside, so water enters the cells, causing turgor and mass increase. In concentrated sucrose solutions, the water potential outside is lower, so water leaves the cells, leading to plasmolysis and mass loss.
解释为什么土豆在低浓度溶液中质量增加，在高浓度溶液中质量减少。水通过渗透作用，从高水势区域穿过部分透性膜向低水势区域移动。在蒸馏水中，细胞外的水势高于细胞内，水进入细胞，细胞变得坚挺，质量增加。在浓蔗糖溶液中，外界水势较低，水离开细胞，导致质壁分离和质量损失。
Estimate the water potential of the potato tissue by finding the concentration where mass does not change. Plot the data and find where the line crosses zero. A line of best fit shows the isotonic point is near 0.28 mol/dm³. This is a key skill for the required practical on osmosis.
通过找出质量不变的浓度来估算土豆组织的水势。绘制数据并找出线条与零点的交点。最佳拟合线显示等渗点大约在 0.28 mol/dm³。这是渗透作用必做实验中的关键技能。
4. Chemistry Case Study: Electrolysis of Aqueous Solutions | 化学案例：水溶液电解
A student set up an electrolysis of aqueous copper(II) chloride using inert graphite electrodes. At the cathode, a pink‑brown solid appeared; at the anode, a greenish gas was produced. Identify the products and write the half equations.
一名学生用惰性石墨电极电解氯化铜(II)水溶液。阴极出现了粉褐色固体；阳极上产生了黄绿色气体。请指出产物并写出半反应式。
The pink‑ the anode: 2Cl⁻ → Cl₂ + 2e⁻. Copper is less reactive than hydrogen, so copper ions are discharged in preference to hydrogen ions. Chloride ions are halide ions, so they are discharged in preference to hydroxide ions from water.
粉褐色固体是金属铜，因为 Cu²⁺ 离子在阴极被还原：Cu²⁺ + 2e⁻ → Cu。黄绿色气体是氯气，因为 Cl⁻ 离子在阳极被氧化：2Cl⁻ → Cl₂ + 2e⁻。铜的活泼性低于氢，因此铜离子优先于氢离子放电。氯离子属于卤素离子，所以比水中的氢氧根离子更易放电。
Always check the reactivity series and the rules for aqueous electrolysis. At the cathode, if the metal is more reactive than hydrogen (e.g. sodium), hydrogen gas is produced. At the anode, if the solution contains halide ions, the halogen is produced; otherwise oxygen is formed from hydroxide ions.
要始终检查金属活动性顺序和水溶液电解规则。在阴极，如果金属比氢活泼（如钠），则生成氢气。在阳极，如果溶液含有卤素离子，则生成相应的卤素；否则由氢氧根离子生成氧气。
5. Chemistry Case Study: Rate of Reaction and Surface Area | 化学案例：反应速率与表面积
A reaction between marble chips (calcium carbonate) and hydrochloric acid was monitored by recording the volume of carbon dioxide produced every 20 seconds. Two runs were performed: with small chips and with large chips of the same mass. The results are shown in the table.
大理石碎片（碳酸钙）与盐酸的反应通过每 20 秒记录生成的二氧化碳体积来监测。进行了两次实验：使用相同质量的小碎片和大碎片。结果见下表。
| Time / s | Volume of CO₂ (small chips) / cm³ | Volume of CO₂ (large chips) / cm³ |
|---|---|---|
| 0 | 0 | 0 |
| 20 | 38 | 14 |
| 40 | 68 | 28 |
| 60 | 85 | 40 |
| 80 | 94 | 51 |
| 100 | 98 | 62 |
Explain why the rate is faster with small chips than with large chips. Small chips have a larger total surface area for the same mass. This means more calcium carbonate particles are exposed at the surface, so there are more frequent successful collisions between reactant particles per second for the same concentration of acid. The surface area increases the frequency of collisions, hence raising the rate of reaction.
解释为什么小碎片的反应速率比大碎片快。相同质量下，小碎片的总表面积更大，这意味着更多的碳酸钙颗粒暴露于反应表面，因此在相同盐酸浓度下，每秒钟反应物粒子间发生成功碰撞的次数更多。表面积增大了碰撞频率，从而提升了反应速率。
You can calculate the initial rate from the gradient of the volume‑ e initial rate is approximately 1.9 cm³/s; for large chips, about 0.7 cm³/s. This directly supports the collision theory explanation.
你可以借助体积‑时间图在 t = 0 处的梯度来计算初始速率。小碎片的初始速率约为 1.9 cm³/s，大碎片约为 0.7 cm³/s。这直接支持了碰撞理论的解释。
6. Physics Case Study: Investigating Resistance | 物理案例：探究电阻
as measured at different currents. Data: current = 0.10 A, voltage = 0.45 V; 0.20 A, 0.90 V; 0.30 A, 1.35 V; 0.40 A, 1.80 V; 0.50 A, 2.25 V. Explain the relationship and calculate the resistance.
一名学生搭建了包含电池、电流表、可变电阻和一段电阻丝的串联电路。在不同电流下测量了电阻丝两端的电压。数据：电流 0.10 A，电压 0.45 V；0.20 A，0.90 V；0.30 A，1.35 V；0.40 A，1.80 V；0.50 A，2.25 V。说明两者的关系并计算电阻。
The voltage across the wire is directly proportional to the current. Using Ohm' is obtained for all data pairs, so the wire is an ohmic conductor at constant temperature. This confirms that resistance is constant as long as temperature does not change significantly.
电阻丝两端的电压与电流成正比。根据欧姆定律 V = I × R，可以算出 R：R = V ÷ I = 0.45 V ÷ 0.10 A = 4.5 Ω。所有数据组都得出了相同的值，说明该电阻丝在恒温条件下是欧姆导体。这证实只要温度没有显著变化，电阻就保持恒定。
If the student had used a filament lamp, the voltage‑current graph would be a curve, showing that resistance increases with current due to heating. In a case study, always check whether temperature is controlled, because resistance in metals increases when ions vibrate more.
如果学生用的是白炽灯，电压‑电流图将是一条曲线，表明电阻随着电流增大而增加，这是由发热所致。在案例分析中，始终要检查温度是否得到控制，因为金属中离子振动加剧时电阻会增大。
7. Physics Case Study: Energy Transfers in a Roller Coaster | 物理案例：过山车中的能量转移
12 m high. Assume negligible friction. Calculate the speed of the car at point B and point C. Take g = 9.8 N/kg.
一辆质量为 500 kg 的过山车从距地面 30 m 的 A 点静止出发，下滑至地面的 B 点，然后上升至 12 m 高的 C 点。忽略摩擦。计算过山车在 B 点和 C 点的速度。取 g = 9.8 N/kg。
ergy (Eₖ = ½ × m × v²). So ½ × 500 × v² = 147 000 J. Solving gives v² = (147 000 × 2) ÷ 500 = 588, thus v = √588 ≈ 24.2 m/s.
在 A 点，过山车具有重力势能 Eₚ = m × g × h = 500 × 9.8 × 30 = 147 000 J。当它滑向 B 点时，这部分能量全部转化为动能 Eₖ = ½ × m × v²。因此 ½ × 500 × v² = 147 000 J，解得 v² = (147 000 × 2) ÷ 500 = 588，于是 v = √588 ≈ 24.2 m/s。
potential energy. Eₚ at C = 500 × 9.8 × 12 = 58 800 J. Remaining kinetic energy = 147 000 – ≈ 18.8 m/s. The conservation of energy principle makes such calculations straightforward.
在 C 点，部分动能已重新转化为重力势能。C 点的 Eₚ = 500 × 9.8 × 12 = 58 800 J。剩余的动能 = 147 000 – 58 800 = 88 200 J。因此 v² = (88 200 × 2) ÷ 500 = 352.8，v ≈ 18.8 m/s。能量守恒原理让这类计算变得十分简单。
8. Data Analysis Techniques | 数据分析技巧
Many case study marks depend on how you handle numerical data. First, always calculate the mean of repeated readings, excluding any anomalous results. Anomalies are values that lie far from the general pattern and may result from measurement error. State clearly why you are excluding an outlier.
案例分析中有不少分值取决于你如何处理数字数据。首先，始终计算多次读数的平均值，并排除异常值。异常值是那些远离总体模式的值，可能源于测量误差。要清楚说明你为什么剔除某个离群值。
When drawing a graph, label axes with quantity and unit, use an appropriate scale, and plot points with small crosses. Draw a line or curve of best fit; do not simply join dots. If the data suggest a directly proportional relationship, the best‑fit line must pass through the origin. Use a ruler for straight lines.
绘制图表时，用物理量和单位标注坐标轴，选用合适的坐标刻度，用小十字标出数据点。画出最佳拟合线或曲线，不要简单地把点连起来。如果数据表明是正比关系，最佳拟合线必须通过原点。直线部分应使用直尺。
For any calculation, show all working steps. In the Physics case above, writing down the formula, substituting values and showing the rearrangement earns method marks even if the final answer contains an arithmetic slip.
任何计算都要展示完整的步骤。在上面的物理案例中，写下公式、代入数值并展示变形过程，即便最终答案出现算术失误，也能拿到方法分。
9. Drawing Conclusions and Evaluating Evidence | 得出结论与评估证据
A conclusion must refer back to the original hypothesis and be supported by the data. Do not overclaim – use words such as 'the results suggest' or 'there is a strong positive correlation' rather than 'this proves'. In science, evidence supports or refutes a hypothesis, but rarely provides absolute proof.
结论必须回扣原假设，并有数据支持。不要过度断言——使用“结果提示”、“两者间存在强烈的正相关”等措辞，而不是“这证明了”。在科学中，证据通常支持或否定假设，但极少提供绝对证明。
Evaluation involves identifying limitations and suggesting realistic improvements. Common points: more repeat readings to improve repeatability, using a more precise measuring instrument (e.g. a digital thermometer instead of a liquid‑in‑glass one), controlling another variable such as room temperature, or extending the range of the independent variable to see the full pattern.
评估包括找出实验的局限性并提出切实可行的改进建议。常见要点：增加重复读数以提高可重复性，使用更精密的测量仪器（例如电子温度计取代液体温度计），控制另一个变量如室温，或者扩大自变量范围以便看清整体规律。
Always link your evaluation to the specific experiment in the case study. Saying 'do the experiment more carefully' is too vague. Instead, write: 'Use a water bath to keep the temperature constant at 25 °C because enzyme activity varies with temperature, which could affect the time taken for starch to disappear.'
评估时一定要紧密结合案例中的具体实验。写“更仔细地做实验”太过笼统。不妨这样写：“使用恒温水浴将温度恒定在 25 °C，因为酶活性会随温度变化，这可能会影响淀粉消失的时间。”
10. Common Pitfalls and How to Avoid Them | 常见错误及如何避免
Pitfall 1: Misidentifying variables. If the question asks 'What type of variable is the concentration of acid?', the answer is independent variable, not control variable, because it is the factor you deliberately change. Fix this by labelling the 'I change, I measure, I keep the same' strategy in your planning.
常见错误一：弄错变量类型。如果题目问“酸的浓度属于哪种变量？”，答案应是自变量而非控制变量，因为它是你刻意改变的因子。解决方法是记住“我要改变、我要测量、我要保持不变”的规划策略。
Pitfall 2: Writing a description when an explanation is required. If the command word is 'explain', you must use scientific ideas like collision theory, enzyme active sites, or energy transfers. Simply repeating the data trend will not earn full marks.
常见错误二：需要解释时却写出了描述。如果指令词是 explain，你务必运用碰撞理论、酶活性位点或能量转移等科学概念。仅仅重复数据趋势不能获得满分。
Pitfall 3: Forgetting units in calculations and graph axes. A number without a unit is meaningless in science. Double‑check that every plotted axis has a unit in brackets and every calculated value has the correct unit, such as cm³/s or J.
常见错误三：计算和图表坐标轴上忘记写单位。在科学中，没有单位的数值毫无意义。仔细检查每个坐标轴上有没有括号标注单位，每个计算结果是否带有正确的单位，如 cm³/s 或 J。
11. Practice Exercise: Combined Science Scenario | 练习：综合科学场景
Scenario: A student investigates how the concentration of sodium chloride solution affects the current passing through it, using graphite electrodes and a fixed voltage of 6 V. The current increases with concentration. At low concentrations, water molecules are abundant and ion movement is easier, yet the current is smaller because there are fewer charge carriers. At high concentrations, more ions are available to carry charge, so the current rises. However, above 0.5 mol/dm³, the current levels off because the electrodes become saturated.
场景：一名学生研究氯化钠溶液浓度对通过其的电流的影响，使用石墨电极和固定的 6 V 电压。电流随浓度增大而增大。低浓度时，水分子充足，离子移动较容易，但由于电荷载体少，电流较小。高浓度时，有更多离子可用来搬运电荷，于是电流上升。然而，当浓度超过 0.5 mol/dm³ 后，电流趋于平缓，因为电极出现饱和。
Identify the independent variable (concentration of NaCl), dependent variable (current), and suggest why carbon electrodes are suitable (inert, so they do not react with the electrolyte). Explain the trend using the idea of ion mobility and charge carriers. Evaluate by noting that temperature was not controlled; heating could increase current as ions move faster, which would introduce a systematic error.
请找出自变量（NaCl 浓度）、因变量（电流），并说明为什么碳电极合适（惰性，不与电解液反应）。运用离子迁移率和电荷载体的概念解释变化趋势。进行评估：温度未加控制；升温可能使离子运动加快而增大电流，这会引入系统误差。
Write a conclusion for this case study. "As the concentration of sodium chloride increases, the current increases up to a point, due to the greater number of ions carrying charge. Beyond 0.5 mol/dm³, the electrode surface becomes saturated, limiting further current rise. The evidence supports the relationship, but uncontrolled variables such as temperature reduce certainty."
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