📚 KS3 OCR Science: Case Study Practical Exercises | KS3 OCR 科学：案例分析实战演练
Practising case studies is a vital part of KS3 OCR Science, helping students apply scientific concepts to real-world scenarios. In this revision guide, we present a series of practical case study exercises covering biology, chemistry and physics. Each case study includes a scenario, experimental data or observations, and guided analysis to build problem-solving skills, together with exam-style questions and model answers written in the language that examiners expect to see.
案例分析练习是 KS3 OCR 科学课程的重要组成部分，能帮助学生把科学概念应用到真实情境中。在这份复习指南中，我们提供一系列涵盖生物、化学和物理的实践案例分析练习。每个案例都包含情境、实验数据或观察记录，并通过引导性分析来培养解决问题的能力，同时配有考试风格的问答题以及按评分标准语言撰写的示范答案。
1. How to Use These Case Study Exercises | 如何使用这些案例分析练习
Each case study follows the same three-part structure: a short scenario, a set of data or observations, and a guided analysis. Work through the scenario and data first, write your own answer, and only then read the analysis section. This active sequence forces you to recall and apply knowledge rather than simply recognise it.
每个案例都采用相同的三段式结构：简短情境、一组数据或观察记录、以及引导性分析。请先阅读情境与数据，自己写出答案，然后再看分析部分。这种主动学习顺序会迫使你回忆并应用知识，而不只是“看懂”知识。
In KS3 OCR assessments, marks are awarded for three things: correct scientific knowledge, correct use of data from the resource, and clear links between the two. A sentence such as "the stem bent because auxin moved" earns one mark; "auxin moved to the shaded side, so cells there elongated more, which bent the stem towards the light" earns three.
在 KS3 OCR 的测评中，得分点主要来自三个方面：正确的科学知识、正确使用材料中的数据、以及两者之间清晰的关联。像 "the stem bent because auxin moved" 这样的句子只能拿到 1 分；而 "auxin moved to the shaded side, so cells there elongated more, which bent the stem towards the light" 则可以拿到 3 分。
Always keep a data table open beside your answer. Examiners frequently reward a direct quotation of a value, a calculated mean, or a stated trend, so writing "the bending angle increased from 30° to 40°" is stronger than writing "the bending angle changed".
作答时请始终把数据表放在旁边。评分者经常会给直接引用数值、计算平均值或描述趋势的答案加分，因此写 "the bending angle increased from 30° to 40°" 比写 "the bending angle changed" 更有力。
2. Case Study: Phototropism in Seedlings | 案例分析：幼苗的向光性
A student placed three bean seedlings in a box with a single hole on one side, allowing light to enter. After five days, she measured the angle of stem bending towards the light. The results are shown in the table below.
一位学生将三株豆苗放入一个只有一侧有一个小孔的盒子里，让光线进入。五天后，她测量了茎朝光源弯曲的角度。结果如下表所示。
| Seedling | Initial height (cm) | Final height (cm) | Bending angle (°) |
|---|---|---|---|
| A | 4.0 | 7.2 | 35 |
| B | 3.8 | 6.5 | 40 |
| C | 4.2 | 7.0 | 30 |
Question: Explain why the stems bent towards the light, using the concept of auxin distribution.
问题：请利用生长素分布的概念，解释为什么茎会向光弯曲。
Analysis: Auxin is a plant hormone that promotes cell elongation. In the seedlings, light causes auxin to move to the shaded side of the stem. This uneven distribution leads to more cell growth on the side away from light, causing the stem to bend towards the light source. The measured angles confirm a positive phototropic response.
分析：生长素是一种促进细胞伸长的植物激素。在幼苗中，光照导致生长素向茎的背光侧移动。这种不均匀的分布使得背光侧的细胞生长更快，从而使茎向光源弯曲。测量的角度证实了正向光性反应。
Key vocabulary you should use in this answer: auxin, hormone, cell elongation, shaded side, uneven distribution, positive phototropism. Using precise vocabulary is one of the fastest ways to raise your mark in KS3 biology questions.
这道题你应当使用的关键术语包括：生长素（auxin）、激素（hormone）、细胞伸长（cell elongation）、背光侧（shaded side）、不均匀分布（uneven distribution）、正向光性（positive phototropism）。在 KS3 生物题目中，使用精确术语是提高分数最快捷的方法之一。
3. Interpreting the Phototropism Data | 向光性数据的解读
2.8 cm. All three seedlings grew, so light was sufficient for growth as well as for the bending response.
第一步：计算每株幼苗的生长量，用最终高度减去初始高度。A：7.2 − 4.0 = 3.2 cm。B：6.5 − 3.8 = 2.7 cm。C：7.0 − 4.2 = 2.8 cm。三株幼苗都生长了，说明光照既足以支持生长，也足以引发弯曲反应。
Step 2: calculate the mean bending angle. Add the three angles and divide by three.
第二步：计算平均弯曲角度。将三个角度相加后除以三。
Mean bending angle = (35° + 40° + 30°) ÷ 3 = 105° ÷ 3 = 35°
Step 3: describe the trend. The mean angle of 35° shows a clear, consistent positive phototropic response in all three seedlings, even though individual values vary between 30° and 40°.
第三步：描述趋势。平均角度为 35°，说明三株幼苗都出现了明显且一致的向光弯曲反应，尽管各株的个体数值在 30° 至 40° 之间波动。
Step 4: evaluate the method. The experiment used only three seedlings, which is a very small sample, so the mean may not be reliable. Ideally the student would repeat the investigation with at least ten seedlings of the same species, age and height, then calculate a new mean. Variables to control include light intensity, distance from the hole, temperature, water volume and soil type.
第四步：评价实验方法。本实验只用了三株幼苗，样本量很小，因此平均值的可靠性有限。理想情况下，学生应使用至少十株品种、苗龄和高度相同的幼苗重复实验，再重新计算平均值。需要控制的变量包括光强、与孔的距离、温度、浇水量和土壤类型。
Independent variable: the direction of the light source. Dependent variable: the bending angle of the stem in degrees. Control variables: light intensity, temperature, water volume, seedling species and starting height.
自变量：光源方向。因变量：茎的弯曲角度（以度为单位）。控制变量：光强、温度、水量、幼苗品种和初始高度。
4. Case Study: Acid–Base Titration Analysis | 案例分析：酸碱滴定分析
A student titrated 25.0 cm³ of hydrochloric acid (HCl) with sodium hydroxide (NaOH) solution using phenolphthalein indicator. She recorded the pH after each addition of NaOH. Some data are shown below.
一位学生用酚酞作指示剂，用氢氧化钠（NaOH）溶液滴定 25.0 cm³ 的盐酸（HCl）。她记录每次加入 NaOH 后的 pH。部分数据如下表所示。
| Volume of NaOH added (cm³) | pH | Colour of indicator |
|---|---|---|
| 0.0 | 1.0 | colourless |
| 10.0 | 1.5 | colourless |
| 20.0 | 2.3 | colourless |
| 24.0 | 3.5 | colourless |
| 24.8 | 6.0 | pale pink |
| 25.0 | 7.0 | pink |
| 26.0 | 10.5 | deep pink |
| 30.0 | 11.5 | deep pink |
Question 1: Describe the change in pH as NaOH is added. Question 2: Explain why phenolphthalein is a suitable indicator for this titration.
问题 1：描述加入 NaOH 过程中 pH 的变化。问题 2：解释为什么酚酞适合用于该滴定实验。
Analysis: Between 0.0 cm³ and 24.0 cm³ the pH rises very slowly, from 1.0 to 3.5, because the added hydroxide ions are being neutralised by the excess acid. Between 24.0 cm³ and 26.0 cm³ the pH rises sharply from 3.5 to 10.5; this is the end point region, where the acid and alkali have reacted in exactly the right proportions. After 26.0 cm³ the pH changes only slightly again because alkali is now in excess.
分析：在 0.0 cm³ 至 24.0 cm³ 之间，pH 上升非常缓慢，从 1.0 升到 3.5，因为加入的氢氧根离子被过量的酸中和。在 24.0 cm³ 至 26.0 cm³ 之间，pH 从 3.5 急剧上升到 10.5，这是终点区域，此时酸与碱恰好按正确的比例完全反应。超过 26.0 cm³ 之后，pH 的变化又变得很小，因为此时碱已过量。
Phenolphthalein is suitable because it changes from colourless to pink over the pH range 8.2 to 10.0, which lies inside the sharp vertical section of the pH curve. A single drop therefore produces a clear, permanent colour change at the end point.
酚酞适合的原因在于：它在 pH 8.2 至 10.0 之间由无色变为粉红色，而这一范围正好落在 pH 曲线急剧上升的竖直部分内。因此只需一滴就能在终点产生明显且稳定的颜色变化。
5. Titration Calculations and Errors | 滴定计算与误差分析
mole of hydrochloric acid reacts with one mole of sodium hydroxide, so the ratio is 1 : 1.
该反应的方程式为：HCl + NaOH → NaCl + H₂O。1 摩尔盐酸与 1 摩尔氢氧化钠反应，因此比例为 1 : 1。
If the hydrochloric acid concentration is 0.100 mol/dm³ and the mean titre of NaOH is 24.8 cm³, the concentration of the alkali can be found as follows.
若盐酸浓度为 0.100 mol/dm³，NaOH 的平均滴定体积为 24.8 cm³，则碱的浓度可按如下方式计算。
c(HCl) × V(HCl) = c(NaOH) × V(NaOH)
0.100 × 25.0 = c(NaOH) × 24.8
c(NaOH) = (0.100 × 25.0) ÷ 24.8 = 0.101 mol/dm³
to divide by 1000 is the single most common error in KS3 and GCSE titration calculations.
注意单位换算：25.0 cm³ = 25.0 ÷ 1000 = 0.0250 dm³。忘记除以 1000 是 KS3 和 GCSE 滴定计算中最常见的错误。
Sources of error worth mentioning in an evaluation: reading the burette from the top rather than at eye level with the meniscus; using too much indicator, which itself is a weak acid; not swirling the flask, so the end point is passed; and using a single rough titration instead of repeating until concordant results (within 0.10 cm³) are obtained.
在评价部分值得提到的误差来源有：从上方而不是与凹液面齐平读数滴定管；加入过多指示剂（指示剂本身是弱酸）；未摇匀锥形瓶，导致越过终点；以及只做一次粗略滴定，而没有重复滴定直到获得吻合结果（相差在 0.10 cm³ 以内）。
6. Case Study: Rates of Reaction with Marble Chips | 案例分析：大理石与酸的反应速率
produced every 20 seconds using a gas syringe. The equation is: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂.
一位学生把 5.0 g 大理石碎块（碳酸钙，CaCO₃）加入 50 cm³ 浓度为 1.0 mol/dm³ 的盐酸中，并用气体注射器每 20 秒测量一次生成的二氧化碳体积。方程式为：CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂。
| Time (s) | Volume of CO₂ (cm³) | Rate (cm³/s) |
|---|---|---|
| 0 | 0 | — |
| 20 | 16 | 0.80 |
| 40 | 28 | 0.60 |
| 60 | 36 | 0.40 |
| 80 | 40 | 0.20 |
| 100 | 40 | 0.00 |
Question: Explain why the rate decreases during the reaction, and predict the effect of using 5.0 g of powdered calcium carbonate instead of chips.
问题：解释反应过程中速率为什么会下降，并预测改用 5.0 g 粉末状碳酸钙代替碎块会有什么影响。
Analysis: The rate is highest at the start because the concentration of hydrochloric acid is at its maximum, so collisions between H⁺ ions and CaCO₃ particles are most frequent. As the acid is used up, its concentration falls, the frequency of successful collisions decreases, and the rate drops. When the graph becomes horizontal at 40 cm³, the reaction has stopped because the acid is the limiting reactant — all of it has reacted.
分析：反应开始时速率最高，因为盐酸浓度最大，H⁺ 离子与 CaCO₃ 颗粒之间的碰撞也最频繁。随着酸被消耗，其浓度下降，有效碰撞的频率降低，速率随之下降。当曲线在 40 cm³ 处变为水平时，反应已停止，因为酸是限量反应物——它已全部反应完。
Powdered calcium carbonate has a much larger surface area to volume ratio than chips. More CaCO₃ particles are exposed to the acid at any moment, so collisions occur more frequently and the initial rate is faster. However, the total volume of CO₂ produced at the end stays the same (40 cm³), because the mass of calcium carbonate and the amount of acid are unchanged.
粉末状碳酸钙的比表面积远大于碎块。任意时刻都有更多 CaCO₃ 颗粒暴露在酸中，因此碰撞更频繁，初始速率更快。但最终生成的 CO₂ 总体积仍为 40 cm³，因为碳酸钙的质量和酸的量都没有改变。
7. Case Study: Food Tests and Enzyme Activity | 案例分析：食物检验与酶活性
A student tested four food samples using Benedict's solution, iodine solution, Biuret reagent and ethanol emulsion test. She then investigated the effect of temperature on amylase, which breaks down starch into maltose.
一位学生用本尼迪克特试剂、碘液、双缩脲试剂和乙醇乳化实验检测了四种食物样品。随后她研究了温度对淀粉酶的影响——淀粉酶会把淀粉分解为麦芽糖。
| Test | Reagent | Positive result | Nutrient detected |
|---|---|---|---|
| Benedict's | Benedict's solution + water bath | blue → green → yellow → orange → brick red | reducing sugar |
| Iodine | iodine solution | yellow-brown → blue-black | starch |
| Biuret | Biuret reagent | blue → purple | protein |
| Emulsion | ethanol then water | clear → cloudy white | lipid |
Question: Explain why the enzyme-controlled reaction is fastest at about 37 °C and stops above 60 °C.
问题：解释为什么该酶促反应在约 37 °C 时最快，而在 60 °C 以上停止。
Analysis: Amylase is a protein with a specific active site. At 37 °C the molecules have enough kinetic energy for frequent successful collisions between starch and the active site, so the rate is at its maximum — this is the optimum temperature, close to human body temperature. Above about 60 °C, the bonds holding the enzyme's tertiary structure break, the active site changes shape, and the substrate no longer fits. The enzyme is denatured, and the reaction stops.
分析：淀粉酶是一种具有特定活性位点的蛋白质。在 37 °C 时，分子具有足够的动能，使淀粉与活性位点之间发生频繁的有效碰撞，因此速率达到最大值——这是最适温度，接近人体体温。当温度升到约 60 °C 以上时，维持酶三级结构的键断裂，活性位点形状改变，底物无法再与之匹配。酶发生变性，反应随即停止。
Denaturation is permanent. Cooling a denatured enzyme back to 37 °C does not restore its activity, whereas an enzyme that has merely been cooled and is inactive will work again when warmed.
变性是不可逆的。把已变性的酶冷却回 37 °C 并不能恢复其活性；而仅仅因为低温而暂时失活的酶，在重新升温后仍能恢复工作。
8. Case Study: Circuits, Resistance and Ohm's Law | 案例分析：电路、电阻与欧姆定律
A student built a circuit with a cell, an ammeter, a voltmeter and a fixed resistor. She increased the number of cells and recorded the current through the resistor each time.
一位学生搭建了包含电源、电流表、电压表和定值电阻的电路。她逐步增加电源（电池）的数量，并记录每次通过电阻的电流。
| Potential difference (V) | Current (A) | Resistance (Ω) |
|---|---|---|
| 1.5 | 0.15 | 10 |
| 3.0 | 0.30 | 10 |
| 4.5 | 0.45 | 10 |
| 6.0 | 0.60 | 10 |
Question: Use the data to show that the resistor obeys
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