📚 Year 11 CAIE Science: Case Study Practice Drills | CAIE 11 年级科学：案例分析实战演练
Case studies in CAIE IGCSE Science require you to move beyond recalling facts. You must apply core concepts to unfamiliar scenarios, interpret data, evaluate experimental designs, and construct logical conclusions. This article delivers a series of practical drills covering Biology, Chemistry and Physics. Each drill mimics real exam material and builds the analytical skills needed to excel in Paper 4 or Paper 6, whether you are following Combined Science 0653, Co-ordinated Sciences 0654, or the separate sciences. Work through them systematically, and you will learn to think like a scientist.
CAIE IGCSE 科学中的案例研究要求你不仅仅是复述事实。你必须将核心概念应用到陌生情境中，解读数据，评估实验设计，并构建合乎逻辑的结论。本文提供一系列涵盖生物、化学和物理的实战演练。每个演练都模拟真实的考试材料，培养你在 Paper 4 或 Paper 6 中取得佳绩所需的分析技能，不论你修读 Combined Science 0653、Co-ordinated Sciences 0654 还是单科科学。系统性地完成这些训练，你将学会像科学家一样思考。
1. Deconstructing a Science Case Study: Variables and Hypotheses | 解构科学案例：变量与假设
Every scientific investigation begins with a clear research question. When you read a case study, underline the aim and identify what is being changed, measured and kept the same. For instance, a case might describe an experiment to find out how the concentration of sodium chloride solution affects the number of germinating radish seeds over five days. The independent variable is the sodium chloride concentration; the dependent variable is the number of germinated seeds; control variables include temperature, volume of solution, light exposure and type of seed. Formulating a precise hypothesis-a testable statement linking independent and dependent variables-is essential.
每个科学调查都始于一个清晰的研究问题。阅读案例时，划出目的并确定正在改变、测量和保持不变的量。例如，一个案例可能描述一项实验，研究氯化钠溶液的浓度如何影响萝卜种子在五天内的发芽数量。自变量是氯化钠浓度；因变量是发芽种子的数量；控制变量包括温度、溶液体积、光照和种子类型。提出一个精确的假设——一个连接自变量和因变量的可检验陈述——至关重要。
Practice tip: Write the hypothesis in an “If … then …” format. For the radish case: “If the concentration of sodium chloride solution increases, then the number of germinated seeds will decrease because high salt concentration lowers the water potential outside the seed, reducing water uptake.” This prediction integrates biological knowledge of osmosis and prepares you for data interpretation.
练习技巧：用“如果……那么……”的格式写假设。对于萝卜案例：“如果氯化钠溶液的浓度增加，那么发芽的种子数量会减少，因为高盐浓度降低了种子外部的水势，从而减少了水分吸收。”这一预测整合了渗透作用的生物学知识，并为你解读数据做好准备。
2. Photosynthesis Experiment: Controlling Variables and Reading Data | 光合作用实验：控制变量与数据解读
A common case study involves using aquatic pondweed, such as Elodea, to measure the rate of photosynthesis via oxygen bubble production. The apparatus often lists a lamp moved to different distances, a beaker of water, and a stopwatch. You need to recognise that the independent variable is light intensity, linked to distance from the lamp, and the dependent variable is the number of bubbles per minute. A key control variable is carbon dioxide concentration, maintained by adding sodium hydrogencarbonate to the water. Temperature is another control, often stabilised by a water bath.
一个常见的案例研究使用水生植物，如伊乐藻，通过氧气气泡的产生来测量光合作用速率。实验装置通常列出一盏可移动不同距离的灯、一个烧杯和一只秒表。你需要认识到自变量是光强度，它与灯的距离相关，因变量是每分钟的气泡数量。一个关键的控制变量是二氧化碳浓度，通过向水中加入碳酸氢钠来维持。温度是另一个控制变量，通常用水浴来稳定。
Carefully read data tables. If results show 42 bubbles min⁻¹ at 10 cm, 28 bubbles min⁻¹ at 20 cm, and 12 bubbles min⁻¹ at 40 cm, you can deduce that as light intensity decreases, the rate of photosynthesis falls. However, look for anomalies, such as a reading that does not fit the trend, and suggest explanations-a miscount, a bubble sticking, or a fluctuation in temperature. Also calculate the rate using the reciprocal of distance (1/d) to see a proportional relationship.
仔细阅读数据表。如果结果显示在 10 厘米处每分钟 42 个气泡，在 20 厘米处每分钟 28 个气泡，在 40 厘米处每分钟 12 个气泡，你可以推断，随着光照强度降低，光合作用速率下降。但是，要寻找不符合趋势的异常值，并提出解释——计数错误、气泡粘附或温度波动。还可以用距离的倒数（1/d）计算速率，以观察比例关系。
3. Rate of Reaction: Analysing the Effect of Concentration | 化学反应速率：分析浓度影响
A case study often presents the reaction between sodium thiosulfate and hydrochloric acid: Na₂S₂O₃ + 2HCl → S + SO₂ + H₂O + 2NaCl. The cross-under-the-flask method is used to measure the time until the precipitate obscures a marked cross. The instruction typically varies the concentration of sodium thiosulfate while keeping the volume of acid and total volume constant. You must identify that rate ∝ 1/time, and so a graph of 1/time against concentration will produce a straight line through the origin, confirming a direct proportionality.
案例研究常呈现硫代硫酸钠与盐酸的反应：Na₂S₂O₃ + 2HCl → S + SO₂ + H₂O + 2NaCl。使用烧杯下十字法，测量沉淀物遮住标记的十字所需的时间。实验指导通常改变硫代硫酸钠的浓度，同时保持酸的体积和总体积不变。你必须知道速率 ∝ 1/时间，因此 1/时间对浓度作图将得到一条过原点的直线，从而确认正比关系。
An evaluative question might ask: ‘Explain why the total volume must be kept constant in each trial.’ The answer relates to controlling the depth of solution, which affects the path length of viewing the cross, ensuring a fair test. Furthermore, the student must explain why repeating the experiment improves reliability and how to identify outliers. A table combining concentration, time and rate (1/time) with proper sign of units reinforces quantitative analysis.
评估性问题可能会问：“解释为什么每次试验必须保持总体积不变。”答案涉及控制溶液深度，这会影响观察十字的光路长度，以确保公平测试。此外，学生必须解释为什么重复实验能提高可靠性以及如何识别异常值。将浓度、时间和速率（1/时间）以及正确的单位符号结合起来的表格能强化定量分析。
4. Electrical Circuit Fault Diagnosis: Applying Ohm’s Law | 电路故障诊断：应用欧姆定律
Case studies in electricity often present a circuit with a lamp, a variable resistor and a cell, and provide voltmeter and ammeter readings. A fault might be a damaged resistor or a loose connection. The core principle is V = I × R. Suppose a data set shows that as the resistance of the variable resistor is increased, the ammeter reading drops while the voltmeter reading across the resistor rises slightly but across the lamp falls. You would identify that the total resistance of the circuit has increased, reducing current, so the lamp becomes dimmer.
电学案例常呈现一个包含灯泡、可变电阻和电池的电路，并提供电压表和电流表的读数。故障可能是一个损坏的电阻或松动连接。核心原理是 V = I × R。假设一组数据显示，随着可变电阻的阻值增加，电流表读数下降，而电阻两端的电压表读数略微上升，但灯泡两端的电压下降。你会判断电路的总电阻增加了，电流减小，因此灯泡变暗。
In another scenario, a case might describe that the ammeter reads zero but the voltmeter across the component reads full source voltage. This indicates an open circuit, perhaps due to a broken filament. You need to link symptoms to component behaviour. Draw the circuit and annotate potential drops. Practise converting energy: E = V × I × t, and use these relationships to explain why a short circuit generates excessive heat, linking to power dissipation P = I²R.
在另一种场景下，案例可能描述电流表读数为零，但元件两端的电压表读数为电源总电压。这表明发生了断路，可能是灯丝断裂。你需要将症状与元件行为联系起来。画出电路并标注电势降。练习能量转换：E = V × I × t，并用这些关系来解释为什么短路会产生过多热量，联系到功率耗散 P = I²R。
5. Genetics Pedigree Analysis: Predicting Inheritance Patterns | 遗传学谱系分析：预测遗传病概率
A typical biology case study provides a family pedigree showing the occurrence of a genetic disease, like cystic fibrosis or Huntington’s disease. You must determine whether the allele is dominant or recessive, and whether it is autosomal or sex-linked. For CF, the pattern is autosomal recessive: unaffected parents can have an affected child, and the trait often skips generations. Use genetic diagrams with symbols and punnet squares to calculate probabilities. If both parents are heterozygous (Ff), the chance of a child being homozygous recessive (ff) and affected is 25%.
一个典型的生物案例研究提供显示遗传病发生情况的家谱，如囊性纤维化或亨廷顿舞蹈症。你必须确定该等位基因是显性还是隐性，以及是常染色体还是性连锁遗传。对于 CF，其遗传模式是常染色体隐性：未患病的父母可能会生出患病的孩子，且该性状常隔代出现。使用符号和庞纳特方格绘制遗传图解以计算概率。如果父母双方都是杂合子（Ff），则孩子为纯合隐性（ff）且患病的概率为 25%。
Another case could ask you to explain why a daughter is a carrier of haemophilia when her father is unaffected and her mother is a carrier. The allele is X-linked recessive: father X^HY, mother X^HX^h. Sons have a 50% chance of being affected, daughters 50% chance of being carriers. Use the pedigree symbols (circles for female, squares for male, shading for affected) to trace inheritance. Pay attention to ethical considerations, such as genetic counselling, often examined in later parts of the question.
另一个案例可能要求你解释为什么一个女儿是血友病携带者，而她的父亲未患病，母亲是携带者。该等位基因为 X 连锁隐性：父亲 X^HY，母亲 X^HX^h。儿子有 50% 概率患病，女儿有 50% 概率为携带者。使用家谱符号（圆圈代表女性、方块代表男性、涂色代表患病）来追溯遗传。注意伦理考量，如遗传咨询，常在后段题目中考查。
6. Separation Techniques Case: Choosing Between Distillation and Chromatography | 分离技术案例：蒸馏与层析的选择
A practical case might describe a mixture of coloured inks or a sample of impure ethanol. The method selection depends on the mixture’s properties. For ethanol and water, fractional distillation is chosen because the two liquids have different boiling points (78°C and 100°C). A fractionating column provides a temperature gradient, improving separation. The case study could supply a graph of temperature against volume of distillate, asking you to identify the melting and boiling points of the pure components. The plateau at 78°C indicates ethanol distilling over.
一个实践案例可能描述有色墨水的混合物或一份不纯的乙醇样本。方法的选择取决于混合物的性质。对于乙醇和水，选择分馏，因为这两种液体具有不同的沸点（78°C 和 100°C）。分馏柱提供温度梯度，改善分离效果。案例研究可能提供温度对馏出液体积的图表，要求你识别纯组分的熔点和沸点。78°C 处的平台期表明乙醇正在蒸馏出来。
Alternatively, if the case involves separating pigments from spinach leaves, paper chromatography is used. The leaf extract is spotted on the baseline. As the solvent (ethanol or propanone) ascends, pigments travel different distances because of their different solubilities and adhesion to the paper. Calculate Rf = distance travelled by substance ÷ distance travelled by solvent. An unknown dye with Rf 0.45 can be identified by comparing with known values. This type of case also tests your understanding of why the baseline must be drawn in pencil (pencil is insoluble) and the solvent level kept below the spots.
反之，如果案例涉及从菠菜叶中分离色素，则使用纸色谱法。将叶提取物点样在基线上。随着溶剂（乙醇或丙酮）上升，色素因其不同的溶解度和对纸张的附着能力而移动不同的距离。计算 Rf = 物质移动距离 ÷ 溶剂移动距离。一个 Rf 值为 0.45 的未知染料可通过与已知值进行比较来鉴定。这类案例还考查你对以下理解：为什么基线必须用铅笔画（铅笔不溶于溶剂），以及溶剂液面要保持在斑点以下。
7. Radioactivity and Half-Life: Calculations and Risk Assessment | 放射性与半衰期：计算与风险评估
Case studies involving radioactive decay present a graph of count rate (corrected for background) against time for a source such as iodine-131. The half-life is read from the graph: the time taken for the count rate to fall from 80 counts/s to 40 counts/s, then from 40 to 20 counts/s, confirming it is constant. You may be asked to calculate the remaining activity after several half-lives. With an initial activity of 1200 Bq and a half-life of 8 days, after 24 days (three half-lives) the activity falls to 1200 × (½)³ = 150 Bq.
涉及放射性衰变的案例研究会呈现一个经过本底校正后的计数率对时间的曲线，例如碘-131。半衰期可从图中读出：计数率从 80 次/秒降到 40 次/秒，再从 40 次/秒降到 20 次/秒所需的时间，从而证实半衰期是恒定的。你可能需要计算经过几个半衰期后剩余的活度。若初始活度为 1200 Bq，半衰期为 8 天，24 天后（三个半衰期）活度降为 1200 × (½)³ = 150 Bq。
Beyond calculation, the exam may explore safety precautions. For instance, handling a gamma source requires lead shielding, long-handled tongs and limited exposure time. You may need to explain why alpha sources are more dangerous when ingested. Alpha particles are highly ionising and cause severe local damage, although they are stopped by skin. Designing a safe experiment involves choosing a beta source for thickness monitoring-allowing penetration through paper but not aluminium. These applications link directly to industrial radiography and medical tracers.
除计算之外，考试可能会探讨安全预防措施。例如，处理伽马源需要使用铅屏蔽、长柄钳，并限制暴露时间。你可能需要解释为什么阿尔法源被摄入后更危险。阿尔法粒子电离能力极强，会造成严重的局部损伤，尽管它们可被皮肤阻挡。设计一个安全实验涉及选择贝塔源进行厚度监测——它能穿透纸张但不能穿透铝。这些应用与工业射线照相和医学示踪剂直接相关。
8. Energy Transfers in Mechanics: The Roller Coaster Case | 力学中的能量转化：过山车案例
A physics case study might present a roller coaster car of mass 500 kg at a height of 30 m. Calculate the gravitational potential energy (GPE) at the top using GPE = mgh = 500 × 10 × 30 = 150,000 J. As the car descends, GPE converts to kinetic energy (KE). Assuming negligible friction, KE at the bottom equals GPE at the top, so 150,000 = ½ × 500 × v², giving v = √(600) ≈ 24.5 m/s. The case adds a loop: you must determine the minimum speed at the top of the loop so the centripetal force equals weight.
一个物理案例研究可能给出一个质量为 500 公斤、高度为 30 米的过山车车厢。计算顶部的重力势能 (GPE)：GPE = mgh = 500 × 10 × 30 = 150,000 焦耳。当车厢下降时，重力势能转化为动能 (KE)。假设摩擦力可以忽略，底部的动能等于顶部的重力势能，因此 150,000 = ½ × 500 × v²，解得 v = √(600) ≈ 24.5 米/秒。案例增加了一个回环：你必须确定回环顶部的最小速度，使得向心力等于重力。
Realistically, friction is present. The case provides a graph of speed versus distance or an energy loss per metre. You must use work done against friction = force × distance to find the actual speed. A subsequent question may ask to improve the design: lubricate wheels, streamline the car, or reduce mass (though mass cancels in ideal calculations). This case practises energy conservation, work and power, all linked to safety regulations in amusement parks.
现实中存在摩擦。案例提供了速度对距离的图表或每米的能量损失。你必须使用克服摩擦力所做的功 = 力 × 距离来求出实际速度。后续问题可能要求改进设计：润滑车轮、使车厢流线型化，或减轻质量（尽管质量在理想计算中会约掉）。这个案例练习了能量守恒、功和功率，并全部与游乐园的安全规程相关联。
9. Ecosystems and Food Webs: Modelling Energy Transfer | 生态系统中的食物网：能量传递建模
A case study may show a pyramid of energy or biomass for a grassland ecosystem. Producers (grass) contain 20,000 kJ m⁻², primary consumers (grasshoppers) contain 2,000 kJ m⁻², and secondary consumers (frogs) contain 200 kJ m⁻². Calculate the percentage energy transfer between trophic levels: (2,000 / 20,000) × 100% = 10%. This 10% rule is a common pattern. You need to explain energy losses: respiration, excretion, uneaten parts and heat loss. The case might challenge you to predict the impact of removing a top predator on the rest of the web.
一个案例研究可能会展示一个草原生态系统的能量金字塔或生物量金字塔。生产者（草）含有 20,000 千焦/平方米，初级消费者（蚱蜢）含有 2,000 千焦/平方米，次级消费者（青蛙）含有 200 千焦/平方米。计算营养级之间的能量传递百分比：(2,000 / 20,000) × 100% = 10%。这一 10% 法则是一个常见模式。你需要解释能量损失：呼吸作用、排泄、未吃掉的部分以及散热。案例可能会让你预测移除顶级捕食者对食物网其余部分的影响。
Build quantitative models. If the producer biomass is 4000 kg and the dry mass conversion efficiency to primary consumer is 12%, then primary consumer biomass is 4000 × 0.12 = 480 kg. Also discuss bioaccumulation: if the grass contained 0.01 ppm of a pesticide, the grasshopper might concentrate it to 0.15 ppm, and the frog to 2.5 ppm. Such case studies integrate ecology, pyramids, and human impact, preparing you for extended answer questions on sustainability and pollution.
建立量化模型。如果生产者的生物量为 4000 公斤，转化为初级消费者的干质量效率为 12%，那么初级消费者的生物量为 4000 × 0.12 = 480 公斤。还要讨论生物累积：如果草含有 0.01 ppm 的杀虫剂，蚱蜢可能会将其浓缩至 0.15 ppm，而青蛙则浓缩至 2.5 ppm。这类案例研究整合了生态学、金字塔和人类影响，为你在可持续发展与污染方面的扩展作答问题做好准备。
10. Environmental Chemistry: Origin and Impact of Acid Rain | 环境化学：酸雨的成因与影响
A case study might present a table of rainwater pH collected from different sites over a decade. The pH drops from 5.6 to 4.2 in an industrial zone. The cause: burning fossil fuels releases sulfur dioxide (SO₂) and nitrogen oxides (NOₓ). Sulfur dioxide oxidises to sulfur trioxide in the atmosphere: 2SO₂ + O₂ → 2SO₃, then reacts with water: SO₃ + H₂O → H₂SO₄. Similarly, NO₂ forms nitric acid. You must write these equations and identify the acid responsible for the lowering of pH. Pure rain has a natural pH of about 5.6 due to dissolved CO₂ forming weak carbonic acid.
一个案例研究可能呈现一张不同地点十年间雨水的 pH 数据表。一个工业区的 pH 值从 5.6 降到 4.2。原因：燃烧化石燃料释放二氧化硫 (SO₂) 和氮氧化物 (NOₓ)。二氧化硫在大气中被氧化成三氧化硫：2SO₂ + O₂ → 2SO₃，然后与水反应：SO₃ + H₂O → H₂SO₄。类似地，NO₂ 会形成硝酸。你必须写出这些方程式，并确定导致 pH 降低的酸。纯雨由于溶解的 CO₂ 形成弱碳酸，其天然 pH 约为 5.6。
Evaluation of control measures is key. Flue gas desulfurisation (scrubbers using CaCO₃ slurry) removes SO₂: CaCO₃ + SO₂ → CaSO₃ + CO₂. Catalytic converters in cars reduce NOₓ emissions. A longer case may ask you to compare the economic and environmental costs of installing scrubbers versus switching to renewable energy. Use data on limestone erosion, fish kill statistics, and forest damage to support arguments. These extended responses test your ability to weigh evidence and justify a position.
控制措施的评估是关键。烟气脱硫（使用 CaCO₃ 浆液的洗涤器）可以去除 SO₂：CaCO₃ + SO₂ → CaSO₃ + CO₂。汽车中的催化转化器可减少 NOₓ 排放。一个较长的案例可能会要求你比较安装洗涤器与转向可再生能源的经济和环境成本。使用石灰岩侵蚀数据、鱼类死亡统计数据和森林损害数据来支持论点。这些扩展作答测试你权衡证据和论证立场的能力。
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