Case Study Practical Drills | 案例分析实战演练

📚 Case Study Practical Drills | 案例分析实战演练

In CCEA Year 12 Science, case study questions move beyond recalling facts—they assess your ability to think like a scientist. You may be asked to design an investigation, interpret unfamiliar data, evaluate the reliability of evidence, or apply scientific ideas to a real-world scenario. These practical drills are designed to sharpen those skills. Each case below presents a short scenario followed by tasks that mirror exam-style challenges. Work through them carefully and treat them as active revision. The more you practise applying knowledge in context, the more confident you become at handling the unseen material that often appears in Unit assessments and the final examination.

在 CCEA 12 年级科学课程中，案例分析题远不止是简单回忆知识点——它们考查的是像科学家一样思考的能力。你可能需要设计一项探究、解读陌生数据、评估证据的可靠性，或将科学理念应用于真实情境。以下这些实战演练正是为了打磨这些技能而设计的。每一个案例都给出一个简短的情景和与考试题型相似的练习任务。请仔细思考、主动复习。越是多练习在具体情境中运用知识，你就越能自信地应对单元评估和终考中常常出现的陌生材料。

1. Sampling a Woodland Habitat | 林地生境取样

A student wants to estimate the population of daisies in a large school field. She places a 1 m² quadrat at ten random locations and records the number of daisies in each quadrat. Her results: 4, 7, 0, 12, 5, 8, 3, 6, 9, 6. The field has a total area of 500 m². Calculate the mean number of daisies per quadrat, estimate the total population, and explain why random sampling is important. What might cause an overestimate?

一名学生想要估计学校大片草地上雏菊的种群数量。她把一个 1 m² 的样方随机放置在十处位置，记录每个样方内雏菊的数量。结果如下：4, 7, 0, 12, 5, 8, 3, 6, 9, 6。草地总面积为 500 m²。请计算每个样方雏菊的平均数量、估算总种群数量，并解释随机取样的重要性。哪些因素可能导致估算值偏高？

First calculate the mean: (4+7+0+12+5+8+3+6+9+6) ÷ 10 = 60 ÷ 10 = 6 daisies per quadrat. Estimated total population = mean × (total area / quadrat area) = 6 × (500 / 1) = 3000 daisies. Random sampling avoids bias—if you only sample where daisies look abundant, the estimate will be too high. An overestimate could also occur if the quadrat placements accidentally hit unusually dense patches or if smaller plants are double-counted. The main skill here is selecting and justifying sampling techniques, a common requirement in CCEA ecology questions.

首先计算平均值：(4+7+0+12+5+8+3+6+9+6) ÷ 10 = 60 ÷ 10 = 6 朵/样方。估算总种群数量 = 平均值 × (总面积 / 样方面积) = 6 × (500 / 1) = 3000 朵雏菊。随机取样避免了人为偏差——如果只选择雏菊看起来茂盛的地方取样，估算值就会偏高。如果样方意外落在密度异常高的斑块，或者小型植株被重复计数，也可能导致估算值偏高。这里的关键技能是选择和论证取样方法，这在 CCEA 生态学题型中经常出现。

2. Investigating Enzyme Activity | 探究酶活性

A group of learners investigates the effect of pH on catalase activity using potato discs and hydrogen peroxide. They measure the volume of oxygen produced in 2 minutes at pH 4, 7, and 10. The table shows mean oxygen volumes: pH 4 → 12 cm³, pH 7 → 28 cm³, pH 10 → 8 cm³. Identify the independent and dependent variables, suggest two controlled variables, and describe why pH 7 gives the highest rate. How would you improve reliability?

一组学生使用土豆片和过氧化氢探究 pH 对过氧化氢酶活性的影响。他们测量了在 pH 4、7 和 10 条件下 2 分钟内产生的氧气体积。表格显示平均氧气体积：pH 4 → 12 cm³，pH 7 → 28 cm³，pH 10 → 8 cm³。请指出自变量和因变量，提出两个控制变量，并说明为何 pH 7 下反应速率最高。如何提高实验的信度？

Independent variable: pH. Dependent variable: volume of oxygen produced (or rate of reaction). Controlled variables could include temperature (keep at room temperature using a water bath), size and freshness of potato discs, concentration and volume of hydrogen peroxide, and timing accuracy. Catalase works optimally near neutral pH; extreme pH denatures the enzyme, altering the active site shape so the substrate no longer fits. To improve reliability, repeat each pH at least three times, calculate means, and exclude anomalies. Use a water bath to control temperature strictly, as temperature also affects enzyme activity.

自变量：pH。因变量：产生的氧气体积（或反应速率）。控制变量可以包括温度（用水浴保持室温）、土豆片的大小和新旧程度、过氧化氢的浓度和体积，以及计时准确性。过氧化氢酶在中性 pH 附近活性最佳；极端 pH 会使酶变性，改变活性位点形状，使底物无法契合。要提高信度，每个 pH 至少重复三次，计算平均值，并剔除异常值。使用水浴严格控制温度，因为温度也会影响酶活性。

3. Interpreting a Disease Graph | 解读疾病传播图

Figure 1 (not shown) plots the number of new measles cases per week in a city over six months. The curve starts low, rises steeply to a peak at week 10, then falls gradually to near zero by week 20. Explain the shape of the curve using terms like ‘susceptible hosts’, ‘herd immunity’, and ‘vaccination’. What does the falling phase tell you about the population?

图 1（未显示）绘制了某城市六个月内每周新增麻疹病例数。曲线开始时很低，到第 10 周急剧上升至峰值，然后逐渐下降，第 20 周时趋近于零。请用“易感宿主”、“群体免疫”和“疫苗接种”等术语解释曲线形状。下降阶段说明了人群的什么特征？

The initial low number reflects the time needed for the pathogen to spread among susceptible individuals. As infected people pass the virus to those not immune, cases rise exponentially—the steep climb shows a lack of herd immunity. After the peak, fewer susceptible hosts remain because many have gained immunity through infection or vaccination efforts. The falling phase suggests that an increasing proportion of the population becomes immune, eventually stopping transmission when the threshold for herd immunity is reached. CCEA marks often reward linking graph trends to biological explanations like transmission dynamics and immunity levels.

初期病例数较低，反映了病原体在易感人群中传播需要时间。随着感染者将病毒传给无免疫力者，病例数呈指数级上升——急剧攀升表明群体免疫不足。峰值过后，易感宿主减少，因为许多人通过感染或疫苗接种获得了免疫力。下降阶段说明越来越多人获得了免疫，当达到群体免疫阈值时，传播便停止了。CCEA 评分常奖励将图形趋势与传染动力学和免疫水平等生物学解释联系起来的答案。

4. Testing a Food Sample | 检测食物成分

A student performs biochemical tests on an unknown liquid food sample. Benedict’s test gives a brick-red precipitate when heated. Iodine solution remains yellow-brown. Biuret reagent turns purple. Ethanol emulsion test produces a cloudy white layer. Identify the nutrients present and explain the colour changes. Why must a control tube be used? How does this relate to a balanced diet?

一名学生对未知液体食物样本进行生化检测。本尼迪克特试剂加热后产生砖红色沉淀。碘液保持黄褐色。双缩脲试剂变为紫色。乙醇乳化试验出现白色浑浊层。请判断样本中含有哪些营养素，并解释颜色变化。为什么必须使用对照管？这与均衡饮食有何关联？

Brick-red precipitate with Benedict’s indicates reducing sugar, such as glucose. Iodine staying yellow-brown means starch is absent. Purple with Biuret confirms protein. Cloudy white emulsion points to lipids (fats). A control tube with distilled water should be run alongside each test to ensure reagents are not contaminated and that any colour change is genuinely due to the sample. Understanding food composition helps design diets that supply enough energy and nutrients for growth and repair—a key link to the ‘Keeping Healthy’ topic in CCEA.

本尼迪克特试验产生砖红色沉淀说明含有还原糖，如葡萄糖。碘液保持黄褐色意味着不存在淀粉。双缩脲变紫色证实有蛋白质。白色浑浊乳化层表明含有脂类（脂肪）。每次试验都应设置一支加入蒸馏水的对照管，以确保试剂未被污染，且任何颜色变化确实由样本引起。了解食物组成有助于设计能提供足够能量和营养素以促进生长和修复的饮食——这与 CCEA “保持健康”主题紧密相关。

5. Force, Mass, and Acceleration | 力、质量与加速度

Using a dynamics trolley and a light gate, a class investigates Newton’s second law. They keep the total mass of the system constant but vary the pulling force by transferring slotted masses from the trolley to the hook. The acceleration is calculated from the light gate data. Explain why transferring masses instead of adding extra masses keeps a key variable constant. Sketch the expected graph of acceleration against force. How would you find the mass of the system from the gradient?

全班使用动力学小车和光门探究牛顿第二定律。他们保持系统的总质量不变，但通过将槽码从小车转移到挂钩上来改变拉力。加速度由光门数据计算得出。解释为什么转移质量而不是额外增加质量能保持某个关键变量不变。画出加速度与力的关系图的预期形状。如何根据斜率求出系统的质量？

By moving masses from the trolley to the hook, the total mass being accelerated (trolley + hanging masses) stays the same, so any change in acceleration is only due to the changing force. This controls the variable ‘mass’ without needing to recalculate for different total masses. The graph of acceleration (a) versus force (F) should be a straight line through the origin, because a ∝ F when mass is constant. The gradient = 1 / mass, so mass = 1 / gradient. CCEA practical-based questions often ask you to justify a procedure and extract information from slopes, so practise rearranging a = F / m confidently.

将质量从小车转移到挂钩，保证了被加速的总质量（小车 + 悬挂的槽码）不变，因此加速度的任何改变完全是由力的变化引起的。这样就能在不重新计算不同总质量的情况下控制“质量”这个变量。加速度 a 与力 F 的关系图应为一条过原点的直线，因为当质量恒定时 a ∝ F。斜率 = 1 / 质量，所以系统质量 = 1 / 斜率。CCEA 的实验题经常会要求你论证操作步骤并从斜率中提取信息，因此要熟练掌握 a = F / m 的变形。

6. Titration and Error Analysis | 滴定与误差分析

In an acid-base titration, a student uses phenolphthalein indicator and records three titre volumes: 24.1 cm³, 24.3 cm³, 24.2 cm³. Concordant results are those within 0.2 cm³. Calculate the mean titre. The student noticed an air bubble in the burette tip before the first trial; that trial gave 25.8 cm³. Should this value be included? Describe the effect of the bubble on the calculated concentration of the unknown acid.

在一项酸碱滴定中，学生使用酚酞指示剂，记录到三个滴定体积：24.1 cm³、24.3 cm³ 和 24.2 cm³。符合要求的结果要求彼此相差在 0.2 cm³ 以内。计算平均滴定体积。学生在第一次滴定前发现滴定管尖端有气泡；那次滴定得到 25.8 cm³。该数值应包含在内吗？描述气泡对未知酸浓度计算结果的影响。

The three concordant results are 24.1, 24.3, 24.2 cm³; mean = (24.1+24.3+24.2) ÷ 3 = 24.2 cm³. The 25.8 cm³ trial is an outlier caused by the air bubble. Air occupies volume that is later replaced by solution, making the recorded delivered volume larger than the actual volume of acid used. This would lead to a calculated acid concentration that is higher than the true value. Therefore the outlier must be excluded, and the titration repeated if necessary. CCEA mark schemes reward identifying ‘anomalous result due to procedural error’ and linking it correctly to the final calculation error.

三个符合要求的结果是 24.1、24.3 和 24.2 cm³；平均值 = (24.1+24.3+24.2) ÷ 3 = 24.2 cm³。25.8 cm³ 这一数值是气泡导致的异常值。气泡占据的体积随后被溶液取代，使得记录到的滴出体积大于实际所用酸的体积。这将导致计算出的酸浓度高于真实值。因此异常值必须剔除，必要时需重新进行滴定。CCEA 的评分标准鼓励你识别“由操作失误引起的异常结果”并正确将其与最终计算误差联系起来。

7. Renewable Energy Evaluation | 可再生能源评估

A council plans to install a wind farm on a coastal site. Wind speed data show an average of 8 m/s year-round. The turbines generate 2.5 MW each at full capacity, but output fluctuates. Local residents are concerned about noise and visual impact. Evaluate the advantages and disadvantages of this wind farm. Use the ideas of reliability, environmental impact, and energy security. Suggest how to manage intermittency.

某地方议会计划在沿海场地建设一座风电场。风速数据显示全年平均风速为 8 m/s。每台风机满负荷时发电功率为 2.5 MW，但输出功率有波动。当地居民担心噪音和景观影响。评价该风电场的优、缺点。请使用可靠性、环境影响和能源安全等概念。建议如何管理间歇性问题。

Advantages: wind is a renewable resource, reducing reliance on fossil fuels and improving energy security. The coastal site has high average wind speed, so capacity factor should be good. No greenhouse gas emissions during operation. Disadvantages: output is intermittent—turbines only generate when wind blows within operating range. This reduces reliability and requires backup storage (e.g. batteries or pumped hydro) or a diverse energy mix. Onshore wind farms create noise and affect landscapes; careful siting and community engagement can mitigate opposition. CCEA often asks you to weigh up socio-scientific aspects, so use balanced arguments and suggest practical solutions like smart grids to handle variability.

优点：风能是可再生资源，有助于减少对化石燃料的依赖并提高能源安全性。沿海场地平均风速高，因此容量因子较为理想。运行过程中不产生温室气体排放。缺点：输出功率具有间歇性——风机仅在风速处于工作范围内时才能发电。这降低了可靠性，需要配备后备储能（如电池或抽水蓄能）或多元化的能源结构。陆上风电场会产生噪音并影响景观；精心选址和社区沟通可以缓解反对意见。CCEA 经常要求你权衡社会-科学层面的问题，因此应给出平衡的论点并提出如智能电网这类解决波动性的实用方案。

8. Climate Data Analysis | 气候数据分析

A table shows atmospheric CO₂ concentration and average global temperature anomaly for 1980, 1990, 2000, 2010, and 2020. CO₂ rises steadily from 338 ppm to 414 ppm. Temperature anomaly increases from +0.25 °C to +0.98 °C. Describe the correlation and suggest whether it proves causation. What other evidence would strengthen the argument that rising CO₂ drives temperature rise? Explain the greenhouse effect.

一张表格列出了 1980、1990、2000、2010 和 2020 年的大气 CO₂ 浓度和全球平均气温距平。CO₂ 浓度从 338 ppm 稳步上升至 414 ppm。气温距平从 +0.25 °C 升至 +0.98 °C。描述两者之间的相关性，并说明这是否能证明因果关系。哪些其他证据能加强“CO₂ 升高导致气温上升”的论证？解释温室效应。

There is a strong positive correlation: as CO₂ concentration increases, temperature anomaly also increases. However, correlation alone does not prove causation; a third factor could be involved. To strengthen the causal link, scientists use laboratory measurements of CO₂’s infrared absorption, climate models that isolate CO₂’s effect, and ice core data showing past CO₂-temperature coupling. The greenhouse effect: solar radiation passes through the atmosphere and warms the Earth’s surface. The surface emits infrared radiation, which greenhouse gases like CO₂ absorb and re-radiate in all directions, trapping heat. This is a classic CCEA explanation question requiring precise use of scientific vocabulary.

两者存在强正相关：随着 CO₂ 浓度增加，气温距平也增加。然而，相关性本身不能证明因果关系；可能存在第三个因素影响。为了加强因果联系的论证，科学家利用实验室测量 CO₂ 对红外辐射的吸收、使用气候模型隔离 CO₂ 的作用，以及冰芯数据展示过去 CO₂ 与温度的相关变化。温室效应：太阳辐射穿过大气层并加热地球表面。地表向外发射红外辐射，CO₂ 等温室气体将其吸收并向各个方向重新辐射，从而将热量圈闭。这是 CCEA 典型的解释题型，要求精准使用科学词汇。

9. Modelling Genetic Inheritance | 模拟遗传规律

A couple are both carriers of the recessive allele for cystic fibrosis (F = normal, f = cystic fibrosis). Draw a Punnett square to show the possible genotypes of their children. What is the probability of a child having cystic fibrosis? Discuss why embryo screening raises ethical issues, even though it can identify affected embryos.

一对夫妇都是囊性纤维化隐性等位基因的携带者（F = 正常，f = 囊性纤维化）。画出庞纳特方格，展示他们孩子可能的基因型。孩子患囊性纤维化的概率是多少？讨论胚胎筛查尽管能识别出患病胚胎，却为何会引发伦理问题。

Punnett square: Ff × Ff → ¼ FF, ½ Ff, ¼ ff

The probability of inheriting two recessive alleles (ff) and having cystic fibrosis is 25% or 1 in 4. Embryo screening can detect the ff genotype before implantation during IVF. While this may prevent serious suffering, ethical concerns include the destruction of affected embryos, the possibility of ‘designer babies’, and views about the sanctity of life. CCEA marks not only biological accuracy but also the ability to present balanced arguments on such ‘Science and Society’ issues.

庞纳特方格：Ff × Ff → ¼ FF，½ Ff，¼ ff

孩子遗传到两个隐性等位基因（ff）并患囊性纤维化的概率为 25% 或 1/4。胚胎筛查可在体外受精（IVF）植入前检测出 ff 基因型。虽然这可以避免严重病痛，但伦理问题包括：销毁患病胚胎、可能出现“设计婴儿”，以及对生命神圣性的不同看法。CCEA 评分不仅关注生物学上的准确性，也看重你在此类“科学与社会”议题上展现平衡论述的能力。

10. Evaluating an Experiment on Transpiration | 评估蒸腾作用实验

A student places a leafy shoot in a potometer under a lamp at three distances: 10 cm, 30 cm, and 50 cm. She records the distance moved by an air bubble in 5 minutes. Results: at 10 cm → 4.2 cm, 30 cm → 2.8 cm, 50 cm → 1.9 cm. Identify the trend. Explain how light intensity affects transpiration rate. What two other environmental factors could be tested, and how would you control them?

一名学生将带叶枝条插入蒸腾计中，分别放在距光源 10 cm、30 cm 和 50 cm 处。她记录 5 分钟内气泡移动的距离。结果：10 cm → 4.2 cm，30 cm → 2.8 cm，50 cm → 1.9 cm。指出变化趋势。解释光照强度如何影响蒸腾速率。还可以测试哪两个环境因素？如何控制它们？

The trend: as the distance from the lamp increases (light intensity decreases), the distance the bubble moves decreases, indicating slower transpiration. Light stimulates stomata to open wider, allowing more water vapour to diffuse out; it also provides energy for evaporation. Two other factors are temperature and air movement (wind). To test temperature, you could use a thermostatically controlled room or water bath; to test wind, use a fan at constant speed. In CCEA practical questions, being able to suggest further valid investigations earns extra marks. Always relate your answers to the collision and diffusion of water vapour molecules.

变化趋势：随着与光源距离增加（光照强度降低），气泡移动的距离减少，说明蒸腾减慢。光照促使气孔张开更大，让更多水蒸气扩散出去；同时为蒸发作用提供能量。另外两个因素为温度和空气流动（风）。要测试温度，可使用恒温室或水浴；要测试风，可使用恒速风扇。在 CCEA 实验题中，能够提出进一步的有效探究计划会赢得额外分数。回答时务必将答案与水蒸气分子的碰撞和扩散联系起来。

11. Bringing It All Together | 融会贯通

Case study success in CCEA Science depends on recognising the type of enquiry being tested—whether it is data analysis, experimental design, graph interpretation, or evaluation of socio-scientific issues. Always read the context carefully, highlight command words (describe, explain, evaluate, suggest), and structure your answer to match. Use correct scientific terminology and show your working for calculations. When evaluating, give at least one advantage and one disadvantage, and always link back to the data or scenario provided. Practise these drills under timed conditions, and for each case, write a short ‘exam tip’ summary in your own words.

在 CCEA 科学考试中，案例分析的成功取决于识别所考查的探究类型——是数据分析、实验设计、图表解读，还是对社会科学议题的评价。务必仔细阅读背景信息，圈出指令词（描述、解释、评价、建议），并据此组织答案。使用正确的科学术语，计算时写出步骤。进行评价时，至少给出一个优点和一个缺点，并始终与给出的数据或情景挂钩。在限时条件下反复练习这些案例，为每个案例用你自己的话写一条简短的“应试技巧”总结。

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Case Study Practical Drills | 案例分析实战演练