📚 Year 12 CCEA Biology: Case Study Practice Drills | Year 12 CCEA 生物：案例分析实战演练
Case study questions are a key part of the CCEA Year 12 Biology assessment, requiring you to apply knowledge from multiple topics to unfamiliar scenarios. This article walks you through realistic practice drills – from enzyme kinetics to population genetics – so you can sharpen your ability to interpret data, evaluate experimental design and draw evidence-based conclusions. Each section presents a sample case paired with a systematic thinking framework.
案例分析题是 CCEA 12 年级生物测评的关键组成部分，要求你将多个专题的知识迁移到陌生情境中。本文通过酶动力学、种群遗传学等贴近真题的实战演练，带你系统练习解读数据、评估实验设计以及基于证据得出结论。每个小节都提供一个样本案例并配有结构化的思考框架。
1. Enzyme Activity and pH | 酶活性与 pH
An investigation measured the volume of oxygen produced by catalase when exposed to different pH buffers. The initial rate of reaction was calculated from the linear portion of each progress curve. The results are shown below:
一项实验测量了过氧化氢酶在不同 pH 缓冲液中产生氧气的体积，并根据每条进程曲线的线性部分计算了初始反应速率。结果如下：
| pH | 2.0 | 4.0 | 6.0 | 7.0 | 8.0 | 10.0 | 12.0 |
|---|---|---|---|---|---|---|---|
| Initial rate / cm³ min⁻¹ | 0.1 | 1.8 | 3.5 | 4.2 | 4.0 | 0.9 | 0.0 |
When tackling such a case, begin by identifying the independent variable (pH) and the dependent variable (initial rate). Plot the data mentally: rate rises to a peak near pH 7 and falls sharply after pH 8. The optimum pH for this catalase is around 7, which matches its intracellular location. Above pH 8, the enzyme begins to denature – hydrogen and ionic bonds in the tertiary structure break, altering the active site shape so the substrate can no longer bind. Below pH 4, excess H⁺ ions similarly disrupt these bonds. Notice that at pH 12 the rate drops to zero, indicating complete denaturation.
遇到此类案例时，先明确自变量（pH）和因变量（初始速率）。在头脑中绘制数据图：速率在 pH 7 附近升至峰值，pH 8 之后急剧下降。该过氧化氢酶的最适 pH 约为 7，与其细胞内定位相符。pH 8 以上酶开始变性——三级结构中的氢键和离子键断裂，活性位点形状改变，底物无法结合。pH 4 以下过多的 H⁺ 同样破坏了这些键。注意在 pH 12 时速率降为零，表明酶已完全变性。
- Always check if the data shows a symmetrical bell-shaped curve, typical of pH-activity profiles.
- Link the molecular interactions (ionic bonds, hydrogen bonds) to the shape of the active site.
- For CCEA exam questions, be prepared to calculate the percentage drop in activity between two pH values.
- 始终检查数据是否呈现对称的钟形曲线，这是 pH-活性曲线的典型特征。
- 将分子间相互作用（离子键、氢键）与活性位点形状关联起来。
- CCEA 考题中常要求计算两个 pH 值之间活性的下降百分比，要提前准备。
2. Osmosis in Plant Tissue | 植物组织中的渗透作用
A student placed potato cylinders of identical mass into sucrose solutions of varying concentration. After 30 minutes the cylinders were blotted dry and reweighed. The percentage change in mass was calculated:
一名学生将等质量的土豆圆柱放入不同浓度的蔗糖溶液中，30 分钟后吸干表面水分并再次称重，计算质量变化百分比：
| Sucrose concentration / mol dm⁻³ | 0.0 | 0.2 | 0.4 | 0.6 | 0.8 | 1.0 |
|---|---|---|---|---|---|---|
| Change in mass / % | +12.5 | +7.0 | +0.8 | -6.4 | -13.2 | -18.5 |
lls. In this case, the isotonic concentration is approximately 0.42 mol dm⁻³. Below this concentration, water enters the cells by osmosis, causing the mass to increase and the cells to become turgid. Above the isotonic point, water leaves the cells, the protoplast shrinks and the tissue loses mass – r sucrose, C is the isotonic concentration, R is the pressure constant and T is temperature in Kelvin).
最佳拟合线与质量变化为零的交点对应土豆细胞的水势。本例中等渗浓度约为 0.42 mol dm⁻³。低于该浓度时，水通过渗透作用进入细胞，质量增加，细胞变得坚硬。高于等渗点时，水离开细胞，原生质体收缩，组织质量下降——这一过程称为质壁分离。若用水势（Ψ）表示，可使用 Ψ = -iCRT（蔗糖 i = 1，C 为等渗浓度，R 为压力常数，T 为开氏温度）。
- Plot percentage change, not absolute change, to correct for variations in initial mass.
- State clearly whether the tissue gains or loses water at each concentration and why.
- Identify the isotonic point from the graph and relate it to the cell's solute potential.
- 绘制的是质量变化百分比而非绝对变化，以校正初始质量差异。
- 明确说明组织在每个浓度下是吸水还是失水，并解释原因。
- 从图中找出等渗点，并将其与细胞的溶质势关联起来。
3. Interpreting Pedigree Charts | 解读系谱图
A family pedigree shows that two unaffected parents have three children: two unaffected daughters and one affected son. The affected son marries an unaffected woman and they have an affected daughter. Use this case to practise determining the inheritance pattern.
某家族系谱显示，一对未患病的父母育有三个孩子：两个未患病的女儿和一个患病的儿子。患病的儿子与一位未患病的女性结婚，生下一个患病的女儿。请利用此案例练习判断遗传方式。
The key observation is that unaffected parents produced an affected child, which rules out a dominant trait (unless a new mutation occurred, but you assume none). The condition is therefore recessive. Since the father is unaffected and the son is affected, the gene cannot be on the Y chromosome. It could be autosomal recessive or X-linked recessive. To distinguish, look at the next generation: the affected male passes the allele to his daughter, who is affected. If it were X-linked recessive, a daughter would receive one X from her affected father (carrying the allele) and one X from her unaffected mother. She would be affected only if the mother is a carrier, but in this case the mother is unaffected and we have no carrier information. However, an affected daughter from an affected father and unaffected mother is possible in autosomal recessive if the mother is a heterozygous carrier. For X-linked recessive, an affected female must receive the recessive allele on both X chromosomes, so the father would necessarily be affected and the mother at least a carrier. The pedigree given could fit either pattern, so you need to discuss probabilities and suggest biochemical tests. In CCEA exams, you are often asked to justify why the condition is most likely autosomal recessive if sons and daughters are affected in equal proportions over several generations.
关键观察是未患病的父母生出了患病的孩子，这排除了显性性状（除非发生新突变，但假定没有）。因此该病症为隐性。由于父亲未患病而儿子患病，基因不可能在 Y 染色体上。它可能是常染色体隐性或 X 连锁隐性。为加以区分，可观察下一代：患病男性将等位基因传给了女儿，女儿患病。如果是 X 连锁隐性，女儿会从患病的父亲那里获得一条带致病等位基因的 X 染色体，从母亲那里获得另一条 X。只有母亲也是携带者时，女儿才会患病，但本案例中母亲未患病且未提供携带信息。然而，对于常染色体隐性遗传，只要母亲是杂合携带者，患病的父亲与未患病的母亲就可能生下患病的女儿。对于 X 连锁隐性，患病女性的两条 X 染色体都必须携带隐性等位基因，因此父亲必然患病，母亲至少是携带者。题述系谱两种可能均存在，因此需要讨论概率并建议进行生化检测。在 CCEA 考试中，若若干代内儿子和女儿的患病比例大致相同，通常会被要求论证该病症很可能是常染色体隐性。
- Label each individual with deduced genotypes where possible.
- Use Punnett squares to calculate the probability that a specific child inherits the condition.
- Always consider whether an X-linked pattern can be ruled out by the data.
- 尽可能为每个个体标注推导出的基因型。
- 使用庞纳特方格计算特定孩子患病的概率。
- 始终考虑数据是否能排除 X 连锁的遗传方式。
4. Vaccination Coverage and Herd Immunity | 疫苗接种覆盖率与群体免疫
A public health study recorded measles incidence per 100,000 population and MMR vaccination coverage in five regions. The data are summarised in the table:
一项公共卫生研究记录了五个地区中每 10 万人的麻疹发病率和 MMR 疫苗接种覆盖率。数据汇总如下：
| Region | Vaccination coverage / % | Incidence per 100,000 |
|---|---|---|
| A | 96 | 1.2 |
| B | 89 | 8.5 |
| C | 77 | 24.3 |
| D | 62 | 67.9 |
| E | 51 | 134.0 |
T), the chain of transmission is broken. For measles, which has a basic reproduction number R₀ ≈ 12–18, the HIT is calculated as 1 – (1/R₀). Using R₀ = 15, HIT = 1 – (1/15) ≈ 0.933, i.e. roughly 93% of the population must be immune to prevent sustained spread. Region A, with 96% coverage, sits above the threshold and records very few cases. Region C, at 77% g outbreaks. In a case study, you may be asked to calculate HIT, predict the effect of a 5% uggest reasons why incidence is not zero even above HIT (imported cases, vaccine failure).
疫苗接种覆盖率与麻疹发病率之间存在明显的负相关。当覆盖率超过群体免疫阈值（HIT）时，传播链就会被切断。麻疹的基本传染数 R₀ 约为 12–18，HIT 的计算公式为 1 – (1/R₀)。若取 R₀ = 15，则 HIT = 1 – (1/15) ≈ 0.933，即大约 93% 的人口必须具有免疫力才能阻止持续传播。地区 A 的覆盖率为 96%，高于阈值，病例极少。地区 C 为 77%，远低于阈值，导致暴发。在案例分析中，你可能要计算 HIT，预测接种率下降 5% 的影响，或者解释为何即使高于 HIT 发病率仍不为零（例如输入性病例、疫苗失效等）。
- Use the equation HIT = (R₀ - 1) / R₀ to show the proportion needed.
- Link the immunological function of memory B cells to long-term protection.
- Discuss ethical issues surrounding mandatory vaccination in the ‘evaluate’ questions.
- 利用公式 HIT = (R₀ - 1) / R₀ 展示所需的比例。
- 将记忆 B 细胞的免疫功能与长期保护联系起来。
- 在“评价”类问题中讨论围绕强制接种的伦理议题。
5. Antibiotic Resistance and Natural Selection | 抗生素耐药性与自然选择
A culture of Staphylococcus aureus was grown on agar plates with antibiotic discs. The diameter of the clear zone (inhibition zone) around each disc was measured on day 1, day 7 and day 14. The results for two antibiotics are shown:
在含有抗生素纸片的琼脂平板上培养金黄色葡萄球菌，于第 1 天、第 7 天和第 14 天测量纸片周围清晰区域（抑菌圈）的直径。两种抗生素的结果如下：
| Day | Antibiotic P diameter / mm | Antibiotic Q diameter / mm |
|---|---|---|
| 1 | 24 | 26 |
| 7 | 18 | 25 |
| 14 | 10 | 24 |
Antibiotic P shows a marked reduction in inhibition zone, indicating that resistant bacteria have been selected. The population originally contained a few cells carrying a resistance gene, probably on a plasmid. When exposed to antibiotic P, susceptible cells died, but resistant cells survived and divided, passing the resistance allele to their offspring. This is directional selection driven by the antibiotic as the selection pressure. By day 14, the majority of the population is resistant, so the inhibition zone shrinks. Antibiotic Q remains effective, so the resistance allele is not prevalent (or the mutations conferring resistance to Q are rare). To control spread, a doctor would prescribe antibiotic Q. In an extended case study, you may be given data on conjugation rates and asked to explain how horizontal gene transfer accelerates resistance.
抗生素 P 的抑菌圈明显缩小，说明耐药菌被选择了出来。原始菌群中存在少数携带耐药基因（可能在质粒上）的细胞。抗生素 P 处理时，敏感细胞死亡，而耐药细胞存活并分裂，将耐药等位基因传递给子代。这是由抗生素作为选择压力驱动的定向选择。到第 14 天，大多数细菌已耐药，抑菌圈因此缩小。抗生素 Q 仍保持有效，说明耐药等位基因不普遍（或赋予 Q 耐药性的突变极为罕见）。为控制传播，医生会选用抗生素 Q。在拓展案例分析中，可能会给出接合率的数据，要求解释水平基因转移如何加速耐药。
- State clearly the variation that initially existed in the population.
- Explain why the resistance allele increases in frequency, not just 'bacteria become immune'.
- Discuss strategies to slow resistance, such as finishing the full antibiotic course and using narrow-spectrum antibiotics.
- 清晰说明群体中最初存在的变异。
- 解释为何耐药等位基因的频率会升高，而不仅仅是“细菌产生免疫力”。
- 讨论延缓耐药的策略，例如完成全疗程抗生素和使用窄谱抗生素。
6. Population Growth and Carrying Capacity | 种群增长与环境容纳量
A laboratory culture of Paramecium was sampled daily. The number of cells per ml was recorded:
每天对实验室培养的草履虫进行取样，记录每毫升细胞数：
| Day | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| Cells / ml × 10² | 2 | 6 | 17 | 45 | 95 | 155 | 210 | 240 | 245 |
Plotting these figures yields a classic sigmoid (S-shaped) growth curve. Day 0– production occurs at near maximum rate because resources are plentiful. After day 6 the growth rate declines as competition intensifies; waste products accumulate and resources become limiting. By day 8 the population plateaus around 245 × 10² cells/ml, indicating the carrying capacity of the closed system. To analyse the data, you can calculate the per capita growth rate during the log phase using the formula: growth rate = (ln N₂ - ln N₁) / (t₂ - t₁). For CCEA, be ready to identify the phase from growth data, estimate carrying capacity, and explain the role of density-dependent factors (food shortage, toxic waste) in slowing growth.
绘制这些数据可得到典型的 S 形生长曲线。第 0–1 天是延滞期，细胞适应培养基并合成酶。第 2–5 天种群处于指数（对数）生长期，此时资源充足，繁殖速率接近最大。第 6 天后，竞争加剧导致生长速率下降；代谢废物累积，资源成为限制因子。到第 8 天，种群在 245 × 10² 细胞/毫升附近趋于平稳，表明封闭系统的环境容纳量。分析数据时，可使用公式：增长率 = (ln N₂ - ln N₁) / (t₂ - t₁) 计算对数期的人均增长率。CCEA 考试中，要能根据生长数据判断所处阶段、估算环境容纳量，并解释密度制约因子（食物短缺、有毒废物）在减缓生长中的作用。
- Label lag, log, stationary and (if present) death phases on any graph you draw.
- Link the stationary phase to the concept of limiting factors – space, oxygen, food.
- Contrast this closed-system growth with natural populations that experience predation and disease.
- 在绘制的任何图形上标注延滞期、对数期、稳定期和（如存在）死亡期。
- 将稳定期与限制因子（空间、氧气、食物）的概念联系起来。
- 对比这种封闭系统生长与经历捕食和疾病的自然种群。
7. Photosynthesis and Limiting Factors | 光合作用与限制因素
A standard elodea experiment measured the volume of oxygen produced per minute at different light intensities. Temperature was kept at 20°C and bicarbonate concentration was 0.2%.
一项标准伊乐藻实验在不同光照强度下测量了每分钟产生的氧气体积，温度保持在 20°C，碳酸氢盐浓度为 0.2%。
| Light intensity / arbitrary units | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
|---|---|---|---|---|---|---|---|
| O₂ produced / cm³ min⁻¹ | 0.8 | 1.6 | 2.4 | 3.0 | 3.3 | 3.4 | 3.4 |
Below 40 units, the rate of oxygen production increases linearly with light intensity. In this region, light is the limiting factor because the Calvin cycle is receiving insufficient ATP and reduced NADP to fix CO₂. At around 40–50 units the curve begins to level off; now another factor, probably CO₂ concentration (0.2% bicarbonate is relatively low) is limiting. If the experiment were repeated at a higher CO₂ concentration (e.g. 0.4%) but the same temperature, the plateau would occur at a higher rate. To identify which factor is limiting, look at whether increasing the factor leads to a proportional increase in rate. In case study questions you may be asked to suggest improvements: use a thermostatically controlled water bath to rule out temperature fluctuations, count oxygen bubbles after steady state, and use a range of bicarbonate concentrations to
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