📚 Case Study Practice for Pre-U CIE Biology | Pre-U CIE 生物：案例分析实战演练
Case studies and data-response questions are the backbone of the Pre-U CIE Biology assessment. They require you to move beyond recall and apply your knowledge to unfamiliar scenarios, interpret graphs and tables, and justify conclusions with precise biological reasoning. This article walks you through ten targeted practice cases covering enzyme kinetics, genetics, physiology, ecology, and molecular biology. Each case includes a problem setup, guided analysis, and the key principles you must master. Work through them systematically to sharpen your exam technique.
案例分析与数据应答是 Pre-U CIE 生物考试的核心。它要求你超越死记硬背，将知识应用于陌生情境，解读图表，并用准确的生物学推理来论证结论。本文将通过十个针对性的实战案例，覆盖酶动力学、遗传学、生理学、生态学和分子生物学。每个案例包含问题设置、引导分析和必须掌握的关键原理。请系统性地完成它们，以打磨你的考试技巧。
1. Enzyme Kinetics: Identifying Inhibition Type | 酶动力学：鉴定抑制类型
An experiment measured the initial rate (v₀) of an enzyme-catalysed reaction at different substrate concentrations [S]. The assays were performed in the absence of inhibitor and in the presence of a fixed concentration of inhibitor X. The results are shown below.
一项实验在不同底物浓度 [S] 下测定了酶催化反应的初始速率 (v₀)。实验分别在不含抑制剂和含有固定浓度抑制剂 X 的条件下进行。结果如下表所示。
| [S] (mmol L⁻¹) | v₀ without inhibitor (µmol min⁻¹) | v₀ with inhibitor X (µmol min⁻¹) |
|---|---|---|
| 0.10 | 0.33 | 0.17 |
| 0.20 | 0.50 | 0.29 |
| 0.50 | 0.71 | 0.50 |
| 1.00 | 0.83 | 0.67 |
| 2.00 | 0.91 | 0.80 |
Identify the type of inhibition displayed by inhibitor X and justify your answer by estimating the apparent Kₘ and Vₘₐₓ values from the data.
请判断抑制剂 X 表现出的抑制类型，并通过从数据中估算表观 Kₘ 和 Vₘₐₓ 值来证明你的答案。
The maximum velocity (Vₘₐₓ) in the absence of inhibitor approaches approximately 1.0 µmol min⁻¹. Half of this Vₘₐₓ is 0.5 µmol min⁻¹, which occurs at a substrate concentration of around 0.20 mmol L⁻¹. Therefore, the uninhibited Kₘ is approximately 0.20 mmol L⁻¹. In the presence of inhibitor X, the rate reaches 0.5 µmol min⁻¹ when [S] is close to 0.50 mmol L⁻¹, giving an apparent Kₘ of around 0.50 mmol L⁻¹. The Vₘₐₓ, however, still approaches 1.0 µmol min⁻¹ at high substrate concentrations.
无抑制剂时，最大速率 (Vₘₐₓ) 大约接近 1.0 µmol min⁻¹。此 Vₘₐₓ 的一半为 0.5 µmol min⁻¹，所对应的底物浓度约为 0.20 mmol L⁻¹。因此，未被抑制的 Kₘ 约为 0.20 mmol L⁻¹。在抑制剂 X 存在下，速率达到 0.5 µmol min⁻¹ 时，[S] 接近 0.50 mmol L⁻¹，表观 Kₘ 约为 0.50 mmol L⁻¹。然而，Vₘₐₓ 在高底物浓度下仍然趋近 1.0 µmol min⁻¹。
Since the apparent Kₘ increases while Vₘₐₓ remains unchanged, inhibitor X behaves as a competitive inhibitor. A competitive inhibitor binds to the active site of the enzyme, competing directly with the substrate. This effect can be overcome by increasing the substrate concentration, which explains why the maximum rate is eventually restored.
由于表观 Kₘ 增加而 Vₘₐₓ 保持不变，抑制剂 X 表现为竞争性抑制剂。竞争性抑制剂结合于酶的活性位点，与底物直接竞争。这种效应可以通过增加底物浓度来克服，这就解释了为何最大速率最终得以恢复。
2. Pedigree Analysis for X-Linked Recessive Traits | X连锁隐性性状的系谱分析
The pedigree below traces the inheritance of an X-linked recessive disorder in a family. Filled symbols indicate affected individuals. Squares represent males; circles represent females.
下面的系谱图追踪了一个家族中 X 连锁隐性遗传病的遗传。实心符号表示患病个体。方块表示男性；圆圈表示女性。
(Image description: Generation I: unaffected male (square) and carrier female (circle, not filled). Generation II: one affected male (filled square), one unaffected male, one unaffected female, one unaffected female (all from same parents). Generation III: an unaffected male married to the unaffected female from gen II; they have one affected son and one unaffected daughter.) For simplicity, use the described pattern.
（图像描述：第 I 代：正常男性与携带者女性；第 II 代：一个患病男性，一个正常男性，两个正常女性；第 III 代：一个正常男性与第 II 代中一位正常女性婚配，育有一个患病儿子和一个正常女儿。）
Using the pedigree, deduce the probability that the unaffected daughter in Generation III is a carrier of the disease allele. Support your reasoning with genotypes.
利用系谱图，推断第 III 代中正常女儿是疾病等位基因携带者的概率。用基因型支持你的推理。
Let the disease allele be represented by Xᵃ and the normal allele by Xᴬ. Males with XᵃY are affected. In Generation I, the mother must be a carrier XᴬXᵃ because she has an affected son (XᵃY) and an unaffected partner (Xᴬ eneration III must herself be a carrier. She inherited one X chromosome from her father (XᴬY → Xᴬ) and one from her carrier mother, with a 50% g Xᵃ. Her carrier status is confirmed because she has an affected son. Therefore, she is XᴬXᵃ. Her husband is an unaffected male (XᴬY). Their daughter is unaffected, so she cannot be XᵃXᵃ. The daughter could be either XᴬXᴬ or Xᴬ ility of inheriting the Xᵃ from her carrier mother is 1/2, and she must inherit her father's Xᴬ. Therefore, the probability that the daughter is a carrier is 0.5 or 50%.
设疾病等位基因为 Xᵃ，正常等位基因为 Xᴬ。携带 XᵃY 的男性患病。第 I 代中的母亲必定是携带者 XᴬXᵃ，因为她有患病的儿子 (XᵃY)，而她的伴侣正常 (XᴬY)。第 II 代中婚配入第 III 代的正常女性必定是携带者。她从父亲处获得 Xᴬ，从携带者母亲处有 50% 几率获得 Xᵃ；她生下了患病儿子，确认为携带者 XᴬXᵃ。她的丈夫是正常男性 (XᴬY)。他们的女儿表现正常，因此不可能是 XᵃXᵃ。女儿可能是 XᴬXᴬ 或 XᴬXᵃ。从母亲处遗传到 Xᵃ 的概率是 1/2，且她一定从父亲处获得 Xᴬ。因此，该女儿是携带者的概率为 0.5 或 50%。
3. Photosynthesis and Limiting Factors | 光合作用与限制因素
A student investigated the effect of carbon dioxide concentration on the rate of photosynthesis in spinach leaf discs. The discs were floated in a bicarbonate solution under constant light intensity and temperature. The rate of oxygen production (µmol O₂ m⁻² s⁻¹) was measured. At low CO₂ concentrations (0.01-0.03%), the rate increased linearly; between 0.03% and 0.08%, the rate rose more slowly; above 0.08%, the rate reached a plateau.
一名学生研究了二氧化碳浓度对菠菜叶圆片光合速率的影响。叶圆片漂浮在碳酸氢盐溶液中，保持恒定的光照强度和温度。测定了氧气产生速率（µmol O₂ m⁻² s⁻¹）。在低 CO₂ 浓度（0.01-0.03%）时，速率线性增加；在 0.03% 至 0.08% 之间，速率上升变缓；超过 0.08% 后，速率达到平台。
Explain the shape of the graph in terms of the Calvin cycle and the role of the enzyme Rubisco. Identify the primary limiting factor in each region.
请从卡尔文循环和 Rubisco 酶的作用来解释图形的形状，并指出每个区域的主要限制因素。
In the linear region, CO₂ concentration is the main limiting factor. The rate of the Calvin cycle is directly governed by the availability of CO₂ for carbon fixation. Rubisco catalyses the addition of CO₂ to ribulose bisphosphate (RuBP), and when CO₂ is scarce, the turnover of the cycle is slow. As CO₂ increases, more 3-phosphoglycerate (GP) is produced, accelerating the cycle.
在线性区域，CO₂ 浓度是主要的限制因素。卡尔文循环的速率直接受碳固定所需的 CO₂ 可用性支配。Rubisco 催化 CO₂ 与核酮糖-1,5-二磷酸 (RuBP) 的加成反应，当 CO₂ 稀缺时，循环的周转很慢。随着 CO₂ 增加，产生更多的 3-磷酸甘油酸 (GP)，加快了循环。
In the intermediate region, the rate of increase declines because the active sites of Rubisco are becoming saturated. Other factors, such as the supply of ATP and NADPH from the light-dependent reactions, begin to exert some limitation. The enzyme is no longer the sole bottleneck. Finally, when the plateau is reached, light intensity or the regeneration of RuBP becomes the limiting factor. At this point, increasing CO₂ further has no effect because the light reactions cannot produce enough ATP and NADPH to support a faster Calvin cycle.
在中间区域，速率增幅下降是因为 Rubisco 的活性位点逐渐饱和。来自光反应的 ATP 和 NADPH 供应等其他因素开始施加限制。酶不再是唯一的瓶颈。最后，当达到平台时，光照强度或 RuBP 的再生成为限制因素。此时，再增加 CO₂ 没有效果，因为光反应无法产生足够的 ATP 和 NADPH 来支持更快的卡尔文循环。
4. Hardy-Weinberg Application to a Genetic Disorder | 哈代-温伯格定律在遗传病中的应用
Cystic fibrosis is an autosomal recessive condition. In a population of 10,000 individuals, 1 in 2,500 newborns is affected. Assume the population is in Hardy-Weinberg equilibrium regarding this locus.
囊性纤维化是一种常染色体隐性遗传病。在一个由 10,000 个个体组成的群体中，每 2,500 名新生儿中有 1 名患病。假设该群体在该位点上处于哈代-温伯格平衡。
Calculate the frequency of the disease allele, the frequency of the normal allele, and the percentage of carriers in the population. Provide all steps.
计算致病等位基因的频率、正常等位基因的频率以及群体中携带者的百分比。请给出所有步骤。
ncy of affected individuals (homozygous recessive) is q² = 1/2500 = 0.0004. Therefore, q = √0 - q = 0.98. Carrier frequency is 2pq = 2 × 0.98 × 0.02 = 0.0392. Thus, approximately 3.92% rates why recessive alleles can persist at low frequencies without high disease incidence.
设致病等位基因频率为 q，正常等位基因频率为 p，且 p + q = 1。患病个体（隐性纯合子）的频率为 q² = 1/2500 = 0.0004。因此，q = √0.0004 = 0.02。正常等位基因的频率 p = 1 - q = 0.98。携带者频率为 2pq = 2 × 0.98 × 0.02 = 0.0392。所以，约 3.92% 的群体是携带者。在一个 10,000 人的群体中，预计约有 392 名携带者。这个案例说明了为何隐性等位基因可以在不引发高发病率的情况下以低频率持续存在。
5. Interpreting an Action Potential Trace | 解读动作电位轨迹
The diagram shows a typical action potential recorded from a mammalian neuron. The resting membrane potential is -70 mV. The trace rises to +30 mV (peak), then repolarises and briefly hyperpolarises before returning to resting value.
该图显示了一个从哺乳动物神经元记录的典型动作电位。静息膜电位为 -70 mV。轨迹上升至 +30 mV（峰值），然后复极化，并短暂超极化，最终回到静息值。
Explain the ionic events underlying the depolarisation, repolarisation, and hyperpolarisation phases. Include the roles of voltage-gated sodium and potassium channels.
解释去极化、复极化和超极化阶段背后的离子事件，包括电压门控钠通道和钾通道的作用。
Depolarisation: A stimulus opens some voltage-gated Na⁺ channels, causing Na⁺ influx. When the threshold (around -55 mV) is reached, a positive feedback loop opens more Na⁺ channels, rapidly raising the membrane potential towards the Na⁺ equilibrium potential (+30 mV). The inside of the axon becomes transiently positive relative to the outside.
去极化：刺激打开部分电压门控 Na⁺ 通道，引起 Na⁺ 内流。当达到阈电位（约 -55 mV）时，正反馈环路使更多的 Na⁺ 通道开放，膜电位迅速向 Na⁺ 平衡电位 (+30 mV) 升高。轴突内部相对外部暂时变为正电。
Repolarisation: At the peak, Na⁺ channels inactivate and voltage-gated K⁺ channels open more slowly. K⁺ efflux out of the axon removes positive charge, driving the membrane potential back toward the resting level. The falling phase is due to the combined effect of Na⁺ channel inactivation and K⁺ outflow.
复极化：在峰值时，Na⁺ 通道失活，而电压门控 K⁺ 通道开放较慢。K⁺ 外流带出正电荷，使膜电位回到静息水平。下降相是由于 Na⁺ 通道失活和 K⁺ 外流的共同作用。
Hyperpolarisation: The voltage-gated K⁺ channels close slowly, allowing more K⁺ to leave the cell than is needed to restore resting potential. The membrane potential becomes more negative than the resting value (e.g., -80 mV). The Na⁺-K⁺ pump and leak channels then restore the resting ionic distribution.
超极化：电压门控 K⁺ 通道关闭缓慢，导致过多的 K⁺ 离开细胞，使膜电位变得比静息值更负（如 -80 mV）。随后 Na⁺-K⁺ 泵和漏通道恢复静息离子分布。
6. Primary and Secondary Immune Responses | 初次与再次免疫应答
A child was vaccinated with two doses of the measles vaccine, administered four weeks apart. Blood antibody concentration against measles virus was monitored over time. The graph showed a small, gradual rise after the first dose, peaking at week 3 and then declining. After the second dose, there was a much faster and higher peak, reaching nearly ten times the concentration of the first response.
一名儿童接种了两剂麻疹疫苗，间隔四周。随时间监测血液中抗麻疹病毒抗体浓度。图形显示第一次接种后抗体浓度缓慢小幅上升，在第 3 周达峰后下降。第二次接种后，峰值出现得更快、更高，达到第一次应答浓度的近十倍。
Explain the immunological basis for the difference between the primary and secondary responses. Refer to memory B and T lymphocytes, clonal selection, and the speed of plasma cell differentiation.
解释初次和再次应答差异的免疫学基础。提及记忆 B 和 T 淋巴细胞、克隆选择以及浆细胞分化的速度。
During the primary response, the immune system encounters the vaccine antigens for the first time. Specific B cells are activated, undergo clonal expansion, and differentiate into plasma cells that secrete antibodies. This process takes several days, hence the lag phase and modest antibody peak. Some activated B and T cells become long-lived memory cells.
在初次应答中，免疫系统首次遇到疫苗抗原。特异性 B 细胞被激活，进行克隆扩增并分化为分泌抗体的浆细胞。该过程需要数天，因此出现滞后期和适中的抗体峰值。部分激活的 B 细胞和 T 细胞成为长寿记忆细胞。
Upon secondary exposure, memory cells quickly recognise the antigen. Memory B cells proliferate rapidly and differentiate into plasma cells without the need for full T helper cell priming. This clonal expansion is faster and more extensive, producing a much higher antibody titre in a shorter time. The antibodies also tend to have higher affinity due to previous affinity maturation. This is why booster doses generate a rapid and powerful protective response.
再次接触抗原时，记忆细胞迅速识别。记忆 B 细胞快速增殖并分化为浆细胞，无需完整的 T 辅助细胞致敏过程。这种克隆扩增更快、规模更大，在更短时间内产生高得多的抗体滴度。由于之前的亲和力成熟，抗体亲和力也往往更高。这便是加强剂量能产生快速且强大的保护性应答的原因。
7. Energy Flow Through an Ecosystem | 生态系统中的能量流动
In a grassland food chain, the gross primary production (GPP) of the grass is 20,000 kJ m⁻² year⁻¹. The grass loses 8,000 kJ m⁻² year⁻¹ in respiration. Rabbits consume 2,000 kJ m⁻² year⁻¹ of the grass's net production, of which 800 kJ is lost in faeces and 600 kJ is used in respiration. Foxes consume 200 kJ m⁻² year⁻¹ of rabbit biomass.
在草原食物链中，草的初级生产总量 (GPP) 为 20,000 kJ m⁻² 年⁻¹。草的呼吸作用损失 8,000 kJ m⁻² 年⁻¹。兔子消耗了草净生产量中的 2,000 kJ m⁻² 年⁻¹，其中 800 kJ 随粪便损失，600 kJ 用于呼吸。狐狸消耗了兔子生物量中的 200 kJ m⁻² 年⁻¹。
Calculate the net primary production (NPP) of the grass, the ecological efficiency of energy transfer from grass to rabbit, and from rabbit to fox. Suggest two reasons why the efficiency is low at each trophic level.
计算草的净初级生产量 (NPP)、能量从草传递到兔子以及从兔子传递到狐狸的生态效率。并为每个营养级的低效率提出两个原因。
estion - faeces = 2,000 - 800 = 1,200 kJ m⁻² year⁻¹. The efficiency of transfer from producer to primary consumer (grass to rabbit) is often given as (assimilation / NPP) × 100% = (1,200 / 12,000) × 100 = 10% fox: assimilation by fox = 200 kJ (assuming negligible faeces loss for simplicity), rabbit' ion - respiration = 1,200 - 600 = 600 kJ m⁻² year⁻¹. Efficiency = (200 / 600) × 100 ≈ 33.3% er, ecological efficiency is typically low; here we use the production efficiency concept.
草的 NPP = GPP - 呼吸 = 20,000 - 8,000 = 12,000 kJ m⁻² 年⁻¹。兔子摄入的能量为 2,000 kJ；同化能量 = 摄入 - 粪便 = 2,000 - 800 = 1,200 kJ m⁻² 年⁻¹。从生产者到初级消费者（草至兔子）的传递效率常用（同化量 / NPP）×100% = (1,200 / 12,000) × 100 = 10%。从兔子到狐狸：狐狸同化量 = 200 kJ，兔子的可传递生产量 = 同化量 - 呼吸 = 1,200 - 600 = 600 kJ m⁻² 年⁻¹。效率 = (200 / 600) × 100 ≈ 33.3%。
Reasons for low efficiency: At the producer level, much light energy is not absorbed by chlorophyll, strikes non-photosynthetic surfaces, or is lost as heat. For herbivores
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