Pre-U AQA Biology: Case Study Practical Exercises | 案例分析实战演练

📚 Pre-U AQA Biology: Case Study Practical Exercises | 案例分析实战演练

Case study analysis is a cornerstone of the AQA Pre-U Biology course, demanding that students apply their knowledge to novel scenarios, interpret data, and evaluate experimental evidence. This article provides a series of carefully designed practical exercises that mirror the style and depth expected in the examination, covering key topics from cellular energetics to population dynamics. Each case is structured to sharpen analytical skills, reinforce mechanistic understanding, and build confidence in tackling open-ended biological problems.

案例分析是 AQA Pre-U 生物课程的核心，它要求学生将已有知识应用于新情境、解读数据并评估实验证据。本文提供一系列精心设计的实战演练，模拟考试中要求的风格与深度，涵盖从细胞能量学至种群动态的关键主题。每个案例旨在锻炼分析技能、强化机制理解，并提升应对开放性生物学问题的信心。

1. Cellular Respiration Case Study | 细胞呼吸案例研究

A 28-year-old patient presents with progressive muscle weakness and exercise intolerance. A muscle biopsy reveals abnormally low activity of Complex I (NADH dehydrogenase) in the electron transport chain. Suspension of isolated mitochondria is supplied with pyruvate and malate as substrates; oxygen consumption is measured polarographically before and after addition of ADP.

一名 28 岁患者出现进行性肌无力和运动不耐受。肌肉活检显示电子传递链复合体 I（NADH 脱氢酶）活性异常低下。分离的线粒体悬液以丙酮酸和苹果酸为底物，用极谱法测定加入 ADP 前后的氧消耗。

Condition	O₂ consumption (nmol min⁻¹ mg protein⁻¹)
Normal, state 3 (with ADP)	58
Patient, state 3 (with ADP)	12

Question: Calculate the percentage decrease in respiratory capacity and explain why a defect in Complex I leads to reduced ATP synthesis, linking your answer to proton motive force and chemiosmosis.

问题：计算呼吸能力下降的百分比，并解释复合体 I 缺陷为何导致 ATP 合成减少，将答案与质子动力势和化学渗透联系起来。

Analysis: The patient’s oxygen consumption is only 21% of normal (12/58 × 100), a 79% decrease. Complex I receives electrons from NADH generated during pyruvate oxidation and the citric acid cycle. Its dysfunction means fewer electrons enter the electron transport chain, so fewer protons are pumped across the inner mitochondrial membrane by complexes I, III, and IV. The resulting proton gradient (Δp) is weaker, dissipating the proton motive force that ordinarily drives H⁺ back through ATP synthase. Consequently, ATP synthase operates at a much lower rate, decreasing the overall ATP yield per glucose molecule.

分析：患者的氧消耗仅为正常的 21%（12/58×100），下降了 79%。复合体 I 接收丙酮酸氧化和柠檬酸循环产生的 NADH 的电子。其功能障碍意味着进入电子传递链的电子减少，因此经复合体 I、III 和 IV 泵出线粒体内膜的质子减少。形成的质子梯度（Δp）较弱，削弱了驱动 H⁺ 回流经过 ATP 合酶的质子动力势。于是 ATP 合酶以低得多的速率工作，降低了每分子葡萄糖的 ATP 总产额。

2. Genetic Pedigree and Hardy–Weinberg | 遗传系谱与哈迪–温伯格平衡

Cystic fibrosis (CF) is an autosomal recessive condition caused by mutations in the CFTR gene. In a geographically isolated human population of 40 000, a hospital records 16 live births affected by CF in a single year. Assuming the population is in Hardy–Weinberg equilibrium for this locus, deduce the frequency of the CF-causing allele and the proportion of phenotypically normal carriers in this community.

囊性纤维化（CF）是一种由 CFTR 基因突变引起的常染色体隐性遗传病。在一个与外界隔绝的 40 000 人口中，某医院单年录得 16 例患有 CF 的活产婴儿。假设该基因座符合哈迪–温伯格平衡，推导该人群 CF 致病等位基因的频率及表型正常携带者的比例。

First, calculate the observed frequency of affected homozygotes (q²): 16/40 000 = 0.0004. Hence q = √0.0004 = 0.02. The frequency of the normal allele (p) is 1 − q = 0.98. Carrier frequency is given by 2pq = 2 × 0.98 × 0.02 = 0.0392, approximately 3.92%. Thus about one in 25 individuals is a carrier despite being healthy. This quantitative approach highlights why genetic counselling often targets carrier screening in populations where recessive disorders show elevated incidence, and it reinforces the importance of the Hardy–Weinberg assumptions—large population size, random mating, and absence of mutation, migration, or selection.

首先计算受累纯合子（q²）的观察频率：16/40 000 = 0.0004。因此 q = √0.0004 = 0.02。正常等位基因频率（p）为 1 − q = 0.98。携带者频率为 2pq = 2×0.98×0.02 = 0.0392，约 3.92%。因此约每 25 人中就有一名表型健康的携带者。这一量化方法表明，为何在隐性遗传病发病率较高的人群中，遗传咨询常建议进行携带者筛查，并强化了哈迪–温伯格假设的重要性——大群体、随机交配、无突变、无迁移和无选择。

3. Energy Flow in a Freshwater Ecosystem | 淡水生态系统能量流动

Productivity data were collected from a small eutrophic lake. The figures below represent annual energy values in kJ m⁻² y⁻¹.

以下数据来自一个小型富营养化湖泊，数值代表年能量值，单位 kJ m⁻² yr⁻¹。

Trophic level	Energy input (kJ m⁻² y⁻¹)	Energy lost as heat/respiration
Phytoplankton GPP	29 400	11 900
Zooplankton ingestion	8 200	6 400
Small fish ingestion	1 500	1 200

Question: Calculate (a) the net primary production (NPP) of phytoplankton, (b) the ecological efficiency of energy transfer from phytoplankton to zooplankton, assuming assimilation efficiency is 45% for zooplankton, and (c) suggest two reasons why so little energy reaches the fish trophic level.

问题：计算 (a) 浮游植物的净初级生产量（NPP），(b) 从浮游植物到浮游动物的生态传递效率（假设浮游动物同化效率为 45%），以及 (c) 提出两个原因解释为何到达鱼类营养级的能量如此少。

Solution: (a) NPP = GPP − respiration = 29 400 − 11 900 = 17 500 kJ m⁻² y⁻¹. (b) Zooplankton assimilated energy = ingestion × assimilation efficiency = 8 200 × 0.45 = 3 690 kJ m⁻² y⁻¹. Efficiency from NPP to assimilated energy = (3 690 / 17 500) × 100 ≈ 21.1%. (c) Causes include: much NPP is not consumed by grazers but enters detritus food chain; heat loss from respiration at each trophic level is substantial; and fecal egesta plus excretion further reduce available energy. This exercise reinforces the principle that biomass pyramids are limited by the low efficiency of energy transfer.

解答：(a) NPP = GPP − 呼吸量 = 29 400 − 11 900 = 17 500 kJ m⁻² yr⁻¹。(b) 浮游动物同化能 = 摄食量 × 同化效率 = 8 200×0.45 = 3 690 kJ m⁻² yr⁻¹。从 NPP 到同化能的效率 = (3 690 / 17 500)×100 ≈ 21.1%。(c) 原因包括：大量 NPP 未被牧食者取食而进入碎屑食物链；各营养级呼吸散热量大；粪便及排泄物进一步减少可用能。本练习强化了生物量金字塔受限于低能量传递效率的原则。

4. Neuromuscular Junction and Botulinum Toxin | 神经肌肉接头与肉毒杆菌毒素

Botulinum toxin type A is produced by Clostridium botulinum and can cause flaccid paralysis. The toxin cleaves SNARE proteins required for docking and fusion of synaptic vesicles with the presynaptic membrane at cholinergic nerve terminals. An experimental set-up uses a frog neuromuscular preparation bathed in Ringer solution. Stimulation of the motor nerve normally evokes end-plate potentials (EPPs) of 35 mV. After exposure to botulinum toxin, the EPP amplitude falls to 3 mV, which is below the threshold for generating a muscle action potential.

A 型肉毒杆菌毒素由肉毒梭菌产生，可导致松弛性瘫痪。该毒素能切割胆碱能神经末梢突触囊泡与突触前膜对接与融合所必需的 SNARE 蛋白。在一项实验装置中，蛙神经肌肉标本浸浴于任氏液。刺激运动神经通常激起 35 mV 的终板电位（EPP）。暴露于肉毒杆菌毒素后，EPP 振幅降至 3 mV，低于触发肌动作电位的阈值。

Question: Explain how the molecular action of botulinum toxin accounts for the reduction in EPP amplitude and the resultant paralysis. Predict whether increasing extracellular Ca²⁺ concentration would overcome the paralysis and justify your answer.

问题：解释肉毒杆菌毒素的分子作用如何导致 EPP 振幅降低及随之而来的瘫痪。预测提高细胞外 Ca²⁺ 浓度能否克服瘫痪并给出理由。

Response: Botulinum toxin prevents neurotransmitter release by dismantling the fusion machinery. Without functional SNARE complexes, synaptic vesicles cannot fuse with the presynaptic membrane even when Ca²⁺ enters the terminal via voltage-gated channels. Consequently, little or no acetylcholine is released into the synaptic cleft, so postsynaptic nicotinic receptors are insufficiently activated. The small remaining EPP likely arises from spontaneous quantal release, but it fails to depolarise the muscle fibre to threshold, preventing an action potential and contraction. Elevating extracellular Ca²⁺ will not rescue the process because the molecular target of the toxin lies downstream of calcium entry—the physical docking step itself is disabled. Therefore paralysis persists regardless of calcium concentration.

回答：肉毒杆菌毒素通过破坏融合装置阻止神经递质释放。缺乏功能性的 SNARE 复合物，即使 Ca²⁺ 经电压门控通道进入末梢，突触囊泡也无法与突触前膜融合。于是极少量或无乙酰胆碱释放至突触间隙，突触后烟碱型受体激活不足。微小的残余 EPP 可能来自自发量子释放，但无法使肌纤维去极化至阈值，从而阻碍动作电位和收缩。提高细胞外 Ca²⁺ 浓度无法挽救该过程，因为毒素的分子靶点位于钙内流的下游——物理对接步骤本身已受损。因此无论钙浓度如何，瘫痪持续存在。

5. Immune Response and Vaccination Strategies | 免疫反应与疫苗接种策略

An influenza vaccination programme monitors seroconversion rates (≥4-fold rise in haemagglutinin-inhibition antibody titre) in three age cohorts. Six weeks post-vaccination the data are:
20–40 years: 78% seroconversion; 60–75 years: 43% seroconversion; 80+ years: 31% seroconversion.

一项流感疫苗接种计划监测三个年龄组的血清转化率（血凝抑制抗体滴度升高 ≥4 倍）。接种后六周的数据如下：20–40 岁 78% 转化；60–75 岁 43% 转化；80 岁以上 31% 转化。

Question: Biologically, why does the vaccine efficacy decline so markedly with age? Propose an immunological strategy that might improve protection in the elderly.

问题：从生物学角度解释疫苗效力为何随年龄增长显著下降？提出一种可能改善老年人保护力的免疫策略。

Explanation: Ageing of the immune system, termed immunosenescence, involves thymic involution, reduced naive T-cell output, and accumulation of memory cells directed against previously encountered pathogens. The elderly mount weaker primary and secondary antibody responses because their follicular helper T cells and germinal centre B cells are less responsive. Furthermore, the receptor repertoire dims, making it harder to recognise drifted influenza epitopes. A high-dose or adjuvanted vaccine can partially overcome this deficit by providing stronger innate immune stimulation, thereby recruiting more antigen-presenting cells and enhancing CD4⁺ T-cell help. This approach has been adopted in some countries to boost protective titres in the over-65 population.

解释：免疫系统老化，即免疫衰老，包括胸腺退化、初始 T 细胞输出减少，以及针对既往病原的记忆细胞积累。老年人产生初级和次级抗体反应较弱，因为滤泡辅助 T 细胞和生发中心 B 细胞反应性降低。此外，受体谱系变窄，更难识别漂变的流感抗原表位。高剂量或含佐剂疫苗可通过提供更强的天然免疫刺激部分弥补这一缺陷，从而招募更多抗原呈递细胞并增强 CD4⁺ T 细胞辅助。一些国家已采用这一方法以提高 65 岁以上人群的保护性滴度。

6. Photosynthesis Limiting Factors Investigation | 光合作用限制因素研究

Elodea canadensis is immersed in a buffer containing sodium hydrogen carbonate. The rate of photosynthesis is measured by counting oxygen bubbles evolved per minute under different light intensities and CO₂ concentrations at 25°C.

加拿大伊乐藻浸于含碳酸氢钠的缓冲液中。在不同光照强度和 CO₂ 浓度下于 25°C 测量氧气泡每分钟释放数，以此作为光合速率。

Light intensity (μmol photons m⁻² s⁻¹)	Rate at 0.05% CO₂ (bubbles min⁻¹)	Rate at 0.15% CO₂ (bubbles min⁻¹)
0	0 (respiration 2)	0 (respiration 2)
50	3	5
200	8	15
500	9	22
1000	9	24

Question: Identify the limiting factors operating at 50 and 500 μmol photons m⁻² s⁻¹, respectively, under low CO₂. Explain why at 1000 μmol m⁻² s⁻¹ with 0.15% CO₂ the rate reaches a plateau.

问题：分别指出低 CO₂ 条件下在 50 和 500 μmol photons m⁻² s⁻¹ 时哪些限制因素在起作用。解释为什么在 0.15% CO₂、1000 μmol m⁻² s⁻¹ 时速率达到平台。

Interpretation: At 50 μmol m⁻² s⁻¹ and 0.05% CO₂, the rate is directly proportional to light intensity, indicating that light is the primary limiting factor because CO₂ supply is relatively adequate. At 500 μmol m⁻² s⁻¹ low CO₂, raising light intensity does not increase the rate; here CO₂ concentration limits the Calvin cycle, specifically the RuBisCO carboxylation step. With ample CO₂ (0.15%), the rate rises with light intensity up to about 500–1000 μmol m⁻² s⁻¹, after which saturation of electron transport carriers and enzyme processing rates—RuBP regeneration or ATP/NADPH supply—becomes limiting. Temperature (25°C) may also set an upper ceiling because enzyme kinetics reach maximum velocity under these conditions.

解读：在 50 μmol m⁻² s⁻¹ 和 0.05% CO₂ 下，速率与光照强度成正比，表明光是主要限制因素，因为 CO₂ 供应相对充足。在 500 μmol m⁻² s⁻¹ 低 CO₂ 条件下，提高光照强度不再使速率增加；此时 CO₂ 浓度限制了卡尔文循环，特别是 RuBisCO 的羧化步骤。在充足 CO₂（0.15%）下，速率随光照强度增加直至约 500–1000 μmol m⁻² s⁻¹，此后电子传递载体饱和以及酶促加工速率——RuBP 再生或 ATP/NADPH 供应——成为限制因素。温度 25°C 也可能设定上限，因为酶动力学在该条件下已达最大速率。

7. Predator–Prey Population Dynamics | 捕食者–猎物种群动态

Records of snowshoe hare (Lepus americanus) and Canada lynx (Lynx canadensis) pelts purchased by the Hudson’s Bay Company from 1845 to 1935 show regular oscillations with a period of approximately 10 years. Hare numbers peak slightly ahead of lynx peaks. In a simple laboratory microcosm, the same pattern can be modelled with the Lotka–Volterra equations, but in the boreal forest several alternative factors are proposed.

哈德逊湾公司 1845–1935 年间收购的雪兔 (Lepus americanus) 和加拿大猞猁 (Lynx canadensis) 毛皮记录显示约 10 年的规则振荡。雪兔数量的峰值略早于猞猁的峰值。在简单的实验室微宇宙中，可用洛特卡–沃尔泰拉方程模拟同一模式，但在北方森林中提出了多种替代因素。

Question: Discuss the extent to which the lynx–hare cycle demonstrates a classic predator–prey interaction, and evaluate the evidence that food availability for hares and external climatic factors could also drive the oscillation.

问题：论述猞猁–雪兔周期在多大程度上展示了经典的捕食者–猎物互作，并评估雪兔食物可得性和外界气候因素也可能驱动振荡的证据。

Discussion: The Lotka–Volterra model predicts coupled oscillations where predator abundance lags behind prey, matching the pelt records superficially. However, field studies reveal that hare populations cycle even on islands lacking lynx, partly because hare browsing can deplete winter forage (deciduous shrubs), leading to density-dependent starvation. Moreover, heavy snow years or excessive cold stress independent of predation can reduce hare survival. Predation from lynx acts as an amplifier, sharpening the cycles rather than being the sole driver. Thus the interaction is more complex than a single predator–prey feedback: it involves a tri-trophic cascade (vegetation–hare–lynx) and bottom-up (food limitation) as well as top-down (predation) controls. Students should recognise that real ecosystems rarely conform to simple mathematical models and must consider multifactorial hypotheses.

讨论：洛特卡–沃尔泰拉模型预测捕食者丰度落后于猎物的耦合振荡，与毛皮记录表面吻合。然而野外研究显示，即使在没有猞猁的岛屿上雪兔种群也会周期性波动，部分原因在于雪兔啃食可耗尽冬季饲料（落叶灌木），导致密度依赖性饥饿。此外，与捕食无关的强降雪年或极度寒冷胁迫也可降低雪兔存活率。猞猁的捕食起到放大器作用，加剧了振荡，但并非唯一驱动力。因此这一互作比单一捕食者–猎物反馈更复杂：涉及三级营养级联（植被–雪兔–猞猁）和自下而上（食物限制）以及自上而下（捕食）调控。学生应认识到真实生态系统很少符合简单的数学模型，必须考虑多因素假说。

8. Gene Expression and Cancer | 基因表达与癌症

BRCA1 is a tumour suppressor gene located on chromosome 17. Its protein product participates in DNA double-strand break repair by homologous recombination. A woman inherits a frameshift mutation in one BRCA1 allele. Surveillance data indicate her cumulative risk of breast cancer by age 70 is 82%, compared to a population risk of 12%. Examination of tumour tissue after prophylactic surgery reveals that the remaining wild-type BRCA1 allele has become inactivated by promoter hypermethylation.

BRCA1 是一个位于 17 号染色体的抑癌基因，其蛋白产物参与同源重组修复 DNA 双链断裂。一位女性遗传了 BRCA1 一个等位基因的移码突变。监测数据表明她 70 岁以前累积乳腺癌风险为 82%，而人群风险为 12%。预防性手术后肿瘤组织检查显示，余下的野生型 BRCA1 等位基因因启动子高甲基化而失活。

Question: Apply Knudson’s two-hit hypothesis to explain her cancer predisposition, and suggest how BRCA1 deficiency leads to genomic instability. Additionally, discuss why p53 mutation is often found alongside BRCA1 loss in invasive tumours.

问题：运用诺德森二次打击假说解释其癌症易感性，并说明 BRCA1 缺陷如何导致基因组不稳定性。同时讨论为何在侵袭性肿瘤中 p53

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Pre-U AQA Biology: Case Study Practical Exercises | 案例分析实战演练