📚 Year 13 CIE Biology: Case Study Masterclass | CIE生物案例分析实战演练
Case study questions in CIE A-Level Biology go beyond simple recall — they demand application, analysis, and evaluation. This masterclass presents real-world scenarios spanning genetics, physiology, ecology, and molecular biology. Each section provides a guided walkthrough, highlighting the step-by-step reasoning that examiners expect. Practise these approaches to sharpen your problem-solving toolkit for Papers 4 and 5.
CIE生物考试中的案例分析题远不只是简单记忆——它要求应用、分析和评价。本实战演练涵盖遗传学、生理学、生态学和分子生物学等真实场景。每个部分提供引导性的分析步骤，突出考官期望的推理过程。通过练习这些方法，让你在试卷4和5中更加游刃有余。
1. Pedigree Analysis: Autosomal Recessive Disorder | 系谱分析：常染色体隐性遗传病
A couple, both unaffected, have three children. Their first child is a daughter affected by cystic fibrosis. The second is an unaffected son, and the third is an unaffected daughter. Draw the pedigree and use symbols to determine the genotypes of all family members. Then calculate the probability that the unaffected son is a carrier.
一对表型正常的夫妇有三个孩子。第一个女儿患囊性纤维化，第二个儿子正常，第三个女儿正常。画出系谱图，用符号确定所有家庭成员的基因型，然后计算该正常儿子是携带者的概率。
Cystic fibrosis is inherited in an autosomal recessive manner. Affected individuals are homozygous recessive (ff). Since the affected daughter has the genotype ff, both parents must be heterozygous carriers (Ff). The unaffected son's phenotype is normal, so his genotype could be FF or Ff. Using a Punnett square for two Ff parents, the genotypic ratio among offspring is 1 FF : 2 Ff : 1 ff. Excluding the affected child, the probability that a normal child is a carrier (Ff) is 2/3. Therefore, the chance the unaffected son carries the recessive allele is 2/3 or about 66.7%.
囊性纤维化是常染色体隐性遗传。患者为隐性纯合子(ff)。既然患病女儿基因型为ff，父母双方必定都是杂合携带者(Ff)。正常儿子的表型正常，其基因型可能为FF或Ff。对两个Ff亲本使用旁氏表，子代基因型比例为1 FF : 2 Ff : 1 ff。排除患病个体后，一个正常孩子是携带者的概率为2/3。因此，该正常儿子携带隐性等位基因的概率为2/3（约66.7%）。
2. Hardy–Weinberg and Evolutionary Change | 哈迪-温伯格与进化变化
In a population of 5000 snails, 800 exhibit the homozygous recessive banded shell pattern (bb). The remaining snails are unbanded (BB or Bb). Assuming the population was in Hardy–Weinberg equilibrium, calculate the allele frequencies and the expected number of heterozygotes. After a severe drought, a follow-up study finds 1200 banded snails. Has the population evolved? Explain.
在一个有5000只蜗牛的种群中，800只表现为隐性纯合条纹壳(bb)，其余为非条纹(BB或Bb)。假设种群处于哈迪-温伯格平衡，计算等位基因频率和预期的杂合子数量。一场严重干旱后，后续研究发现1200只条纹蜗牛。种群进化了吗？请解释。
The frequency of the recessive genotype bb is q² = 800/5000 = 0.16, so q = √0 0.48 × 5000 = 2400 heterozygotes. After the drought, the new q² = 1200/5000 = 0.24, so q = √0 as increased from 0.4 to 0.49, indicating a change in the gene pool. Since allele frequencies have shifted across generations, the population has evolved — likely due to natural selection against the unbanded phenotype during drought conditions.
隐性基因型bb频率q² = 800/5000 = 0.16, 故q = √0.16 = 0.4。若种群处于平衡，p = 1 - q = 0.6。预期杂合子频率2pq = 2×0.6×0.4 = 0.48, 数量为0.48×5000 = 2400。干旱后新q² = 1200/5000 = 0.24，q = √0.24 ≈ 0.49。等位基因b的频率从0.4升至0.49，表明基因库发生变化。等位基因频率在世代间发生改变，因此种群已经进化——很可能是因为干旱条件下自然选择对抗非条纹表型。
3. Limiting Factors in Photosynthesis | 光合作用的限制因素
An experiment measured the rate of oxygen production in Elodea at varying light intensities and two CO₂ concentrations (0.04% ambient and 0.1% enriched). The data show that at low light, both treatments produce similar rates, but at high light the enriched CO₂ treatment gives a significantly higher rate. Explain the pattern using the concept of limiting factors.
一项实验测量了不同光强和两种CO₂浓度（0.04%环境浓度和0.1%富集浓度）下伊乐藻的产氧速率。数据表明，在低光强下两种处理的速率相似，但在高光强下富集CO₂处理的速率明显更高。运用限制因素的概念解释该模式。
At low light intensity, light is the limiting factor for the light-dependent reactions — increasing CO₂ cannot accelerate photosynthesis because the Calvin cycle is limited by the supply of ATP and reduced NADP. Both curves therefore rise together as light increases. Once light reaches a saturating level, the light-dependent reactions produce ample ATP and reduced NADP. Now CO₂ concentration becomes the limiting factor. The 0.1% CO₂ treatment allows a faster Calvin cycle, so the photosynthetic rate plateaus at a higher value than the ambient treatment. This classic pattern illustrates that when one factor is lifted, another becomes limiting.
在低光强下，光是光反应的限制因素——增加CO₂不能加快光合作用，因为卡尔文循环受限于ATP和还原型NADP的供应。因此两条曲线随光强上升而同步升高。一旦光强达到饱和，光反应产生充足的ATP和还原型NADP，此时CO₂浓度成为限制因素。0.1% CO₂处理允许卡尔文循环更快运行，因此光合速率在更高水平达到平台期。这一经典模式说明当一个限制因素被解除时，另一因素即成为限制。
4. Respiratory Quotient and Metabolic Fuel | 呼吸商与代谢燃料
A germinating seed respires in a respirometer, consuming 50 cm³ of oxygen and producing 35 cm³ of carbon dioxide over 24 hours. Calculate its respiratory quotient (RQ). Deduce the likely respiratory substrate and explain what happens to the remaining oxygen volume.
一个发芽的种子在呼吸计中呼吸，24小时内消耗50 cm³氧气，产生35 cm³二氧化碳。计算其呼吸商(RQ)。推断最可能的呼吸底物，并解释剩余氧气的去向。
, whereas a lipid gives approximately 0.7. Therefore the seed is predominantly respiring lipids. Lipids contain fewer oxygen atoms relative to hydrogen compared with carbohydrates, so more oxygen is required for complete oxidation. The 'remaining' oxygen ends up in metabolic water — removed from the substrate to form H₂O, accounting for the O₂ that does not appear as CO₂.
RQ = 产生的CO₂体积 / 消耗的O₂体积 = 35/50 = 0.7。纯碳水化合物的RQ为1.0，而脂质约0.7。因此种子主要利用脂质呼吸。脂质中氧含量相对氢原子较少，完全氧化需要更多氧气。剩余的氧气最终进入代谢水——每分子O₂从底物中移除两个氢原子生成H₂O，这部分氧气不以CO₂形式出现。
5. Kidney Function and Osmoregulation in Diabetes | 肾脏功能与糖尿病渗透调节
A patient with untreated type 1 diabetes has a high blood glucose concentration (18 mmol dm⁻³) and produces a large volume of dilute urine. Use your knowledge of ultrafiltration and selective reabsorption to explain why glucose appears in the urine and why urine output is increased.
一名未治疗的1型糖尿病患者血糖浓度高达18 mmol dm⁻³，并产生大量稀释尿液。运用超滤和选择性重吸收的知识，解释为何尿中出现葡萄糖以及尿量增加。
In the Bowman’s capsule, glucose is freely filtered into the glomerular filtrate. Normally, the proximal convoluted tubule reabsorbs 100 % of glucose via co-transporter proteins, which have a maximum transport rate (Tm). When blood glucose exceeds the renal threshold (~11 mmol dm⁻³), the Tm is saturated; glucose remains in the filtrate. This unabsorbed glucose lowers the water potential of the filtrate, reducing the osmotic gradient for water reabsorption in the collecting duct. Consequently, less water is reabsorbed into the blood and more is excreted, producing a large volume of urine (polyuria) that tests positive for glucose.
在鲍曼氏囊中，葡萄糖自由滤入肾小球滤液。正常情况下，近曲小管通过协同转运蛋白重吸收100%的葡萄糖，这些蛋白有最大转运速率(Tm)。当血糖超过肾糖阈(约11 mmol dm⁻³)时，Tm饱和，葡萄糖残留在滤液中。未被吸收的葡萄糖降低了滤液的水势，减小了集合管重吸收水的渗透梯度。因此重吸收入血的水减少，排出增多，产生大量尿液（多尿症），且尿糖检测呈阳性。
6. Action Potential Propagation: Ion Channel Dynamics | 动作电位传导：离子通道动力学
The diagram shows an action potential with labelled phases: resting potential, depolarisation, repolarisation, and hyperpolarisation. Explain how voltage-gated Na⁺ and K⁺ channels contribute to each phase. Why does the action potential travel in one direction along an unmyelinated axon?
图中显示一个动作电位，标出了静息电位、去极化、复极化和超极化各阶段。解释电压门控Na⁺和K⁺通道如何参与每个阶段。为什么动作电位在无髓鞘轴突中单向传导？
At resting potential (-70 mV), voltage-gated Na⁺ channels are closed. A threshold stimulus opens some Na⁺ channels, causing Na⁺ influx and depolarisation. This triggers a positive-feedback opening of more Na⁺ channels, driving the membrane potential toward +40 mV. Then Na⁺ channels inactivate, and voltage-gated K⁺ channels open slowly, allowing K⁺ efflux, which repolarises the membrane. K⁺ channels close slowly, so the potential overshoots to hyperpolarisation before the sodium-potassium pump restores resting potential. The action potential propagates in one direction because the region behind is in its absolute refractory period — inactivated Na⁺ channels cannot reopen immediately, so depolarisation can only spread forward.
静息电位(-70 mV)时，电压门控Na⁺通道关闭。阈刺激打开部分Na⁺通道，Na⁺内流引起去极化，进而以正反馈方式打开更多Na⁺通道，膜电位冲向+40 mV。随后Na⁺通道失活，电压门控K⁺通道缓慢打开，K⁺外流使膜复极化。K⁺通道关闭较慢，电位超射至超极化，然后钠钾泵恢复静息电位。动作电位单向传导是因为后方区域处于绝对不应期——失活的Na⁺通道无法立即再开放，因此去极化只能向前传播。
7. Plant Hormones and Phototropism: Experimental Design | 植物激素与向光性：实验设计
Seedlings grown in unidirectional light bend toward the light. Design a controlled experiment using agar blocks and coleoptile tips to show that a diffusible chemical (auxin) from the tip causes differential growth. Include the control setup and predict the results.
在单向光下生长的幼苗会向光弯曲。设计一个利用琼脂块和胚芽鞘尖的对照实验，证明来自尖端的可扩散化学物质（生长素）导致差异性生长。包括对照设置并预测结果。
Cut the tips off several coleoptiles and place them on agar blocks for two hours, allowing auxin to diffuse into the agar. Then place the agar blocks asymmetrically on decapitated coleoptiles — half on the shaded side only, half on the illuminated side. A control group receives plain agar blocks without tip extract. The coleoptiles with auxin-containing agar on the shaded side should bend toward the light, as the higher auxin concentration promotes cell elongation on that side. Those with auxin on the illuminated side bend away from light. Control coleoptiles show no bending. This demonstrates that auxin redistributes to the shaded side under unidirectional light, causing asymmetric growth and phototropism.
切下几个胚芽鞘尖端，置于琼脂块上两小时，使生长素扩散到琼脂中。然后将琼脂块不对称地放在去尖的胚芽鞘上——一半仅置于背光侧，一半仅置于向光侧。对照组接受无尖端提取物的普通琼脂块。背光侧含生长素琼脂块的处理组应弯向光源，因为该侧较高的生长素浓度促进细胞伸长。向光侧含生长素者则弯离光源。对照组无弯曲。这证明单向光下生长素重新分布到背光侧，引起不对称生长和向光性。
8. Predator–Prey Dynamics and Population Oscillations | 捕食者-猎物动态与种群波动
Data from a classic snowshoe hare–lynx trapping record shows cyclic oscillations with an approximately 10-year period. The hare population peaks just before the lynx population. Explain the lag and the factors that prevent either population from growing unchecked.
经典的雪鞋兔-猞猁捕猎数据显示出约10年周期的循环波动，兔种群在猞猁种群之前达到顶峰。解释这一滞后现象以及阻止任一群体无限制增长的因素。
The hare population increases when resources are abundant and predation pressure is low. As hare numbers rise, the lynx find more prey, which supports a higher lynx reproduction rate — but the lynx population increase lags because gestation and development take time. Eventually, higher predation drives the hare population down. With fewer hares, the lynx experience food shortage, and their numbers decline. The hare population then recovers. Density-dependent factors such as food supply and disease also regulate the hare population, preventing uncontrolled growth. The predator–prey interaction produces the linked oscillations with a temporal offset.
当资源丰富且捕食压力低时，兔种群增长。随着兔数量上升，猞猁获得更多猎物，促进了猞猁的繁殖率——但猞猁种群的增长滞后，因为妊娠和发育需要时间。最终，更高的捕食压力使兔种群下降。兔减少后，猞猁面临食物短缺，数量随之下降。兔种群随后恢复。食物供应和疾病等密度制约因素也调节兔种群，防止无限制增长。捕食者-猎物相互作用产生相互关联的周期性波动，且存在时间上的错位。
9. Antibiotic Resistance and Natural Selection | 抗生素耐药性与自然选择
A hospital isolates Staphylococcus aureus from patients before and after a course of methicillin. The proportion of methicillin-resistant S. aureus (MRSA) rises from 4 % to 22 %. Explain how natural selection can account for this change and discuss why completing the full antibiotic course matters.
一家医院在使用甲氧西林前后从患者体内分离出金黄色葡萄球菌。耐甲氧西林金黄色葡萄球菌(MRSA)比例从4%上升至22%。解释自然选择如何导致这一变化，并讨论完成完整抗生素疗程的重要性。
Within the original bacterial population, random mutations confer varying degrees of antibiotic resistance. Methicillin administration acts as a strong selection pressure: susceptible bacteria are killed, whereas those carrying resistance genes survive and reproduce. Over the course of treatment, the resistant allele frequency increases in the population. If a patient stops the antibiotic prematurely, the concentration may not eliminate semi-resistant mutants, allowing them to proliferate and pass on resistance. Completing the full course helps ensure the remaining bacteria — including partially resistant ones — are eradicated, reducing the chance of a resistant strain becoming dominant in the body and spreading to others.
在原始菌群中，随机突变赋予不同程度的抗生素耐药性。甲氧西林的使用构成强大的选择压力：敏感菌被杀死，携带抗性基因的细菌存活并繁殖。在治疗过程中，耐药等位基因的频率在种群中上升。如果患者提前停药，药物浓度可能无法清除半耐药突变体，使其继续增殖并传递耐药性。完成整个疗程有助于确保残留细菌——包括部分耐药菌——被清除，降低耐药菌株在体内成为优势并传播给他人的风险。
10. Primary and Secondary Immune Response | 初次与二次免疫应答
Graph antibody concentration after two exposures to the same antigen, spaced four weeks apart. The second response is faster, higher, and dominated by IgG. Explain the cellular events underlying this immunological memory.
图中显示两次接触同一抗原后的抗体浓度变化，间隔四周。第二次应答更快、更高，且以IgG为主。解释免疫记忆背后的细胞事件。
Upon first exposure, specific B lymphocytes are activated by the antigen and T helper cells. They undergo clonal selection, proliferating into plasma cells that secrete mainly IgM, and memory B cells that remain in the body. After the antigen is cleared, the memory cells persist. When the same antigen enters again, the memory B cells divide rapidly into plasma cells without needing T helper activation again. This produces a much faster and larger secondary response, with abundant IgG — the class-switched, high-affinity antibody that is more effective at neutralising the pathogen. The rapidity prevents the individual from developing symptoms.
首次接触时，特异性B淋巴细胞被抗原和辅助T细胞活化。它们经历克隆选择，增殖为分泌IgM的浆细胞和留存在体内的记忆B细胞。抗原清除后，记忆细胞持续存在。当同一抗原再次进入时，记忆B细胞迅速分裂为浆细胞，无需再次借助辅助T细胞。这产生更快、更大的二次应答，大量出现IgG——经类别转换的高亲和力抗体，更有效地中和病原体。快速应答使个体不出现病症。
11. Lac Operon: Gene Regulation in Prokaryotes | 乳糖操纵子：原核生物的基因调控
E. coli is grown in a medium containing glucose only, then shifted to a medium containing lactose only. Explain the molecular events that activate the lac operon, and why the switch is not immediate if glucose was present beforehand.
大肠杆菌先在仅含葡萄糖的培养基中生长，然后换到仅含乳糖的培养基。解释激活乳糖操纵子的分子事件，并说明如果预先存在葡萄糖，为什么切换不是即刻发生的。
In the absence of lactose, the lac repressor protein binds to the operator, blocking RNA polymerase from transcribing the lacZ, lacY and lacA genes. When lactose is available, its isomer allolactose binds to the repressor, changing its shape so it detaches from the operator. Simultaneously, if glucose is absent, cAMP levels rise, allowing cAMP–CAP complex to bind near the promoter, which facilitates RNA polymerase binding. This dual control ensures high transcription only when lactose is present and glucose is scarce. The initial lag occurs because allolactose must accumulate, and the glucose-lowered cAMP must recover, before full operon expression can proceed.
无乳糖时，lac阻遏蛋白结合于操纵基因，阻止RNA聚合酶转录lacZ、lacY和lacA基因。当乳糖存在时，其异构体异乳糖与阻遏蛋白结合，改变其构象从而从操纵基因上脱落。同时，若无葡萄糖，cAMP水平上升，cAMP–CAP复合物结合于启动子附近，促进RNA聚合酶结合。这种双重控制确保只有在乳糖存在且葡萄糖匮乏时才进行高效转录。初始滞后是因为异乳糖必须积累，且受葡萄糖抑制的cAMP必须恢复，之后操纵子才能充分表达。
12. Energy Flow and Ecological Efficiency | 能量流动与生态效率
In a grassland ecosystem, the producers fix 8000 kJ m⁻² yr⁻¹ of energy in gross primary production (GPP). Respiration consumes 4800 kJ m⁻² yr⁻¹. Primary consumers ingest 1200 kJ m⁻² yr⁻¹, of which 600 kJ is lost in faeces and 360 kJ is used in respiration. Calculate the net primary production (NPP), the assimilation efficiency, and the net production of primary consumers. Suggest why so little energy reaches the secondary consumer level.
在一个草地生态系统中，生产者总初级生产量(GPP)固定了8000 kJ m⁻² yr⁻¹ 的能量，呼吸消耗4800 kJ m⁻² yr⁻¹。初级消费者摄入1200 kJ m⁻² yr⁻¹，其中600 kJ随粪便损失，360 kJ用于呼吸。计算净初级生产量(NPP)、同化效率和初级消费者的净生产量。解释为什么到达次级消费者的能量如此之少。
mary consumers = (ingested - faeces) / ingested = (1200 - 600)/1200 = 600/1200 = 0.5 or 50 % ssimilated energy - respiration = 600 - 360 = 240 kJ m⁻² yr⁻¹. This is only 240/3200 = 7.5 % through respiration and as waste. Energy transfer between trophic levels is usually only 5–20 % tle remains for secondary consumers, explaining why food chains rarely exceed five levels.
NPP = GPP - 呼吸 = 8000 - 4800 = 3200 kJ m⁻² yr⁻¹。初级消费者的同化效率 = (摄入 - 粪便)/摄入 = (1200 - 600)/1200 = 600/1200 = 0.5 或50%。其净生产量 = 同化能量 - 呼吸 = 600 - 360 = 240 kJ m⁻² yr⁻¹，仅为NPP的240/3200 = 7.5%。损失巨大是因为并非所有NPP均被采食（部分成为碎屑），而且被采食的能量中，大部分以呼吸热能和废物形式损失。营养级间能量传递效率通常仅5–20%，因此留给次级消费者的极少，这解释了食物链很少超过五个环节。
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