CIE A-Level Biology: Worked Example Questions Explained | CIE A-Level 生物：典型例题详解

📚 CIE A-Level Biology: Worked Example Questions Explained | CIE A-Level 生物：典型例题详解

Exam questions in CIE A-Level Biology are designed to test not just recall, but the ability to apply knowledge, analyse data, and evaluate experimental evidence. This article goes through carefully selected worked examples from core topics, breaking down the command words, marking points, and common pitfalls. Each example models the depth of answer expected at this level, helping you build confidence before the final papers.

CIE A-Level 生物的考试题目不仅考查记忆，更着重于应用知识、分析数据和评估实验证据的能力。本文精选了核心主题中的典型例题，逐一拆解指令词、得分点和常见失分点。每个例题都展示了这一级别所期望的答案深度，帮助你在最终试卷前建立信心。

1. Enzyme Kinetics: Interpreting a Lineweaver–Burk Plot | 酶动力学：解读莱恩威弗–伯克图

The Lineweaver–Burk plot is a double-reciprocal graph (1/V₀ against 1/[S]) that frequently appears in CIE multiple‑choice and structured questions. It allows determination of Vₘₐₓ and Kₘ, and is particularly useful for distinguishing competitive and non‑competitive inhibition.

莱恩威弗–伯克图是双倒数图（1/V₀ 对 1/[S]），频繁出现在CIE的选择题和结构化问题中。它能用来求算 Vₘₐₓ 和 Kₘ，尤其适用于区分竞争性抑制和非竞争性抑制。

Example question: The graph shows two lines: one for an enzyme‑catalysed reaction without inhibitor (line A) and one in the presence of inhibitor I (line B). Both lines intersect the y‑axis at the same point. Identify the type of inhibition and explain your reasoning.

典型例题：图中显示两条直线：一条代表无抑制剂时的酶促反应（线A），另一条代表存在抑制剂I时的反应（线B）。两条线与y轴的交点相同。请判断抑制类型并解释理由。

Since the y‑intercept (1/Vₘₐₓ) is unchanged, the Vₘₐₓ remains the same. The x‑intercept (–1/Kₘ) moves closer to the origin, meaning Kₘ has increased. This is the hallmark of competitive inhibition: the inhibitor competes with the substrate for the active site, so a higher substrate concentration is needed to reach half‑Vₘₐₓ. A perfect answer would also mention that the effect can be overcome at very high substrate concentrations.

由于y轴截距（1/Vₘₐₓ）未变，说明 Vₘₐₓ 保持不变。x轴截距（–1/Kₘ）向原点移动，意味着Kₘ值增大。这正是竞争性抑制的特征：抑制剂与底物竞争活性位点，因此需要更高的底物浓度才能达到半最大反应速率。完美的回答还应提到，在极高的底物浓度下抑制作用可以被克服。

2. Oxidative Phosphorylation: The Chemiosmotic Model | 氧化磷酸化：化学渗透模型

One of the most essay‑style tasks in CIE Paper 4 is to describe how ATP is synthesised in the inner mitochondrial membrane. Answers must integrate electron transport, proton pumping, the proton motive force, and ATP synthase.

在CIE卷四中，最典型的论述题任务之一就是描述线粒体内膜上ATP是如何合成的。答案必须整合电子传递、质子泵、质子动力势和ATP合酶。

Example: Explain how the transfer of electrons along the electron transport chain leads to the production of ATP. [8 marks]

例题：解释电子沿电子传递链的传递如何导致ATP的产生。[8分]

Reduced NAD and reduced FAD donate electrons to the chain; electrons pass through a series of carriers (complex I → coenzyme Q → complex III → cytochrome c → complex IV) with progressively lower energy. At complexes I, III, and IV, the energy released pumps H⁺ from the matrix into the intermembrane space, establishing a proton gradient (high H⁺ concentration in intermembrane space). This creates a proton motive force. H⁺ ions flow back into the matrix through ATP synthase (chemiosmosis), causing the enzyme to rotate and catalyse the phosphorylation of ADP to ATP (oxidative phosphorylation). Oxygen acts as the final electron acceptor, forming water. The total yield is about 2.5 ATP per NADH and 1.5 ATP per FADH₂.

还原态的NAD和还原态的FAD将电子传递给传递链；电子通过一系列能量递减的载体（复合物I → 辅酶Q → 复合物III → 细胞色素c → 复合物IV）传递。在复合物I、III和IV处，释放的能量将H⁺从基质泵入膜间隙，建立起质子梯度（膜间隙中H⁺浓度高）。由此产生质子动力势。H⁺离子通过ATP合酶流回基质（化学渗透），导致该酶旋转并催化ADP磷酸化生成ATP（氧化磷酸化）。氧气作为最终电子受体，生成水。每个NADH约产生2.5个ATP，每个FADH₂约产生1.5个ATP。

3. Calvin Cycle: Fixation, Reduction, Regeneration | 卡尔文循环：固定、还原、再生

Questions on the Calvin cycle often ask for the three main stages and the products that leave the cycle. Students often lose marks by not specifying the number of carbon atoms or by confusing RuBP with G3P.

有关卡尔文循环的问题常要求描述三个主要阶段以及离开循环的产物。学生丢分往往是因为没有明确碳原子数量，或混淆了RuBP与G3P。

Example: Outline the reactions of the Calvin cycle that occur in the stroma of a chloroplast. [6 marks]

例题：概述发生在叶绿体基质中的卡尔文循环反应。[6分]

1. Carbon fixation: CO₂ combines with ribulose bisphosphate (RuBP, a 5‑carbon sugar) catalysed by RuBisCO, forming an unstable 6‑carbon intermediate that immediately splits into two molecules of glycerate‑3‑phosphate (GP, 3‑carbon).

1. 碳固定：CO₂ 与核酮糖‑1,5‑二磷酸（RuBP，五碳糖）在RuBisCO催化下结合，形成不稳定的六碳中间产物，该中间产物立即分裂为两分子甘油酸‑3‑磷酸（GP，三碳）。

2. Reduction: GP is phosphorylated by ATP and reduced by reduced NADP (from the light‑dependent reactions) to form glyceraldehyde‑3‑phosphate (GALP, also called triose phosphate).

2. 还原：GP由ATP磷酸化并被还原态NADP（来自光反应）还原，生成甘油醛‑3‑磷酸（GALP，也称磷酸丙糖）。

3. Regeneration: Most GALP molecules are used to regenerate RuBP, using ATP. Some GALP molecules leave the cycle to be synthesised into glucose, sucrose, starch, amino acids, or lipids.

3. 再生：大多数GALP分子用于再生RuBP，此过程需要ATP。部分GALP分子离开循环，用于合成葡萄糖、蔗糖、淀粉、氨基酸或脂质。

For six turns of the cycle, six CO₂ are fixed, producing twelve GALP; ten of them (30‑carbon) regenerate six RuBP (30‑carbon), while two GALP can form one hexose sugar.

循环六次可固定六分子CO₂，生成十二分子GALP；其中十分子（30碳）再生六分子RuBP（30碳），剩余两分子GALP可形成一分子己糖。

4. Genetics: Dihybrid Cross and Epistasis | 遗传学：双因子杂交与上位效应

Epistasis questions appear almost every year in CIE. The key is to recognise when the expected 9:3:3:1 phenotypic ratio becomes 9:3:4, 12:3:1, or 9:7, and to explain the underlying gene interaction.

上位效应的问题在CIE考试中几乎每年都出现。关键在于识别预期的9:3:3:1表型比例何时变为9:3:4、12:3:1或9:7，并解释其背后的基因互作。

Example: In sweet peas, two genes are involved in flower colour. Gene A produces an enzyme that converts a colourless precursor to a purple intermediate. Gene B converts the purple intermediate to blue pigment. If both steps are necessary for blue flowers, predict the phenotypic ratio in the F₂ generation from a cross between AABB (blue) and aabb (white).

例题：香豌豆的花色涉及两个基因。基因A编码一种酶，将无色前体转化为紫色中间产物。基因B将紫色中间产物转化为蓝色色素。如果蓝色花朵需要这两步都完成，预测AABB（蓝色）与aabb（白色）杂交F₂代的表型比例。

This is a case of recessive epistasis (complementary gene action). The F₁ are AaBb (blue). In the F₂, any genotype with at least one dominant A and one dominant B (A_B_) will be blue: 9/16. Genotypes with homozygous recessive aa (aaB_ and aabb) will block the pathway at step one, giving white: 3/16 + 1/16 = 4/16. Genotypes with homozygous recessive bb (A_bb) will have the purple intermediate but cannot convert it to blue, so purple: 3/16. Thus the ratio is 9 blue : 3 purple : 4 white (recessive epistasis, ratio 9:3:4). Many students incorrectly call it dominant epistasis because they see 9:3:4; always trace the pathway.

这是隐性上位（互补基因作用）的情况。F₁ 代均为 AaBb（蓝色）。在F₂中，任何具有至少一个显性A和一个显性B的基因型（A_B_）都会呈现蓝色：9/16。隐性纯合aa（aaB_ 和 aabb）会在第一步阻断通路，表现为白色：3/16 + 1/16 = 4/16。隐性纯合bb（A_bb）会产生紫色中间产物但不能转化为蓝色，因此为紫色：3/16。所以比例为 9 蓝色 : 3 紫色 : 4 白色（隐性上位，比例 9:3:4）。许多学生误以为是显性上位，只因看到9:3:4；务必追溯通路进行分析。

5. Immunology: B Cell and T Cell Activation | 免疫学：B细胞和T细胞的激活

Clarifying the difference between humoral and cell‑mediated immunity is a classic CIE focus. Students must link antigen‑presenting cells, helper T cells, clonal selection, and differentiation into plasma and memory cells.

厘清体液免疫和细胞免疫的区别是CIE的经典考点。学生必须将抗原呈递细胞、辅助T细胞、克隆选择和分化为浆细胞及记忆细胞联系起来。

Example: Describe the role of helper T cells in the immune response. [5 marks]

例题：描述辅助T细胞在免疫应答中的作用。[5分]

Helper T cells have CD4 receptors that bind to antigens displayed on MHC class II molecules of antigen‑presenting cells (macrophages, dendritic cells, B cells). Once activated, they secrete cytokines such as interleukins. These cytokines stimulate: (1) clonal expansion of cytotoxic T cells that recognise the same antigen; (2) proliferation and differentiation of B cells into antibody‑producing plasma cells and memory B cells; and (3) enhanced activity of macrophages. Thus, helper T cells are central to both cell‑mediated and humoral responses. HIV destroys helper T cells, leading to immunodeficiency.

辅助T细胞拥有CD4受体，能够与抗原呈递细胞（巨噬细胞、树突状细胞、B细胞）的MHC II类分子上展示的抗原结合。一旦激活，它们便分泌细胞因子，如白细胞介素。这些细胞因子刺激：（1）识别相同抗原的细胞毒性T细胞的克隆增殖；（2）B细胞增殖并分化为产生抗体的浆细胞和记忆B细胞；（3）增强巨噬细胞的活性。因此，辅助T细胞是细胞免疫和体液免疫的核心。HIV破坏辅助T细胞，导致免疫缺陷。

6. Ecology: Mark‑Release‑Recapture Calculation | 生态学：标记-释放-重捕法计算

The Lincoln index (N = M × C / R) is a straightforward quantitative skill tested almost every session. The challenge lies in stating assumptions and recognising sources of error, not just plugging in numbers.

林肯指数（N = M × C / R）是一项基础定量技能，几乎每次考试都会涉及。挑战在于陈述假设条件并识别误差来源，而不仅仅是代入数字。

Example: In a woodland, 40 field mice were captured, marked, and released. One week later, 50 mice were captured, of which 10 were marked. Estimate the population size and explain two assumptions necessary for this estimate to be valid.

例题：在林地里，捕获了40只田鼠，标记后释放。一周后，再捕获50只，其中10只已标记。估算种群大小，并解释该估算有效所必需的两个假设条件。

Population estimate N = (40 × 50) / 10 = 200 mice. Acceptable assumptions: (1) The marked individuals have had enough time to mix randomly with the unmarked population. (2) The mark is not lost or removed, and does not affect survival (no increased predation). (3) No immigration, emigration, births, or deaths between the two sampling events. (4) The marking does not alter the behaviour of the mice. Errors leading to over‑estimation occur if marked individuals are more easily predated (R becomes smaller) or if the mark washes off. Under‑estimation occurs if marked individuals become trap‑happy (recapture rate higher than expected).

种群估算值 N = (40 × 50) / 10 = 200只。可接受的假设条件：（1）已标记的个体有充足时间与未标记种群随机混合。（2）标记不脱落、不被清除，且不影响存活（不增加被捕食风险）。（3）两次采样之间没有迁入、迁出、出生或死亡。（4）标记不改变田鼠的行为。如果已标记个体更易被捕食（R变小）或标记脱落，会导致估算值偏高。如果已标记个体变得“喜陷阱”（重捕率高于预期），则估算值偏低。

7. DNA Replication: Semi‑Conservative Mechanism | DNA复制：半保留机制

The Meselson–Stahl experiment and the molecular details of DNA replication are commonly assessed together. Students must name enzymes, explain the directionality of synthesis, and differentiate between the leading and lagging strands.

梅塞尔森-斯塔尔实验以及DNA复制的分子细节常被一起考查。学生必须列出酶的名称，解释合成方向性，并区分前导链与滞后链。

Example: Explain why DNA replication is described as semi‑conservative and discontinuous. [7 marks]

例题：解释为什么DNA复制被描述为半保留和不连续的过程。[7分]

Semi‑conservative means each new DNA molecule consists of one original (parental) strand and one newly synthesised strand. The Meselson–Stahl experiment using ¹⁵N and ¹⁴N density‑gradient centrifugation confirmed this: after one generation in ¹⁴N, DNA showed a single intermediate band; after two generations, one intermediate and one light band. Discontinuous replication refers to the lagging strand synthesis: DNA polymerase III can only add nucleotides in the 5′ → 3′ direction. On the template strand that runs 5′ → 3′ overall, short Okazaki fragments are synthesised in the opposite direction and later joined by DNA ligase. Key enzymes: DNA helicase unwinds the double helix; primase adds RNA primers; DNA polymerase III extends the chain; DNA polymerase I replaces RNA primers with DNA; ligase seals nicks.

半保留意味着每一个新的DNA分子都由一条原有的（亲代）链和一条新合成的链组成。梅塞尔森和斯塔尔使用¹⁵N和¹⁴N进行密度梯度离心的实验证实了这一点：在¹⁴N中培养一代后，DNA显示单一条中间带；两代后，出现一条中间带和一条轻带。不连续复制指的是滞后链的合成：DNA聚合酶III只能以5’→3’方向添加核苷酸。在整体方向为5’→3’的模板链上，短的冈崎片段以相反方向合成，随后由DNA连接酶连接。关键酶：DNA解旋酶解开双螺旋；引物酶添加RNA引物；DNA聚合酶III延伸链；DNA聚合酶I将RNA引物置换为DNA；连接酶封闭切口。

8. Transpiration: Cohesion–Tension Theory | 蒸腾作用：内聚力-张力理论

Questions on water transport in xylem test understanding of water potential gradients, adhesion, cohesion, and the tension generated by transpiration. Diagrams of a potometer are also a common feature.

关于木质部水分运输的问题，考查对水势梯度、附着力、内聚力以及由蒸腾作用产生的张力的理解。蒸腾计（potometer）的图示也是常见内容。

Example: Explain how water moves from the soil into the root hair and up to the leaves according to the cohesion–tension theory. [8 marks]

例题：根据内聚力–张力理论，解释水分如何从土壤进入根毛并向上运输到叶片。[8分]

1. Soil water has a higher water potential than the root hair cell, so water enters by osmosis down the water potential gradient. This continues into the root cortex via the apoplast and symplast pathways until reaching the endodermis, where the Casparian strip forces water into the symplast. Mineral ions are actively transported into the xylem, lowering its water potential; water follows by osmosis, generating root pressure.

1. 土壤水的水势高于根毛细胞，因此水分通过渗透作用顺着水势梯度进入。水通过质外体和共质体途径继续进入根皮层，直至到达内皮层，在这里凯氏带迫使水分进入共质体。矿质离子被主动运输到木质部，降低其水势；水分通过渗透作用跟随进入，产生根压。

2. In the leaf, water evaporates from the surfaces of mesophyll cells into air spaces and diffuses out through stomata (transpiration). This lowers the water potential in the cell walls of mesophyll cells, drawing water out of the xylem. The loss of water from the xylem creates tension (negative pressure) that is transmitted down the column of water due to the strong cohesion between water molecules (hydrogen bonds). Adhesion of water to xylem walls (capillarity) helps maintain the column. The continuous column of water in the xylem is pulled up under tension—hence the cohesion–tension theory.

2. 在叶片中，水分从叶肉细胞表面蒸发进入气室，并通过气孔扩散出去（蒸腾作用）。这降低了叶肉细胞壁中的水势，从而从木质部中吸取水分。木质部中水分的流失产生张力（负压），由于水分子之间强大的内聚力（氢键），张力沿水柱向下传递。水对木质部壁的附着力（毛细管作用）有助于维持水柱的连续。木质部中连续的水柱在张力下被向上拉——这就是内聚力–张力理论。

No energy is required from the plant because transpiration is a passive process driven by the sun’s energy. The theory explains how water can rise to the top of tall trees, often over 100 m.

植物无需消耗能量，因为蒸腾作用是一个由太阳能驱动的被动过程。该理论解释了水分如何能够上升到100多米高的树顶。

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