📚 A-Level Biology: Typical Example Questions Explained | A-Level 生物：典型例题详解
This article provides detailed walkthroughs of common A-Level Biology exam questions, covering topics from enzyme kinetics to population genetics. Each worked example is designed to reinforce key concepts and problem-solving techniques.
本文详细解析A-Level生物考试中的典型例题，涵盖酶动力学、种群遗传学等多个主题。每个例题精讲旨在巩固核心概念和解题技巧。
1. Enzyme Kinetics: Interpreting Michaelis–Menten Data | 酶动力学：解读米氏方程数据
A student measures the initial rate of an enzyme‑catalysed reaction at different substrate concentrations. The results are shown in the table below. Calculate the approximate Vmax and Km for this enzyme.
一名学生测定了不同底物浓度下酶促反应的初始速率。结果如下表所示。请计算该酶的大致 Vmax 和 Km。
| [S] (mmol dm⁻³) | Rate (μmol min⁻¹) |
|---|---|
| 0.1 | 12 |
| 0.2 | 20 |
| 0.5 | 35 |
| 1.0 | 48 |
| 2.0 | 55 |
x is just above 55 μmol min⁻¹. To estimate Km, find the substrate concentration at which the rate is half of Vmax. Take Vmax ≈ 58 μmol min⁻¹; half‑maximal rate is 29 μmol min⁻¹.
随着底物浓度增加，反应速率上升但逐渐趋于平稳。最大速率 (Vmax) 在高底物浓度时接近；数据显示 Vmax 略高于 55 μmol min⁻¹。取 Vmax ≈ 58 μmol min⁻¹，则半最大速率为 29 μmol min⁻¹。
Interpolating roughly gives an apparent Km of approximately 0.35 mmol dm⁻³. More precise values would be obtained using a Lineweaver–Burk plot, but this estimation is often sufficient.
从数据来看，速率为 29 时对应的底物浓度介于 0.2 和 0.5 mmol dm⁻³ 之间。粗略插值得到表观 Km 约为 0.35 mmol dm⁻³。使用双倒数作图可以得到更精确的值，但这种估算通常已经足够。
2. Monohybrid Cross and Test Cross | 单基因杂交与测交
In pea plants, tall stem (T) is dominant over dwarf (t). A tall plant is crossed with a dwarf plant and produces 45 tall and 41 dwarf offspring. Determine the genotype of the tall parent.
在豌豆中，高茎 (T) 对矮茎 (t) 为显性。一株高茎植株与一株矮茎植株杂交，产生 45 株高茎和 41 株矮茎后代。请确定高茎亲本的基因型。
The dwarf parent must be homozygous recessive (tt). It can only contribute a t allele to the offspring. The offspring show a ratio of roughly 1:1 (45 tall : 41 dwarf). For a dwarf offspring (tt) to appear, the tall parent must have supplied a t allele. Therefore the tall parent must be heterozygous (Tt).
矮茎亲本必为隐性纯合子 (tt)，只能向后代提供 t 等位基因。子代表现型比例约 1:1 (45 高 : 41 矮)。要出现矮茎后代 (tt)，高茎亲本必须提供 t 等位基因，因此高茎亲本必为杂合子 (Tt)。
| Gametes (Tt parent) | t |
|---|---|
| t (from dwarf) | tt (dwarf) |
| t (from dwarf) | Tt (tall) |
The cross is a test cross: breeding an organism of dominant phenotype with a homozygous recessive to reveal its genotype. A 1:1 ratio indicates that the dominant‑phenotype parent was heterozygous.
该杂交属于测交：将显性表现型个体与隐性纯合子杂交以揭示其基因型。1:1 的分离比表明显性亲本为杂合子。
3. Respiration: ATP Yield from One Glucose Molecule | 细胞呼吸：一分子葡萄糖的 ATP 产量
Calculate the net ATP yield from the complete oxidation of one glucose molecule in aerobic respiration. State the contribution of each stage and the assumptions made for NADH shuttles.
计算有氧呼吸中一分子葡萄糖完全氧化所产生的净 ATP 数量。说明各阶段的贡献以及对 NADH 穿梭的假设。
Glycolysis produces 2 ATP (substrate‑level) and 2 NADH. The link reaction and Krebs cycle produce 2 ATP (GTP), 8 NADH, and 2 FADH₂ in total per glucose. Most ATP is generated by oxidative phosphorylation, where electrons from reduced coenzymes pass along the electron transport chain.
糖酵解产生 2 分子 ATP（底物水平磷酸化）和 2 分子 NADH。连接反应和克雷布斯循环总共每个葡萄糖分子产生 2 分子 ATP (GTP)、8 分子 NADH 和 2 分子 FADH₂。绝大多数 ATP 由氧化磷酸化产生，还原性辅酶的电子经电子传递链传递。
Conventionally, each NADH yields 2.5 ATP and each FADH₂ yields 1.5 ATP. The 2 glycolytic NADH may yield either 1.5 or 2.5 ATP depending on the shuttle. Using 2.5:
通常每个 NADH 产生 2.5 个 ATP，每个 FADH₂ 产生 1.5 个 ATP。糖酵解的 2 个 NADH 根据穿梭机制可能产生 1.5 或 2.5 个 ATP。按 2.5 计算：
Glycolysis: 2 ATP + 2 NADH × 2.5 = 2 + 5 = 7 ATP
糖酵解：2 ATP + 2 NADH × 2.5 = 2 + 5 = 7 ATP
Link/Krebs: 2 ATP + 8 NADH × 2.5 + 2 FADH₂ × 1.5 = 2 + 20 + 3 = 25 ATP
连接/克雷布斯：2 ATP + 8 NADH × 2.5 + 2 FADH₂ × 1.5 = 2 + 20 + 3 = 25 ATP
Total net ATP = 7 + 25 = 32 ATP per glucose
If the glycerol‑phosphate shuttle is used, glycolytic NADH gives only 1.5 ATP each, reducing the total to 30 ATP. Both 30 and 32 are acceptable answers depending on context.
如果采用甘油磷酸穿梭，糖酵解的 NADH 每个仅产 1.5 ATP，总数降为 30 ATP。根据题目背景，30 或 32 均可接受。
4. Photosynthesis: Limiting Factors Graph | 光合作用：限制因素曲线
The graph below shows the rate of photosynthesis of a plant at two different carbon dioxide concentrations as light intensity increases. Explain the shape of the curves.
下图表示两种不同二氧化碳浓度下植物光合速率随光强增加的变化。解释曲线的形状。
(Imagine a graph: at low light intensity both curves rise steeply and overlap; at higher light intensities the curve for low CO₂ levels off earlier, while the high CO₂ curve continues to rise until it too reaches a plateau.)
（想象图形：低光强下两条曲线陡升且重合；光强增高后，低浓度 CO₂ 的曲线先趋于平缓，高浓度 CO₂ 的曲线继续上升直至也到达平台。）
At low light intensity, light is the limiting factor. Increasing light provides more energy for the light‑dependent reactions, boosting ATP and reduced NADP production. Both CO₂ concentrations show the same rate because CO₂ is not limiting.
在低光强下，光是限制因素。增加光照为光反应提供更多能量，生成更多 ATP 和还原型 NADP。由于 CO₂ 不成为限制，两种浓度下光合速率相同。
As light intensity rises, at a certain point the low CO₂ concentration becomes rate‑limiting. The Calvin cycle cannot fix carbon fast enough, so the rate levels off. For the high CO₂ treatment, rate continues to increase until light or enzyme capacity (e.g., RuBisCO activity) becomes limiting, eventually producing a plateau.
随着光强升高，在某一点低浓度 CO₂ 成为速率限制因子。卡尔文循环无法足够快地固定碳，因此速率趋于平稳。对于高 CO₂ 处理，速率继续增加，直到光或酶容量（如 RuBisCO 活性）成为限制，最终出现平台。
5. Immunity: Antibody Structure and Vaccination | 免疫：抗体结构与疫苗接种
Describe how the structure of an antibody is related to its function. Use a labelled diagram in your explanation.
阐述抗体结构如何与其功能相适应。请用标注图辅助解释。
An antibody (immunoglobulin) is a Y‑shaped protein made of four polypeptide chains – two identical heavy chains and two identical light chains – held together by disulfide bonds. Each arm of the Y contains a variable region with a specific antigen‑binding site. The variable regions differ between antibodies and allow recognition of diverse antigens.
抗体（免疫球蛋白）是一种 Y 形蛋白质，由四条多肽链组成——两条相同的重链和两条相同的轻链——通过二硫键连接。Y 的每个臂含有一个可变区，带有特异的抗原结合位点。不同抗体可变区不同，从而能够识别多种抗原。
The constant region of the heavy chains determines the antibody class (IgG, IgM etc.) and binds to receptors on phagocytes and other immune cells, facilitating opsonisation and complement activation. The hinge region confers flexibility, allowing both arms to bind antigens simultaneously.
重链的恒定区决定抗体类别（IgG、IgM 等），并与吞噬细胞等免疫细胞上的受体结合，促进调理作用和补体激活。铰链区赋予抗体灵活性，使两个臂能够同时结合抗原。
In vaccination, a harmless form of the pathogen (or antigen) is introduced. This triggers a primary immune response, producing memory B cells. Upon subsequent exposure to the real pathogen, memory cells rapidly divide, giving a faster, stronger secondary response that often prevents disease.
在疫苗接种中，引入无害形式的病原体（或抗原）。这引发初次免疫应答，产生记忆 B 细胞。当再次接触真实病原体时，记忆细胞迅速分裂，产生更快、更强的二次应答，常能阻止发病。
6. Ecology: Mark‑Release‑Recapture | 生态学：标记重捕法
A biologist captures 30 woodlice in a log pile, marks their shells with non‑toxic paint, and releases them back. The next day, she captures 50 woodlice from the same area and finds that 10 of them are marked. Estimate the total population size of woodlice in this habitat.
一名生物学家在木桩堆中捕获 30 只潮虫，用无毒颜料标记它们的壳后放回。第二天，她从同一区域捕获 50 只潮虫，发现其中 10 只带有标记。估算该栖息地潮虫的总种群大小。
Lincoln index: N = (M × C) / R
mber captured in second sample (50), and R = number of marked individuals recaptured (10).
其中 M = 首次标记数 (30)，C = 第二次捕获数 (50)，R = 重捕的标记个体数 (10)。
N = (30 × 50) / 10 = 150. The estimated population size is 150 woodlice.
N = (30 × 50) / 10 = 150。估算种群大小为 150 只潮虫。
This method assumes: marked individuals mix randomly with the population, marks do not wear off or affect survival, no immigration or emigration occurs, and the population is sampled randomly. Violations of these assumptions can lead to over‑ or underestimation.
该方法假设：标记个体与种群随机混合，标记不脱落且不影响生存，没有迁入或迁出，且种群被随机取样。若违反这些假设，可能导致估算值偏高或偏低。
7. Nervous System: Changes During an Action Potential | 神经系统：动作电位期间的变化
Explain the changes in membrane permeability of a neurone during an action potential. Refer to the roles of voltage‑gated sodium and potassium channels.
解释动作电位过程中神经元膜通透性的变化。涉及电压门控钠通道和钾通道的作用。
At resting potential (−70 mV), the membrane is polarised. A stimulus depolarises the membrane to threshold, triggering voltage‑gated Na⁺ channels to open. The membrane becomes highly permeable to Na⁺, which rushes in, making the inside more positive (depolarisation).
在静息电位 (−70 mV) 时，膜处于极化状态。刺激使膜去极化到阈值，触发电压门控 Na⁺ 通道开放。膜对 Na⁺ 通透性大增，Na⁺ 大量内流，使膜内变得更正（去极化）。
At the peak (+30 to +40 mV), Na⁺ channels inactivate and voltage‑gated K⁺ channels open. The membrane becomes highly permeable to K⁺, which diffuses out, restoring the negative internal charge (repolarisation). Sometimes the membrane briefly becomes more negative than the resting potential (hyperpolarisation) because K⁺ channels close slowly.
在电位峰值 (+30 至 +40 mV) 时，Na⁺ 通道失活，电压门控 K⁺ 通道开放。膜对 K⁺ 通透性大增，K⁺ 外流，使膜内恢复负电（复极化）。有时因 K⁺ 通道关闭较慢，膜电位会暂时比静息电位更负（超极化）。
The sodium‑potassium pump then restores the original ion distributions, using ATP to pump Na⁺ out and K⁺ in. The refractory period ensures that impulses travel in one direction.
随后，钠‑钾泵利用 ATP 将 Na⁺ 泵出、K⁺ 泵入，恢复离子分布。不应期确保神经冲动单向传导。
8. Hardy–Weinberg Equilibrium | 哈代–温伯格平衡
In a population of 500 individuals, 16% display a recessive trait caused by a recessive allele a. Assuming Hardy–Weinberg conditions, calculate the frequency of the recessive allele and the percentage of heterozygous carriers in the population.
在一个 500 人的群体中，16% 的个体表现出由隐性等位基因 a 控制的隐性性状。假设符合哈代–温伯格条件，计算隐性等位基因的频率和群体中杂合子携带者的百分比。
p + q = 1 and p² + 2pq + q² = 1
where q = frequency of allele a, q² = frequency of homozygous recessive genotype.
其中 q = 等位基因 a 的频率，q² = 隐性纯合子基因型的频率。
Given q² = 0.16, so q = √0 hen p = 1 − 0.4 = 0.6. The frequency of heterozygotes is 2pq = 2 × 0.6 × 0.4 = 0.48, or 48%. In a population of 500, approximately 240 individuals are carriers.
已知 q² = 0.16，则 q = √0.16 = 0.4。p = 1 − 0.4 = 0.6。杂合子的频率为 2pq = 2 × 0.6 × 0.4 = 0.48，即 48%。在 500 人中，约有 240 人是携带者。
The Hardy–Weinberg principle requires: no mutation, no natural selection, large population size, random mating, and no migration. Deviations from these conditions can cause microevolution.
哈代–温伯格原理的前提是：无突变、无自然选择、群体足够大、随机交配、无迁移。偏离这些条件可能导致微进化。
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