Typical CCEA A-Level Biology Worked Examples | A-Level CCEA 生物：典型例题详解

📚 Typical CCEA A-Level Biology Worked Examples | A-Level CCEA 生物：典型例题详解

Mastering CCEA A‑Level Biology requires more than memorising facts; it demands the ability to apply concepts to unfamiliar data, interpret experimental results, and perform precise calculations. This article presents a selection of typical exam‑style worked examples that cover key topics from the specification. Each example is broken down into clear steps with paired English and Chinese explanations, helping you build confidence for the exam.

掌握 CCEA A‑Level 生物学不仅仅需要记忆知识点；它还要求能将概念应用于陌生的数据、解释实验结果并进行精确计算。本文精选了一系列典型的考试风格例题，涵盖了课程大纲中的重点主题。每道例题都分解为明确的步骤，并配有中英文对照解析，帮助你增强考试信心。

1. Membrane Transport and Osmolarity | 膜转运与渗透浓度

A student placed identical pieces of potato tissue in a range of sucrose solutions. After 30 minutes, the percentage change in mass was recorded. The results are shown in the table below:

一名学生将相同大小的马铃薯组织块放入一系列不同浓度的蔗糖溶液中。30 分钟后，记录质量变化的百分比。结果如下表所示：

Sucrose concentration / mol dm⁻³	0.0	0.2	0.4	0.6	0.8	1.0
% change in mass	+12.0	+5.0	-2.0	-9.5	-16.0	-22.5

Determine the approximate water potential of the potato tissue and explain the pattern of mass change.

请确定马铃薯组织的大致水势，并解释质量变化的规律。

Step 1: Identify the point of no net water movement. The solution in which mass change is zero represents isotonic conditions. By plotting or interpolating the data, the sucrose concentration giving 0% change lies between 0.2 and 0.4 mol dm⁻³, approximately 0.35 mol dm⁻³.

步骤 1：确定无净水移动的点。质量变化为零的溶液代表等渗条件。通过绘图或内插数据，质量变化为 0% 对应的蔗糖浓度介于 0.2 与 0.4 mol dm⁻³ 之间，约为 0.35 mol dm⁻³。

Step 2: Convert concentration to water potential. Water potential of a solution (ψₛ) is given by ψₛ = -iCRT. For sucrose, i=1, R=0.00831 kPa m³ mol⁻¹ K⁻¹, T=298 K. At 0.35 mol m⁻³ (equivalent to 350 mol m⁻³), ψₛ = -1 × 350 × 0.00831 × 298 ≈ -866 kPa. Therefore, the potato water potential is approximately -866 kPa.

步骤 2：将浓度转换为水势。溶液的水势（ψₛ）由 ψₛ = -iCRT 计算。对蔗糖而言 i=1，R=0.00831 kPa m³ mol⁻¹ K⁻¹，T=298 K。在 0.35 mol dm⁻³（相当于 350 mol m⁻³）时，ψₛ = -1 × 350 × 0.00831 × 298 ≈ -866 kPa。因此，马铃薯的水势约为 -866 kPa。

Step 3: Explain the pattern. In solutions with higher water potential (less negative, lower sucrose concentration), water enters the cells by osmosis, increasing mass. In solutions with lower water potential (more negative), water leaves, decreasing mass.

步骤 3：解释变化规律。在水势较高（负值较小、蔗糖浓度较低）的溶液中，水分通过渗透作用进入细胞，质量增加。在水势较低（负值更大）的溶液中，水分流失，质量减少。

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2. Enzyme Kinetics: Michaelis-Menten Calculation | 酶动力学：米氏方程计算

An enzyme has a V_max of 120 μmol min⁻¹ and a K_m of 0.5 mM. What is the initial reaction rate when the substrate concentration is 2.0 mM? Express your answer as a percentage of V_max.

一种酶的 V_max 为 120 μmol min⁻¹，K_m 为 0.5 mM。当底物浓度为 2.0 mM 时，初始反应速率是多少？请将结果表示为 V_max 的百分比。

Step 1: Recall the Michaelis-Menten equation: v = V_max × [S] / (K_m + [S]). This allows calculation of initial reaction rate at any substrate concentration.

步骤 1：回顾米氏方程：v = V_max × [S] / (K_m + [S])。该方程可用于计算任意底物浓度下的初始反应速率。

Step 2: Substitute the values. V_max = 120 μmol min⁻¹, [S] = 2.0 mM, K_m = 0.5 mM. v = 120 × 2.0 / (0.5 + 2.0) = 240 / 2.5 = 96 μmol min⁻¹.

步骤 2：代入数值。V_max = 120 μmol min⁻¹，[S] = 2.0 mM，K_m = 0.5 mM。v = 120 × 2.0 / (0.5 + 2.0) = 240 / 2.5 = 96 μmol min⁻¹。

Step 3: Calculate the percentage of V_max. Percentage = (96 / 120) × 100% = 80%. This indicates that at 2.0 mM substrate, the enzyme is operating at 80% of its maximum capacity.

步骤 3：计算 V_max 的百分比。百分比 = (96 / 120) × 100% = 80%。这表明在 2.0 mM 底物浓度下，酶正以其最大能力的 80% 运行。

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3. DNA Semi‑Conservative Replication and Isotope Labelling | DNA 半保留复制与同位素标记

A sample of E. coli was grown for many generations in a medium containing ¹⁵N. The bacteria were then transferred to a ¹⁴N‑containing medium and allowed to replicate twice. DNA was extracted and centrifuged in a CsCl gradient. Describe the expected pattern of DNA bands and calculate the proportion of DNA molecules that contain only ¹⁴N after two generations.

一批大肠杆菌在含有 ¹⁵N 的培养基中培养了许多代。随后将细菌转移到含 ¹⁴N 的培养基中，并让其复制两次。提取 DNA 并在 CsCl 梯度中离心。描述预期的 DNA 带型，并计算在两次复制后，仅含 ¹⁴N 的 DNA 分子所占的比例。

Step 1: Understand the starting condition. After growth in ¹⁵N, all DNA molecules are ‘heavy’, with both strands containing ¹⁵N (¹⁵N—¹⁵N).

步骤 1：理解起始条件。在 ¹⁵N 中生长后，所有 DNA 分子都是“重链”，双链均含 ¹⁵N（¹⁵N—¹⁵N）。

Step 2: After the first replication in ¹⁴N medium, each heavy strand serves as a template. The new complementary strand is synthesised using ¹⁴N, so all DNA molecules become hybrid (¹⁵N—¹⁴N). Centrifugation would show a single intermediate band.

步骤 2：在 ¹⁴N 培养基中第一次复制后，每条重链作为模板。新合成的互补链使用 ¹⁴N，因此所有 DNA 分子都变成杂合链（¹⁵N—¹⁴N）。离心会显示出单一中间条带。

Step 3: After the second replication, the hybrid molecules separate. Each strand of the hybrid acts as a template. The ¹⁵N strand generates a new hybrid (¹⁵N—¹⁴N), while the ¹⁴N strand generates a light molecule (¹⁴N—¹⁴N). Thus, 50% of the molecules are hybrid and 50% are light. After two generations, 50% of the DNA molecules contain only ¹⁴N.

步骤 3：第二次复制后，杂合分子分开。杂合分子的每条链都作为模板。¹⁵N 链产生新的杂合分子（¹⁵N—¹⁴N），而 ¹⁴N 链产生轻链分子（¹⁴N—¹⁴N）。因此，50% 的分子为杂合链，50% 为轻链。经过两代后，仅含 ¹⁴N 的 DNA 分子占 50%。

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4. Protein Synthesis: Determining Polypeptide Length | 蛋白质合成：确定多肽长度

A mature mRNA molecule consists of 1200 nucleotides, including the start codon (AUG) and the stop codon (UAA). How many amino acids does the translated polypeptide contain, assuming no introns are present?

一个成熟的 mRNA 分子由 1200 个核苷酸组成，包含起始密码子 (AUG) 和终止密码子 (UAA)。假设没有内含子，翻译出的多肽含多少个氨基酸？

Step 1: Convert nucleotide count to codons. The genetic code is read in triplets, so the total number of codons is 1200 / 3 = 400 codons.

步骤 1：将核苷酸数量转换为密码子。遗传密码以三联体形式读取，所以密码子总数为 1200 ÷ 3 = 400 个密码子。

Step 2: Account for the stop codon. The stop codon does not code for an amino acid and terminates translation. Therefore, the number of amino acid‑specifying codons is 400 – 1 = 399.

步骤 2：考虑终止密码子。终止密码子不编码氨基酸并终止翻译。因此，编码氨基酸的密码子数量为 400 – 1 = 399 个。

Step 3: The start codon (AUG) codes for methionine, which is included in the polypeptide. Thus, the final polypeptide consists of 399 amino acids.

步骤 3：起始密码子 (AUG) 编码甲硫氨酸，该氨基酸包含在多肽中。因此，最终的多肽含有 399 个氨基酸。

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5. Pedigree Analysis: Autosomal Recessive Inheritance | 系谱分析：常染色体隐性遗传

The pedigree below shows the inheritance of a rare autosomal recessive condition. Affected individuals are shaded. II‑3 and II‑4 are planning a child. What is the probability that their child will be affected?

下面的系谱图显示了一种罕见的常染色体隐性遗传病的遗传情况。患者用阴影表示。II‑3 和 II‑4 正计划生育一个孩子。他们的孩子患病的概率是多少？

[Assume a simple pedigree: I‑1 and I‑2 are normal and have an affected son (II‑2). II‑1 is normal female, II‑2 affected male, II‑3 normal female (sister of affected), marries II‑4 who is normal but unrelated.]

[假设一个简单系谱：I‑1 和 I‑2 表型正常，育有患病儿子 (II‑2)。II‑1 为正常女性，II‑2 为患病男性，II‑3 为正常女性（患者的姐妹），与未患病的非近亲 II‑4 结婚。]

Step 1: Deduce genotypes of parents I‑1 and I‑2. Since they have an affected child (aa), both must be heterozygous (Aa).

步骤 1：推断亲代 I‑1 和 I‑2 的基因型。由于他们生有患病孩子 (aa)，两人必定都是杂合子 (Aa)。

Step 2: Determine the probability that II‑3 is a carrier. II‑3 is normal, so her genotype could be AA or Aa. From an Aa × Aa cross, the normal offspring ratio is 1 AA : 2 Aa. Thus, the probability that II‑3 is a carrier (Aa) is 2/3.

步骤 2：确定 II‑3 是携带者的概率。II‑3 表型正常，因此她的基因型可能是 AA 或 Aa。在 Aa × Aa 的婚配中，正常后代的比例为 1 AA : 2 Aa。所以，II‑3 为携带者 (Aa) 的概率为 2/3。

Step 3: Assess the risk from II‑4. The condition is rare, so the probability that an unrelated normal individual is heterozygous is very low; we assume II‑4 is homozygous normal (AA) unless there is evidence otherwise. Thus, the probability that II‑4 is a carrier is approximated as 0.

步骤 3：评估 II‑4 的风险。该疾病是罕见的，因此一个无亲缘关系的正常个体是杂合子的概率非常低；除非有证据表明，我们假设 II‑4 是纯合正常 (AA)。因此，II‑4 是携带者的概率近似为 0。

Step 4: Calculate the child’s risk. For the child to be affected, both parents must contribute a recessive allele. Since II‑4 is assumed AA, he cannot pass on a recessive allele. Hence the probability of an affected child is 0.

步骤 4：计算孩子患病风险。孩子要患病，父母双方均需提供隐性等位基因。由于假定 II‑4 为 AA，他不可能传递隐性等位基因。因此，孩子患病的概率为 0。

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6. Hardy-Weinberg Equilibrium: Allele and Genotype Frequencies | 哈迪‑温伯格平衡：等位基因与基因型频率

In a population of 500 individuals, 20 individuals exhibit a recessive phenotype (aa). Assuming the population is in Hardy‑Weinberg equilibrium, calculate the frequency of heterozygous individuals.

在一个 500 个体的种群中，有 20 个个体表现出隐性表型 (aa)。假设该种群处于哈迪‑温伯格平衡，计算杂合子个体的频率。

Step 1: Determine the frequency of the recessive genotype. q² = number of aa individuals / total population = 20 / 500 = 0.04.

步骤 1：确定隐性基因型频率。q² = aa 个体数 / 总种群数 = 20 / 500 = 0.04。

Step 2: Calculate the frequency of the recessive allele. q = √q² = √0.04 = 0.2.

步骤 2：计算隐性等位基因频率。q = √q² = √0.04 = 0.2。

Step 3: Calculate the frequency of the dominant allele. Since p + q = 1, p = 1 – 0.2 = 0.8.

步骤 3：计算显性等位基因频率。由于 p + q = 1，p = 1 – 0.2 = 0.8。

Step 4: Find the heterozygous frequency. The frequency of heterozygotes (2pq) is 2 × 0.8 × 0.2 = 0.32. Therefore, 32% of the population are heterozygous carriers.

步骤 4：计算杂合子频率。杂合子频率 (2pq) 为 2 × 0.8 × 0.2 = 0.32。因此，32% 的种群是杂合携带者。

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7. Ecological Efficiency: Energy Transfer in a Food Chain | 生态效率：食物链中的能量传递

In a marine ecosystem, phytoplankton fix 12,000 kJ m⁻² yr⁻¹ of energy. Zooplankton ingest 4,500 kJ m⁻² yr⁻¹ of phytoplankton energy, of which 1,500 kJ is lost in faeces, 1,200 kJ is used in respiration, and the remainder is converted to new biomass. Calculate the net production of zooplankton and the ecological efficiency between the two trophic levels.

在一个海洋生态系统中，浮游植物固定了 12,000 kJ m⁻² yr⁻¹ 的能量。浮游动物摄入了 4,500 kJ m⁻² yr⁻¹ 的浮游植物能量，其中 1,500 kJ 通过粪便流失，1,200 kJ 用于呼吸作用，剩余部分转化为新的生物量。计算浮游动物的净生产量以及这两个营养级之间的生态效率。

Step 1: Calculate the assimilated energy (A) of zooplankton. A = Ingestion (I) – loss in faeces (F). A = 4,500 – 1,500 = 3,000 kJ m⁻² yr⁻¹.

步骤 1：计算浮游动物的同化能量 (A)。A = 摄入量 (I) – 粪便流失 (F)。A = 4,500 – 1,500 = 3,000 kJ m⁻² yr⁻¹。

Step 2: Calculate net production (N). N = assimilated energy – respiratory loss (R). N = 3,000 – 1,200 = 1,800 kJ m⁻² yr⁻¹. This is the energy available to the next trophic level.

步骤 2：计算净生产量 (N)。N = 同化能量 – 呼吸消耗 (R)。N = 3,000 – 1,200 = 1,800 kJ m⁻² yr⁻¹。这是可供下一营养级利用的能量。

Step 3: Determine ecological efficiency. Efficiency = (Net production of zooplankton / Net production of phytoplankton) × 100%. Phytoplankton net production is 12,000 kJ m⁻² yr⁻¹. Efficiency = (1,800 / 12,000) × 100% = 15%.

步骤 3：确定生态效率。效率 = (浮游动物净生产量 / 浮游植物净生产量) × 100%。浮游植物净生产量为 12,000 kJ m⁻² yr⁻¹。效率 = (1,800 / 12,000) × 100% = 15%。

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8. Microscopy Calibration and Cell Measurement | 显微镜校准与细胞测量

A student calibrated an eyepiece graticule using a stage micrometer with a known scale of 10 µm per small division. At ×400 magnification, 50 divisions on the eyepiece graticule aligned with 20 divisions on the stage micrometer. The student then measured the diameter of a cheek cell at 12 eyepiece graticule divisions. Calculate the actual diameter of the cell in micrometres.

一名学生使用已知每小格 10 µm 的物镜测微尺校准目镜测微尺。在 ×400 放大倍数下，目镜测微尺的 50 格与物镜测微尺的 20 格对齐。然后，该学生测量得一个口腔上皮细胞直径为目镜测微尺的 12 格。计算该细胞的实际直径（以微米计）。

Step 1: Determine the actual length corresponding to one eyepiece graticule division. First, find the total actual length of the aligned stage micrometer divisions: 20 divisions × 10 µm/division = 200 µm.

步骤 1：确定目镜测微尺每格对应的实际长度。首先，计算对齐的物镜测微尺格数的总实际长度：20 格 × 10 µm/格 = 200 µm。

Step 2: Calculate the calibration factor. This total length of 200 µm corresponds to 50 eyepiece graticule divisions. Therefore, 1 eyepiece division = 200 µm / 50 = 4 µm.

步骤 2：计算校准因子。200 µm 的总长度对应 50 个目镜测微尺格。因此，1 个目镜格 = 200 µm ÷ 50 = 4 µm。

Step 3: Measure the cell. The cell spans 12 divisions on the eyepiece graticule, so its actual diameter = 12 × 4 µm = 48 µm.

步骤 3：测量细胞。细胞占据目镜测微尺的 12 格，因此其实际直径 = 12 × 4 µm = 48 µm。

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