Worked Examples in A-Level Edexcel Biology | A-Level Edexcel 生物：典型例题详解

📚 Worked Examples in A-Level Edexcel Biology | A-Level Edexcel 生物：典型例题详解

Mastering A-Level Edexcel Biology requires not only memorising content but also applying knowledge to unfamiliar scenarios. This article presents a collection of typical exam-style questions with detailed step-by-step solutions. By working through these examples, you will develop your analytical skills, data interpretation and ability to structure answers clearly – all essential for top marks.

掌握A-Level Edexcel生物不仅需要记忆内容，还需将知识应用于陌生情境。本文精选一批典型考题，并提供详细的分步解析。通过这些例题的训练，你将提升分析技巧、数据解读能力和清晰组织答案的能力——这些都是夺取高分的关键。

1. Monohybrid Inheritance and Test Cross | 单基因遗传与测交

A homozygous recessive short pea plant (tt) is crossed with a pea plant of unknown genotype that shows the dominant tall phenotype. All offspring are tall. Determine the genotype of the unknown parent.

一株纯合隐性的矮生豌豆(tt)与一株表型为显性高茎但基因型未知的豌豆杂交。子代全部为高茎。判断未知亲本的基因型。

Key principle: A test cross with a homozygous recessive reveals the unknown genotype. If the unknown parent were heterozygous (Tt), we would expect a 1 : 1 tall : short ratio. Since all offspring are tall, the unknown parent must be homozygous dominant (TT).

关键原理：与纯合隐性的测交可以揭示未知基因型。若未知亲本是杂合子(Tt)，预期将出现高茎与矮茎1:1的比例。子代全为高茎，说明未知亲本必然是纯合显性(TT)。

Punnett square for Tt × tt: gametes T and t combine with t to give Tt (tall) and tt (short). The observed outcome (all tall) contradicts this; therefore the cross was TT × tt, producing only Tt offspring.

Tt × tt的庞纳特方格：配子T和t与t结合，得到Tt(高茎)和tt(矮茎)。实际观察结果(全为高茎)与此矛盾；因此杂交应为TT × tt，只产生Tt子代。

2. Sex-linked Inheritance (Haemophilia A) | 伴性遗传（血友病A）

Haemophilia A is an X-linked recessive condition. A carrier female (XHXh) and a normal male (XHY) have a son. What is the probability that this son will have haemophilia?

血友病A是一种X连锁隐性遗传病。一名携带者女性(XHXh)与一名正常男性(XHY)育有一子。该儿子患血友病的概率是多少？

Males inherit their X chromosome from the mother and the Y chromosome from the father. The mother’s gametes are XH and Xh with equal probability. The father contributes a Y chromosome to a son.

男性的X染色体来自母亲，Y染色体来自父亲。母亲的配子为XH和Xh，各占一半概率。父亲为儿子提供Y染色体。

Son’s possible genotype: XHY (normal) or XhY (haemophiliac). The probability of receiving Xh from the carrier mother is 1/2. Therefore, the probability the son has haemophilia = 50%.

儿子可能的基因型：XHY(正常)或XhY(患者)。从携带者母亲获得Xh的概率为1/2。因此，儿子患血友病的概率为50%。

3. Codominance and Multiple Alleles (ABO Blood Groups) | 共显性与复等位基因（ABO血型）

A man with blood group A (genotype IAi) and a woman with blood group B (genotype IBi) have a child. Determine the possible blood groups of their child.

一名A型血男性（基因型IAi）与一名B型血女性（基因型IBi）育有一子。判断孩子可能的血型。

IA and IB are codominant; both are expressed when present together. The i allele is recessive. Parental gametes: father produces IA and i; mother produces IB and i.

IA和IB为共显性，同时存在时二者皆表达。等位基因i为隐性。亲本配子：父亲产生IA和i；母亲产生IB和i。

Gametes	IB	i
IA	IAIB (AB)	IAi (A)
i	IBi (B)	ii (O)

Genotypic ratio: 1 IAIB : 1 IAi : 1 IBi : 1 ii. Phenotypic ratio: 1 AB : 1 A : 1 B : 1 O. The child can have any of the four blood groups, each with a 25% chance.

基因型比例：1 IAIB : 1 IAi : 1 IBi : 1 ii。表现型比例：1 AB型 : 1 A型 : 1 B型 : 1 O型。孩子可能拥有四种血型中的任意一种，每种概率为25%。

4. Enzyme Kinetics and Competitive Inhibition | 酶动力学与竞争性抑制

An enzyme-catalysed reaction was studied with varying substrate concentrations in the absence and presence of a competitive inhibitor. Sketch the expected Lineweaver–Burk plots and explain how Vmax and Km are affected.

在不同底物浓度下研究了某酶促反应，并分别在不添加和添加竞争性抑制剂条件下进行。试绘出预期的Lineweaver–Burk图，并解释Vmax与Km如何受影响。

In competitive inhibition, the inhibitor resembles the substrate and binds to the active site. Increasing substrate concentration can outcompete the inhibitor, so Vmax remains unchanged. However, a higher substrate concentration is needed to reach half Vmax, so Km (apparent) increases.

在竞争性抑制中，抑制剂与底物结构相似，结合于活性位点。增加底物浓度可以竞争性排除抑制剂，因此Vmax不变。但达到½ Vmax需要更高的底物浓度，因而表观Km值增大。

Lineweaver–Burk plot (1/V vs 1/[S]): the inhibited line rotates around the y‑axis intercept, producing the same 1/Vmax intercept but a less negative x‑intercept (‑1/Km becomes closer to zero).

Lineweaver–Burk图（1/V对1/[S]）：抑制线绕y轴截距旋转，1/Vmax截距相同，但x截距（‑1/Km）负值变小（更接近零）。

Competitive inhibition: Vmax unchanged, Km increased

竞争性抑制：Vmax不变，Km增大

5. Limiting Factors in Photosynthesis (Light Intensity) | 光合作用限制因子（光照强度）

A data table shows the rate of oxygen production by Elodea at different light intensities, under saturating CO₂ and constant temperature. Interpret the shape of the curve and identify the limiting factors in each region.

数据表展示了伊乐藻在不同光照强度下的氧气产生速率，CO₂饱和且温度恒定。解释曲线形态，并确定各区间内的限制因子。

At low light intensity, the rate increases linearly with light intensity – light is the limiting factor. As light intensity continues to rise, the rate plateaus; light is no longer limiting, and another factor (e.g. CO₂ concentration or temperature) becomes limiting.

在低光照强度下，速率随光照线性增加——此时光照是限制因子。随着光照继续增强，速率达到平台期；光照不再是限制因子，另一个因子（如CO₂浓度或温度）成为限制因子。

If the experiment were repeated at a higher CO₂ level, the plateau would rise, indicating that CO₂ had been limiting. Typical exam questions require identifying the limiting factor from a graph and predicting the effect of changing conditions.

如果在更高CO₂浓度下重复实验，平台会升高，表明原本CO₂是限制因子。典型考题要求从图中识别限制因子，并预测条件改变的影响。

6. Mark-Release-Recapture Estimating Population Size | 标记重捕法估算种群大小

In a woodland, students captured 40 woodlice, marked them with a spot of paint, and released them. The next day they captured 50 woodlice, of which 10 were marked. Estimate the population size and discuss assumptions.

在一片林地中，学生捕获了40只潮虫，用油漆斑点标记后释放。次日他们捕获了50只潮虫，其中10只带有标记。估算种群大小并讨论假设条件。

The Lincoln Index formula:

N = (M × C) / R

where M = number marked initially (40), C = number captured on second visit (50), R = number of marked recaptures (10).

林肯指数公式：

N = (M × C) / R

其中M = 初次标记数(40)，C = 第二次捕获数(50)，R = 重捕中的标记数(10)。

Plugging in: N = (40 × 50) / 10 = 200. Estimated population size = 200 woodlice.

代入：N = (40 × 50) / 10 = 200。估算种群大小为200只潮虫。

Assumptions: no migration, no births or deaths between samplings, marks are not lost and do not affect survival, marked individuals mix randomly. Violation of these leads to over‑ or underestimation.

假设条件：采样间无迁移、无出生或死亡，标记不脱落且不影响生存，标记个体随机混合。违背这些假设将导致高估或低估。

7. Cardiac Cycle and Pressure Changes | 心动周期与压力变化

The diagram shows pressure changes in the left atrium, left ventricle and aorta during one cardiac cycle. Identify the points when the atrioventricular (AV) valves close and the semilunar valves open.

图示一个心动周期中左心房、左心室和主动脉的压力变化。标出房室(AV)瓣关闭和半月瓣打开的时刻。

At the start of ventricular systole, the ventricular pressure rises rapidly and exceeds atrial pressure – the AV valves (bicuspid) close. Shortly after, ventricular pressure overcomes aortic pressure, forcing the semilunar (aortic) valve open.

心室收缩初期，心室内压迅速升高并超过心房压力——房室瓣（二尖瓣）关闭。稍后，心室压超过主动脉压，迫使半月瓣（主动脉瓣）打开。

On a pressure‑time graph, AV valve closure corresponds to the point where ventricular pressure crosses above atrial pressure. Semilunar valve opening occurs when ventricular pressure exceeds aortic pressure. These events mark the boundaries of the phases of the cardiac cycle.

在压力‑时间图上，房室瓣关闭对应心室压刚刚超过心房压的时刻。半月瓣打开则发生在心室压超过主动脉压之时。这些事件划分了心动周期的各时相界限。

8. Hardy-Weinberg Equilibrium Calculation | 哈代-温伯格平衡计算

Cystic fibrosis is caused by a recessive allele. In a population of 10,000 individuals, 1 in 2,500 is born with the disease. Estimate the frequency of the carrier genotype, assuming Hardy‑Weinberg equilibrium.

囊性纤维化由隐性等位基因引起。在一个10,000人的群体中，每2,500个新生儿中有1人患病。假设哈代‑温伯格平衡，估算携带者基因型的频率。

Let q² = frequency of homozygous recessive (disease) genotype. q² = 1/2,500 = 0.0004, so q = √0.0004 = 0.02.

设q² = 纯合隐性(患病)基因型频率。q² = 1/2,500 = 0.0004，故q = √0.0004 = 0.02。

p = 1 – q = 0.98. The carrier genotype is heterozygous (2pq). Calculate 2pq = 2 × 0.98 × 0.02 = 0.0392.

p = 1 – q = 0.98。携带者基因型为杂合子(2pq)。计算2pq = 2 × 0.98 × 0.02 = 0.0392。

Thus, approximately 3.92% of the population are carriers, equivalent to about 392 individuals in this town. The calculation assumes random mating, large population, no mutation, no selection and no gene flow.

因此，约3.92%的群体为携带者，相当于该城镇约392人。此计算假设随机交配、大群体、无突变、无选择和无基因流。

9. PCR and DNA Amplification | PCR与DNA扩增

A single molecule of double‑stranded DNA is amplified by PCR for 30 cycles. Ignoring limitations of primers and enzyme, calculate the theoretical number of DNA copies produced.

一个双链DNA分子经过PCR扩增30个循环。忽略引物和酶的限制，计算产生的理论DNA拷贝数。

Each PCR cycle doubles the amount of DNA. After n cycles, the number of copies = 2ⁿ. For n = 30, copies = 2³⁰.

每个PCR循环使DNA量翻倍。经过n个循环，拷贝数 = 2ⁿ。当n = 30时，拷贝数 = 2³⁰。

2³⁰ = 1,073,741,824 copies

2³⁰ = 1,073,741,824 个拷贝

In practice, after many cycles the amplification plateaus due to reagent depletion and enzyme denaturation. However, for theoretical exam questions, the exponential formula is expected. Real‑time PCR (qPCR) curves show the exponential phase followed by a plateau.

实际上，多次循环后因试剂耗竭和酶变性，扩增会进入平台期。但理论考题要求使用指数公式。实时PCR（qPCR）曲线展示指数期以及后续的平台期。

10. Osmosis and Water Potential | 渗透作用与水势

A plant cell with a water potential (Ψ) of –400 kPa is placed in a sucrose solution with Ψ = –200 kPa. Predict the net movement of water and describe the effect on the cell.

一个水势(Ψ)为–400 kPa的植物细胞放入Ψ = –200 kPa的蔗糖溶液中。预测水的净移动方向，并描述对细胞的影响。

Water moves from a region of higher (less negative) water potential to a region of lower (more negative) water potential. Here, –200 kPa > –400 kPa, so water will enter the cell from the solution.

水从水势较高（负值较小）的区域流向水势较低（负值较大）的区域。此处–200 kPa > –400 kPa，因此水将从溶液进入细胞。

The influx of water increases the cell’s turgor pressure. The cell will become turgid; the cell wall prevents bursting. If the solution were more negative (e.g. –600 kPa), water would leave the cell, causing plasmolysis.

水的流入使细胞膨压增大。细胞将变得硬挺；细胞壁可防止破裂。若溶液水势更负（如–600 kPa），水将离开细胞，导致质壁分离。

Using the formula Ψ = Ψs + Ψp, students may also be asked to calculate solute potential or pressure potential in a given scenario. Remember that at incipient plasmolysis, Ψp = 0.

利用公式Ψ = Ψs + Ψp，学生也可能被要求计算特定情境下的溶质势或压力势。切记在初始质壁分离时，Ψp = 0。

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Worked Examples in A-Level Edexcel Biology | A-Level Edexcel 生物：典型例题详解