📚 A-Level WJEC Biology: Typical Exam Question Walkthrough | A-Level WJEC 生物：典型例题详解
Welcome to this targeted WJEC A-Level Biology revision guide. Through ten carefully constructed example questions covering microscopy, enzymes, osmosis, genetics, ecology, circulation, neurobiology, photosynthesis, molecular biology and energetics, you will learn how to break down typical exam problems, apply key formulas and structure your answers for maximum marks.
欢迎阅读这篇专门针对 WJEC A-Level 生物学的复习指南。我们精心设计了十道典型例题，涵盖显微镜测量、酶学、渗透作用、遗传学、生态学、循环系统、神经生物学、光合作用、分子生物学和能量平衡。你将学会如何剖析典型考题、运用核心公式并规范作答，从而在考试中稳操胜券。
1. Example 1: Calculating Magnification and Actual Size from Microscope Images | 例题1：根据显微图像计算放大倍数与实际大小
A student observed a human cheek epithelial cell using a light microscope with an eyepiece magnification of ×10 and an objective lens of ×40. The image of the cell on the graticule measured 3.6 mm in diameter. Calculate the actual diameter of this cell in micrometres (µm). Show your working.
一名学生使用目镜放大倍数为 ×10、物镜放大倍数为 ×40 的光学显微镜观察人口腔上皮细胞。测微尺显示该细胞的影像直径为 3.6 mm。计算该细胞的实际直径，以微米（µm）表示，并写出计算过程。
Total magnification = Eyepiece magnification × Objective magnification
总放大倍数 = 目镜放大倍数 × 物镜放大倍数
10 × 40 = 400
Actual size = Image size ÷ Total magnification
实际大小 = 影像大小 ÷ 总放大倍数
Actual size = 3.6 mm ÷ 400 = 0.009 mm
Convert millimetres to micrometres: 1 mm = 1000 µm. So, 0.009 mm = 0.009 × 1000 = 9 µm.
将毫米换算为微米：1 mm = 1000 µm，因此 0.009 mm = 0.009 × 1000 = 9 µm。
The actual diameter of the cheek cell is 9 µm. Always remember to use the correct units and to show the conversion step clearly.
该口腔上皮细胞的实际直径为 9 µm。务必使用正确单位并清晰呈现换算步骤。
2. Example 2: Interpreting Enzyme Kinetics and Inhibition Data | 例题2：解读酶动力学与抑制作用数据
An investigation measured the initial rate of an enzyme-catalysed reaction at different substrate concentrations, both without an inhibitor and in the presence of a fixed concentration of inhibitor X. The key results are summarised in the table below.
某实验测定了不存在抑制剂以及存在固定浓度抑制剂 X 的条件下，不同底物浓度对应的酶促反应初始速率。主要结果归纳于下表中。
| Substrate concentration / mmol dm⁻³ | Rate without inhibitor / a.u. | Rate with inhibitor X / a.u. |
|---|---|---|
| 0.2 | 20 | 10 |
| 0.5 | 40 | 20 |
| 1.0 | 58 | 30 |
| 2.0 | 70 | 38 |
| 4.0 | 75 | 42 |
| 8.0 | 78 | 45 |
Use the data to identify the type of inhibition displayed by inhibitor X. Explain your reasoning with reference to the apparent Km and Vmax.
利用上述数据判断抑制剂 X 的抑制类型。参考表观 Km 与 Vmax 解释你的判断依据。
In the absence of inhibitor, the maximum velocity (Vmax) is approached at around 78 a.u. In the presence of inhibitor X, the maximum rate reached is about 45 a.u., showing that Vmax is significantly reduced. However, the substrate concentration needed to reach half of the uninhibited Vmax – the apparent Km – remains similar; half of 45 a.u. (~22.5) is reached at approximately 0.5 mmol dm⁻³, which is close to the half-rate concentration for the uninhibited reaction (half of 78 a.u. ≈ 39, reached near 0.5 mmol dm⁻³). This pattern – a lower Vmax with no major change in Km – is characteristic of non-competitive inhibition. The inhibitor binds not to the active site but to an allosteric site, reducing the number of functional enzyme molecules but not affecting the affinity of remaining active sites for the substrate.
在无抑制剂时，最大反应速率（Vmax）趋近于 78 a.u.。而存在抑制剂 X 时，最高速率仅约 45 a.u.，说明 Vmax 显著下降。然而，达到无抑制反应 Vmax 一半（约 39 a.u.）所需的底物浓度与达到有抑制 Vmax 一半（约 22.5 a.u.）所需的浓度均约为 0.5 mmol dm⁻³，即表观 Km 几乎未变。这种 Vmax 降低而 Km 基本不变的规律是非竞争性抑制的典型特征。抑制剂并非与活性位点结合，而是结合于别构部位，减少了功能酶分子的数量，但不影响余下活性位点对底物的亲和力。
If inhibitor X were competitive, we would see an increased apparent Km while Vmax remained unchanged. The data clearly show the opposite.
若抑制剂 X 为竞争性抑制剂，则表观 Km 会增大而 Vmax 不变，但数据明显呈现相反规律。
3. Example 3: Osmosis and Determining Water Potential | 例题3：渗透作用与细胞水势的测定
A student placed potato cylinders of equal mass into a series of sucrose solutions ranging from 0.0 mol dm⁻³ to 0.8 mol dm⁻³. After 60 minutes, she recorded the percentage change in mass. The cylinder in 0.35 mol dm⁻³ sucrose showed no net change in mass. The water potential (ψ) of 0.35 mol dm⁻³ sucrose at 20 °C is known to be −860 kPa. What is the water potential of the potato tissue? Explain the result in terms of water movement and ψ.
一名学生将等质量的土豆圆柱分别浸泡在一系列浓度从 0.0 mol dm⁻³ 到 0.8 mol dm⁻³ 的蔗糖溶液中。60 分钟后记录质量变化的百分比。其中处于 0.35 mol dm⁻³ 蔗糖液中的圆柱体质量无净变化。已知该浓度蔗糖溶液在 20 °C 下的水势（ψ）为 −860 kPa。请问该土豆组织的水势是多少？从水分移动与水势角度解释这一结果。
At equilibrium, when there is no net movement of water, the water potential of the potato tissue equals the water potential of the surrounding solution. Therefore, the potato tissue has a water potential of −860 kPa.
当水分净移动为零时，体系达到渗透平衡，此时土豆组织的水势与外界溶液的水势相等。因此，该土豆组织的水势为 −860 kPa。
Water always moves from a region of higher (less negative) water potential to a region of lower (more negative) water potential. In solutions more dilute than 0.35 mol dm⁻³, the external water potential is higher than −860 kPa, so water enters the potato cells, causing the mass to increase. In solutions more concentrated than 0.35 mol dm⁻³, the external water potential is lower, so water leaves the cells, and mass decreases. The point of incipient plasmolysis for these cells corresponds closely to the water potential of the 0.35 mol dm⁻³ solution.
水分总是从水势较高（负值较小）的区域移向水势较低（负值较大）的区域。在比 0.35 mol dm⁻³ 更稀的溶液中，外界水势高于 −860 kPa，水分子进入土豆细胞，导致质量增加；在更高浓度溶液中，外界水势更低，水分子离开细胞，质量减少。这些细胞的初始质壁分离点非常接近 0.35 mol dm⁻³ 溶液的水势。
ψ (potato) = ψ (solution at no net mass change) = −860 kPa
4. Example 4: Genetic Crosses and Probability (ABO Blood Groups) | 例题4：遗传杂交与概率计算（ABO 血型系统）
A man with blood group A, whose mother had blood group O, marries a woman with blood group B, whose father had blood group O. The ABO blood group system is controlled by three alleles: Iᴬ, Iᴮ, and i. Alleles Iᴬ and Iᴮ are codominant, and both are dominant to i. What are the possible blood groups of their children and the probability of each?
一名血型为 A 型的男性（其母亲血型为 O 型）与一名血型为 B 型的女性（其父亲血型为 O 型）结婚。ABO 血型系统由三个等位基因控制：Iᴬ、Iᴮ 和 i。其中 Iᴬ 与 Iᴮ 为共显性，且两者对 i 均为显性。他们的子女可能出现哪些血型，各自的概率是多少？
The man has blood group A but must have inherited an i allele from his group O mother. Therefore his genotype is Iᴬi. The woman has blood group B must have inherited an i allele from her group O father, so her genotype is Iᴮi.
该男性为 A 型血，但其母亲为 O 型（ii），他必定从母亲处获得了一个 i 等位基因，故其基因型为 Iᴬi。该女性为 B 型血，其父亲为 O 型，她从父亲处获得了 i 基因，故基因型为 Iᴮi。
Cross: Iᴬi × Iᴮi. Construct a Punnett square:
杂交组合：Iᴬi × Iᴮi。构建庞纳特方格：
| Gametes | Iᴮ | i |
|---|---|---|
| Iᴬ | Iᴬ Iᴮ (group AB) | Iᴬ i (group A) |
| i | Iᴮ i (group B) | ii (group O) |
Each genotype occurs with a 25% probability. Thus the possible blood groups are: AB (25%), A (25%), B (25%) and O (25%).
每种基因型的出现概率均为 25%。因此子女可能的血型及概率为：AB 型（25%）、A 型（25%）、B 型（25%）和 O 型（25%）。
5. Example 5: Estimating Population Size Using Mark-Release-Recapture | 例题5：运用标记重捕法估算种群数量
In a meadow habitat, a biologist captured 80 woodlice in her first sample. She marked them with a dot of non-toxic paint and released them back into the habitat. Two days later, she captured a second sample of 100 woodlice and found that 20 of them carried the paint mark. Estimate the total population size of woodlice in that meadow and state one assumption made when using this method.
在一片草地生境中，一位生物学家第一次取样捕获了 80 只鼠妇，用无毒颜料标记后放回。两天后她第二次取样捕获了 100 只鼠妇，发现其中 20 只带有标记。请估算这片草地鼠妇种群的总数量，并说明使用该方法的一个前提假设。
The Lincoln index formula:
林肯指数公式：
N = (M × C) / R
where M = number of individuals marked in first sample (80), C = total number caught in second sample (100), and R = number of marked individuals recaptured (20).
其中 M = 首次标记数（80），C = 第二次捕获总数（100），R = 重捕中带标记的个体数（20）。
N = (80 × 100) ÷ 20 = 8000 ÷ 20 = 400
Estimated population size = 400 woodlice. A key assumption is that the marked individuals have mixed completely and randomly with the rest of the population between samples, and that marking does not affect survival or probability of recapture. Other assumptions include a closed population with no births, deaths, immigration or emigration.
估算的种群数量为 400 只鼠妇。一个关键假设是：两次取样之间，标记个体已与种群其他成员充分、随机地混合，且标记行为不影响存活率或重捕概率。其他假设还包括种群封闭，即没有出生、死亡、迁入或迁出。
6. Example 6: Cardiac Output and Blood Pressure Calculations | 例题6：心输出量与血压的相关计算
At rest, an athlete has a stroke volume of 80 cm³ and a heart rate of 60 beats per minute. During intense exercise, her heart rate rises to 180 bpm and her stroke volume increases to 130 cm³. Calculate her cardiac output (CO) both at rest and during exercise. Express your answers in dm³ min⁻¹. Explain how the increase in CO affects mean arterial blood pressure.
某运动员静息时每搏输出量为 80 cm³，心率为 60 次/分钟。剧烈运动时其心率升至 180 次/分钟，每搏输出量增至 130 cm³。分别计算其静息时与运动时的心输出量（CO），结果以 dm³ min⁻¹ 表示。并解释心输出量上升如何影响平均动脉血压。
Cardiac output = Stroke volume × Heart rate. First, convert cm³ to dm³: 1 dm³ = 1000 cm³.
心输出量 = 每搏输出量 × 心率。首先将 cm³ 换算为 dm³：1 dm³ = 1000 cm³。
CO (rest) = 80 cm³ × 60 min⁻¹ = 4800 cm³ min⁻¹ = 4.8 dm³ min⁻¹
CO (exercise) = 130 cm³ × 180 min⁻¹ = 23400 cm³ min⁻¹ = 23.4 dm³ min⁻¹
Mean arterial pressure (MAP) = Cardiac output × Total peripheral resistance. During exercise, despite a drop in total peripheral resistance due to vasodilation in muscles, the massive increase in cardiac output typically causes a moderate rise in MAP. This ensures adequate perfusion of active tissues.
平均动脉压（MAP）= 心输出量 × 总外周阻力。运动时，虽然骨骼肌血管舒张导致总外周阻力下降，但心输出量的大幅上升通常仍会使得平均动脉压适度升高，从而保证活跃组织的充足灌注。
7. Example 7: Nerve Impulse Transmission and the Action Potential | 例题7：神经冲动传导与动作电位
Explain the changes in membrane permeability to Na⁺ and K⁺ ions that produce the depolarisation and repolarisation phases of an action potential. Describe how the refractory period ensures unidirectional propagation of the impulse.
解释动作电位去极化与复极化阶段中，膜对 Na⁺ 和 K⁺ 的通透性变化。说明不应期如何确保神经冲动单向传导。
At resting potential (−70 mV), voltage-gated Na⁺ channels are closed and voltage-gated K⁺ channels are mostly closed. Upon stimulus, some Na⁺ channels open; if threshold is reached, many voltage-gated Na⁺ channels open, allowing Na⁺ to rush in. This depolarises the membrane to about +40 mV. Then, Na⁺ channels inactivate, and voltage-gated K⁺ channels open, allowing K⁺ to efflux, which repolarises the membrane. A slight overshoot (hyperpolarisation) occurs before the resting potential is restored by the Na⁺/K⁺ pump.
静息电位（约 −70 mV）时，电压门控 Na⁺ 通道关闭，电压门控 K⁺ 通道也大部分关闭。刺激使部分 Na⁺ 通道开放；一旦达到阈值，大量电压门控 Na⁺ 通道开放，
Published by TutorHao | A-Level Biology Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply