Pre-U CCEA Biology: Unit Test Mock Exam Analysis | Pre-U CCEA 生物：单元测试模拟卷解析

📚 Pre-U CCEA Biology: Unit Test Mock Exam Analysis | Pre-U CCEA 生物：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for CCEA Pre-U Biology. Each question is analysed to reinforce key concepts, highlight common pitfalls, and model effective exam technique. Use this review to sharpen your understanding and boost your confidence before the real assessment.

本文逐题解析一套 CCEA Pre-U 生物单元模拟卷。通过深入剖析每个问题，强化核心概念，指出常见失分点，并示范高效答题技巧。希望你在真实考试前用这份解析巩固知识、提升信心。

1. Cell Membrane Structure and Function | 细胞膜的结构与功能

Question: Describe how the fluid mosaic model accounts for the selective permeability of the plasma membrane. Include the roles of phospholipids, cholesterol, and proteins.

问题： 说明流动镶嵌模型如何解释质膜的选择透过性。请提及磷脂、胆固醇和蛋白质的作用。

The fluid mosaic model explains that the membrane is a dynamic bilayer of phospholipids, with hydrophobic tails facing inward and hydrophilic heads outward. This arrangement restricts free passage of ions and large polar molecules, allowing only small, non-polar substances to diffuse directly through the lipid layer. Cholesterol inserts between phospholipids, reducing membrane fluidity and permeability to very small water-soluble molecules. Proteins serve as selective channels, carriers, or pumps for specific substances, enabling facilitated diffusion and active transport. Glycoproteins also contribute to cell recognition, indirectly influencing transport regulation.

流动镶嵌模型指出，生物膜由磷脂双分子层构成，疏水尾向内、亲水头向外，这种排列限制离子和大极性分子自由通过，只允许小而非极性的物质直接扩散过膜。胆固醇插在磷脂之间，降低膜的流动性，减小对极小的水溶性分子的通透性。蛋白质作为选择性的通道、载体或泵，驱动特定物质的协助扩散和主动运输；糖蛋白还参与细胞识别，间接调节跨膜运输。

2. Enzyme Kinetics and Temperature | 酶动力学与温度

Question: Explain why the rate of an enzyme-catalysed reaction rises with temperature up to an optimum, then falls sharply. Use the concept of activation energy and tertiary structure in your answer.

问题： 解释为何酶促反应速率随温度升高直至最适温度，随后急剧下降。回答需应用活化能和三级结构的概念。

As temperature increases, substrate and enzyme molecules gain kinetic energy. This leads to more frequent and forceful collisions, raising the proportion of collisions that exceed the activation energy barrier. Consequently, the rate of reaction increases up to an optimum. Beyond this point, the thermal energy disrupts the weak bonds (hydrogen bonds, ionic interactions, hydrophobic forces) maintaining the enzyme’s tertiary structure. The active site loses its complementary shape, the substrate can no longer bind, and the enzyme denatures irreversibly, causing a rapid drop in activity.

温度升高时，底物和酶分子获得更多动能，碰撞频率和力度增大，超过活化能壁垒的碰撞比例增加，因此反应速率上升至最适温度。超出最适温度后，热能破坏维持酶三级结构的弱键（氢键、离子键、疏水作用），活性位点失去互补构象，底物无法结合，酶发生不可逆变性，导致活性急剧下降。

3. DNA Replication | DNA 复制

Question: Compare the roles of DNA polymerase III and DNA ligase in the replication of the lagging strand.

问题： 比较 DNA 聚合酶 III 和 DNA 连接酶在后随链复制中的作用。

On the lagging strand, DNA polymerase III synthesises short Okazaki fragments in the 5′ to 3′ direction, using RNA primers as starting points. It adds DNA nucleotides complementary to the template strand and proofreads as it goes. DNA ligase then seals the nicks between adjacent Okazaki fragments by catalysing the formation of phosphodiester bonds, creating a continuous sugar-phosphate backbone. Unlike polymerase III, ligase does not synthesise new DNA; it only joins pre-existing fragments.

在后随链上，DNA 聚合酶 III 以 RNA 引物为起点，沿 5’→3′ 方向合成短冈崎片段，按模板链互补添加脱氧核苷酸并边合成边校读。DNA 连接酶则催化相邻冈崎片段之间形成磷酸二酯键，填补缺口，形成连续的糖-磷酸骨架。与聚合酶 III 不同，连接酶不合成新 DNA，只连接已有片段。

4. Monohybrid Inheritance and Probability | 单基因遗传与概率

Question: In pea plants, tall (T) is dominant over dwarf (t). Two heterozygous tall plants are crossed. Calculate the probability that an offspring is tall. Show your Punnett square.

问题： 在豌豆中，高茎(T)对矮茎(t)显性。两株杂合高茎豌豆杂交，计算子代为高茎的概率，并画出庞尼特方格。

Parental genotypes: Tt × Tt. The Punnett square yields combinations: TT, Tt, Tt, tt. Among the four equally probable outcomes, three (TT, Tt, Tt) produce tall phenotype. Thus, probability of tall offspring = 3/4 or 75%. The dwarf phenotype appears only when the genotype is homozygous recessive (tt).

亲本基因型: Tt × Tt。庞尼特方格显示配子组合为 TT、Tt、Tt、tt。四者概率相等，其中三种（TT, Tt, Tt）表现高茎，因此高茎概率 = 3/4 即 75%。只有基因型为纯合隐性(tt)时才表现矮茎。

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

Question: A red blood cell is placed in pure water. Describe the changes observed and explain them in terms of water potential gradients.

问题： 将红细胞置于纯水中，描述所观察到的变化，并用水势梯度解释。

Pure water has a water potential (Ψ) of zero, which is higher than the cytoplasm of a red blood cell (negative Ψ due to solutes). Water moves by osmosis from the region of higher water potential to the lower water potential, so it enters the cell. The cell swells and eventually bursts (haemolysis) because the membrane cannot withstand the increasing internal pressure. This occurs in the absence of a cell wall to resist turgor.

纯水的水势(Ψ)为零，高于红细胞细胞质的水势（因含溶质而为负值）。水通过渗透作用从高水势区域流向低水势区域，因此水进入细胞。细胞膨胀并最终破裂（溶血），因为质膜无法承受不断增加的内压。红细胞无细胞壁，故无法抵抗膨压。

6. Immune Response: B and T Cells | 免疫应答：B 细胞与 T 细胞

Question: Distinguish between the roles of B lymphocytes and T helper cells in the humoral immune response.

问题： 区分 B 淋巴细胞和辅助性 T 细胞在体液免疫应答中的作用。

B lymphocytes recognise specific antigens via surface immunoglobulins. Upon activation – often with help from T helper cells – they differentiate into plasma cells that secrete large quantities of antibodies. These antibodies neutralise pathogens and mark them for destruction. T helper cells do not produce antibodies. Instead, they bind to antigen-presenting cells displaying antigens on MHC class II molecules, become activated, and release cytokines that stimulate B cell proliferation and differentiation. Thus, T helper cells act as central regulators of the humoral response.

B 淋巴细胞通过表面免疫球蛋白识别特定抗原，被激活后（通常需辅助性 T 细胞协助）分化为浆细胞，大量分泌抗体。抗体中和病原体并标记其供清除。辅助性 T 细胞不产生抗体，而是与抗原呈递细胞表面 MHC II 类分子呈递的抗原结合，被激活后释放细胞因子，刺激 B 细胞增殖分化。因此，辅助性 T 细胞在体液免疫中起核心调控作用。

7. Photosynthesis: Light-Dependent Stage | 光合作用：光反应阶段

Question: Outline the key events of the light-dependent stage of photosynthesis, identifying where they occur within the chloroplast.

问题： 概述光合作用光反应阶段的关键事件，并指出它们在叶绿体中的发生部位。

Light-dependent reactions take place in the thylakoid membranes. Chlorophyll absorbs light energy, exciting electrons to a higher energy level. These electrons are passed along an electron transport chain, generating ATP via chemiosmosis as protons are pumped into the thylakoid space and flow back through ATP synthase. Simultaneously, water is split (photolysis) to release electrons, protons, and oxygen. The electrons replace those lost by chlorophyll, while the protons contribute to the formation of reduced NADP (NADPH). Thus, the products are ATP, reduced NADP, and O₂.

光反应发生在类囊体膜上。叶绿素吸收光能，激发电子跃迁到高能级。电子沿电子传递链传递，同时质子被泵入类囊体腔内，再通过 ATP 合酶回流，经化学渗透偶联生成 ATP。水发生光解，释放电子、质子和氧气。电子补充叶绿素失去的电子，质子参与形成还原型 NADP (NADPH)。最终产物为 ATP、还原型 NADP 和 O₂。

8. Population Genetics: Hardy-Weinberg Principle | 群体遗传学：哈代-温伯格原理

Question: In a population, 16% of individuals show a recessive phenotype (aa). Assuming Hardy-Weinberg equilibrium, calculate the frequency of the dominant allele A and the percentage of heterozygous carriers.

问题： 某群体中，16% 个体表现隐性性状(aa)。假设群体符合哈代-温伯格平衡，计算显性等位基因 A 的频率和杂合携带者的百分比。

Frequency of aa (q²) = 0.16, so q = √0.16 = 0.4. Since p + q = 1, p = 1 − 0.4 = 0.6. Frequency of heterozygous carriers (2pq) = 2 × 0.6 × 0.4 = 0.48, or 48%. Therefore, the dominant allele frequency is 0.6, and nearly half the population are carriers.

隐性纯合子频率 (q²) = 0.16，故 q = √0.16 = 0.4。由 p + q = 1 得 p = 1 − 0.4 = 0.6。杂合携带者频率 = 2pq = 2 × 0.6 × 0.4 = 0.48，即 48%。因此显性等位基因频率为 0.6，近一半个体为携带者。

9. Nervous System: Action Potential | 神经系统：动作电位

Question: Explain how voltage-gated sodium and potassium channels generate the rising and falling phases of an action potential.

问题： 解释电压门控钠通道和钾通道如何产生动作电位的上升相与下降相。

At resting potential, both channel types are closed. Depolarisation to threshold opens voltage-gated Na⁺ channels; Na⁺ rushes in, driving the membrane potential towards +40 mV (rising phase). Inactivation of Na⁺ channels and delayed opening of voltage-gated K⁺ channels then allow K⁺ to exit the cell, repolarising the membrane (falling phase). A transient hyperpolarisation (undershoot) may occur before K⁺ channels close and resting potential is restored by Na⁺/K⁺ pumps.

静息电位时，两种通道均关闭。去极化达到阈电位使电压门控 Na⁺ 通道开放，Na⁺ 快速内流，膜电位趋向 +40 mV（上升相）。随后 Na⁺ 通道失活，同时电压门控 K⁺ 通道延迟开放，K⁺ 外流使膜复极化（下降相）。在 K⁺ 通道关闭前可能出现短时超极化，最终由 Na⁺/K⁺ 泵恢复静息电位。

10. Ecological Sampling and Chi-Squared Test | 生态取样与卡方检验

Question: A student records the distribution of two plant species in two habitats using quadrats. Explain how the chi-squared test can determine whether there is a significant association between species and habitat.

问题： 学生用样方法记录两种植物在两个生境的分布。解释如何用卡方检验判断物种与生境之间是否存在显著关联。

First, the observed frequencies are entered into a contingency table. Expected frequencies are calculated assuming no association (null hypothesis): for each cell, expected = (row total × column total) / grand total. The chi-squared value is computed as χ² = Σ (O − E)² / E. Degrees of freedom = (number of rows − 1) × (number of columns − 1). Comparing the calculated χ² to a critical value at p = 0.05 determines significance. If χ² exceeds the critical value, the null hypothesis is rejected, indicating a significant association between species and habitat.

首先将观测频数填入列联表，并在零假设（无关联）前提下计算预期频数：每格预期值 = (行总和 × 列总和) / 总频数。卡方值 χ² = Σ (O − E)² / E。自由度 = (行数−1) × (列数−1)。将计算所得 χ² 与 p=0.05 的临界值比较，若大于临界值则拒绝零假设，表明物种与生境间存在显著关联。

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Pre-U CCEA Biology: Unit Test Mock Exam Analysis | Pre-U CCEA 生物：单元测试模拟卷解析