📚 Year 13 Edexcel Biology: Unit Test Mock Paper Walkthrough | Edexcel 生物高三单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Year 13 Edexcel International A-Level Biology, covering Units 4, 5 and relevant practical skills. Each question is modelled after real exam style, with step-by-step explanations to help you consolidate key concepts and improve exam technique.
本文详细解析一份针对 Edexcel 国际 A-Level 生物高三的单元测试模拟卷，涵盖单元 4、5 及相关实验技能。每道题都模仿真实考试风格，配有逐步解释，帮助你巩固核心概念并提高应试技巧。
1. Light-dependent Reactions: Role of Water | 光合作用光反应：水的作用
Question: Describe the role of water in the light-dependent reactions of photosynthesis. (4 marks)
问题：描述水在光合作用光反应中的作用。（4分）
Water is split by light energy at Photosystem II in a process called photolysis.
水在光系统II中被光能分解，该过程称为光解作用。
Photolysis releases electrons, which replace those lost by the chlorophyll a (P680) reaction centre when it passes excited electrons to the electron transport chain.
光解释放出电子，用于替代叶绿素a（P680）反应中心将激发电子传递给电子传递链时失去的电子。
Protons (H⁺) are generated and released into the thylakoid lumen, building up a proton gradient that drives ATP synthesis via chemiosmosis.
产生质子（H⁺）并释放到类囊体腔内，积累质子梯度，通过化学渗透驱动 ATP 合成。
Oxygen gas (O₂) is released as a by-product, originating from the splitting of water.
氧气（O₂）作为副产品释放，来源于水的分解。
2. Calvin Cycle: Role of RuBisCO | 卡尔文循环：RuBisCO 的作用
Question: Explain the role of RuBisCO in the Calvin cycle and why it is considered the most abundant enzyme on Earth. (4 marks)
问题：解释 RuBisCO 在卡尔文循环中的作用，以及为何它被认为是地球上最丰富的酶。（4分）
RuBisCO catalyses the fixation of carbon dioxide (CO₂) to ribulose bisphosphate (RuBP), the first major step of the Calvin cycle.
RuBisCO 催化二氧化碳（CO₂）与核酮糖-1,5-二磷酸（RuBP）的固定反应，这是卡尔文循环的第一个主要步骤。
The reaction produces an unstable 6‑carbon intermediate which immediately splits into two molecules of glycerate 3‑phosphate (GP).
该反应生成不稳定的6碳中间物，随即分解为两分子甘油酸-3-磷酸（GP）。
RuBisCO has a low turnover number and can also bind oxygen (photorespiration), so plants produce very large amounts to maintain sufficient carbon fixation.
RuBisCO 的转换数低，且还能结合氧气（光呼吸），因此植物大量生产该酶以维持足够的固碳量。
It accounts for up to 50% of soluble leaf protein, making it the most abundant protein on Earth.
它可占叶片可溶性蛋白的50%，这使它成为地球上最丰富的蛋白质。
3. Aerobic Respiration: Coenzymes | 有氧呼吸：辅酶的作用
Question: Outline the roles of coenzymes NAD⁺ and FAD in aerobic respiration. (4 marks)
问题：概述辅酶 NAD⁺ 和 FAD 在有氧呼吸中的作用。（4分）
NAD⁺ and FAD act as hydrogen carriers, accepting hydrogen atoms (protons and electrons) during oxidation reactions in glycolysis, the link reaction and the Krebs cycle.
NAD⁺ 和 FAD 作为氢载体，在糖酵解、连接反应和克雷布斯循环中的氧化反应中接受氢原子（质子和电子）。
Reduced NAD (NADH) and reduced FAD (FADH₂) deliver high‑energy electrons to the electron transport chain on the inner mitochondrial membrane.
还原型 NAD (NADH) 与还原型 FAD (FADH₂) 将高能电子传递到线粒体内膜上的电子传递链。
As electrons pass along the chain, energy is released to pump protons across the membrane, generating a proton gradient for ATP synthesis by oxidative phosphorylation.
电子沿传递链传递时释放能量，将质子泵过膜，形成质子梯度，用于氧化磷酸化合成 ATP。
NAD⁺ is the main coenzyme during the link reaction and Krebs cycle, while FAD is specifically involved in one step of the Krebs cycle (succinate to fumarate), yielding fewer ATP molecules per hydrogen carrier.
NAD⁺ 是连接反应和克雷布斯循环中的主要辅酶，而 FAD 仅参与克雷布斯循环中的一个步骤（琥珀酸至延胡索酸），每个氢载体产生的 ATP 分子较少。
4. Muscle Contraction: Sliding Filament | 肌肉收缩：滑动丝模型
Question: Describe the sliding filament mechanism of muscle contraction, referring to the roles of Ca²⁺ ions and ATP. (6 marks)
问题：描述肌肉收缩的滑动丝模型，提及 Ca²⁺ 离子和 ATP 的作用。（6分）
An action potential arrives at the neuromuscular junction and depolarises the sarcolemma, spreading along T‑tubules and triggering the release of Ca²⁺ from the sarcoplasmic reticulum.
动作电位到达神经肌肉接头，使肌膜去极化，沿 T 管传导并触发肌质网释放 Ca²⁺。
Ca²⁺ binds to troponin, causing tropomyosin to move and expose the myosin‑binding sites on the actin filament.
Ca²⁺ 与肌钙蛋白结合，导致原肌球蛋白移动，暴露肌动蛋白丝上的肌球蛋白结合位点。
Myosin heads bind to actin forming cross‑bridges; ATP is hydrolysed to ADP and phosphate (Pi) to provide energy, allowing the myosin head to pivot and pull actin (power stroke).
肌球蛋白头与肌动蛋白结合形成横桥；ATP 水解为 ADP 和磷酸 (Pi) 提供能量，使肌球蛋白头摆动并拉动肌动蛋白（动力冲程）。
A new ATP molecule binds to the myosin head, causing detachment from actin; ATP is also required for the active transport of Ca²⁺ back into the sarcoplasmic reticulum to allow relaxation.
新的 ATP 分子与肌球蛋白头结合，使其与肌动蛋白分离；ATP 还用于将 Ca²⁺ 主动转运回肌质网，促使肌肉松弛。
Repeated attach‑pull‑detach cycles shorten the sarcomere, leading to muscle contraction as actin and myosin filaments slide past each other.
反复的附着‑拉动‑分离循环使肌节缩短，促使肌动蛋白与肌球蛋白丝彼此滑动，实现肌肉收缩。
5. Action Potential Propagation | 动作电位的传播
Question: Explain how a nerve impulse is propagated along a non‑myelinated axon. (4 marks)
问题：解释神经冲动如何沿无髓鞘轴突传播。（4分）
An action potential at one region of the axon causes the membrane to depolarise, opening voltage‑gated Na⁺ channels and allowing Na⁺ to enter.
轴突某一区域产生动作电位，使膜去极化，打开电压门控 Na⁺ 通道，Na⁺ 内流。
This creates a local current as positive ions spread to the adjacent resting region, depolarising that region to threshold.
这产生局部电流，正离子扩散到相邻的静息区域，使其去极化达到阈值。
Voltage‑gated Na⁺ channels in the new region open, generating a new action potential.
新区域的电压门控 Na⁺ 通道打开，产生新的动作电位。
The refractory period of the previous region (inactivated Na⁺ channels) ensures the impulse moves in one direction only.
前一区域的不应期（Na⁺ 通道失活）确保冲动仅沿一个方向传播。
This sequential depolarisation allows the action potential to travel continuously along the non‑myelinated axon.
这种顺序去极化使动作电位沿着无髓鞘轴突连续传播。
6. Synaptic Transmission | 突触传递
Question: Describe the sequence of events at a cholinergic synapse. (5 marks)
问题：描述胆碱能突触中的事件顺序。（5分）
An action potential arrives at the presynaptic knob, causing voltage‑gated Ca²⁺ channels to open and Ca²⁺ to diffuse in.
动作电位到达突触前末梢，使电压门控 Ca²⁺ 通道开放，Ca²⁺ 扩散进入。
Ca²⁺ causes synaptic vesicles containing acetylcholine (ACh) to fuse with the presynaptic membrane, releasing ACh into the synaptic cleft by exocytosis.
Ca²⁺ 导致含有乙酰胆碱 (ACh) 的突触小泡与突触前膜融合，通过胞吐将 ACh 释放到突触间隙。
ACh diffuses across the cleft and binds to nicotinic receptors on the postsynaptic membrane, opening ligand‑gated Na⁺ channels.
ACh 扩散过间隙，与突触后膜上的烟碱受体结合，打开配体门控 Na⁺ 通道。
Na⁺ influx depolarises the postsynaptic membrane, generating an excitatory postsynaptic potential (EPSP); if threshold is reached, an action potential is initiated.
Na⁺ 内流使突触后膜去极化，产生兴奋性突触后电位 (EPSP)；若达到阈值，则引发动作电位。
ACh is swiftly broken down by acetylcholinesterase to prevent continuous stimulation, and the products are re‑uptaken for recycling.
ACh 迅速被乙酰胆碱酯酶分解以防止持续刺激，分解产物被重摄取以循环利用。
7. Kidney Function: Ultrafiltration and Reabsorption | 肾脏功能：超滤与重吸收
Question: Explain how the nephron carries out ultrafiltration and selective reabsorption. (6 marks)
问题：解释肾单位如何进行超滤和选择性重吸收。（6分）
In the Bowman’s capsule, high hydrostatic pressure in the glomerular capillaries forces water, glucose, ions and urea out of the blood into the capsule, forming glomerular filtrate; large molecules such as proteins and blood cells are retained.
在鲍曼囊中，肾小球毛细血管内的高静水压迫使水、葡萄糖、离子和尿素从血液进入囊腔形成肾小球滤液；大分子如蛋白质和血细胞被截留。
The filtrate passes into the proximal convoluted tubule (PCT), where about 80% of the filtrate is reabsorbed.
滤液进入近曲小管 (PCT)，约80%的滤液在此被重吸收。
PCT cells have microvilli and many mitochondria, enabling active transport of sodium ions out of the cells, which drives co‑transport of glucose and amino acids back into the blood.
近曲小管细胞具有微绒毛和大量线粒体，能够主动转运钠离子出细胞，驱动葡萄糖和氨基酸的协同转运回血液。
Water follows by osmosis, and some solutes (e.g., urea) are partially reabsorbed by diffusion.
水通过渗透跟随，部分溶质（如尿素）通过扩散被重吸收。
The loop of Henle and distal convoluted tubule fine‑tune water and ion balance under hormonal control (ADH and aldosterone).
亨勒袢和远曲小管在激素（抗利尿激素和醛固酮）控制下精细调节水和离子平衡。
8. Specific Immune Response: B and T Cells | 特异性免疫应答：B 细胞与 T 细胞
Question: Compare the roles of B lymphocytes and T lymphocytes in the specific immune response. (6 marks)
问题：比较 B 淋巴细胞和 T 淋巴细胞在特异性免疫应答中的作用。（6分）
B cells are involved in humoral immunity; they produce antibodies that bind to specific antigens on pathogens, neutralising them or marking them for destruction.
B 细胞参与体液免疫；它们产生抗体，与病原体上的特定抗原结合，使其失活或标记以被破坏。
When a B cell encounters its complementary antigen, it is activated by helper T cells and undergoes clonal expansion, differentiating into plasma cells and memory B cells.
当 B 细胞遇到其互补抗原时，被辅助 T 细胞激活，进行克隆扩增，分化为浆细胞和记忆 B 细胞。
T cells are responsible for cell‑mediated immunity; helper T cells (CD4⁺) release cytokines that activate B cells, cytotoxic T cells and macrophages.
T 细胞负责细胞介导免疫；辅助 T 细胞 (CD4⁺) 释放细胞因子，激活 B 细胞、细胞毒性 T 细胞和巨噬细胞。
Cytotoxic T cells (CD8⁺) directly kill virus‑infected or cancerous cells by releasing perforin and granzymes, which induce apoptosis.
细胞毒性 T 细胞 (CD8⁺) 通过释放穿孔素和颗粒酶直接杀死病毒感染细胞或癌细胞，诱导细胞凋亡。
Both B and T cells produce memory cells that provide long‑term immunity, enabling a faster, stronger secondary response.
B 细胞和 T 细胞都产生记忆细胞，提供长期免疫力，产生更快更强的二次应答。
9. Primary Succession | 原生演替
Question: Describe the main stages of primary succession from bare rock to climax community. (5 marks)
问题：描述从裸岩到顶级群落的原生演替的主要阶段。（5分）
Pioneer species such as lichens and mosses colonise bare rock; they break down rock by weathering and add organic matter when they die, forming a thin, nutrient‑poor soil.
先锋物种如地衣和苔藓首先在裸岩上定居；它们通过风化分解岩石，死亡后添加有机质，形成稀薄、贫瘠的初始土壤。
These changes make the environment less hostile, allowing small grasses and herbs to establish, which out‑compete the pioneers.
这些变化使环境变得不那么严酷，允许小草和草本植物建群，它们会竞争淘汰先锋物种。
Shrubs and fast‑growing trees then colonise, increasing soil depth, nutrient content and water retention.
接着灌木和速生树木定居，增加土壤深度、养分含量和保水能力。
Over time, larger tree species dominate, forming a stable climax community with high biodiversity that is in equilibrium with the climate.
随着时间的推移，高大的树种成为优势种，形成与气候平衡的稳定顶级群落，具有高生物多样性。
The climax community persists until a significant disturbance initiates secondary succession.
顶级群落持续存在，直到重大干扰引发次生演替。
10. Genetic Drift vs Natural Selection | 遗传漂变与自然选择
Question: Explain how genetic drift differs from natural selection in shaping allele frequencies. (4 marks)
问题：解释遗传漂变与自然选择在塑造等位基因频率上有何不同。（4分）
Genetic drift is a random change in allele frequency due to chance events, not driven by environmental pressures.
遗传漂变是由于随机事件导致的等位基因频率变化，并非由环境压力驱动。
It has a more pronounced effect in small populations, where random loss or fixation of alleles can occur independently of fitness.
在小种群中影响更为显著，等位基因的随机丢失或固定与适合度无关。
Natural selection, in contrast, is a non‑random process where alleles that confer a survival or reproductive advantage increase in frequency over generations.
相比之下，自然选择是非随机过程，提供生存或繁殖优势的等位基因频率会逐代增加。
Selection leads to adaptation, whereas drift can lead to loss of genetic diversity and differentiation between populations by chance.
选择导致适应，而漂变可因偶然因素导致遗传多样性的丧失和种群间的分化。
11. Data Analysis: Chi‑squared Test | 数据分析：卡方检验
Question: A student investigated a genetic cross and observed the following phenotypes: 82 tall, 24 dwarf. The expected ratio is 3:1. Use the chi‑squared (χ²) test to determine whether the difference is significant at the 5% level. The critical value for 1 degree of freedom is 3.84. (5 marks)
问题：一名学生研究了遗传杂交，观察到以下表型：82 高茎，24 矮茎。预期比例为3:1。使用卡方（
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