Year 13 CIE Biology: Unit Test Mock Exam Walkthrough | CIE 生物单元测试模拟卷解析

📚 Year 13 CIE Biology: Unit Test Mock Exam Walkthrough | CIE 生物单元测试模拟卷解析

This article provides a comprehensive walkthrough of a mock unit test designed for Year 13 students following the CIE A Level Biology (9700) syllabus. We break down a series of exam‑style questions covering core A2 topics, including respiration, photosynthesis, neural communication, gene expression, evolution, and ecosystems. Each section analyses the mark scheme, highlights common errors, and reinforces the key concepts that examiners expect you to master.

本文为Year 13学生提供一份按照CIE A Level生物（9700）大纲设计的单元测试模拟卷的全方位解析。我们逐题拆解呼吸作用、光合作用、神经传导、基因表达、进化与生态系统等A2核心主题，结合评分标准分析常见错误，并强化考官期望你掌握的关键概念。

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1. Question 1: Oxidative Phosphorylation & Chemiosmosis | 第1题：氧化磷酸化与化学渗透

Question brief: Explain how the electron transport chain (ETC) and chemiosmosis lead to the synthesis of ATP in aerobic respiration. (8 marks)

题目要求：解释电子传递链和化学渗透如何导致有氧呼吸中ATP的合成。（8分）

To gain full marks, you must link the flow of electrons to the pumping of protons and the formation of a proton gradient, and finally to the action of ATP synthase. Start with reduced NAD and FAD delivering electrons to the ETC on the inner mitochondrial membrane.

要拿到这8分，必须将电子传递、质子泵出、质子梯度的建立以及ATP合酶的作用清晰地串联起来。从还原型NAD和FAD将电子传递给位于线粒体内膜上的电子传递链开始。

Marking points	评分要点
1. Reduced NAD (and reduced FAD) donate electrons to the ETC / are oxidised.	1. 还原型NAD（和还原型FAD）将电子提供给电子传递链 / 被氧化。
2. Electrons pass along a series of electron carriers (cytochromes) with progressively lower energy levels.	2. 电子沿着一系列能量水平逐渐降低的电子载体（细胞色素）传递。
3. Energy released from electron transfer is used to pump protons (H⁺) from the matrix into the intermembrane space.	3. 电子传递释放的能量被用来将质子（H⁺）从线粒体基质泵入膜间隙。
4. A proton gradient is established – higher H⁺ concentration in the intermembrane space than in the matrix – creating a proton motive force.	4. 质子梯度建立 —— 膜间隙中H⁺浓度高于基质 —— 形成质子动力势。
5. Protons diffuse back into the matrix through ATP synthase (chemiosmosis).	5. 质子通过ATP合酶顺浓度梯度流回基质（化学渗透）。
6. The flow of protons drives the rotation / conformational change of ATP synthase, catalysing ADP + Pᵢ → ATP.	6. 质子流驱动ATP合酶旋转 / 构象变化，催化 ADP + Pᵢ → ATP。
7. Oxygen acts as the final electron acceptor, combining with electrons and H⁺ to form water. Without oxygen, electrons back up and the ETC stops.	7. 氧气作为最终的电子受体，与电子和H⁺结合生成水。没有氧气，电子堆积，电子传递链停止。
8. All the above describes oxidative phosphorylation, producing ~28–34 ATP per glucose molecule (awarded for correct linking).	8. 以上全过程即氧化磷酸化，每分子葡萄糖产生约28–34个ATP（正确关联可得分）。

Common error: students often forget to state that the energy for proton pumping comes from the exergonic transfer of electrons, or they confuse the locations of the intermembrane space and matrix. Also, always name oxygen as the terminal electron acceptor – simply writing ‘O₂’ without stating its role may lose a mark.

常见错误：学生常忘记说明泵出质子的能量来自电子传递的放能过程，或将膜间隙与基质的位置搞混。此外，务必明确指出氧气是最终的电子受体，只写“O₂”而不说明其作用可能丢分。

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2. Question 2: Light‑Dependent and Light‑Independent Reactions | 第2题：光反应与暗反应

Question brief: Compare the locations, energy sources, and products of the light‑dependent reactions and the Calvin cycle in photosynthesis. (6 marks)

题目要求：比较光合作用中光反应和卡尔文循环的发生部位、能量来源和产物。（6分）

A structured table makes this comparison crystal clear. Focus on thylakoid membrane vs stroma, light energy captured by chlorophyll vs chemical energy (ATP and reduced NADP), and initial products such as O₂, ATP, reduced NADP vs triose phosphate and eventually glucose.

用表格作答最为清晰。重点对比类囊体膜与基质，叶绿素捕获的光能与化学能（ATP和还原型NADP），以及O₂、ATP、还原型NADP与磷酸丙糖、最终生成葡萄糖等产物。

Feature	Light‑dependent reactions	Calvin cycle (light‑independent)
Location	Thylakoid membranes of chloroplasts	Stroma of chloroplasts
部位	叶绿体类囊体膜	叶绿体基质
Energy source	Light energy absorbed by chlorophyll	Chemical energy from ATP and reduced NADP (from light‑dependent stage)
能量来源	叶绿素吸收的光能	来自光反应阶段的ATP和还原型NADP的化学能
Key products	ATP, reduced NADP, O₂ (from photolysis of water)	Triose phosphate (which can be converted to glucose, sucrose, starch, etc.); NADP and ADP + Pᵢ are regenerated
主要产物	ATP、还原型NADP、O₂（来自水的光解）	磷酸丙糖（可转化为葡萄糖、蔗糖、淀粉等）；NADP和ADP+Pᵢ被再生
Role of water	Photolysis provides electrons to replace those lost by chlorophyll, and releases H⁺ and O₂	Not directly involved
水的作用	光解提供电子以补充叶绿素失去的电子，并释放H⁺和O₂	不直接参与

Many candidates lose marks by stating that the Calvin cycle ‘produces glucose’ directly. In reality, the immediate product is triose phosphate (a 3‑carbon sugar phosphate); two triose phosphate molecules are needed to form one hexose sugar. Also, the Calvin cycle regenerates NADP and ADP for the light‑dependent reactions to use again – this interdependence is a favourite exam point.

很多考生因声称卡尔文循环直接“产生葡萄糖”而失分。实际上，直接产物是磷酸丙糖（三碳糖磷酸）；需要两个磷酸丙糖分子才能形成一分子己糖。同时，卡尔文循环会再生NADP和ADP供光反应再次使用——这种相互依存关系是常见的考点。

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3. Question 3: Action Potential Generation & Conduction | 第3题：动作电位的产生与传导

Question brief: Describe how an action potential is generated in a neurone and explain how it is propagated along the axon. (7 marks)

题目要求：描述神经元中动作电位如何产生，并解释它如何沿轴突传导。（7分）

Begin by stating the resting potential is around –70 mV, maintained by the sodium‑potassium pump and differential membrane permeability. Stimulus leads to opening of voltage‑gated Na⁺ channels, Na⁺ influx, depolarisation, and if threshold is reached (~ –55 mV), an action potential fires. Repolarisation follows via K⁺ efflux.

首先说明静息电位约为–70 mV，由钠钾泵和膜对不同离子的通透性差异维持。刺激导致电压门控Na⁺通道开放，Na⁺内流，去极化，若达到阈电位（约–55 mV），则爆发动作电位。随后K⁺外流引起复极化。

Then for propagation, emphasise that the depolarisation of one section of axon raises the membrane potential of the adjacent region to threshold, opening voltage‑gated Na⁺ channels there, thereby conducting the impulse. In myelinated axons, saltatory conduction occurs – action potentials ‘jump’ between nodes of Ranvier, dramatically increasing speed.

对于传导，要强调轴突某一段的去极化使相邻区域的膜电位升高至阈值，打开那里的电压门控Na⁺通道，从而传导冲动。在有髓鞘轴突中，发生跳跃式传导——动作电位在郎飞结之间“跳跃”，大大提高了传导速度。

Key sequence: Resting → Depolarisation (Na⁺ in) → Repolarisation (K⁺ out) → Hyperpolarisation → Refractory period

关键序列：静息 → 去极化（Na⁺内流） → 复极化（K⁺外流） → 超极化 → 不应期

Watch out for omission of the absolute and relative refractory periods. The absolute refractory period ensures unidirectional propagation and prevents overlap of action potentials; it results from the inactivation of voltage‑gated Na⁺ channels. The relative refractory period is due to continued K⁺ efflux and requires a larger‑than‑threshold stimulus. Clearly distinguish between the roles of Na⁺/K⁺ pump and the ion channels.

注意不要遗漏绝对不应期和相对不应期。绝对不应期保证了单向传导并防止动作电位叠加，它由电压门控Na⁺通道失活导致。相对不应期是由于K⁺持续外流，需要更大的刺激才能触发。需明确区分钠钾泵和离子通道的作用。

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4. Question 4: Transcription and Translation – Gene Expression | 第4题：转录与翻译——基因表达

Question brief: Outline the roles of transcription factors and post‑transcriptional modification in controlling gene expression in eukaryotes. (5 marks)

题目要求：概述转录因子和转录后修饰在控制真核生物基因表达中的作用。（5分）

Gene expression can be regulated at multiple levels. At the transcriptional level, specific transcription factors bind to promoter regions or enhancers to activate or repress RNA polymerase binding. For example, oestrogen binds to an oestrogen receptor, and the hormone‑receptor complex acts as a transcription factor for certain genes.

基因表达可在多个水平进行调控。在转录水平，特定的转录因子与启动子区域或增强子结合，激活或抑制RNA聚合酶的结合。例如，雌激素与雌激素受体结合，形成的激素‑受体复合物可作为某些基因的转录因子。

Post‑transcriptional control includes the removal of introns and splicing together of exons in pre‑mRNA to form mature mRNA. Alternative splicing allows a single gene to code for several different proteins. Additionally, mRNA may be edited, or its stability and transport from the nucleus regulated.

转录后调控包括前体mRNA中内含子的切除和外显子的剪接，形成成熟mRNA。可变剪接使一个基因能够编码多种不同的蛋白质。此外，mRNA可能被编辑，或者其稳定性和从细胞核的运输受到调节。

Level	Mechanism	Example / detail
Transcriptional	Transcription factors, promoter sequences, enhancers, epigenetic modifications (DNA methylation, histone acetylation)	Oestrogen‑receptor complex; Lac operon in prokaryotes (comparison)
Post‑transcriptional	mRNA processing (5′ capping, 3′ poly‑A tail, splicing), alternative splicing, mRNA stability	Removal of introns; alternative splicing producing different antibody domains
Translational	Regulatory proteins binding to mRNA, miRNA/siRNA interference	miRNA binding to target mRNA and blocking translation

Students often confuse introns (non‑coding, removed) with exons (coding, expressed). Make sure to use the terms ‘transcription factor’ and ‘promoter’ precisely – a common mistake is saying that transcription factors bind directly to RNA polymerase; they typically bind to DNA near the promoter and influence polymerase recruitment.

学生经常混淆内含子（非编码，被切除）和外显子（编码，被表达）。务必精确使用“转录因子”和“启动子”——一个常见错误是说转录因子直接与RNA聚合酶结合；它们通常与启动子附近的DNA结合，影响聚合酶的招募。

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5. Question 5: Natural Selection & Antibiotic Resistance | 第5题：自然选择与抗生素抗性

Question brief: Using antibiotic resistance in bacteria as an example, explain how natural selection leads to evolutionary change. (6 marks)

题目要求：以细菌抗生素抗性为例，解释自然选择如何导致进化改变。（6分）

This is a classic application. Begin by noting that genetic variation exists in bacterial populations, with some individuals possessing alleles conferring resistance (e.g., via random mutation). When an antibiotic is used, it acts as a selective pressure.

这是一个经典应用。首先指出细菌种群中存在遗传变异，一些个体携带赋予抗性的等位基因（例如通过随机突变产生）。当使用抗生素时，它构成了选择压力。

• Resistant bacteria survive and reproduce, while sensitive bacteria are killed.

  抗性细菌存活并繁殖，而敏感菌被杀死。

• The favourable alleles (resistance genes) are passed to the next generation in greater proportion.

  有利等位基因（抗性基因）以更高的比例传递给下一代。

• Over many generations, the allele frequency for antibiotic resistance increases in the population.

  经过多代，抗生素抗性等位基因频率在种群中上升。

• Thus the population evolves to become predominantly resistant. This illustrates evolution by natural selection – a change in allele frequency over time driven by differential reproductive success.

  因此种群进化为主要以抗性菌为主。这体现了自然选择驱动的进化——由差异繁殖成功率引起的等位基因频率随时间改变。

Examiners want precise language: ‘selective pressure’ (the antibiotic), ‘differential survival’, and ‘increase in allele frequency’ are essential. Also, note that mutations occur randomly and not ‘in response’ to the antibiotic. The resistance gene often resides on plasmids and can be transferred horizontally, accelerating the spread – mentioning this can secure the top band.

考官希望看到准确的用语：“选择压力”（抗生素）、“差异存活”和“等位基因频率增加”是必不可少的。还要注意突变是随机发生的，并非对抗生素的“应答”。抗性基因常位于质粒上并可水平转移，加速传播——提及这一点可获得高分。

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6. Question 6: Energy Transfer in Ecosystems | 第6题：生态系统中的能量传递

Question brief: Explain why the efficiency of energy transfer between trophic levels is typically low (around 10%), and calculate the energy available to secondary consumers from 50,000 kJ of producer energy. (4 marks)

题目要求：解释为什么营养级之间的能量传递效率通常很低（约10%），并计算从生产者能量50,000 kJ中次级消费者可获得的能量。（4分）

Energy is lost at each trophic level because not all of an organism is consumed (e.g., roots, bones), not all ingested material is digested and absorbed (lost in faeces), and a large fraction of assimilated energy is used for respiration, generating heat and kinetic energy rather than new biomass. Only the energy stored in new tissues is available to the next trophic level.

每个营养级都有能量损失：生物体并非全部被食用（如根、骨骼），摄入的物质并非全部被消化吸收（随粪便流失），且同化的能量很大一部分用于呼吸作用，产生热能和动能，而不是形成新的生物量。只有储存在新组织中的能量才可供下一营养级利用。

Calculation: Producer (50,000 kJ) → Primary consumer: 10% × 50,000 = 5,000 kJ → Secondary consumer: 10% × 5,000 = 500 kJ

计算：生产者 (50,000 kJ) → 初级消费者：10% × 50,000 = 5,000 kJ → 次级消费者：10% × 5,000 = 500 kJ

The 10% figure is an approximate rule of thumb. Exam questions often ask you to calculate energy transfer efficiency using the formula: Efficiency = (energy at higher trophic level ÷ energy at lower trophic level) × 100. Make sure you can use it both ways. A common mistake is to apply the percentage to the wrong figure, or to treat respiratory loss as being available to decomposers without acknowledging they obtain energy from dead organic matter of all trophic levels.

10%是一个近似的经验法则。考题常要求用公式：效率 = （较高营养级能量 ÷ 较低营养级能量） × 100 计算。务必能灵活运用。常见错误包括将百分比用于错误的数据，或认为呼吸损耗的能量可被分解者利用，却没有认识到分解者从各个营养级的死亡有机质中获取能量。

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7. Question 7: Homeostasis – Blood Glucose Regulation | 第7题：稳态——血糖调节

Question brief: Describe the negative feedback mechanism that regulates blood glucose concentration in mammals, including the roles of insulin and glucagon. (6 marks)

题目要求：描述哺乳动物调节血糖浓度的负反馈机制，包括胰岛素和胰高血糖素的作用。（6分）

Blood glucose concentration is maintained around 90 mg per 100 cm³ by the antagonistic actions of insulin and glucagon, secreted by β‑cells and α‑cells of the pancreatic islets of Langerhans respectively.

血糖浓度通过胰岛素和胰高血糖素的拮抗作用维持在大约90 mg/100 cm³，这两种激素分别由胰岛的β细胞和α细胞分泌。

When glucose rises above the set point (e.g., after a meal), β‑cells detect the increase and release insulin. Insulin stimulates target cells – especially hepatocytes and muscle cells – to increase uptake of glucose (via recruitment of GLUT4 transporters), convert glucose to glycogen (glycogenesis), and increase the rate of respiration using glucose. It also inhibits glycogenolysis and gluconeogenesis. These processes reduce blood glucose back to normal.

当血糖升高超过调定点（如餐后），β细胞感知到变化并释放胰岛素。胰岛素刺激靶细胞——尤其是肝细胞和肌细胞——增加葡萄糖的摄取（通过募集GLUT4转运蛋白），将葡萄糖转化为糖原（糖原生成），并提高利用葡萄糖的呼吸速率。它还抑制糖原分解和糖异生。这些过程使血糖回降到正常水平。

Conversely, when glucose falls below normal (e.g., during fasting), α‑cells secrete glucagon. Glucagon acts primarily on the liver to stimulate glycogenolysis (breakdown of glycogen to glucose) and gluconeogenesis (formation of glucose from non‑carbohydrate sources such as amino acids and glycerol). Glucose is released into the blood, restoring normal levels. This dual‑hormone negative feedback loop prevents wide fluctuations.

反之，当血糖低于正常水平（如禁食时），α细胞分泌胰高血糖素。胰高血糖素主要作用于肝脏，促进糖原分解和糖异生（从氨基酸、甘油等非碳水化合物物质形成葡萄糖）。葡萄糖释放到血液中，恢复至正常水平。这种双激素负反馈回路防止了血糖的大幅波动。

Be sure to use the term ‘negative feedback’ correctly – the response counteracts the initial change. Diagrams are helpful but if writing prose, clearly state that insulin secretion stops as glucose falls, and glucagon secretion stops as glucose rises. A frequent error is to state that insulin converts glucose to glycogen; the enzyme cascade activates glycogen synthase, but the hormone itself does not perform the chemical conversion. Link to second messengers such as cyclic AMP (cAMP) for top marks.

一定要正确使用“负反馈”这个术语——反应抵消了初始变化。画图有帮助，但如果用文字描述，需明确说明随着葡萄糖下降胰岛素分泌停止，随着葡萄糖上升胰高血糖素分泌停止。一个常见错误是说胰岛素将葡萄糖转化为糖原；实际上激素通过酶级联反应激活糖原合酶，本身并不进行化学转化。联系第二信使如环磷酸腺苷(cAMP)可获得高分。

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8. Question 8: Polymerase Chain Reaction (PCR) & Its Applications | 第8题：聚合酶链式反应（PCR）及其应用

Question brief: Outline the principles of the polymerase chain reaction (PCR) and suggest one practical application in medicine or forensics. (5 marks)

题目要求：概述聚合酶链式反应（PCR）的原理，并提及其在医学或法医学中的一个实际应用。（5分）

PCR is an in vitro technique used to amplify a specific region of DNA exponentially. It cycles through three main temperature‑dependent stages: denaturation (~94–96 °C), annealing (typically 50–65 °C), and extension (72 °C, using Taq polymerase). Primers, free DNA nucleotides, and a heat‑stable DNA polymerase are required.

PCR是一种体外扩增特定DNA片段的技术，呈指数增长。它循环三个主要的温度依赖步骤：变性（~94–96 °C）、退火（通常50–65 °C）和延伸（72 °C，使用Taq聚合酶）。需要引物、游离的脱氧核苷酸和耐热DNA聚合酶。

During denaturation, hydrogen bonds between complementary strands break, yielding single strands. During annealing, primers bind to complementary sequences on the target DNA. During extension, Taq polymerase synthesises a new complementary strand from the 3′ end of each primer. Each cycle doubles the amount of target DNA – after n cycles, there can be 2ⁿ copies.

变性时，互补链之间的氢键断裂，产生单链。退火时，引物与目标DNA上的互补序列结合。延伸时，Taq聚合酶从每个引物的3′末端合成新的互补链。每一循环使目标DNA数量加倍—— n个循环后可得到2ⁿ个拷贝。

Applications include diagnosis of genetic disorders (e.g., detecting mutations in the CFTR gene for cystic fibrosis), detection of viral RNA (after reverse transcription, RT‑PCR), DNA fingerprinting in forensic investigations using short tandem repeats (STRs), and amplification of ancient DNA samples. Be specific: ‘forensics’ is too vague; say ‘amplifying DNA from a crime scene to match a suspect’s profile via STR analysis’.

应用包括遗传病诊断（如检测囊性纤维化CFTR基因的突变）、病毒RNA的检测（经逆转录后，RT‑PCR）、利用短串联重复序列（STR）进行法医DNA指纹分析，以及古代DNA样本的扩增。务必具体：“法医学”太笼统；要说“通过STR分析扩增犯罪现场DNA，与嫌疑人图谱进行比对”。

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