📚 Year 13 CIE Biology: Unit Test Simulated Paper Analysis | CIE 生物单元测试模拟卷解析
This article provides a detailed analysis of a simulated unit test for Year 13 CIE Biology, covering core A2 topics such as cell signalling, gene expression, immunity, neurophysiology, homeostasis, photosynthesis, respiration, genetics, speciation and biotechnology. Each section presents a typical exam-style question followed by mark-scheme key points and explanatory comments, with English–Chinese paired paragraphs to consolidate understanding and exam technique.
本文详细解析一份为 CIE A Level 生物 Year 13 设计的单元测试模拟卷，涵盖细胞信号、基因表达、免疫、神经生理、稳态、光合作用、呼吸作用、遗传学、物种形成和生物技术等核心 A2 主题。每个小节展示一道典型考题，提供评分方案要点和解析，采用中英双语分段对应阐述，帮助夯实概念并熟悉答题方法。
1. Cell Signalling: Role of Cyclic AMP | 细胞信号：环腺苷酸的作用
The question asks you to explain how adrenaline binding to a receptor on a liver cell leads to increased glycogenolysis. The mark scheme expects: adrenaline binds to the β-adrenergic receptor, activating a G-protein; the G-protein stimulates adenylyl cyclase, which converts ATP into cyclic AMP (cAMP); cAMP acts as a second messenger, activating protein kinase A; protein kinase A phosphorylates and activates phosphorylase kinase, which in turn activates glycogen phosphorylase; glycogen phosphorylase catalyses the breakdown of glycogen to glucose-1‑phosphate, raising blood glucose concentration.
题目要求解释肾上腺素与肝细胞受体结合后如何促进糖原分解。评分要点包括：肾上腺素与β‑肾上腺素能受体结合，激活 G 蛋白；G 蛋白激活腺苷酸环化酶，将 ATP 转化为环腺苷酸 (cAMP)；cAMP 作为第二信使，激活蛋白激酶 A；蛋白激酶 A 磷酸化并激活磷酸化酶激酶，后者再激活糖原磷酸化酶；糖原磷酸化酶催化糖原分解为葡糖‑1‑磷酸，最终升高血糖浓度。
A common pitfall is confusing the roles of G-protein and adenylyl cyclase. Remember that cAMP is the intracellular secondary messenger that amplifies the hormone signal, triggering a cascade of phosphorylation events. CIE examiners also expect you to state that the system is a signal transduction pathway, showing how a hydrophilic hormone can exert an effect without entering the cell.
常见错误是混淆 G 蛋白与腺苷酸环化酶的功能。要牢记 cAMP 是胞内第二信使，可将激素信号放大，启动磷酸化级联反应。CIE 考官还要求明确说明该系统属于信号转导途径，体现了亲水性激素无需进入细胞即可发挥作用的特点。
2. Gene Expression: Control of the lac Operon | 基因表达：乳糖操纵子的调控
A typical question asks you to outline how the lac operon is regulated in the presence and absence of lactose. Key points: when lactose is absent, the lac repressor protein binds to the operator, blocking RNA polymerase and preventing transcription of the structural genes β‑galactosidase, permease and transacetylase. When lactose is present, it is converted to allolactose, which acts as an inducer by binding to the repressor and changing its shape so it cannot bind to the operator; consequently, RNA polymerase transcribes the structural genes, producing the enzymes needed for lactose metabolism. A bonus mark can be gained by mentioning that the cAMP–CAP complex (catabolite activator protein) positively regulates the operon when glucose is scarce.
典型考题要求概述乳糖存在与否时乳糖操纵子的调控。答案要点：无乳糖时，lac 阻遏蛋白结合于操纵基因，阻碍 RNA 聚合酶，阻止结构基因（β‑半乳糖苷酶、透性酶和乙酰转移酶）的转录。有乳糖时，乳糖转变为别乳糖，作为诱导物结合阻遏蛋白并改变其构象，使阻遏蛋白无法结合操纵基因；于是 RNA 聚合酶转录结构基因，产生代谢乳糖所需的酶。加分点在于提及当葡萄糖缺乏时，cAMP–CAP 复合物（代谢激活蛋白）对操纵子进行正调控。
To earn full marks, be precise with terminology such as ‘operator’, ‘repressor protein’ and ‘inducer’. Do not simply say ‘lactose switches on the gene’, but describe the molecular detail. The lac operon is a classic example of negative control and is frequently examined in CIE papers.
获得满分的关键在于术语准确，例如“操纵基因”、“阻遏蛋白”和“诱导物”。不要只说“乳糖开启基因”，而应详述分子机制。乳糖操纵子是负调控的经典实例，在 CIE 考试中出现频率极高。
3. T Lymphocytes and Cell‑Mediated Immunity | T 淋巴细胞与细胞免疫
You may be asked to describe the role of cytotoxic T lymphocytes (Tc cells) in defence against virus‑infected cells. The mark scheme includes: antigen‑presenting cells (e.g. dendritic cells) display viral peptides on MHC class I molecules; helper T cells (Th) that recognise the same antigen are activated and secrete cytokines; these cytokines activate specific Tc cells with complementary T‑cell receptors; activated Tc cells recognise the foreign peptide–MHC I complex on infected body cells and release perforin and granzymes; perforin forms pores in the target cell membrane, allowing granzymes to enter and induce apoptosis, thereby destroying the infected cell.
考题可能要求描述细胞毒性 T 淋巴细胞 (Tc 细胞) 在防御病毒感染细胞中的作用。评分方案包括：抗原提呈细胞（如树突状细胞）通过 MHC I 类分子呈递病毒肽段；识别同一抗原的辅助 T 细胞 (Th) 被活化并分泌细胞因子；这些细胞因子激活携带互补 T 细胞受体的特异性 Tc 细胞；活化的 Tc 细胞识别受感染细胞表面的外源肽–MHC I 复合物，释放穿孔素和颗粒酶；穿孔素在靶细胞膜上形成孔道，颗粒酶进入并诱导细胞凋亡，从而清除感染细胞。
Make sure you distinguish between humoral and cell‑mediated responses. The Tc cell response is cell‑mediated, directly attacking host cells. Memory T cells are also formed for long‑term immunity, a point examiners like to see.
务必区分体液免疫与细胞免疫。Tc 细胞应答属于细胞免疫，直接攻击宿主细胞。此外，记忆 T 细胞也会形成以获得长期免疫，这一点常是考官的加分项。
4. The Action Potential and Saltatory Conduction | 动作电位与跳跃传导
A structured question on nerve impulses might ask you to explain propagation of an action potential along a myelinated axon. Top‑scoring answers state: at resting potential the axonal membrane is polarised (~‑70 mV); a threshold depolarisation opens voltage‑gated Na⁺ channels, so Na⁺ rushes in (rising phase); K⁺ channels open more slowly, K⁺ efflux restores the membrane potential (falling phase). The local current generated by Na⁺ entry depolarises the adjacent section of the axon, triggering the next action potential. Myelin sheaths of Schwann cells insulate the axon, so ionic exchanges occur only at the nodes of Ranvier; the action potential ‘jumps’ from node to node – saltatory conduction – increasing speed significantly without needing an increase in axon diameter.
神经冲动结构化问题可能要求解释动作电位在髓鞘化轴突上的传播。高分答案应包含：静息电位时轴突膜处于极化状态（约‑70 mV）；去极化达到阈电位时，电压门控 Na⁺ 通道开放，Na⁺ 快速内流（上升相）；K⁺ 通道开放较慢，K⁺ 外流使膜电位恢复（下降相）。Na⁺ 内流形成的局部电流使轴突相邻区域去极化，触发下一个动作电位。施万细胞形成的髓鞘具有绝缘作用，离子交换只发生在郎飞结处；动作电位从一个结“跳跃”至下一个结——跳跃传导——可在不增加轴突直径的情况下大幅提升传导速度。
Examiners look for accurate use of terms such as ‘voltage‑gated’, ‘threshold’ and ‘saltatory conduction’. Emphasise the roles of the myelin sheath and nodes of Ranvier, as this is a key application of the core concept in CIE biology.
考官看重“电压门控”“阈电位”“跳跃传导”等术语的准确使用。重点强调髓鞘与郎飞结的作用，这是 CIE 生物中核心概念的重要应用。
5. Homeostatic Control of Blood Glucose | 血糖的稳态调控
A question on homeostasis frequently asks for an outline of negative feedback control of blood glucose. Marks are given for: rise in blood glucose detected by β‑cells in the islets of Langerhans, which secrete insulin; insulin binds to receptors on liver and muscle cells, stimulating glucose uptake, glycogenesis (glucose → glycogen) and increased respiration; fall in blood glucose below the set‑point is detected by α‑cells, which secrete glucagon; glucagon stimulates glycogenolysis (glycogen → glucose) and gluconeogenesis in the liver, returning glucose to normal. The system is a classic negative feedback loop because the response counteracts the initial change.
稳态考题常要求概述血糖的负反馈调节。得分点包括：血糖升高被胰岛中的 β 细胞感知，分泌胰岛素；胰岛素与肝细胞和肌肉细胞上的受体结合，促进葡萄糖摄取、糖原合成（葡萄糖→糖原）并增强呼吸作用；血糖降至调定点以下时，α 细胞感知并分泌胰高血糖素；胰高血糖素促进肝糖原分解（糖原→葡萄糖）和糖异生，使血糖恢复至正常水平。该系统为典型的负反馈回路，因为响应方向与初始变化相反。
In CIE exams, be specific about cell types (α‑ and β‑cells), target organs and the names of processes. Stating merely ‘insulin lowers blood glucose’ is not sufficient; you must describe how. Drawing a simple flow diagram in your revision is helpful for linking these events.
CIE 考试中，需明确细胞类型（α 和 β 细胞）、靶器官及各过程名称。仅说“胰岛素降低血糖”是不够的，必须描述方式。复习时画出简单流程图有助于将这些事件串联起来。
6. The Calvin Cycle in Photosynthesis | 光合作用中的卡尔文循环
Expect a question asking you to describe the three stages of the Calvin cycle. Mark scheme essentials: (1) Carboxylation – CO₂ combines with ribulose bisphosphate (RuBP) to form two molecules of glycerate‑3‑phosphate (GP), catalysed by Rubisco. (2) Reduction – GP is phosphorylated by ATP and reduced by reduced NADP to form triose phosphate (TP). (3) Regeneration – most TP is used to regenerate RuBP, with ATP required; some TP leaves the cycle to be converted into glucose and other organic molecules. The cycle is light‑independent but requires the products of the light‑dependent reactions, ATP and reduced NADP.
考题可能要求描述卡尔文循环的三个阶段。评分方案要点：(1) 羧化阶段——CO₂ 与核酮糖‑1,5‑二磷酸 (RuBP) 结合，形成两分子甘油酸‑3‑磷酸 (GP)，由 Rubisco 催化。(2) 还原阶段——GP 被 ATP 磷酸化并被还原型 NADP 还原，形成磷酸丙糖 (TP)。(3) 再生阶段——大部分 TP 用于再生 RuBP，此过程需 ATP；部分 TP 离开循环，转化为葡萄糖及其他有机物。该循环虽不直接需光，但依赖光反应产生的 ATP 和还原型 NADP。
Students often lose marks by forgetting to mention ATP and reduced NADP in the right places or by confusing GP and TP. Remember that the Calvin cycle must turn a total of three times to produce one net TP for synthesis of half a hexose sugar. Practising the stoichiometry helps.
学生常因未在恰当步骤提及 ATP 和还原型 NADP 或混淆 GP 与 TP 而失分。牢记卡尔文循环需运转三次才能净生成一分子 TP，用以合成半分子六碳糖。练习化学计量关系有助于掌握。
7. Oxidative Phosphorylation in Respiration | 呼吸作用中的氧化磷酸化
A question on oxidative phosphorylation expects you to describe how the electron transport chain (ETC) and chemiosmosis generate ATP. Key points: reduced NAD and reduced FAD donate electrons to the ETC on the inner mitochondrial membrane; as electrons pass along a series of carriers (complexes I–IV), energy is released to pump protons (H⁺) from the matrix into the intermembrane space, creating an electrochemical gradient; this proton motive force drives H⁺ back through ATP synthase (chemiosmosis), causing the enzyme to rotate and catalyse the synthesis of ATP from ADP and Pi; oxygen acts as the final electron acceptor, combining with electrons and H⁺ to form water, thereby maintaining the flow of electrons along the chain.
有关氧化磷酸化的题目要求描述电子传递链 (ETC) 和化学渗透如何生成 ATP。关键点：还原型 NAD 和还原型 FAD 将电子提供给线粒体内膜上的 ETC；电子经复合体 I–IV 传递时释放的能量将质子 (H⁺) 从基质泵入膜间隙，形成电化学梯度；该质子动力驱动 H⁺ 通过 ATP 合酶回流（化学渗透），使酶旋转并催化 ADP 与 Pi 合成 ATP；氧气作为最终电子受体，结合电子和 H⁺ 生成水，维持电子沿链的流动。
Always clarify the distinction between glycolysis, the Krebs cycle and oxidative phosphorylation. CIE examiners frequently target the role of oxygen and the compartmentalisation within the mitochondrion. Using a labelled diagram in your answer can convey the sequence elegantly, but you must still explain it in words.
务必区分糖酵解、克雷布斯循环和氧化磷酸化。CIE 考官常涉及氧气的作用及线粒体的区室化结构。答题时附上标注简洁的图示很有效，但仍需用文字加以阐释。
8. Genetic Linkage and Recombination Frequency | 遗传连锁与重组频率
A typical test item provides data from a test cross involving two linked genes and asks you to calculate the recombination frequency and deduce the map distance. You must: identify recombinant phenotypes (different from the parental combinations); count the recombinant progeny and divide by the total number of offspring, then multiply by 100 to give a percentage; 1% recombination frequency equals 1 map unit (centimorgan, cM). You may also be asked to explain how crossing over during prophase I of meiosis produces recombinant chromatids, and why double crossovers can lead to underestimation of map distance for genes far apart.
典型试题会提供两个连锁基因测交的数据，要求计算重组频率并推出图距。必须：识别重组表型（与亲本组合不同）；计数重组子代数，除以总子代数，乘以 100 得到百分比；1% 的重组频率相当于 1 个图距单位（厘摩，cM）。也可能要求解释减数分裂前期 I 的交叉互换如何产生重组染色单体，以及双交换为何会导致相距较远的基因图距被低估。
The formula and concept of independent assortment versus linkage are frequently examined. Remember that genes on the same chromosome do not assort independently unless separated by crossing over. Any remaining ambiguity in a calculation usually signals a need to check for double‑crossover events in a three‑point cross.
自由组合与连锁的公式和概念常被考查。记住同一染色体上的基因除非发生互换分离，否则不表现为自由组合。如果计算结果有疑义，往往需要考虑三点测交中的双交换事件。
9. Allopatric Speciation | 异地物种形成
You might need to describe how allopatric speciation results in the formation of new species. The mark scheme would credit: a physical barrier (e.g. a mountain range, river) splits a population, preventing gene flow between the two sub‑populations; different environmental conditions or genetic drift cause the separate gene pools to diverge in allele frequencies; natural selection acts on heritable variations, favouring different adaptations; over many generations, genetic differences accumulate; eventually, even if the barrier is removed, the populations can no longer interbreed to produce fertile offspring – reproductive isolation has occurred, defining the formation of new species.
考题可能要求描述异地物种形成如何导致新物种的产生。评分方案认可：物理屏障（如山脉、河流）分隔种群，阻止两个亚群间的基因交流；不同的环境条件或遗传漂变使得隔离的基因库在等位基因频率上发生分歧；自然选择作用于可遗传变异，选择不同的适应特征；经多代累积，遗传差异加大；最终即使屏障消失，种群也无法交配产生可育后代，生殖隔离形成，标志着新物种的诞生。
Make sure you distinguish allopatric (geographical) speciation from sympatric speciation. CIE often asks for a named example such as Darwin’s finches or island‑based species. Concluding with ‘reproductive isolation’ is essential for full marks.
务必区分异地（地理）物种形成与同地物种形成。CIE 常要求举出实例，如达尔文雀或岛屿物种。答案以“生殖隔离”结尾是获得满分的关键。
10. Polymerase Chain Reaction (PCR) | 聚合酶链式反应 (PCR)
Examiners frequently ask for a description of the PCR process, including the roles of temperature. Model answer: (1) Denaturation at approximately 95 °C breaks hydrogen bonds, separating the double‑stranded DNA into two single strands. (2) Annealing at 50–60 °C allows short DNA primers to bind by complementary base pairing to the ends of the target sequence. (3) Extension at 72 °C, the optimum temperature for Taq DNA polymerase, which synthesises new DNA strands by adding free nucleotides to the 3′ ends of the primers. Each cycle doubles the amount of target DNA, yielding an exponential increase. Taq polymerase is thermostable, thus not denatured at the high denaturation temperature.
考题常要求描述 PCR 过程，包括温度的作用。标准答案：(1) 变性阶段于约 95 °C 进行，打破氢键，使双链 DNA 解离为两条单链。(2) 退火阶段于 50–60 °C 进行，让短 DNA 引物通过互补碱基配对与目标序列末端结合。(3) 延伸阶段于 72 °C 进行，是 Taq DNA 聚合酶的最适温度，该酶在引物 3′ 端添加游离核苷酸合成新 DNA 链。每轮循环使目标 DNA 数量加倍，呈指数增长。Taq 聚合酶具有热稳定性，在高温变性步骤中不会失活。
To gain maximum credit, name the enzyme as Taq polymerase and explicitly state it is obtained from the thermophilic bacterium Thermus aquaticus. Knowing the direction of synthesis (5′ to 3′) and the role of primers as starting points for DNA synthesis demonstrates a more thorough understanding.
赢得高分的要点：明确写出酶名 Taq 聚合酶，并指出其来源于嗜热细菌 Thermus aquaticus。掌握合成方向（5′→3′）以及引物作为 DNA 合成起点的作用，能体现更深入的理解。
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