📚 OCR A-Level Biology June 2023 Paper 1 Exam Practice | OCR A-Level 生物 2023年6月 卷一 真题精练
Welcome to our focused breakdown of OCR A-Level Biology A Paper 1 (June 2023). This session presents curated exam-style questions, model answers marked against the specification, and key examiner tips to help you master Biological Processes. Each explanation is given in English first, followed by its Chinese equivalent, so you can cross-reference terminology and deepen understanding.
欢迎阅读 OCR A-Level 生物 A 卷 1（2023 年 6 月）真题精练。本文精选了试卷中的典型考题，提供符合评分标准的参考答案，并提炼出关键考点与常见失分点。每个解析点先给出英文，再配以中文，帮助双语对照复习，巩固生物过程模块的核心知识。
1. Phospholipid Bilayer and Selective Permeability | 磷脂双分子层与选择透过性
Question: Explain how the structure of the cell-surface membrane enables it to be selectively permeable. (4 marks)
问题：解释细胞表面膜的结构如何使其具有选择透过性。(4分)
The hydrophobic core of the phospholipid bilayer repels ions, charged solutes, and large polar molecules, preventing their free diffusion.
磷脂双分子层的疏水核心排斥离子、带电溶质和大分子极性物质，阻止它们自由扩散。
Small, non-polar molecules such as O₂ and CO₂ can dissolve in the fatty acid tails and pass through by simple diffusion.
小分子、非极性物质如 O₂ 和 CO₂ 可溶于脂肪酸尾部，通过简单扩散穿过膜。
Intrinsic proteins including channel and carrier proteins provide hydrophilic pathways for specific ions and polar molecules like glucose or Na⁺.
通道蛋白和载体蛋白等内在蛋白为特定离子和极性分子（如葡萄糖、Na⁺）提供了亲水通道。
Cholesterol molecules within the bilayer regulate membrane fluidity and reduce permeability to very small water-soluble molecules.
磷脂双分子层中的胆固醇分子调节膜的流动性，并降低对极微水溶性分子的通透性。
Examiner tip: Always mention the hydrophobic barrier role and give named examples of molecules that cross or are blocked.
考官提示：务必点明疏水屏障的作用，并给出能通过与被阻挡分子的具体例子。
2. Enzyme Inhibition: Competitive vs Non-Competitive | 酶抑制：竞争性与非竞争性
Question: Compare the mechanisms of competitive and non-competitive enzyme inhibition and their effects on Vmax. (3 marks)
问题：比较竞争性抑制与非竞争性抑制的机制及其对 Vmax 的影响。(3分)
A competitive inhibitor has a similar shape to the substrate; it occupies the active site, preventing substrate binding. Increasing substrate concentration can overcome this inhibition, so Vmax remains unchanged.
竞争性抑制剂与底物形状相似，占据活性位点，阻止底物结合。增加底物浓度可克服此抑制，因此 Vmax 不变。
A non-competitive inhibitor binds to an allosteric site (away from the active site), altering the enzyme’s tertiary structure so the active site is no longer complementary. This reduces the number of functional enzyme molecules, so Vmax decreases and cannot be restored by adding more substrate.
非竞争性抑制剂结合于别构部位（远离活性位点），改变酶的三级结构，使活性位点不再互补。这减少了功能性酶分子数量，因此 Vmax 降低，且无法通过增加底物恢复。
Common mistake: stating non-competitive inhibitor binds to the active site – it does not.
常见错误：声称非竞争性抑制剂结合于活性位点——事实并非如此。
3. Maths in Biology: Cardiac Output Calculation | 生物学数学：心输出量计算
Question: A person at rest has a heart rate of 72 beats min⁻¹ and a stroke volume of 70 cm³. Calculate the cardiac output and give appropriate units. (2 marks)
问题：静息状态下心率为 72 beats min⁻¹，每搏输出量为 70 cm³，计算心输出量并写出合适单位。(2分)
Cardiac output (Q) = stroke volume (SV) × heart rate (HR) → Q = 70 cm³ × 72 min⁻¹ = 5040 cm³ min⁻¹
心输出量 Q = 每搏输出量 SV × 心率 HR → Q = 70 cm³ × 72 min⁻¹ = 5040 cm³ min⁻¹
One mark is for the correct calculation, and one for the correct unit (cm³ min⁻¹ or dm³ min⁻¹ after conversion). Don’t forget to show your working.
第一分给正确计算，第二分给正确单位（cm³ min⁻¹ 或换算成 dm³ min⁻¹）。别忘了展示运算步骤。
Related skill: Be ready to rearrange the equation to find heart rate if given cardiac output and stroke volume.
关联技能：在给出心输出量和每搏输出量时，要能变换公式求出心率。
4. Haemoglobin: Bohr Shift and Oxygen Transport | 血红蛋白：波尔效应与氧气运输
Question: Describe the Bohr effect and explain how it benefits actively respiring tissues. (3 marks)
问题：描述波尔效应并解释它如何有益于旺盛呼吸的组织。(3分)
The Bohr effect is a decrease in haemoglobin’s affinity for oxygen caused by increased carbon dioxide concentration (or lower pH).
波尔效应是指由于二氧化碳浓度升高（或 pH 降低）导致血红蛋白对氧亲和力下降的现象。
In actively respiring tissues, more CO₂ is produced, which lowers the pH. This shifts the oxyhaemoglobin dissociation curve to the right, meaning haemoglobin releases more oxygen where it is most needed.
在旺盛呼吸的组织中，产生更多 CO₂，pH 降低。这使氧解离曲线右移，意味着血红蛋白在最需要氧的部位释放更多氧气。
The Bohr shift thus ensures that oxygen is unloaded precisely at sites of high metabolic activity, improving the efficiency of respiration.
因此波尔效应确保氧气在代谢活跃部位被精准卸载，提高呼吸效率。
5. Xylem Transport: Cohesion-Tension Theory | 木质部运输：内聚力-张力理论
Question: Explain how water moves up a tall tree from roots to leaves, with reference to the cohesion-tension theory. (4 marks)
问题：根据内聚力-张力理论，解释水如何从根部上升到高大树木的叶片中。(4分)
Water evaporates from mesophyll cells into air spaces during transpiration, lowering the water potential of leaf cells. This draws water out of xylem vessels, creating a tension (negative pressure) at the top of the plant.
蒸腾作用中，水从叶肉细胞蒸发到气腔，降低了叶细胞的水势，从而将木质部导管中的水拉出，在植物顶端形成张力（负压）。
The strong cohesion between water molecules, due to hydrogen bonding, allows the continuous water column to be pulled up without breaking.
水分子间因氢键产生的强大内聚力使得连续水柱在上升时不会断开。
Adhesion of water molecules to the lignin-lined xylem walls also helps to support the column.
水分子对木质化导管壁的附着力也有助于支撑水柱。
Thus, the transpiration stream is maintained by a combination of evaporation, cohesion, and adhesion, moving water and dissolved minerals upwards.
因此，蒸腾流由蒸发、内聚力和附着力共同维持，将水和溶解的矿物质向上运输。
6. Homeostasis: Blood Glucose Control | 稳态：血糖调控
Question: Outline the role of insulin and glucagon in the regulation of blood glucose concentration. (4 marks)
问题：概述胰岛素和胰高血糖素在血糖浓度调控中的作用。(4分)
When blood glucose rises (e.g. after a meal), β-cells in the islets of Langerhans secrete insulin. Insulin binds to receptors on target cells, increasing the permeability of cell membranes to glucose and activating enzymes for glycogenesis – converting glucose to glycogen for storage, mainly in liver and muscle cells.
当血糖升高（如进食后），胰岛 β 细胞分泌胰岛素。胰岛素与靶细胞受体结合，增加细胞膜对葡萄糖的通透性，并激活糖原生成酶——将葡萄糖转化为糖原储存，主要在肝细胞和肌细胞。
When blood glucose falls, α-cells secrete glucagon. Glucagon stimulates glycogenolysis (breakdown of glycogen to glucose) and gluconeogenesis (synthesis of glucose from non-carb sources such as amino acids) in the liver, releasing glucose into the blood.
当血糖降低，α 细胞分泌胰高血糖素。胰高血糖素促进肝糖原分解为葡萄糖以及糖异生（由氨基酸等非糖物质合成葡萄糖），将葡萄糖释放入血。
These negative feedback loops maintain normoglycaemia. A lack of insulin, as in type 1 diabetes, leads to hyperglycaemia.
通过这些负反馈环路维持血糖浓度正常。胰岛素缺乏（如 1 型糖尿病）会导致高血糖。
7. Kidney Function: Ultrafiltration and Reabsorption | 肾功能：超滤与重吸收
Question: Describe how the structure of the glomerulus and Bowman’s capsule facilitates ultrafiltration. (3 marks)
问题：描述肾小球和鲍曼氏囊的结构如何促进超滤。(3分)
Blood enters the glomerulus via a wider afferent arteriole and leaves via a narrower efferent arteriole, generating high hydrostatic pressure in the glomerular capillaries.
血液经较宽的入球小动脉进入肾小球，经较窄的出球小动脉离开，在肾小球毛细血管中产生高静水压。
The capillary wall is fenestrated with pores, and the inner layer of Bowman’s capsule is made of podocytes with filtration slits. Together with the basement membrane, they act as a molecular filter, allowing water, glucose, ions, and urea to pass through but retaining blood cells and large proteins.
毛细血管壁有孔，鲍曼氏囊内层由足细胞构成并带有滤过裂隙。它们与基膜共同充当分子滤器，允许水、葡萄糖、离子和尿素通过，但截留血细胞和大分子蛋白质。
The filtrate collected in Bowman’s capsule is protein-free and enters the renal tubule for selective reabsorption.
鲍曼氏囊中收集的滤液不含蛋白质，进入肾小管进行选择性重吸收。
8. Immune System: Cell-Mediated Response via T Lymphocytes | 免疫系统：T 淋巴细胞介导的细胞免疫
Question: Explain how T helper cells activate B lymphocytes in the humoral response. (3 marks)
问题：解释辅助性 T 细胞如何在体液免疫中激活 B 淋巴细胞。(3分)
T helper cells possess CD4 receptors that bind to the antigen-MHC class II complex presented by antigen-presenting cells such as macrophages.
辅助性 T 细胞拥有 CD4 受体，能与巨噬细胞等抗原提呈细胞表面的抗原-MHC II 类复合物结合。
Once activated, the T helper cell releases cytokines (interleukins) that stimulate specific B lymphocytes to undergo clonal expansion and differentiation into plasma cells and memory cells.
激活后，辅助性 T 细胞释放细胞因子（白细胞介素），刺激特定的 B 淋巴细胞进行克隆扩增并分化为浆细胞和记忆细胞。
Thus, T helper cells are essential for effective antibody production and the establishment of immunological memory.
因此，辅助性 T 细胞对有效的抗体生成和免疫记忆的建立至关重要。
9. Photosynthesis: Non-Cyclic Photophosphorylation | 光合作用：非循环光合磷酸化
Question: Outline the process of non-cyclic photophosphorylation in the light-dependent reactions. (4 marks)
问题：概述光依赖反应中非循环光合磷酸化的过程。(4分)
Light energy is absorbed by Photosystem II, exciting electrons that are passed to an electron transport chain. Photolysis of water replaces these electrons, releasing protons, electrons, and O₂.
光系统 II 吸收光能激发电子，电子传递给电子传递链。水的光解为光系统 II 补充电子，同时释放质子、电子和 O₂。
As electrons move along the electron transport chain, energy is used to pump protons into the thylakoid space, creating a proton gradient. Protons flow back through ATP synthase to generate ATP.
电子沿传递链移动时，能量用于将质子泵入类囊体腔，形成质子梯度。质子通过 ATP 合酶回流，驱动 ATP 合成。
Meanwhile, light energy absorbed by Photosystem I excites electrons that, together with protons in the stroma and NADP, form reduced NADP (NADPH).
同时，光系统 I 吸收光能激发电子，电子与基质中的质子和 NADP 结合，生成还原型 NADP (NADPH)。
This process yields ATP and NADPH but does not return electrons to PSI, hence ‘non-cyclic’.
该过程产生 ATP 和 NADPH，且电子不再返回 PSI，故称为“非循环”。
10. Respiration: Link Reaction and Krebs Cycle | 呼吸作用：连接反应与克雷布斯循环
Question: Describe the events of the link reaction and the Krebs cycle in aerobic respiration, and state the net products from one molecule of glucose. (5 marks)
问题：描述有氧呼吸中连接反应和克雷布斯循环的过程，并说明一分子葡萄糖的净产物。(5分)
For one glucose molecule, glycolysis yields 2 pyruvate. The link reaction occurs in the mitochondrial matrix: each pyruvate is decarboxylated (CO₂ removed) and oxidised by NAD, producing acetyl CoA (a 2C compound) along with 2 reduced NAD per glucose.
一分子葡萄糖经糖酵解产生 2 个丙酮酸。连接反应在线粒体基质中进行：每个丙酮酸脱羧（释放 CO₂）并被 NAD 氧化，生成乙酰辅酶 A（2C 化合物），每分子葡萄糖共产生 2 个还原型 NAD。
Acetyl CoA enters the Krebs cycle by combining with a 4C compound (oxaloacetate) to form citrate (6C). Through a series of decarboxylations, dehydrogenations, and substrate-level phosphorylation, the cycle regenerates oxaloacetate.
乙酰辅酶 A 与 4C 化合物（草酰乙酸）结合形成柠檬酸（6C），进入克雷布斯循环。经过一系列脱羧、脱氢和底物水平磷酸化反应，循环再生草酰乙酸。
Per glucose, the two turns of the Krebs cycle produce 2 ATP, 6 reduced NAD, 2 reduced FAD, and 4 CO₂.
每分子葡萄糖（两轮循环）产生 2 ATP、6 还原型 NAD、2 还原型 FAD 和 4 CO₂。
Many students lose marks by listing products for only one turn – always double for glucose.
许多考生只列出一轮循环的产物而失分——一定要对葡萄糖按两轮计算。
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