📚 Aerobic Respiration: Exam-style Practice | 有氧呼吸：真题精练
Aerobic respiration is a central metabolic pathway that powers nearly every eukaryotic cell. In A‑level biology, exam questions frequently probe the four stages, the role of coenzymes, chemiosmotic ATP synthesis, and the interpretation of respirometer data. This article combines concise topic revision with authentic exam‑style questions and model answers, helping you build both conceptual clarity and exam technique.
有氧呼吸是几乎所有真核细胞供能的核心代谢途径。在 A‑level 生物考试中，真题经常围绕四个阶段、辅酶的作用、化学渗透合成 ATP 以及呼吸计数据的分析展开。本文结合精炼的知识点与真实的真题风格问题和参考答案，帮你同时夯实概念和提升解题能力。
1. Overall Equation and Cellular Location | 总反应式与细胞定位
Aerobic respiration can be summarised by the equation: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy (as ATP). While the net yield is about 30–32 ATP per glucose, the process is stepwise, distributed across the cytoplasm and the mitochondrion.
有氧呼吸可用总反应式概括：C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + 能量（以 ATP 形式）。每分子葡萄糖净生成约 30–32 ATP，但过程是分步进行的，分布在细胞质和线粒体中。
Exam-style Question: List the four major stages of aerobic respiration and identify the precise location of each within the cell. (4 marks)
真题模拟：列举有氧呼吸的四个主要阶段，并逐一指出在细胞内的确切位置。（4分）
Model Answer: Glycolysis – cytoplasm; Link reaction – mitochondrial matrix; Krebs cycle – mitochondrial matrix; Oxidative phosphorylation – inner mitochondrial membrane (cristae).
参考答案：糖酵解——细胞质；连接反应——线粒体基质；克雷布斯循环——线粒体基质；氧化磷酸化——线粒体内膜（嵴）。
2. Glycolysis: Cytosolic Splitting of Glucose | 糖酵解：葡萄糖的胞质分解
Glycolysis breaks one 6‑carbon glucose (C₆H₁₂O₆) into two molecules of 3‑carbon pyruvate (CH₃COCOO⁻). It consists of two phases: energy investment (phosphorylation of glucose, using 2 ATP) and energy pay‑off (oxidation of triose phosphate, yielding 4 ATP and 2 NADH). The net products per glucose are 2 ATP (substrate‑level phosphorylation), 2 NADH, and 2 pyruvate. Importantly, glycolysis does not require oxygen.
糖酵解将一分子 6 碳的葡萄糖（C₆H₁₂O₆）分解为两分子 3 碳的丙酮酸（CH₃COCOO⁻）。它分为两个阶段：能量投入期（葡萄糖磷酸化，消耗 2 ATP）和能量回报期（磷酸丙糖氧化，产生 4 ATP 和 2 NADH）。每分子葡萄糖的净产物为 2 ATP（底物水平磷酸化）、2 NADH 和 2 丙酮酸。重要的是，糖酵解不需要氧气。
Exam-style Question: Explain why the net gain of ATP in glycolysis is 2, despite 4 ATP being produced. (2 marks)
真题模拟：尽管糖酵解中产生了 4 个 ATP，但净增 ATP 为 2，请解释原因。（2分）
Model Answer: Two ATP are used to phosphorylate glucose and fructose‑6‑phosphate during the energy investment phase. These are subtracted from the four produced in the energy pay‑off phase, giving a net gain of 2 ATP.
参考答案：在能量投入期，有两个 ATP 被用于磷酸化葡萄糖和果糖‑6‑磷酸。从能量回报期产生的四个 ATP 中减去这两个，净增益为 2 ATP。
3. Link Reaction: From Pyruvate to Acetyl‑CoA | 连接反应：丙酮酸到乙酰辅酶A
In the mitochondrial matrix, pyruvate undergoes oxidative decarboxylation. Each pyruvate (3C) loses a carbon as CO₂, is oxidised by NAD⁺ to NADH, and combines with coenzyme A to form acetyl‑CoA (2C). Thus, per glucose, two link reactions occur, producing 2 acetyl‑CoA, 2 NADH, and releasing 2 CO₂. No ATP is made directly here.
在线粒体基质中，丙酮酸经历氧化脱羧。每个丙酮酸（3C）失去一个碳以 CO₂ 形式释放，被 NAD⁺ 氧化为 NADH，并与辅酶 A 结合生成乙酰辅酶 A（2C）。因此，每分子葡萄糖发生两次连接反应，生成 2 乙酰辅酶 A、2 NADH，并释放 2 CO₂。此阶段不直接产生 ATP。
Exam-style Question: State precisely what happens to the carbon atoms originally present in one glucose molecule by the end of the link reaction. (3 marks)
真题模拟：准确说明从一分子葡萄糖到最后完成连接反应时，其原有的碳原子去向如何。（3分）
Model Answer: Glucose has 6 carbons. During glycolysis, 2 pyruvate (3C each) are formed. Each pyruvate then loses one carbon as CO₂ in the link reaction. Therefore, at the end of the link reaction, 2 CO₂ have been released, and 4 carbons remain in the 2 molecules of acetyl‑CoA (2C each).
参考答案：葡萄糖有 6 个碳。糖酵解中生成 2 个丙酮酸（各 3C）。每个丙酮酸在连接反应中失去一个碳成为 CO₂。因此，连接反应结束时共释放 2 CO₂，剩余 4 个碳存在于 2 个乙酰辅酶 A（各 2C）中。
4. The Krebs Cycle: Complete Oxidation of Acetyl‑CoA | 克雷布斯循环：乙酰辅酶A的彻底氧化
Acetyl‑CoA (2C) combines with a 4‑carbon acceptor (oxaloacetate) to form citrate (6C). A series of decarboxylations and oxidations then regenerate oxaloacetate. Per turn, the cycle yields 3 NADH, 1 FADH₂, 1 ATP (via substrate‑level phosphorylation), and 2 CO₂. Since two acetyl‑CoA enter the cycle per glucose, the total products are 6 NADH, 2 FADH₂, 2 ATP, and 4 CO₂.
乙酰辅酶 A（2C）与 4 碳受体草酰乙酸结合生成柠檬酸（6C）。通过一系列脱羧和氧化反应重新生成草酰乙酸。每一轮循环产生 3 NADH、1 FADH₂、1 ATP（底物水平磷酸化）和 2 CO₂。由于每分子葡萄糖有两个乙酰辅酶 A 进入循环，总产物为 6 NADH、2 FADH₂、2 ATP 和 4 CO₂。
Exam-style Question: Name the 4‑carbon molecule that accepts acetyl‑CoA and explain why it is considered to be ‘regenerated’ in the cycle. (2 marks)
真题模拟：指出接受乙酰辅酶 A 的 4 碳分子名称，并解释为何说它在循环中被“再生”。（2分）
Model Answer: Oxaloacetate. It is regenerated at the end of each turn of the cycle, thereby allowing the cycle to continue accepting new acetyl‑CoA molecules.
参考答案：草酰乙酸。它在每一轮循环结束时重新生成，从而使循环能够继续接受新的乙酰辅酶 A 分子。
5. Role of NADH and FADH₂ as Electron Carriers | NADH 与 FADH₂ 作为电子载体的作用
The reduced coenzymes NADH and FADH₂ store energy harvested during glycolysis, the link reaction and the Krebs cycle. They donate high‑energy electrons to the electron transport chain on the inner mitochondrial membrane. NADH donates electrons to Complex I, while FADH₂ donates to Complex II. This difference means that fewer H⁺ are pumped for FADH₂, ultimately yielding fewer ATP.
还原型辅酶 NADH 和 FADH₂ 储存了糖酵解、连接反应和克雷布斯循环中捕获的能量。它们向线粒体内膜上的电子传递链提供高能电子。NADH 将电子传递给复合物 I，而 FADH₂ 传递给复合物 II。这一差异意味着 FADH₂ 跨膜泵送的 H⁺ 较少，最终生成的 ATP 也较少。
Exam-style Question: Explain why less ATP is produced per FADH₂ compared with per NADH. (3 marks)
真题模拟：解释为何每个 FADH₂ 产生的 ATP 少于每个 NADH。（3分）
Model Answer: FADH₂ donates electrons to Complex II rather than Complex I. Since Complex II does not pump H⁺ across the membrane, fewer protons are pumped into the intermembrane space when FADH₂ is oxidised. A smaller proton gradient drives chemiosmosis, hence fewer ATP are synthesised per FADH₂.
参考答案：FADH₂ 将电子传递给复合物 II 而不是复合物 I。因为复合物 II 不跨膜泵送 H⁺，所以氧化 FADH₂ 时泵入膜间隙的质子较少。较小的质子梯度驱动化学渗透，因此每个 FADH₂ 合成的 ATP 较少。
6. Electron Transport Chain and Proton Pumping | 电子传递链与质子泵送
The electron transport chain (ETC) consists of a series of carrier proteins (Complexes I–IV) and mobile carriers (ubiquinone, cytochrome c) embedded in the inner mitochondrial membrane. As electrons pass from NADH and FADH₂ through the complexes, energy released is used to pump H⁺ from the matrix into the intermembrane space, creating an electrochemical gradient. Molecular oxygen (O₂) acts as the final electron acceptor, combining with electrons and H⁺ to form H₂O. Without O₂, electrons back up, stopping the chain.
电子传递链（ETC）由一系列嵌入线粒体内膜的载体蛋白（复合物 I–IV）和移动载体（泛醌、细胞色素 c）组成。当电子从 NADH 和 FADH₂ 经过复合物传递时，释放的能量用于将 H⁺ 从基质泵入膜间隙，形成电化学梯度。分子氧（O₂）作为最终电子受体，与电子和 H⁺ 结合生成 H₂O。没有 O₂，电子会堆积，使传递链停止。
Exam-style Question: Describe the role of cytochrome c in the electron transport chain. (2 marks)
真题模拟：描述细胞色素 c 在电子传递链中的作用。（2分）
Model Answer: Cytochrome c is a mobile electron carrier that transfers electrons from Complex III to Complex IV. It picks up an electron at Complex III and shuttles it to Complex IV, contributing to proton pumping at Complex IV.
参考答案：细胞色素 c 是一种移动电子载体，将电子从复合物 III 传递到复合物 IV。它在复合物 III 处接受电子并将其运送到复合物 IV，促进复合物 IV 的质子泵送。
7. Chemiosmosis and ATP Synthase | 化学渗透与ATP合酶
The H⁺ gradient (proton motive force) forces H⁺ to flow back into the matrix through the enzyme ATP synthase. This flow (chemiosmosis) causes the rotor‑like portion of ATP synthase to spin, driving the synthesis of ATP from ADP and inorganic phosphate (Pᵢ). This process is called oxidative phosphorylation and accounts for the vast majority of ATP produced in aerobic respiration.
H⁺ 梯度（质子动力势）迫使 H⁺ 通过 ATP 合酶流回基质。这种流动（化学渗透）使 ATP 合酶的转子部分旋转，驱动由 ADP 和无机磷酸（Pᵢ）合成 ATP。此过程称为氧化磷酸化，占有了氧呼吸产生的绝大多数 ATP。
Exam-style Question: Define the term ‘chemiosmosis’ and link it directly to ATP synthesis. (2 marks)
真题模拟：定义“化学渗透”并直接联系 ATP 合成进行说明。（2分）
Model Answer: Chemiosmosis is the diffusion of H⁺ ions down their electrochemical gradient through ATP synthase. This proton flow drives the conformational changes in ATP synthase that catalyse the phosphorylation of ADP to ATP.
参考答案：化学渗透是 H⁺ 沿电化学梯度通过 ATP 合酶的扩散过程。这一质子流驱动 ATP 合酶的构象变化，催化 ADP 磷酸化为 ATP。
8. Net ATP Yield and Theoretical vs. Actual Values | 净ATP产量与理论值对比实际值
The theoretical maximum ATP yield from one glucose molecule is often quoted as 36–38, derived from 10 NADH and 2 FADH₂. In current A‑level specifications, the accepted figure is typically around 30–32 ATP, because: NADH generated in glycolysis cannot directly enter the mitochondrion and must shuttle electrons (costing ATP), some H⁺ leak back across the membrane, and the exact number of H⁺ pumped per NADH and FADH₂ is not a whole‑number ratio.
一分子葡萄糖的理论最大 ATP 产量通常被引用为 36–38，由 10 NADH 和 2 FADH₂ 推算而来。在现行 A‑level 大纲中，公认的数字通常约为 30–32 ATP，原因是：糖酵解产生的 NADH 不能直接进入线粒体，需要通过穿梭传递电子（消耗 ATP），部分 H⁺ 会漏回膜内，且每个 NADH 和 FADH₂ 泵送的 H⁺ 数目并非整数比。
Exam-style Question: Give two reasons why the actual ATP yield from aerobic respiration is lower than the theoretical maximum. (2 marks)
真题模拟：给出有氧呼吸实际 ATP 产量低于理论最大值的两个原因。（2分）
Model Answer: 1) Some H⁺ leak across the inner mitochondrial membrane without passing through ATP synthase. 2) The energy from the proton gradient is used for other processes, such as transporting pyruvate and ADP into the mitochondrion. (Also accept: the glycerol phosphate shuttle reduces the number of ATP per cytosolic NADH.)
参考答案：1）部分 H⁺ 未经过 ATP 合酶而漏过线粒体内膜。2）质子梯度的能量被用于其他过程，如运输丙酮酸和 ADP 进入线粒体。（同样接受：磷酸甘油穿梭降低了胞质 NADH 对应的 ATP 数量。）
9. Oxygen as the Final Electron Acceptor | 氧气作为最终电子受体
Oxygen is essential for aerobic respiration because it has the highest electronegativity of the components of the ETC, allowing it to accept electrons at the end of the chain. Without O₂, all electron carriers remain reduced, the ETC halts, and no H⁺ gradient is generated. Anaerobic respiration or fermentation becomes the only ATP source.
氧气对有氧呼吸不可或缺，因为它在 ETC 各组分中具有最高的电负性，能够在链末端接受电子。没有 O₂，所有电子载体都将处于还原态，ETC 停止，不再形成 H⁺ 梯度。此时只能依赖无氧呼吸或发酵提供 ATP。
Exam-style Question: Predict the consequence for ATP synthesis if cyanide, which inhibits Complex IV, is applied to respiring mitochondria. (3 marks)
真题模拟：如果向正在呼吸的线粒体施用能抑制复合物 IV 的氰化物，预测对 ATP 合成的影响。（3分）
Model Answer: Cyanide blocks Complex IV, preventing electron transfer to O₂. As a result, electrons cannot be passed along the ETC, so no H⁺ are pumped. The proton gradient dissipates, chemiosmosis stops, and oxidative phosphorylation ceases. ATP synthase cannot produce ATP, so the cell must rely only on glycolysis for ATP.
参考答案：氰化物阻断复合物 IV，阻止电子传递给 O₂。因此电子无法沿 ETC 传递，没有 H⁺ 被泵出。质子梯度消散，化学渗透停止，氧化磷酸化终止。ATP 合酶无法生成 ATP，细胞只能依赖糖酵解产生 ATP。
10. Respiratory Substrates and Respiratory Quotient (RQ) | 呼吸底物与呼吸商
Cells can oxidise carbohydrates, lipids and proteins. The respiratory quotient (RQ) is the ratio of CO₂ produced to O₂ consumed per unit time. RQ for carbohydrates ≈ 1.0, for lipids ≈ 0.7, for proteins ≈ 0.9. RQ values provide clues about the type of substrate being respired and are frequently tested with respirometer data.
细胞可以氧化碳水化合物、脂类和蛋白质。呼吸商（RQ）是单位时间内产生的 CO₂ 与消耗的 O₂ 的体积比。碳水化合物的 RQ ≈ 1.0，脂类 ≈ 0.7，蛋白质 ≈ 0.9。RQ 值能提示正在被呼吸的底物类型，常与呼吸计数据结合考察。
Exam-style Question: A respirometer shows that a germinating seed consumes 25 cm³ of O₂ and produces 22.5 cm³ of CO₂ in 30 minutes. Calculate the RQ and suggest the likely respiratory substrate. (3 marks)
真题模拟：呼吸计显示萌发种子在 30 分钟内消耗 25 cm³ O₂，产生 22.5 cm³ CO₂。计算 RQ 并推测可能的呼吸底物。（3分）
Model Answer: RQ = CO₂ produced / O₂ consumed = 22.5/25 = 0.9. An RQ of 0.9 suggests that the seed is respiring a mixture of substrates, likely including some protein or lipid alongside carbohydrate.
参考答案：RQ = 产生的 CO₂/消耗的 O₂ = 22.5/25 = 0.9。RQ 为 0.9 表明种子正在呼吸混合底物，很可能蛋白质或脂类与碳水化合物同时被利用。
11. Interpreting Respirometer Data and Experimental Design | 呼吸计数据分析与实验设计
A‑level exams often require students to explain the set‑up of a respirometer, including the role of the potassium hydroxide (KOH) solution to absorb CO₂, the use of a manometer to measure pressure changes, and the importance of a thermostatically controlled water bath. Questions also test your ability to calculate respiration rates from pressure or volume changes.
A‑level 考试常要求学生解释呼吸计的装置，包括氢氧化钾（KOH）溶液吸收 CO₂ 的作用、使用压力计测量压力变化，以及恒温水浴的重要性。同时考察根据压力或体积变化计算呼吸速率的能力。
Exam-style Question: A respirometer contains KOH. After 20 minutes, the coloured liquid in the manometer moved 15 mm towards the respirometer chamber. The capillary tube has an internal diameter of 1.0 mm. Calculate the volume of oxygen consumed per minute. (3 marks) (Volume of cylinder = πr² × length; π ≈ 3.14)
真题模拟：一个呼吸计装有 KOH。20 分钟后，压力计中的有色液体向呼吸室方向移动了 15 mm。毛细管的内径为 1.0 mm。计算每分钟消耗的氧气体积。（3分）（圆柱体积 = πr² × 长度；π ≈ 3.14）
Model Answer: Radius = 0.5 mm. Volume change = π × (0.5)² × 15 = 3.14 × 0.25 × 15 = 11.775 mm³. This volume of O₂ was consumed in 20 min, so rate = 11.775 ÷ 20 = 0.58875 mm³ min⁻¹ ≈ 0.59 mm³ min⁻¹.
参考答案：半径 = 0.5 mm。体积变化 = π × (0.5)² × 15 = 3.14 × 0.25 × 15 = 11.775 mm³。此体积的 O₂ 在 20 min 内被消耗，速率 = 11.775 ÷ 20 = 0.58875 mm³ min⁻¹ ≈ 0.59 mm³ min⁻¹。
12. Common Misconceptions and Exam Tips | 常见误区与应试技巧
Many students confuse substrate‑level phosphorylation with oxidative phosphorylation. Remember: substrate‑level phosphorylation occurs in glycolysis and the Krebs cycle when an enzyme transfers a phosphate group directly to ADP. Oxidative phosphorylation requires the ETC and chemiosmosis. Also, do not write that ‘energy is produced’ – instead state that ‘ATP is synthesised’ or ‘chemical energy is transferred to ATP’.
许多学生混淆底物水平磷酸化与氧化磷酸化。记住：底物水平磷酸化发生在糖酵解和克雷布斯循环中，由酶直接将磷酸基团转移给 ADP。氧化磷酸化需要 ETC 和化学渗透。此外，不要写“产生能量”，而应写“合成 ATP”或“化学能转移至 ATP”。
Exam-style Question: A student writes: ‘The Krebs cycle produces energy.’ Identify the error and rewrite the statement correctly. (2 marks)
真题模拟：某学生写道：“克雷布斯循环产生能量。”指出错误并正确重写该句。（2分）
Model Answer: The statement is incorrect because energy is not produced – it is conserved. Correct version: ‘The Krebs cycle transfers chemical energy to reduced coenzymes (NADH and FADH₂) and synthesises a small amount of ATP by substrate‑level phosphorylation.’
参考答案：该表述错误，因为能量不会被创造，而只是被转化。正确说法：“克雷布斯循环将化学能转移至还原型辅酶（NADH 和 FADH₂），并通过底物水平磷酸化合成少量 ATP。”
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