A-Level Biology: Cellular Respiration Complete Guide | A-Level 生物：细胞呼吸完整指南

Cellular respiration is one of the most fundamental processes in biology — and one of the most heavily examined topics in A-Level Biology. Whether you’re studying AQA, OCR, Edexcel, or CIE, a deep understanding of how cells convert glucose into ATP is essential for top marks. This comprehensive guide covers every stage, from glycolysis to oxidative phosphorylation, in clear bilingual format.

细胞呼吸是生物学中最基础的过程之一，也是 A-Level 生物考试中最常出现的主题。无论你学习的是 AQA、OCR、Edexcel 还是 CIE 考试局，深入理解细胞如何将葡萄糖转化为 ATP 都是获得高分的关键。这篇完整指南以清晰的双语格式涵盖了从糖酵解到氧化磷酸化的每一个阶段。

1. What Is Cellular Respiration? | 什么是细胞呼吸？

Cellular respiration is the metabolic pathway that breaks down glucose to release energy in the form of ATP (adenosine triphosphate). The overall equation for aerobic respiration is:

细胞呼吸是分解葡萄糖以释放能量（以 ATP 即三磷酸腺苷的形式）的代谢途径。有氧呼吸的总方程式为：

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + Energy (≈38 ATP)

This process occurs in four main stages: Glycolysis, the Link Reaction, the Krebs Cycle, and Oxidative Phosphorylation (which includes the Electron Transport Chain). Each stage takes place in a specific location within the cell and involves distinct enzymes, coenzymes, and intermediate molecules.

这个过程分为四个主要阶段：糖酵解、连接反应、克雷布斯循环以及氧化磷酸化（包括电子传递链）。每个阶段发生在细胞内的特定位置，涉及不同的酶、辅酶和中间分子。

2. Glycolysis — The Universal First Step | 糖酵解——通用的第一步

Location: Cytoplasm | 位置：细胞质

Glycolysis is the only stage of respiration that occurs in the cytoplasm. It does not require oxygen, making it the universal energy-releasing pathway found in virtually all living organisms — from bacteria to humans.

糖酵解是唯一在细胞质中进行的呼吸阶段。它不需要氧气，因此是几乎所有生物体（从细菌到人类）中都存在的通用能量释放途径。

The Process | 过程

Glycolysis converts one molecule of glucose (a 6-carbon sugar) into two molecules of pyruvate (a 3-carbon compound). The process can be divided into two phases:

糖酵解将一个葡萄糖分子（6碳糖）转化为两个丙酮酸分子（3碳化合物）。该过程可分为两个阶段：

1. Phosphorylation (Energy Investment Phase) | 磷酸化（能量投入阶段）: Glucose is phosphorylated using 2 ATP molecules to form hexose bisphosphate. This makes the glucose more reactive and prevents it from leaving the cell. The hexose bisphosphate is then split into two molecules of triose phosphate (TP).
2. Oxidation (Energy Payoff Phase) | 氧化（能量产出阶段）: Each triose phosphate is oxidised to pyruvate. During this oxidation, NAD⁺ is reduced to NADH (2 NADH total). Four ATP molecules are produced by substrate-level phosphorylation, giving a net gain of 2 ATP (4 produced minus 2 used).

1. 磷酸化（能量投入阶段）：葡萄糖使用 2 个 ATP 分子被磷酸化，形成己糖二磷酸。这使葡萄糖更具反应性并防止其离开细胞。然后己糖二磷酸分裂为两个磷酸丙糖（TP）分子。

2. 氧化（能量产出阶段）：每个磷酸丙糖被氧化为丙酮酸。在此氧化过程中，NAD⁺ 被还原为 NADH（总共 2 个 NADH）。通过底物水平磷酸化产生四个 ATP 分子，净增益为 2 个 ATP（产生 4 个减去消耗 2 个）。

Glycolysis Summary | 糖酵解总结

Input | 输入	Output | 输出
1 Glucose (6C)	2 Pyruvate (3C)
2 ATP (used)	4 ATP (produced) → Net +2 ATP
2 NAD⁺	2 NADH (reduced NAD)
2 ADP + 2 Pi	2 H₂O

3. The Link Reaction — Bridging Glycolysis and the Krebs Cycle | 连接反应——连接糖酵解与克雷布斯循环

Location: Mitochondrial Matrix | 位置：线粒体基质

After glycolysis, pyruvate enters the mitochondria via active transport. Once inside the mitochondrial matrix, each pyruvate molecule undergoes the link reaction (also called the pyruvate oxidation or the transition reaction).

糖酵解之后，丙酮酸通过主动运输进入线粒体。一旦进入线粒体基质，每个丙酮酸分子都会经历连接反应（也称为丙酮酸氧化或过渡反应）。

The Process | 过程

Each pyruvate (3C) molecule is converted into acetyl coenzyme A (acetyl-CoA, 2C) through the following steps:

每个丙酮酸（3C）分子通过以下步骤转化为乙酰辅酶 A（acetyl-CoA，2C）：

1. Decarboxylation: One carbon atom is removed from pyruvate in the form of CO₂.
2. Oxidation: The remaining 2-carbon fragment is oxidised, and NAD⁺ is reduced to NADH.
3. Coenzyme A attachment: The 2-carbon acetyl group combines with coenzyme A (CoA) to form acetyl-CoA, the molecule that enters the Krebs cycle.

1. 脱羧：从丙酮酸中移除一个碳原子，以 CO₂ 的形式释放。

2. 氧化：剩余的 2 碳片段被氧化，NAD⁺ 被还原为 NADH。

3. 辅酶 A 附着：2 碳乙酰基与辅酶 A（CoA）结合形成乙酰辅酶 A，这是进入克雷布斯循环的分子。

Link Reaction Summary (per glucose molecule) | 连接反应总结（每个葡萄糖分子）

Since one glucose produces two pyruvate molecules, the link reaction occurs twice per glucose:

由于一个葡萄糖产生两个丙酮酸分子，连接反应每个葡萄糖发生两次：

• 2 Pyruvate (3C) → 2 Acetyl-CoA (2C)
• 2 CO₂ released | 释放 2 个 CO₂
• 2 NADH produced | 产生 2 个 NADH

Key point: No ATP is produced directly in the link reaction. The energy is captured in the form of reduced NAD (NADH), which will be used later in oxidative phosphorylation.

关键点：连接反应中不直接产生 ATP。能量以还原型 NAD（NADH）的形式被捕获，将在后续的氧化磷酸化中使用。

4. The Krebs Cycle — The Metabolic Hub | 克雷布斯循环——代谢枢纽

Location: Mitochondrial Matrix | 位置：线粒体基质

The Krebs cycle (also called the citric acid cycle or TCA cycle) is a closed loop of enzyme-controlled reactions that oxidises acetyl-CoA completely to CO₂. It is named after Sir Hans Krebs, who discovered the cycle in 1937 and won the Nobel Prize for this work in 1953.

克雷布斯循环（也称为柠檬酸循环或 TCA 循环）是一个由酶控制的闭合反应循环，将乙酰辅酶 A 完全氧化为 CO₂。它以汉斯·克雷布斯爵士命名，他于 1937 年发现了该循环，并因此于 1953 年获得诺贝尔奖。

The Process | 过程

For each turn of the cycle (processing one acetyl-CoA molecule):

循环每转一圈（处理一个乙酰辅酶 A 分子）：

1. Acetyl-CoA (2C) combines with oxaloacetate (4C) to form citrate (6C). CoA is released and recycled.
2. Citrate is decarboxylated and dehydrogenated through a series of steps, releasing 2 CO₂ molecules.
3. During these reactions, 3 NAD⁺ are reduced to 3 NADH, and 1 FAD is reduced to 1 FADH₂.
4. One ATP is produced directly by substrate-level phosphorylation (GTP in some organisms).
5. Oxaloacetate is regenerated, ready to combine with the next acetyl-CoA molecule.

1. 乙酰辅酶 A（2C）与草酰乙酸（4C）结合形成柠檬酸（6C）。CoA 被释放并循环使用。

2. 柠檬酸通过一系列步骤进行脱羧和脱氢，释放 2 个 CO₂ 分子。

3. 在这些反应中，3 个 NAD⁺ 被还原为 3 个 NADH，1 个 FAD 被还原为 1 个 FADH₂。

4. 通过底物水平磷酸化直接产生一个 ATP（在某些生物体中为 GTP）。

5. 草酰乙酸被再生，准备与下一个乙酰辅酶 A 分子结合。

Krebs Cycle Summary (per glucose = 2 turns) | 克雷布斯循环总结（每个葡萄糖 = 2 圈）

Product | 产物	Per Turn | 每圈	Per Glucose | 每个葡萄糖
CO₂	2	4
NADH	3	6
FADH₂	1	2
ATP	1	2

Exam tip: The Krebs cycle is a common source of confusion in exams. Remember that the cycle itself does not use oxygen directly — it is the electron transport chain that requires O₂ as the final electron acceptor. The Krebs cycle is simply a series of oxidation and decarboxylation reactions that strip electrons from carbon compounds.

考试提示：克雷布斯循环是考试中常见的混淆点。记住，循环本身不直接使用氧气——需要 O₂ 作为最终电子受体的是电子传递链。克雷布斯循环仅仅是一系列从碳化合物中剥离电子的氧化和脱羧反应。

5. Oxidative Phosphorylation — The ATP Factory | 氧化磷酸化——ATP 工厂

Location: Inner Mitochondrial Membrane (Cristae) | 位置：线粒体内膜（嵴）

Oxidative phosphorylation is the final and most productive stage of aerobic respiration. It consists of two coupled processes: the Electron Transport Chain (ETC) and Chemiosmosis. Together, these two processes produce the vast majority of ATP — approximately 34 out of the total ~38 ATP molecules per glucose.

氧化磷酸化是有氧呼吸的最终也是产量最高的阶段。它由两个耦合过程组成：电子传递链（ETC）和化学渗透。这两个过程共同产生绝大多数 ATP——每个葡萄糖大约 38 个 ATP 中的约 34 个。

5a. The Electron Transport Chain (ETC) | 电子传递链

The ETC is a series of protein complexes (Complex I, II, III, and IV) and mobile electron carriers (ubiquinone and cytochrome c) embedded in the inner mitochondrial membrane. Reduced NAD (NADH) and reduced FAD (FADH₂) from earlier stages donate their electrons to the chain:

ETC 是嵌入线粒体内膜的一系列蛋白质复合物（复合体 I、II、III 和 IV）以及移动电子载体（泛醌和细胞色素 c）。来自早期阶段的还原型 NAD（NADH）和还原型 FAD（FADH₂）将电子提供给该链：

1. NADH donates electrons to Complex I (NADH dehydrogenase). These electrons are passed to ubiquinone.
2. FADH₂ donates electrons to Complex II (succinate dehydrogenase). These electrons are also passed to ubiquinone.
3. Ubiquinone carries electrons to Complex III (cytochrome bc1 complex).
4. Cytochrome c shuttles electrons from Complex III to Complex IV (cytochrome c oxidase).
5. At Complex IV, electrons are finally transferred to molecular oxygen (O₂), which combines with H⁺ ions to form water: ½O₂ + 2e⁻ + 2H⁺ → H₂O

1. NADH 将电子提供给复合体 I（NADH 脱氢酶）。这些电子被传递给泛醌。

2. FADH₂ 将电子提供给复合体 II（琥珀酸脱氢酶）。这些电子也被传递给泛醌。

3. 泛醌将电子携带到复合体 III（细胞色素 bc1 复合体）。

4. 细胞色素 c 将电子从复合体 III 穿梭到复合体 IV（细胞色素 c 氧化酶）。

5. 在复合体 IV，电子最终被传递给分子氧（O₂），后者与 H⁺ 离子结合形成水：½O₂ + 2e⁻ + 2H⁺ → H₂O

As electrons move through the chain, they lose energy. This energy is used by Complexes I, III, and IV to pump protons (H⁺ ions) from the mitochondrial matrix into the intermembrane space, creating a proton gradient — a higher concentration of H⁺ in the intermembrane space than in the matrix.

随着电子在链中移动，它们失去能量。这些能量被复合体 I、III 和 IV 用来将质子（H⁺ 离子）从线粒体基质泵入膜间隙，产生质子梯度——膜间隙中的 H⁺ 浓度高于基质。

5b. Chemiosmosis — Harnessing the Proton Gradient | 化学渗透——利用质子梯度

The proton gradient created by the ETC represents stored potential energy. The inner mitochondrial membrane is impermeable to H⁺ ions, so they can only flow back into the matrix through a specialised protein channel called ATP synthase (also known as ATP synthetase). As protons flow down their concentration gradient through ATP synthase, the enzyme uses this energy to catalyse the reaction:

ETC 产生的质子梯度代表了储存的势能。线粒体内膜对 H⁺ 离子是不通透的，因此它们只能通过一个称为 ATP 合酶的特殊蛋白质通道流回基质。当质子沿浓度梯度通过 ATP 合酶流动时，该酶利用此能量催化反应：

ADP + Pi → ATP

This coupling of electron transport to ATP synthesis via a proton gradient is called chemiosmosis, a theory proposed by Peter Mitchell in 1961 (Nobel Prize in Chemistry, 1978). It is one of the most elegant examples of energy coupling in biology.

这种通过质子梯度将电子传递与 ATP 合成耦合的过程称为化学渗透，由 Peter Mitchell 于 1961 年提出（1978 年诺贝尔化学奖）。这是生物学中最优雅的能量耦合例子之一。

ATP Yield from NADH vs FADH₂ | NADH 与 FADH₂ 的 ATP 产量

A crucial distinction: NADH donates electrons at Complex I, which pumps protons. This results in approximately 2.5 ATP per NADH. FADH₂ enters at Complex II, which does not pump protons, so it yields only about 1.5 ATP per FADH₂.

一个关键区别：NADH 在复合体 I 处提供电子，该复合体泵送质子。这导致每个 NADH 约产生 2.5 个 ATP。FADH₂ 在复合体 II 处进入，该复合体不泵送质子，因此每个 FADH₂ 仅产生约 1.5 个 ATP。

6. Anaerobic Respiration — When Oxygen Runs Out | 无氧呼吸——当氧气耗尽时

When oxygen is unavailable, the electron transport chain cannot function because there is no final electron acceptor. However, glycolysis can still occur, producing 2 ATP and 2 NADH per glucose. The problem is that NAD⁺ must be regenerated for glycolysis to continue — otherwise, all NAD⁺ would become locked up as NADH.

当氧气不可用时，电子传递链无法运作，因为没有最终电子受体。然而，糖酵解仍然可以进行，每个葡萄糖产生 2 个 ATP 和 2 个 NADH。问题在于 NAD⁺ 必须被再生才能使糖酵解继续进行——否则所有 NAD⁺ 都会被锁定为 NADH。

In Animals: Lactate Fermentation | 在动物中：乳酸发酵

In animal cells (including human muscle cells during intense exercise), pyruvate is reduced to lactate (lactic acid) by the enzyme lactate dehydrogenase. This reaction oxidises NADH back to NAD⁺, allowing glycolysis to continue:

在动物细胞中（包括剧烈运动中的人类肌肉细胞），丙酮酸被乳酸脱氢酶还原为乳酸。该反应将 NADH 氧化回 NAD⁺，使糖酵解得以继续：

Pyruvate + NADH → Lactate + NAD⁺

Lactate can be converted back to pyruvate in the liver when oxygen becomes available again (the Cori cycle). Contrary to popular belief, lactate is not the cause of muscle soreness — that is due to micro-tears in muscle fibres.

当氧气再次可用时，乳酸可以在肝脏中被转化回丙酮酸（Cori 循环）。与普遍认知相反，乳酸不是肌肉酸痛的原因——肌肉酸痛是由于肌纤维的微撕裂。

In Plants and Yeast: Ethanol Fermentation | 在植物和酵母中：乙醇发酵

In plants and microorganisms such as yeast (Saccharomyces cerevisiae), pyruvate undergoes a two-step process to produce ethanol:

在植物和微生物（如酿酒酵母 Saccharomyces cerevisiae）中，丙酮酸经过两步过程产生乙醇：

1. Pyruvate is decarboxylated to ethanal (acetaldehyde), releasing CO₂. Catalysed by pyruvate decarboxylase.
2. Ethanal is reduced to ethanol by alcohol dehydrogenase, oxidising NADH back to NAD⁺.

1. 丙酮酸脱羧为乙醛，释放 CO₂。由丙酮酸脱羧酶催化。

2. 乙醛被乙醇脱氢酶还原为乙醇，将 NADH 氧化回 NAD⁺。

Pyruvate → Ethanal + CO₂ → Ethanol + NAD⁺

This process is the basis of alcoholic beverage production and bread-making. The CO₂ released causes bread to rise, while the ethanol evaporates during baking.

该过程是酒精饮料生产和面包制作的基础。释放的 CO₂ 使面包膨胀，而乙醇在烘焙过程中蒸发。

7. Complete Energy Yield Summary | 完整能量产量总结

Stage | 阶段	Location | 位置	ATP (direct) | ATP（直接）	NADH	FADH₂	CO₂
Glycolysis | 糖酵解	Cytoplasm | 细胞质	2	2	0	0
Link Reaction ×2 | 连接反应×2	Matrix | 基质	0	2	0	2
Krebs Cycle ×2 | 克雷布斯循环×2	Matrix | 基质	2	6	2	4
Oxidative Phosphorylation | 氧化磷酸化	Inner membrane | 内膜	~34	—	—	0
Total | 总计	—	~38	10	2	6

Note: The theoretical maximum of ~38 ATP is rarely achieved in practice. The actual yield is closer to 30-32 ATP per glucose due to proton leak across the mitochondrial membrane and the energy cost of transporting NADH from glycolysis into the mitochondria.

注意：理论上 ~38 个 ATP 的最大值在实践中很少达到。由于线粒体膜的质子泄漏以及将 NADH 从糖酵解转运到线粒体的能量成本，实际产量更接近每个葡萄糖 30-32 个 ATP。

8. Key Coenzymes and Their Roles | 关键辅酶及其作用

Coenzyme | 辅酶	Full Name | 全称	Role | 作用
NAD⁺	Nicotinamide Adenine Dinucleotide | 烟酰胺腺嘌呤二核苷酸	Electron carrier; reduced to NADH | 电子载体；被还原为 NADH
FAD	Flavin Adenine Dinucleotide | 黄素腺嘌呤二核苷酸	Electron carrier; reduced to FADH₂ | 电子载体；被还原为 FADH₂
CoA	Coenzyme A | 辅酶 A	Carries acetyl groups into Krebs cycle | 将乙酰基带入克雷布斯循环

9. Exam Tips for Top Marks | 高分考试技巧

• Use precise terminology: Write “oxidative phosphorylation” not “making ATP with oxygen”. Examiners reward accurate scientific language.
• State locations: Always specify where each stage occurs — cytoplasm, mitochondrial matrix, or inner mitochondrial membrane (cristae).
• Balance your equations: For the overall equation, make sure C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O is correctly balanced.
• Distinguish substrate-level from oxidative phosphorylation: Substrate-level phosphorylation produces ATP directly from a phosphorylated intermediate. Oxidative phosphorylation uses the proton gradient and ATP synthase.
• Explain the role of oxygen: Oxygen is the final electron acceptor in the ETC, not a direct reactant in glycolysis or the Krebs cycle.
• Link structure to function: The highly folded cristae of the inner mitochondrial membrane increase surface area for ETC complexes and ATP synthase, maximising ATP production.
• Compare anaerobic pathways: Be able to explain both lactate fermentation (animals) and ethanol fermentation (plants/yeast), including the enzymes involved and the products formed.

• 使用精确术语：写”氧化磷酸化”而不是”用氧气制造 ATP”。考官奖励准确的科学语言。
• 说明位置：始终指定每个阶段发生的位置——细胞质、线粒体基质或线粒体内膜（嵴）。
• 平衡方程式：对于总方程式，确保 C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O 正确平衡。
• 区分底物水平磷酸化和氧化磷酸化：底物水平磷酸化直接从磷酸化中间体产生 ATP。氧化磷酸化使用质子梯度和 ATP 合酶。
• 解释氧气的作用：氧气是 ETC 中的最终电子受体，而不是糖酵解或克雷布斯循环中的直接反应物。
• 将结构与功能联系起来：线粒体内膜高度折叠的嵴增加了 ETC 复合物和 ATP 合酶的表面积，最大化 ATP 产量。
• 比较无氧途径：能够解释乳酸发酵（动物）和乙醇发酵（植物/酵母），包括涉及的酶和形成的产物。

10. Common Misconceptions | 常见误区

1. “Plants only photosynthesise; they don’t respire” | “植物只进行光合作用，不进行呼吸” — False. Plants respire 24 hours a day. Photosynthesis only occurs in the light. At night, plants rely entirely on respiration for ATP.
2. “The Krebs cycle uses oxygen” | “克雷布斯循环使用氧气” — False. The Krebs cycle itself does not use O₂. Oxygen is only used at the very end of the ETC as the final electron acceptor.
3. “Lactic acid causes muscle soreness” | “乳酸导致肌肉酸痛” — False. DOMS (delayed onset muscle soreness) is caused by micro-damage to muscle fibres, not lactate accumulation.
4. “Each NADH always produces exactly 3 ATP” | “每个 NADH 总是产生恰好 3 个 ATP” — Outdated. Current estimates are ~2.5 ATP per NADH and ~1.5 ATP per FADH₂, because proton pumping and ATP synthase stoichiometries are now better understood.

1. “植物只进行光合作用，不进行呼吸” ——错误。植物每天 24 小时都在呼吸。光合作用仅在光下发生。夜间，植物完全依赖呼吸来获取 ATP。

2. “克雷布斯循环使用氧气” ——错误。克雷布斯循环本身不使用 O₂。氧气仅在 ETC 的最末端作为最终电子受体使用。

3. “乳酸导致肌肉酸痛” ——错误。延迟性肌肉酸痛（DOMS）是由肌纤维的微损伤引起的，而不是乳酸积累。

4. “每个 NADH 总是产生恰好 3 个 ATP” ——过时。目前估计每个 NADH 约 2.5 个 ATP，每个 FADH₂ 约 1.5 个 ATP，因为现在对质子泵送和 ATP 合酶的化学计量有了更好的理解。

Conclusion | 结论

Cellular respiration is a beautifully coordinated sequence of reactions that extracts energy from glucose with remarkable efficiency. Understanding each stage — glycolysis, the link reaction, the Krebs cycle, and oxidative phosphorylation — along with their locations, inputs, outputs, and the roles of key enzymes and coenzymes, will give you a solid foundation for A-Level Biology exams. Remember to practise drawing and labelling the key pathways, and always link structure to function when discussing the mitochondria.

细胞呼吸是一系列精妙协调的反应，以卓越的效率从葡萄糖中提取能量。理解每个阶段——糖酵解、连接反应、克雷布斯循环和氧化磷酸化——以及它们的位置、输入、输出和关键酶与辅酶的作用，将为你的 A-Level 生物考试打下坚实基础。记住练习绘制和标注关键途径，并在讨论线粒体时始终将结构与功能联系起来。

This bilingual guide provides the essential knowledge for A-Level Cellular Respiration. For practice questions, past papers, and exam-style quizzes, visit the resources section at aleveler.com.

本双语指南提供了 A-Level 细胞呼吸的核心知识。如需练习题、历年真题和考试风格的测验，请访问 aleveler.com 的资源专区。