Cellular Respiration: Glycolysis, Krebs Cycle & Oxidative Phosphorylation | A-Level Biology 细胞呼吸全解析

Introduction | 引言

Cellular respiration is one of the most fundamental processes in biology — it is how every living cell extracts energy from organic molecules to power life. For A-Level Biology students, mastering respiration means understanding not just the chemical equations, but the intricate dance of enzymes, membranes, and electron carriers that convert a single molecule of glucose into up to 38 molecules of ATP. This article provides a complete bilingual walkthrough of the four stages of aerobic respiration: Glycolysis, the Link Reaction, the Krebs Cycle, and Oxidative Phosphorylation, followed by a concise treatment of anaerobic respiration.

细胞呼吸是生物学中最基本的过程之一——每一个活细胞都通过它从有机分子中提取能量来维持生命。对于A-Level生物学学生来说，掌握呼吸作用不仅意味着理解化学方程式，还意味着理解酶、膜和电子载体的精妙配合：将一个葡萄糖分子转化为多达38个ATP分子。本文提供有氧呼吸四个阶段的双语完整指南：糖酵解、连接反应、克雷布斯循环和氧化磷酸化，并简要介绍无氧呼吸。

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1. Overview of Respiration | 呼吸作用概述

What is Respiration? | 什么是呼吸作用？

Respiration is the process by which cells release energy from organic molecules (primarily glucose) and transfer it to ATP (adenosine triphosphate). ATP is the universal energy currency of the cell — it powers active transport, muscle contraction, protein synthesis, and virtually every endergonic reaction. The overall equation for aerobic respiration is:

呼吸作用是细胞从有机分子（主要是葡萄糖）中释放能量并将其转移至ATP（三磷酸腺苷）的过程。ATP是细胞的通用能量货币——它为主动运输、肌肉收缩、蛋白质合成以及几乎所有吸能反应提供动力。有氧呼吸的总方程式为：

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

This equation masks enormous complexity. In reality, respiration proceeds through four tightly coupled stages, each occurring in a specific cellular compartment. The table below summarises the key facts you must know for A-Level exams:

这个方程式掩盖了巨大的复杂性。实际上，呼吸作用通过四个紧密耦合的阶段进行，每个阶段发生在特定的细胞区室中。下表总结了A-Level考试必须掌握的关键事实：

Stage | 阶段	Location | 位置	ATP Yield (per glucose) | ATP产量（每葡萄糖）	Coenzymes Produced | 产生的辅酶	O₂ Required? | 需氧？
Glycolysis | 糖酵解	Cytoplasm | 细胞质	2 (net) | 净产2	2 NADH	No | 否
Link Reaction | 连接反应	Mitochondrial matrix | 线粒体基质	0	2 NADH	Yes (indirectly) | 是（间接）
Krebs Cycle | 克雷布斯循环	Mitochondrial matrix | 线粒体基质	2 ATP (GTP)	6 NADH + 2 FADH₂	Yes (indirectly) | 是（间接）
Oxidative Phosphorylation | 氧化磷酸化	Inner mitochondrial membrane | 线粒体内膜	~34	None (NAD⁺ & FAD regenerated) | 无（NAD⁺和FAD再生）	Yes (terminal acceptor) | 是（最终受体）

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2. Glycolysis | 糖酵解

Glycolysis (from Greek glykys = sweet, lysis = splitting) is the first stage of respiration and the only one that occurs in the cytoplasm. It does not require oxygen and is therefore the sole ATP-producing pathway available to anaerobic organisms and to cells temporarily deprived of oxygen (such as muscle cells during intense exercise).

糖酵解（源自希腊语glykys=甜，lysis=分裂）是呼吸作用的第一阶段，也是唯一发生在细胞质中的阶段。它不需要氧气，因此是无氧生物和暂时缺氧细胞（如剧烈运动时的肌肉细胞）唯一可用的ATP产生途径。

Key Steps | 关键步骤

Glycolysis converts one molecule of glucose (6C) into two molecules of pyruvate (3C each). The process consumes 2 ATP in the energy investment phase but produces 4 ATP in the energy payoff phase, yielding a net gain of 2 ATP. Two molecules of NAD⁺ are also reduced to NADH.

糖酵解将一个葡萄糖分子（6C）转化为两个丙酮酸分子（各3C）。该过程在能量投入阶段消耗2个ATP，但在能量回报阶段产生4个ATP，净得2个ATP。两个NAD⁺分子也被还原为NADH。

1. Phosphorylation of glucose | 葡萄糖磷酸化： Glucose is phosphorylated by ATP to form glucose-6-phosphate. This traps glucose inside the cell (the phosphate group prevents it from crossing the plasma membrane) and makes it more reactive. A second phosphorylation by another ATP produces fructose-1,6-bisphosphate.
   葡萄糖被ATP磷酸化形成葡萄糖-6-磷酸。这将葡萄糖困在细胞内（磷酸基团阻止其穿过质膜）并使其更具反应性。另一个ATP的第二次磷酸化产生果糖-1,6-二磷酸。
2. Lysis (splitting) | 裂解（分裂）： Fructose-1,6-bisphosphate is split into two 3-carbon molecules: glyceraldehyde-3-phosphate (GALP) and dihydroxyacetone phosphate (DHAP). DHAP is rapidly isomerised into GALP, so the subsequent steps process two molecules of GALP.
   果糖-1,6-二磷酸被分裂为两个3碳分子：甘油醛-3-磷酸（GALP）和磷酸二羟丙酮（DHAP）。DHAP迅速异构化为GALP，因此后续步骤处理两个GALP分子。
3. Oxidation and ATP synthesis | 氧化与ATP合成： Each GALP is oxidised, reducing NAD⁺ to NADH. The energy released drives the production of ATP via substrate-level phosphorylation — a phosphate group is transferred directly from a substrate molecule to ADP.
   每个GALP被氧化，将NAD⁺还原为NADH。释放的能量通过底物水平磷酸化驱动ATP的产生——磷酸基团直接从底物分子转移至ADP。

Exam Tip | 考试提示： A-Level examiners frequently ask about substrate-level phosphorylation. Remember: it is the direct transfer of a phosphate group from a phosphorylated intermediate to ADP, catalysed by a kinase enzyme. This is distinct from oxidative phosphorylation, which relies on the electron transport chain and chemiosmosis.

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3. The Link Reaction | 连接反应

Pyruvate produced by glycolysis cannot enter the Krebs Cycle directly. It must first be transported into the mitochondrial matrix, where it undergoes oxidative decarboxylation — the Link Reaction. This reaction is catalysed by the multi-enzyme pyruvate dehydrogenase complex.

糖酵解产生的丙酮酸不能直接进入克雷布斯循环。它必须首先被转运到线粒体基质中，在那里经历氧化脱羧——连接反应。该反应由多酶丙酮酸脱氢酶复合体催化。

Pyruvate (3C) + NAD⁺ + CoA → Acetyl-CoA (2C) + CO₂ + NADH

Key points for the exam:

考试关键点：

• Decarboxylation: One carbon atom is removed from pyruvate as CO₂. The molecule is now a 2-carbon acetyl group.
  脱羧：一个碳原子以CO₂形式从丙酮酸中移除。该分子现在是2碳的乙酰基。
• Oxidation: Pyruvate is oxidised, reducing NAD⁺ to NADH.
  氧化：丙酮酸被氧化，将NAD⁺还原为NADH。
• Coenzyme A: The acetyl group is attached to coenzyme A (CoA) to form acetyl-CoA, which enters the Krebs Cycle.
  辅酶A：乙酰基附着在辅酶A（CoA）上形成乙酰辅酶A，进入克雷布斯循环。
• Per glucose: Two pyruvate molecules are produced per glucose, so the Link Reaction occurs twice, producing 2 acetyl-CoA, 2 CO₂, and 2 NADH.
  每葡萄糖：每葡萄糖产生两个丙酮酸分子，因此连接反应发生两次，产生2个乙酰辅酶A、2个CO₂和2个NADH。

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4. The Krebs Cycle | 克雷布斯循环

The Krebs Cycle (also called the citric acid cycle or TCA cycle) takes place in the mitochondrial matrix. It is a cyclic series of enzyme-catalysed reactions that oxidises the acetyl group from acetyl-CoA completely to CO₂, generating reduced coenzymes (NADH and FADH₂) and a small amount of ATP. The cycle was discovered by Sir Hans Krebs in 1937, earning him the 1953 Nobel Prize.

克雷布斯循环（也称为柠檬酸循环或TCA循环）发生在线粒体基质中。它是一系列酶催化的环状反应，将乙酰辅酶A中的乙酰基完全氧化为CO₂，产生还原辅酶（NADH和FADH₂）和少量ATP。该循环由汉斯·克雷布斯爵士于1937年发现，为他赢得了1953年诺贝尔奖。

Outline of one turn of the cycle | 循环一周概述：

1. Acetyl-CoA (2C) + Oxaloacetate (4C) → Citrate (6C): The acetyl group combines with oxaloacetate (a 4-carbon molecule) to form citrate (6C). CoA is released and recycled.
   乙酰辅酶A (2C) + 草酰乙酸 (4C) → 柠檬酸 (6C)：乙酰基与草酰乙酸（4碳分子）结合形成柠檬酸（6C）。辅酶A被释放并循环使用。
2. Decarboxylation and oxidation: Citrate is progressively oxidised and decarboxylated. Two CO₂ molecules are released, and the molecule is reduced back to oxaloacetate (4C). During this process, 3 NAD⁺ are reduced to 3 NADH, 1 FAD is reduced to FADH₂, and 1 ATP is produced by substrate-level phosphorylation (GTP in some organisms).
   脱羧与氧化：柠檬酸逐步被氧化和脱羧。释放两个CO₂分子，分子被还原回草酰乙酸（4C）。在此过程中，3个NAD⁺被还原为3个NADH，1个FAD被还原为FADH₂，并通过底物水平磷酸化产生1个ATP（某些生物中为GTP）。
3. Regeneration of oxaloacetate: The cycle ends with the regeneration of oxaloacetate, ready to accept another acetyl group.
   草酰乙酸的再生：循环以草酰乙酸的再生结束，准备接受另一个乙酰基。

Per glucose molecule (two turns): 2 ATP, 6 NADH, 2 FADH₂, 4 CO₂.
每葡萄糖分子（两轮）：2 ATP、6 NADH、2 FADH₂、4 CO₂。

Exam Tip | 考试提示： You do not need to memorise every intermediate of the Krebs Cycle for most A-Level specifications, but you MUST know the inputs (acetyl-CoA), outputs (CO₂, NADH, FADH₂, ATP), and that oxaloacetate is regenerated. Some exam boards (AQA, Edexcel) expect you to name citrate as the first product and oxaloacetate as the final regenerated molecule.

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5. Oxidative Phosphorylation | 氧化磷酸化

Oxidative phosphorylation is the final and most productive stage of aerobic respiration, accounting for approximately 34 of the ~38 ATP molecules produced per glucose. It consists of two tightly coupled processes: the Electron Transport Chain (ETC) and Chemiosmosis. Both occur on the inner mitochondrial membrane, which is highly folded into cristae to maximise surface area.

氧化磷酸化是有氧呼吸的最终且最高产阶段，约占每葡萄糖产生约38个ATP中的34个。它由两个紧密结合的过程组成：电子传递链（ETC）和化学渗透。两者都发生在线粒体内膜上，内膜高度折叠成嵴以最大化表面积。

5.1 The Electron Transport Chain (ETC) | 电子传递链

The NADH and FADH₂ produced in glycolysis, the Link Reaction, and the Krebs Cycle donate their electrons to the ETC. The chain consists of four protein complexes (I–IV) and two mobile carriers (ubiquinone and cytochrome c) embedded in the inner mitochondrial membrane.

糖酵解、连接反应和克雷布斯循环中产生的NADH和FADH₂将其电子捐赠给ETC。该链由嵌入线粒体内膜的四个蛋白质复合体（I–IV）和两个移动载体（泛醌和细胞色素c）组成。

1. Complex I (NADH dehydrogenase): NADH donates electrons. The electrons pass through the complex and are transferred to ubiquinone (Q). Protons (H⁺) are pumped from the matrix into the intermembrane space.
   复合体I（NADH脱氢酶）：NADH提供电子。电子通过复合体并转移至泛醌（Q）。质子（H⁺）从基质泵入膜间隙。
2. Complex II (Succinate dehydrogenase): FADH₂ donates electrons here. Unlike Complex I, Complex II does NOT pump protons. Electrons are transferred to ubiquinone.
   复合体II（琥珀酸脱氢酶）：FADH₂在此提供电子。与复合体I不同，复合体II不泵送质子。电子转移至泛醌。
3. Complex III (Cytochrome bc1): Electrons from ubiquinone pass through Complex III. More protons are pumped into the intermembrane space.
   复合体III（细胞色素bc1）：来自泛醌的电子通过复合体III。更多质子被泵入膜间隙。
4. Complex IV (Cytochrome c oxidase): Electrons are transferred to the final electron acceptor — molecular oxygen (O₂). Oxygen combines with electrons and protons to form water: ½O₂ + 2e⁻ + 2H⁺ → H₂O. This is why oxygen is essential for aerobic respiration.
   复合体IV（细胞色素c氧化酶）：电子转移至最终电子受体——分子氧（O₂）。氧与电子和质子结合形成水：½O₂ + 2e⁻ + 2H⁺ → H₂O。这就是氧气对有氧呼吸必不可少的原因。

FADH₂ yields fewer ATP: Because FADH₂ enters at Complex II (which does not pump protons), it contributes to a smaller proton gradient than NADH. This is why FADH₂ produces approximately 1.5 ATP compared to NADH’s 2.5 ATP.

FADH₂产生较少ATP：因为FADH₂在复合体II（不泵送质子）进入，它对质子梯度的贡献小于NADH。这就是为什么FADH₂产生约1.5个ATP而NADH产生约2.5个ATP。

5.2 Chemiosmosis | 化学渗透

As electrons pass along the ETC, complexes I, III, and IV pump protons (H⁺) from the mitochondrial matrix into the intermembrane space. This creates:

随着电子沿ETC传递，复合体I、III和IV将质子（H⁺）从线粒体基质泵入膜间隙。这产生了：

• A proton gradient (higher [H⁺] in the intermembrane space, lower [H⁺] in the matrix)
  质子梯度（膜间隙[H⁺]高，基质[H⁺]低）
• An electrochemical gradient (the membrane is more positively charged on the intermembrane side)
  电化学梯度（膜在膜间隙侧带更多正电荷）
• This combined gradient is the proton motive force (PMF)
  这个组合梯度就是质子动力势（PMF）

Protons can only flow back into the matrix through a specialised protein channel called ATP synthase (Complex V). As protons flow down their electrochemical gradient through ATP synthase, the enzyme rotates and catalyses the synthesis of ATP from ADP + Pi. This process is called chemiosmosis, a mechanism proposed by Peter Mitchell (Nobel Prize, 1978).

质子只能通过一种特殊的蛋白质通道——ATP合酶（复合体V）流回基质。当质子沿电化学梯度通过ATP合酶流动时，酶旋转并催化ADP + Pi合成ATP。这个过程称为化学渗透，由彼得·米切尔提出（1978年诺贝尔奖）。

A-Level Definition | A-Level定义： Chemiosmosis is the diffusion of protons (H⁺) down their electrochemical gradient through ATP synthase, coupled to the synthesis of ATP from ADP and inorganic phosphate.

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6. Anaerobic Respiration | 无氧呼吸

When oxygen is unavailable, the ETC cannot function because there is no final electron acceptor. NADH accumulates and NAD⁺ becomes depleted, bringing glycolysis (and all ATP production) to a halt. Anaerobic respiration solves this problem by regenerating NAD⁺ from NADH, allowing glycolysis to continue producing 2 ATP per glucose.

当氧气不可用时，ETC无法运作，因为没有最终电子受体。NADH积累，NAD⁺被耗尽，导致糖酵解（及所有ATP生产）停止。无氧呼吸通过从NADH再生NAD⁺来解决这个问题，使糖酵解能够继续每葡萄糖产生2个ATP。

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

Pyruvate + NADH → Lactate + NAD⁺ (catalysed by lactate dehydrogenase)

丙酮酸 + NADH → 乳酸 + NAD⁺ （由乳酸脱氢酶催化）

This occurs in mammalian muscle cells during strenuous exercise when oxygen delivery cannot keep pace with demand. The lactate can be transported to the liver and converted back to glucose (the Cori Cycle) or, when oxygen becomes available, oxidised back to pyruvate.

这发生在哺乳动物肌肉细胞剧烈运动期间，当氧气供应跟不上需求时。乳酸可被转运至肝脏并转化回葡萄糖（科里循环），或在氧气恢复时被氧化回丙酮酸。

In Yeast and Plants: Alcoholic Fermentation | 酵母和植物中：酒精发酵

Pyruvate → Ethanal + CO₂ → Ethanol + NAD⁺ (catalysed by pyruvate decarboxylase and alcohol dehydrogenase)

丙酮酸 → 乙醛 + CO₂ → 乙醇 + NAD⁺ （由丙酮酸脱羧酶和乙醇脱氢酶催化）

This pathway is exploited commercially in brewing, baking, and biofuel production.

该途径在酿造、烘焙和生物燃料生产中被商业利用。

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7. Respiratory Quotient (RQ) | 呼吸商

The Respiratory Quotient (RQ) is the ratio of CO₂ produced to O₂ consumed during respiration:

呼吸商（RQ）是呼吸过程中产生的CO₂与消耗的O₂之比：

RQ = CO₂ produced / O₂ consumed

Different respiratory substrates give different RQ values, making RQ a useful experimental tool for identifying which substrate an organism is respiring:

不同的呼吸底物给出不同的RQ值，使RQ成为识别生物体正在呼吸哪种底物的有用实验工具：

• Carbohydrate | 碳水化合物： RQ = 1.0 (C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, ratio 6:6)
• Lipid | 脂质： RQ ≈ 0.7 (lipids are more reduced, requiring more O₂ per CO₂ released)
• Protein | 蛋白质： RQ ≈ 0.8–0.9 (varies by amino acid composition)

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8. Common Exam Questions & Model Answers | 常见考试问题与标准答案

Q1: Explain why the Link Reaction and Krebs Cycle cannot occur in the absence of oxygen. | 解释为什么连接反应和克雷布斯循环在缺氧时无法进行。

Answer: In the absence of oxygen, the ETC stops because O₂ is the final electron acceptor. NADH cannot be reoxidised to NAD⁺. The Link Reaction and Krebs Cycle both require NAD⁺ as an electron acceptor. When NAD⁺ is depleted, these pathways halt.
答案：在缺氧情况下，ETC停止因为O₂是最终电子受体。NADH无法被再氧化为NAD⁺。连接反应和克雷布斯循环都需要NAD⁺作为电子受体。当NAD⁺耗尽时，这些途径停止。

Q2: Compare substrate-level phosphorylation and oxidative phosphorylation. | 比较底物水平磷酸化和氧化磷酸化。

Answer: Substrate-level phosphorylation transfers a phosphate group directly from a phosphorylated intermediate to ADP, catalysed by an enzyme (occurs in glycolysis and the Krebs Cycle). Oxidative phosphorylation uses energy from the ETC to create a proton gradient, and ATP synthase uses the proton motive force to synthesise ATP (occurs on the inner mitochondrial membrane).
答案：底物水平磷酸化直接将磷酸基团从磷酸化中间体转移至ADP，由酶催化（发生在糖酵解和克雷布斯循环中）。氧化磷酸化利用ETC的能量产生质子梯度，ATP合酶利用质子动力势合成ATP（发生在线粒体内膜上）。

Q3: Why does FADH₂ produce fewer ATP molecules than NADH? | 为什么FADH₂产生的ATP分子比NADH少？

Answer: FADH₂ donates electrons to Complex II of the ETC, which does NOT pump protons across the membrane. NADH donates electrons to Complex I, which pumps protons. With fewer protons pumped, FADH₂ generates a smaller proton motive force, resulting in fewer ATP produced by chemiosmosis.
答案：FADH₂将电子提供给ETC的复合体II，该复合体不跨膜泵送质子。NADH将电子提供给复合体I，该复合体泵送质子。泵送的质子较少，FADH₂产生较小的质子动力势，导致化学渗透产生的ATP较少。

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Summary | 总结

Respiration is a masterpiece of biochemical engineering — a multi-stage system that extracts energy from glucose with remarkable efficiency. For A-Level success, focus on:

呼吸作用是生物化学工程的杰作——一个多阶段系统，以卓越的效率从葡萄糖中提取能量。要在A-Level中取得成功，请关注：

1. The location of each stage (cytoplasm vs. mitochondria) and whether O₂ is required
2. The ATP yield at each stage and whether it comes from substrate-level or oxidative phosphorylation
3. The role of reduced coenzymes (NADH and FADH₂) as electron carriers
4. The chemiosmotic mechanism and the role of the proton gradient
5. The difference between aerobic and anaerobic pathways and why anaerobic respiration yields far less ATP

1. 每个阶段的位置（细胞质 vs. 线粒体）以及是否需要O₂
2. 每个阶段的ATP产量以及来自底物水平磷酸化还是氧化磷酸化
3. 还原辅酶（NADH和FADH₂）作为电子载体的作用
4. 化学渗透机制和质子梯度的作用
5. 有氧和无氧途径的区别以及为什么无氧呼吸产生的ATP少得多