📚 A-Level Biology Respiration: Glycolysis, Link Reaction, Krebs Cycle and Oxidative Phosphorylation | A-Level生物细胞呼吸：糖酵解、连接反应、三羧酸循环与氧化磷酸化
Cellular respiration is one of the highest-yield topics in A-Level Biology. It links enzymes, membranes, energy transfer and biochemical calculations into a single narrative, and examiners regularly test it through structured data questions as well as extended writing. This guide walks through glycolysis, the link reaction, the Krebs cycle, the electron transport chain and oxidative phosphorylation, then adds the exam technique and calculation practice that push answers from a grade B to a grade A.
细胞呼吸是 A-Level 生物中分值产出最高的话题之一。它把酶、生物膜、能量转换与生化计算串成一条主线，考官既用结构化数据分析题考查它，也用长论述题考查它。本文依次梳理糖酵解、连接反应、三羧酸循环、电子传递链与氧化磷酸化，并补充把答案从 B 提升到 A 所需的答题技巧与计算训练。
1. Why Respiration Matters in the Specification | 呼吸作用在考纲中的位置
Respiration is defined as the controlled release of energy from organic molecules to synthesise ATP. It is not a single reaction but a sequence of enzyme-catalysed steps, each catalysed by a specific enzyme, and each localised to a particular cellular compartment. For A-Level you must be able to name those compartments and explain why the compartmentalisation matters.
呼吸作用被定义为：以可控方式从有机物中释放能量以合成 ATP 的过程。它不是单一反应，而是一系列由特定酶催化的步骤，且每一步都定位在特定的细胞区室中。A-Level 要求你能说出这些区室，并解释区室化为何重要。
The standard ATP yield quoted by most current specifications is 30 or 32 ATP per glucose molecule under aerobic conditions. Older textbooks printed 36 or 38 ATP; exam boards now accept the lower modern figure because the older value ignored the cost of moving reducing power into mitochondria. Always check which convention your specification uses.
目前多数考纲给出的标准数值是：有氧条件下每分子葡萄糖净产 30 或 32 个 ATP。较早的教材印的是 36 或 38 个 ATP；现在考试局接受较低的新数值，因为旧值忽略了把还原力运入线粒体的代价。务必确认你所考考纲采用哪一套数值。
- Glycolysis - cytoplasm / 糖酵解 -- 细胞质基质
- Link reaction and Krebs cycle - mitochondrial matrix / 连接反应与三羧酸循环 -- 线粒体基质
- Oxidative phosphorylation - inner mitochondrial membrane / 氧化磷酸化 -- 线粒体内膜
- Lactate fermentation - cytoplasm / 乳酸发酵 -- 细胞质基质
2. Glycolysis: The Universal First Stage | 糖酵解：通用的第一阶段
Glycolysis occurs in the cytoplasm of every living cell and requires no oxygen and no membrane-bound organelles. Glucose, a hexose, is phosphorylated twice and then split into two triose phosphate molecules. This investment phase costs 2 ATP per glucose.
糖酵解发生在所有活细胞的细胞质中，不需要氧气，也不需要膜性细胞器。葡萄糖是己糖，先被磷酸化两次，随后裂解为两分子丙糖磷酸。这个"投资阶段"每分子葡萄糖消耗 2 个 ATP。
In the pay-off phase, each triose phosphate is oxidised by NAD⁺ and its phosphate groups are transferred to ADP by substrate-level phosphorylation, producing pyruvate. Because two triose phosphates are formed per glucose, all products of the pay-off phase are doubled.
在"回收阶段"，每分子丙糖磷酸被 NAD⁺ 氧化，其磷酸基团通过底物水平磷酸化转移给 ADP，生成丙酮酸。由于每分子葡萄糖产生两分子丙糖磷酸，回收阶段的产物都要乘以 2。
C₆H₁₂O₆ + 2 NAD⁺ + 2 ADP + 2 Pi → 2 C₃H₄O₃ + 2 NADH + 2 H⁺ + 2 ATP + 2 H₂O
Net yield per glucose: 2 pyruvate, 2 NADH, 2 ATP. Note that the 2 ATP produced are the net figure already accounting for the 2 ATP consumed at the start.
每分子葡萄糖净产量：2 丙酮酸、2 NADH、2 ATP。注意这里的 2 ATP 已是扣除起始消耗的 2 ATP 后的净值。
- Phosphorylation of glucose to glucose phosphate, then to fructose bisphosphate / 葡萄糖磷酸化为葡萄糖磷酸，再转为果糖二磷酸
- Lysis into two triose phosphate molecules / 裂解为两分子丙糖磷酸
- Dehydrogenation generating NADH / 脱氢生成 NADH
- Substrate-level phosphorylation generating ATP / 底物水平磷酸化生成 ATP
3. The Link Reaction: Oxidative Decarboxylation | 连接反应：氧化脱羧
Pyruvate is actively transported into the mitochondrial matrix. There, pyruvate dehydrogenase removes a carboxyl group as CO₂ (decarboxylation) and a hydrogen pair as NADH (dehydrogenation). The remaining two-carbon acetyl group is bonded to coenzyme A.
丙酮酸被主动运输进入线粒体基质。在这里，丙酮酸脱氢酶把一个羧基以 CO₂ 形式脱去（脱羧），同时把一对氢以 NADH 形式脱去（脱氢）。剩下的二碳乙酰基与辅酶 A 结合。
Pyruvate + NAD⁺ + CoA → Acetyl CoA + NADH + CO₂
Because glycolysis produces two pyruvate molecules, this reaction happens twice per glucose, so the link reaction contributes 2 NADH and 2 CO₂. The name "link reaction" reflects its role: it connects the cytoplasmic process of glycolysis to the matrix-based Krebs cycle.
由于糖酵解产生 2 分子丙酮酸，该反应每分子葡萄糖进行两次，因此连接反应贡献 2 个 NADH 与 2 个 CO₂。"连接反应"这一名称正体现其定位：把细胞质中的糖酵解与基质中的三羧酸循环连接起来。
4. The Krebs Cycle: Regenerating Oxaloacetate | 三羧酸循环：草酰乙酸的再生
Acetyl CoA delivers its two-carbon acetyl group to the four-carbon oxaloacetate, forming six-carbon citrate. Through a series of eight enzyme-controlled steps, two carbons are removed as CO₂, four dehydrogenations occur, and one direct substrate-level phosphorylation forms ATP (or GTP) by transfer of phosphate from a substrate to ADP.
乙酰辅酶 A 把二碳乙酰基交给四碳的草酰乙酸，生成六碳的柠檬酸。经过八个酶控步骤，两个碳以 CO₂ 形式脱去，发生四次脱氢，同时发生一次直接的底物水平磷酸化，由底物把磷酸转移给 ADP 生成 ATP（或 GTP）。
The cycle is described as a cycle for one reason only: oxaloacetate is regenerated at the end so the process can accept another acetyl group. This is a favourite exam explanation question.
之所以称为"循环"只有一个原因：草酰乙酸在末端被重新生成，以接受下一个乙酰基。这一点是考试中高频的解释类问题。
Yield per acetyl CoA: 3 NADH, 1 FADH₂, 1 ATP, 2 CO₂. Per glucose (multiply by two): 6 NADH, 2 FADH₂, 2 ATP, 4 CO₂.
每分子乙酰辅酶 A 产量：3 NADH、1 FADH₂、1 ATP、2 CO₂。每分子葡萄糖（乘以二）：6 NADH、2 FADH₂、2 ATP、4 CO₂。
5. The Electron Transport Chain on the Cristae | 嵴上的电子传递链
NADH and FADH₂ release hydrogen atoms at the inner mitochondrial membrane. The electrons are passed down a chain of protein complexes embedded in the membrane, while the protons remain in the matrix. Each electron carrier in the chain is at a slightly lower energy level than the previous one, so energy is released in small, usable steps rather than in one destructive burst.
NADH 与 FADH₂ 在线粒体内膜上释放氢原子。电子沿膜内的一系列蛋白复合体向下传递，而质子留在基质中。链中每个电子载体的能级都比前一个略低，因此能量以小而可用的步骤释放，而不是一次猛然释放造成破坏。
The final electron acceptor is oxygen, which combines with electrons and protons to form water. If oxygen is absent, the chain becomes fully reduced, protons can no longer be pumped, and oxidative phosphorylation stops within seconds. This is why oxygen is called the terminal electron acceptor rather than simply a "reactant".
最终的电子受体是氧，它与电子和质子结合生成水。若没有氧，整条链被完全还原，质子无法继续被泵出，氧化磷酸化在数秒内停止。这正是氧被称为"末端电子受体"而不只是"反应物"的原因。
½ O₂ + 2 e⁻ + 2 H⁺ → H₂O
6. Chemiosmosis and ATP Synthase | 化学渗透与 ATP 合酶
As electrons pass along the chain, the energy released pumps protons from the matrix into the intermembrane space. This creates a proton electrochemical gradient: the space becomes more acidic and more positively charged than the matrix. In mammals, the intermembrane space sits at roughly pH 7.0-7.4 while the matrix is close to pH 7.8.
电子沿链传递时，释放的能量把质子从基质泵入膜间隙。这形成质子电化学梯度：膜间隙比基质更酸、更正电。在哺乳动物中，膜间隙 pH 约 7.0-7.4，而基质接近 pH 7.8。
Protons then flow back down the gradient through ATP synthase, a membrane-spanning enzyme with a proton channel and a catalytic head. The flow causes the head to rotate, coupling the exergonic proton movement to the endergonic phosphorylation of ADP. This mechanism is chemiosmosis, proposed by Peter Mitchell in 1961.
随后质子顺浓度梯度经 ATP 合酶回流。ATP 合酶是一种跨膜酶，含有质子通道与催化头部。质子流使头部旋转，把放能的质子移动与吸能的 ADP 磷酸化偶联起来。这一机制即化学渗透，由 Peter Mitchell 于 1961 年提出。
ADP + Pi → ATP
Key exam point: ATP synthase is not part of the electron transport chain. It is a separate enzyme complex that uses the gradient the chain created.
关键考点：ATP 合酶不属于电子传递链。它是独立的酶复合体，利用电子传递链建立的梯度。
7. Calculating the Aerobic ATP Yield | 有氧 ATP 产量的计算
Modern values give approximately 2.5 ATP per NADH and 1.5 ATP per FADH₂ because proton pumping is not perfectly coupled and some protons leak across the membrane. Adding the substrate-level ATP from glycolysis and the Krebs cycle gives roughly 32 ATP per glucose.
现代数值给出约每 NADH 产 2.5 个 ATP、每 FADH₂ 产 1.5 个 ATP，因为质子泵送并非完全偶联，且部分质子会漏过膜。加上糖酵解与三羧酸循环中的底物水平 ATP，总计约每分子葡萄糖 32 个 ATP。
| Stage / 阶段 | ATP / NADH / FADH₂ |
| Glycolysis 糖酵解 | 2 ATP, 2 NADH |
| Link reaction 连接反应 | 0 ATP, 2 NADH |
| Krebs cycle 三羧酸循环 | 2 ATP, 6 NADH, 2 FADH₂ |
| Oxidative phosphorylation 氧化磷酸化 | 10 NADH × 2.5 + 2 FADH₂ × 1.5 ≈ 28 ATP |
| Total 合计 | ≈ 32 ATP（若计入穿梭消耗则为 30） |
If your specification expects 30, the difference comes from the ATP needed to shuttle cytosolic NADH into the mitochondrion using the glycerol phosphate shuttle.
若你的考纲期望 30，差额来自把细胞质 NADH 通过甘油磷酸穿梭运入线粒体所消耗的 ATP。
8. Anaerobic Respiration: Keeping Glycolysis Running | 无氧呼吸：让糖酵解持续运转
Without oxygen the electron transport chain stops, NADH accumulates and NAD⁺ is depleted. Glycolysis would halt because it needs NAD⁺ to oxidise triose phosphate. Anaerobic pathways solve this by regenerating NAD⁺ from NADH, allowing a small but continuous ATP supply.
没有氧时电子传递链停止，NADH 累积而 NAD⁺ 被耗尽。糖酵解将因缺少氧化丙糖磷酸所需的 NAD⁺ 而中断。无氧途径通过把 NADH 重新氧化为 NAD⁺ 来解决这一问题，从而维持少量但持续的 ATP 供应。
In mammals, pyruvate is reduced to lactate by lactate dehydrogenase, catalysing the reaction below. In yeast and plants, pyruvate is decarboxylated to ethanal, which is then reduced to ethanol. Both pathways release no additional CO₂ in the case of lactate, but yeast fermentation produces CO₂, which is why it is used in bread making.
在哺乳动物中，丙酮酸被乳酸脱氢酶还原为乳酸，反应如下。在酵母与植物中，丙酮酸先脱羧生成乙醛，再被还原为乙醇。乳酸途径不额外释放 CO₂，而酵母发酵产生 CO₂，这正是它被用于面包制作的原因。
Pyruvate + NADH + H⁺ → Lactate + NAD⁺
Pyruvate → Ethanal + CO₂; Ethanal + NADH + H⁺ → Ethanol + NAD⁺
The net ATP yield of anaerobic respiration is only 2 per glucose, roughly 6 percent of the aerobic yield. This is why intense exercise cannot be sustained indefinitely and why lactate accumulates in muscle, lowering pH and inhibiting enzymes.
无氧呼吸每分子葡萄糖净产仅 2 个 ATP，约为有氧产量的 6%。这就是剧烈运动无法长期持续、乳酸在肌肉中累积并降低 pH、抑制酶活性的原因。
9. Respiratory Substrates and the Respiratory Quotient | 呼吸底物与呼吸商
Carbohydrates are the preferred substrate, but lipids and amino acids can also be respired. Lipids are hydrolysed to glycerol and fatty acids; glycerol enters glycolysis while fatty acids undergo beta-oxidation to acetyl CoA. Proteins are deaminated first, with the amino group removed as ammonia and the carbon skeleton entering as pyruvate or Krebs cycle intermediates.
糖类是首选底物，但脂质与氨基酸也可被呼吸。脂质水解为甘油与脂肪酸；甘油进入糖酵解，脂肪酸经 β-氧化生成乙酰辅酶 A。蛋白质先脱氨，氨基以氨的形式脱去，碳骨架以丙酮酸或三羧酸循环中间产物形式进入。
The respiratory quotient (RQ) compares carbon dioxide released with oxygen consumed and reveals which substrate is being respired.
呼吸商（RQ）比较释放的 CO₂ 与消耗的 O₂，可揭示正在被呼吸的底物种类。
RQ = CO₂ produced ÷ O₂ consumed
- Carbohydrate: RQ ≈ 1.0 / 糖类：RQ ≈ 1.0
- Protein: RQ ≈ 0.8-0.9 / 蛋白质：RQ ≈ 0.8-0.9
- Lipid: RQ ≈ 0.7 / 脂质：RQ ≈ 0.7
- Mixed diet: RQ ≈ 0.85 / 混合膳食：RQ ≈ 0.85
Lipids give a low RQ because they are highly reduced and hydrogen-rich, so they consume far more oxygen per carbon dioxide released.
脂质 RQ 较低，是因为它们高度还原、富含氢，因此每释放一个 CO₂ 需要消耗更多氧。
10. Measuring Respiration Rate in the Laboratory | 实验室中测量呼吸速率
A respirometer measures oxygen uptake by small organisms such as germinating seeds or woodlice. The organisms are placed on a platform above a carbon dioxide absorbent, usually soda lime or potassium hydroxide solution, in a sealed tube connected to a manometer or capillary U-tube.
呼吸计通过萌发种子或潮虫等小型生物测量耗氧量。生物置于平台上，下方放二氧化碳吸收剂（通常是钠石灰或氢氧化钾溶液），整体密封在管内，并与压力计或毛细 U 形管相连。
As oxygen is consumed, the CO₂ released is absorbed, so the gas volume falls and coloured fluid moves along the capillary. A control tube containing the same mass of dead or boiled organisms corrects for changes in temperature and atmospheric pressure.
随着氧被消耗，释放的 CO₂ 被吸收，气体体积下降，有色液柱沿毛细管移动。对照管放入等量死亡或煮沸过的生物，用于校正温度与大气压的变化。
Key practical points: use a water bath to keep temperature constant and to allow investigation of temperature effects; equilibrate for five to ten minutes before taking readings; read the fluid level at eye level; calculate rate as distance moved divided by time.
关键实验要点：用水浴保持温度恒定并便于探究温度影响；读数前先平衡 5 至 10 分钟；平视读取液面；速率以移动距离除以时间计算。
11. Common Exam Errors and Answer Technique | 常见失分点与答题技巧
Many marks are lost not through lack of knowledge but through imprecise language. Examiners want causal chains with the words "because", "so that" and "therefore", and they want named locations for every process.
许多分数并非因知识缺失而丢失，而是因为语言不精确。考官要求使用"因为""从而使""因此"等词构成因果链，并要求每个过程都点名发生位置。
- Saying glycolysis occurs in the mitochondrion - it is in the cytoplasm / 误说糖酵解在线粒体发生 -- 实际在细胞质基质
- Saying "ATP is made by the electron transport chain" - ATP is made by ATP synthase using the proton gradient / 误说"ATP 由电子传递链制造" -- ATP 由 ATP 合酶利用质子梯度制造
- Forgetting to double yields per glucose / 忘记每分子葡萄糖产量要乘二
- Writing "energy is released as heat" without specifying ATP synthesis / 只写"能量以热形式释放"而未提 ATP 合成
- Saying oxygen "reacts with glucose" - oxygen is the terminal electron acceptor at the end of the chain / 误说氧"与葡萄糖反应" -- 氧是链末端的最终电子受体
A strong six-mark answer typically names the stage, names the enzyme or carrier, states the location, quantifies the yield and explains the consequence of inhibition.
一道优秀的 6 分题答案通常会点出阶段、酶或载体名称、发生位置、量化产量，并解释抑制后的后果。
12. Uncouplers, Inhibitors and Real-World Links | 解偶联剂、抑制剂与现实联系
Uncoupling agents such as 2,4-dinitrophenol carry protons across the inner membrane without passing through ATP synthase. The gradient collapses, ATP synthesis falls, and the energy is released as heat instead. This is exactly what uncoupling protein 1 does in brown adipose tissue of hibernating mammals and newborn babies.
2,4-二硝基苯酚等解偶联剂可携带质子穿过内膜而不经过 ATP 合酶。梯度瓦解，ATP 合成下降，能量改以热的形式释放。冬眠哺乳动物与新生儿的棕色脂肪中的解偶联蛋白 1 正是如此发挥作用。
Respiratory inhibitors act at specific points: cyanide blocks cytochrome oxidase, the final complex, so the whole chain backs up. Oligomycin blocks the proton channel of ATP synthase, causing the gradient to build and electron transport to slow. Understanding these gives precise answers in data-analysis questions.
呼吸抑制剂作用于特定环节：氰化物阻断末端复合体细胞色素氧化酶，使整条链堵塞；寡霉素阻断 ATP 合酶的质子通道，使梯度积累、电子传递减慢。理解这些机制能在数据分析题中给出精确答案。
Finally, remember the wider biological context. Respiration connects to photosynthesis through the shared intermediates of the Calvin cycle and the Krebs cycle, links to homeostasis through thermoregulation, and underpins every calculation of biomass, productivity and metabolic rate you will meet in the ecology and physiology papers.
最后，记住更宏观的生物学语境。呼吸通过卡尔文循环与三羧酸循环的共享中间产物与光合作用相连，通过体温调节与稳态相连，并支撑你在生态学与生理学试卷中会遇到的一切关于生物量、生产力与代谢速率的计算。
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