A-Level Biology: Cellular Respiration Complete Guide | 细胞呼吸完全指南

掌握细胞呼吸：从糖酵解到氧化磷酸化 — A-Level 生物高分指南

Cellular respiration is one of the most heavily examined topics in A-Level Biology — and one of the most misunderstood. Whether you’re studying CIE, AQA, Edexcel, or OCR, you will face questions that test not just your ability to recall the stages, but your deep understanding of why each step happens and how they connect. This guide walks you through the entire process in both English and Chinese, with clear diagrams, exam-focused explanations, and the most common pitfalls to avoid.

细胞呼吸是 A-Level 生物考试中最常考、也最容易被误解的章节之一。无论你学的是 CIE、AQA、Edexcel 还是 OCR 考纲，你面对的考题不仅要求你记住各个阶段，更考验你对为什么每一步会发生、以及它们如何相互连接的深刻理解。本文用中英双语带你走完整条细胞呼吸路径，配合清晰的解释和最常见的考试陷阱分析。

1. The Big Picture: What Is Cellular Respiration? | 什么是细胞呼吸？

Cellular respiration is the process by which cells break down organic molecules — primarily glucose — to release energy in the form of ATP (adenosine triphosphate). This energy is then used to power every cellular process, from active transport to protein synthesis. The overall equation is deceptively simple:

细胞呼吸是细胞分解有机分子（主要是葡萄糖）以释放 ATP（三磷酸腺苷） 形式能量的过程。这些能量随后用于驱动从主动运输到蛋白质合成的每一个细胞过程。总方程式看似简单：

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

But behind this simple equation lies a beautifully orchestrated sequence of four major stages: Glycolysis, the Link Reaction, the Krebs Cycle, and Oxidative Phosphorylation. Each stage takes place in a specific location within the cell and involves a cascade of enzyme-controlled reactions. Let’s explore each one.

但在这个简单方程背后，隐藏着一个精妙编排的四阶段序列：糖酵解、连接反应、克雷布斯循环和氧化磷酸化。每个阶段都在细胞内特定位置发生，并涉及一系列酶控制的反应。让我们逐一探索。

2. Stage 1: Glycolysis — The Universal First Step | 第一阶段：糖酵解

Location: Cytoplasm (cytosol)
Oxygen requirement: None (occurs in both aerobic and anaerobic conditions)

位置：细胞质（胞质溶胶）
需氧情况：不需要（有氧和无氧条件下均可发生）

The Process | 过程

Glycolysis literally means “sugar splitting”. A single molecule of glucose (a 6-carbon sugar) is split into two molecules of pyruvate (a 3-carbon compound). This happens through a series of 10 enzyme-catalysed reactions, which can be divided into two phases:

糖酵解的字面意思是”糖的裂解”。一分子葡萄糖（6碳糖）被分解为两分子丙酮酸（3碳化合物）。这通过一系列10个酶催化反应完成，可分为两个阶段：

• Energy Investment Phase (磷酸化阶段): 2 ATP molecules are used to phosphorylate glucose. This makes the glucose molecule more reactive and prevents it from leaving the cell. The phosphorylated glucose (fructose-1,6-bisphosphate) is then split into two 3-carbon molecules — glyceraldehyde-3-phosphate (GALP).
• Energy Payoff Phase (能量释放阶段): Each GALP molecule is oxidised through a series of steps. NAD⁺ acts as a hydrogen carrier and is reduced to NADH (reduced NAD). Substrate-level phosphorylation produces ATP directly — 4 ATP molecules are generated per glucose (2 per GALP).

• 能量投入阶段： 消耗 2 个 ATP 分子来磷酸化葡萄糖。这使葡萄糖分子更具反应活性并阻止其离开细胞。磷酸化的葡萄糖（果糖-1,6-二磷酸）随后被裂解为两个3碳分子——甘油醛-3-磷酸（GALP）。
• 能量回报阶段： 每个 GALP 分子通过一系列步骤被氧化。NAD⁺ 作为氢载体被还原为 NADH（还原型 NAD）。底物水平磷酸化直接产生 ATP——每分子葡萄糖净生成 4 个 ATP（每个 GALP 产生 2 个）。

Net Products of Glycolysis (per glucose) | 糖酵解的净产物（每分子葡萄糖）

Product 产物	Amount 数量
Pyruvate 丙酮酸	2
ATP (net) ATP（净值）	2
NADH (reduced NAD) 还原型 NAD	2

💡 Exam Tip: Many students forget to say “net” when stating the ATP yield of glycolysis (4 produced, 2 used, net = 2). Always specify that the net gain is 2 ATP. Also remember that glycolysis occurs in the cytoplasm, not in the mitochondria — a common trick in multiple-choice questions.

💡 考试提示： 许多学生在陈述糖酵解的 ATP 产量时忘记说”净”。请一定明确净收益是 2 个 ATP（产生 4 个，消耗 2 个）。还要记住糖酵解发生在细胞质中，而非线粒体——这是选择题中常见的陷阱。

3. Stage 2: The Link Reaction — Bridging Glycolysis and the Krebs Cycle | 第二阶段：连接反应

Location: Mitochondrial matrix
Oxygen requirement: Indirect (NAD⁺ regeneration requires O₂ in the electron transport chain)

位置：线粒体基质
需氧情况：间接需要（NAD⁺ 的再生需要电子传递链中的 O₂）

Before pyruvate can enter the Krebs cycle, it must be converted to acetyl coenzyme A (acetyl-CoA). This is the Link Reaction, and it happens as pyruvate crosses from the cytoplasm into the mitochondrial matrix. For each pyruvate molecule:

在丙酮酸进入克雷布斯循环之前，它必须转化为乙酰辅酶 A（acetyl-CoA）。这就是连接反应，发生在丙酮酸从细胞质穿过线粒体膜进入基质时。对每个丙酮酸分子：

1. Decarboxylation (脱羧): A carboxyl group (COO⁻) is removed, releasing CO₂.
2. Dehydrogenation (脱氢): Two hydrogen atoms are removed. NAD⁺ is reduced to NADH.
3. Coenzyme A attachment (辅酶A结合): The remaining 2-carbon acetyl group binds to coenzyme A, forming acetyl-CoA.

1. 脱羧： 移除羧基（COO⁻），释放 CO₂。
2. 脱氢： 移除两个氢原子，NAD⁺ 被还原为 NADH。
3. 辅酶A结合： 剩余的2碳乙酰基与辅酶A结合，形成乙酰辅酶A。

Key point: Since one glucose produces two pyruvate molecules, the Link Reaction occurs twice per glucose. This yields: 2 acetyl-CoA, 2 CO₂, and 2 NADH per glucose molecule.

关键点： 由于一分子葡萄糖产生两分子丙酮酸，连接反应每分子葡萄糖发生两次。这产生：每分子葡萄糖生成 2 分子乙酰辅酶A、2 分子 CO₂ 和 2 分子 NADH。

💡 Exam Tip: The Link Reaction is often a “hidden” mark in longer questions. Students jump straight from glycolysis to the Krebs cycle and lose marks. Always include it explicitly. Also note that the CO₂ released here is the first time carbon dioxide appears in aerobic respiration.

💡 考试提示： 连接反应在长答题中常常是”隐藏”的得分点。学生往往直接从糖酵解跳到克雷布斯循环从而丢分。请务必明确写出连接反应。还要注意，这里释放的 CO₂ 是有氧呼吸中二氧化碳首次出现的位置。

4. Stage 3: The Krebs Cycle — The Metabolic Hub | 第三阶段：克雷布斯循环

Location: Mitochondrial matrix
Also known as: The Citric Acid Cycle, The Tricarboxylic Acid (TCA) Cycle

位置：线粒体基质
别称：柠檬酸循环、三羧酸（TCA）循环

The Krebs cycle is a closed loop of enzyme-controlled reactions that oxidises acetyl-CoA completely to CO₂. For each turn of the cycle (one acetyl-CoA molecule):

克雷布斯循环是一个酶控制的闭环反应序列，将乙酰辅酶A完全氧化为 CO₂。循环每运转一圈（一分子乙酰辅酶A）：

• Acetyl-CoA (2C) combines with oxaloacetate (4C) to form citrate (6C). Coenzyme A is released and recycled.
• Citrate undergoes a series of rearrangements and oxidations, releasing 2 CO₂ molecules through decarboxylation.
• 3 NAD⁺ are reduced to 3 NADH through dehydrogenation.
• 1 FAD is reduced to 1 FADH₂ (another hydrogen carrier, though it carries electrons to a different point in the electron transport chain).
• 1 ATP is produced directly via substrate-level phosphorylation (actually GTP, which is readily converted to ATP).
• Oxaloacetate is regenerated, ready to accept the next acetyl-CoA.

• 乙酰辅酶A（2C）与草酰乙酸（4C）结合形成柠檬酸（6C）。辅酶A被释放并循环利用。
• 柠檬酸经过一系列重排和氧化反应，通过脱羧释放 2 分子 CO₂。
• 3 个 NAD⁺ 通过脱氢被还原为 3 个 NADH。
• 1 个 FAD 被还原为 1 个 FADH₂（另一种氢载体，但其电子进入电子传递链的位置与 NADH 不同）。
• 通过底物水平磷酸化直接产生 1 个 ATP（实际是 GTP，但可轻易转化为 ATP）。
• 草酰乙酸被再生，准备接受下一个乙酰辅酶A。

Products Per Glucose (Two Turns of the Cycle) | 每分子葡萄糖的产物（循环运转两圈）

Product 产物	Per Turn 每圈	Per Glucose 每分子葡萄糖
CO₂	2	4
NADH	3	6
FADH₂	1	2
ATP (GTP)	1	2

💡 Exam Tip: AQA and OCR exam boards particularly love asking about the role of coenzymes (NAD, FAD, CoA) in respiration. Be able to name each one, state what it carries, and where it fits in the process. Also, don’t confuse FAD with FADH₂ — FAD is the oxidised form, FADH₂ is the reduced form.

💡 考试提示： AQA 和 OCR 考纲特别喜欢考察辅酶（NAD、FAD、CoA）在呼吸中的作用。要能够说出每种辅酶的名称、携带什么、以及在过程中处于什么位置。另外，不要混淆 FAD 和 FADH₂——FAD 是氧化型，FADH₂ 是还原型。

5. Stage 4: Oxidative Phosphorylation — Where Most ATP Is Made | 第四阶段：氧化磷酸化

Location: Inner mitochondrial membrane (cristae)
Oxygen requirement: Essential — O₂ is the final electron acceptor

位置：线粒体内膜（嵴）
需氧情况：必需 — O₂ 是最终电子受体

Oxidative phosphorylation is where the real ATP payoff happens. It consists of two linked processes: the Electron Transport Chain (ETC) and Chemiosmosis.

氧化磷酸化是真正大量产生 ATP 的阶段。它由两个相连的过程组成：电子传递链（ETC） 和 化学渗透。

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

The reduced coenzymes NADH and FADH₂, produced in the earlier stages, now donate their electrons to a series of protein complexes embedded in the inner mitochondrial membrane:

前期产生的还原型辅酶 NADH 和 FADH₂ 现在将其电子传递给嵌入线粒体内膜的一系列蛋白质复合体：

1. NADH donates electrons to Complex I (NADH dehydrogenase). The electrons pass through a series of redox carriers — including ubiquinone (coenzyme Q) and cytochrome c — each at progressively lower energy levels.
2. FADH₂ donates electrons to Complex II (succinate dehydrogenase), entering the chain at a slightly lower energy level than NADH. This is why FADH₂ produces fewer ATP molecules than NADH.
3. At each transfer, the energy released is used to pump protons (H⁺) from the mitochondrial matrix into the intermembrane space. This creates an electrochemical gradient — a high concentration of protons in the intermembrane space relative to the matrix.
4. Finally, electrons are passed to Complex IV (cytochrome c oxidase), where they combine with oxygen (O₂) and protons to form water (H₂O). This is why oxygen is essential — without it, electrons back up along the chain and the whole process halts.

1. NADH 向复合体 I（NADH脱氢酶）提供电子。 电子经过一系列氧化还原载体——包括泛醌（辅酶Q）和细胞色素c——每个载体的能量水平逐渐降低。
2. FADH₂ 向复合体 II（琥珀酸脱氢酶）提供电子， 进入链的能量水平略低于 NADH。这就是 FADH₂ 产生的 ATP 比 NADH 少的原因。
3. 每次转移释放的能量用于将质子（H⁺）从线粒体基质泵入膜间空间。这创建了一个电化学梯度——膜间空间中的质子浓度远高于基质。
4. 最终，电子传递到复合体 IV（细胞色素c氧化酶），在此与氧气（O₂）和质子结合形成水（H₂O）。这就是氧气必不可少的原因——没有它，电子会在链上堆积，整个过程停止。

5.2 Chemiosmosis | 化学渗透

The proton gradient built up by the ETC represents stored potential energy. Protons can only flow back into the matrix through a specialised channel: ATP synthase (also called ATP synthetase). As protons flow down their electrochemical gradient through this enzyme, the energy released drives the synthesis of ATP from ADP and inorganic phosphate (Pi):

ETC 建立的质子梯度代表了储存的势能。质子只能通过一个特殊通道流回基质：ATP合酶。当质子沿其电化学梯度流经此酶时，释放的能量驱动 ADP 与无机磷酸（Pi）合成 ATP：

ADP + Pi → ATP

This coupling of electron transport to ATP synthesis via a proton gradient is the chemiosmotic theory, proposed by Peter Mitchell in 1961 — for which he won the Nobel Prize in Chemistry in 1978.

这种通过质子梯度将电子传递与 ATP 合成耦合的机制就是化学渗透学说，由 Peter Mitchell 于 1961 年提出——他因此获得了 1978 年诺贝尔化学奖。

ATP Yield: The Full Accounting | ATP 产量：完整核算

The theoretical maximum ATP yield from one glucose molecule in aerobic respiration is approximately 30–32 ATP (the exact number varies because the proton yield from NADH and FADH₂ depends on which shuttle system transports cytoplasmic NADH into the mitochondria):

一分子葡萄糖在有氧呼吸中的理论最大 ATP 产量约为 30–32 ATP（确切数字会变化，因为 NADH 和 FADH₂ 的质子产量取决于将细胞质 NADH 转运到线粒体的穿梭系统）：

Stage 阶段	ATP Produced (approx.) ATP 产量（约）
Glycolysis 糖酵解	2 (net)
Link Reaction 连接反应	0 (only produces NADH)
Krebs Cycle 克雷布斯循环	2 (GTP → ATP)
Oxidative Phosphorylation 氧化磷酸化	~26–28 (from NADH + FADH₂)
Total 总计	~30–32

💡 Exam Tip: NADH produced in glycolysis (in the cytoplasm) yields slightly fewer ATP molecules than NADH produced in the mitochondrial matrix. This is because the NADH must be shuttled across the mitochondrial membrane, which costs energy. In eukaryotic cells, cytoplasmic NADH typically yields ~1.5 ATP (via the glycerol-3-phosphate shuttle) rather than the ~2.5 ATP from mitochondrial NADH. FADH₂ yields ~1.5 ATP.

💡 考试提示： 在糖酵解（细胞质）中产生的 NADH 比在线粒体基质中产生的 NADH 产出略少的 ATP。这是因为 NADH 必须通过穿梭系统跨线粒体膜转运，这需要消耗能量。在真核细胞中，细胞质 NADH 通常产生约 1.5 ATP（通过甘油-3-磷酸穿梭），而非线粒体 NADH 的约 2.5 ATP。FADH₂ 产生约 1.5 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 operate (there is no final electron acceptor). However, glycolysis can still proceed — but only if NAD⁺ is available. In anaerobic conditions, cells regenerate NAD⁺ through fermentation pathways:

当氧气不可用时，电子传递链无法运作（没有最终电子受体）。然而，糖酵解仍然可以进行——但前提是 NAD⁺ 可用。在无氧条件下，细胞通过发酵途径再生 NAD⁺：

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

Pyruvate + NADH → Lactate + NAD⁺

The enzyme lactate dehydrogenase catalyses this reaction. NAD⁺ is regenerated, allowing glycolysis to continue producing 2 ATP per glucose — far less than aerobic respiration, but enough for short-term survival. Lactate can be converted back to pyruvate in the liver when oxygen becomes available again (the Cori cycle).

乳酸脱氢酶催化此反应。NAD⁺ 被再生，使糖酵解得以继续，每分子葡萄糖产生 2 ATP——远少于有氧呼吸，但足以短期维持生存。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）。

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

Pyruvate → Ethanal + CO₂ (decarboxylation, catalysed by pyruvate decarboxylase)
Ethanal + NADH → Ethanol + NAD⁺ (catalysed by alcohol dehydrogenase)

丙酮酸 → 乙醛 + CO₂（脱羧，由丙酮酸脱羧酶催化）
乙醛 + NADH → 乙醇 + NAD⁺（由乙醇脱氢酶催化）

💡 Exam Tip: A common mistake is stating that anaerobic respiration produces CO₂ in animals — it doesn’t! Only the ethanol fermentation pathway in plants and yeast releases CO₂. Also, remember that the sole purpose of anaerobic respiration is to regenerate NAD⁺ so that glycolysis can continue. The ATP yield is only 2 per glucose (from glycolysis alone).

💡 考试提示： 常见错误是声称无氧呼吸在动物中产生 CO₂——事实并非如此！只有植物和酵母中的乙醇发酵途径才释放 CO₂。此外，请记住无氧呼吸的唯一目的是再生 NAD⁺ 以使糖酵解得以继续。ATP 产量仅为每分子葡萄糖 2 个（仅来自糖酵解）。

7. Key Coenzymes Summary | 关键辅酶总结

Coenzyme 辅酶	Role 角色	Reduced Form 还原型	Stage(s) Involved 涉及阶段
NAD⁺	Hydrogen/electron carrier 氢/电子载体	NADH	Glycolysis, Link, Krebs
FAD	Hydrogen/electron carrier 氢/电子载体	FADH₂	Krebs Cycle
Coenzyme A	Acetyl group carrier 乙酰基载体	Acetyl-CoA	Link Reaction

8. Common Exam Pitfalls | 常见考试陷阱

1. Confusing location: Glycolysis = cytoplasm, NOT mitochondria. The Krebs cycle = mitochondrial matrix, NOT cristae. Oxidative phosphorylation = inner mitochondrial membrane (cristae).
2. Forgetting the Link Reaction: Always include it between glycolysis and the Krebs cycle when describing the full pathway.
3. “Energy is produced”: Energy is not “produced” or “created” — it is released from chemical bonds and transferred to ATP. Use precise language.
4. CO₂ in anaerobic respiration: Only ethanol fermentation (plants/yeast) releases CO₂. Lactate fermentation (animals) does NOT.
5. Role of oxygen: Oxygen is the final electron acceptor in the ETC, not a direct reactant in glycolysis or the Krebs cycle. Say “oxygen accepts electrons” rather than “oxygen is used to break down glucose”.
6. NAD vs NADH: NAD⁺ is the oxidised form; NADH is the reduced form. Don’t mix them up — examiners look for this.
7. ATP counting: State whether you are giving gross or net ATP figures. For glycolysis, always specify “net gain of 2 ATP”.

1. 混淆位置： 糖酵解 = 细胞质，不是线粒体。克雷布斯循环 = 线粒体基质，不是嵴。氧化磷酸化 = 线粒体内膜（嵴）。
2. 忘记连接反应： 在描述完整通路时，一定要将连接反应放在糖酵解和克雷布斯循环之间。
3. “能量被产生”： 能量不是”产生”或”创造”的——它是从化学键中释放并转移到 ATP 中的。使用精确的语言。
4. 无氧呼吸中的 CO₂： 只有乙醇发酵（植物/酵母）释放 CO₂。乳酸发酵（动物）不释放。
5. 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。应该说”氧气接受电子”而非”氧气用于分解葡萄糖”。
6. NAD vs NADH： NAD⁺ 是氧化型；NADH 是还原型。不要混淆——考官会检查这一点。
7. ATP 计数： 说明你给的是总产量还是净产量。对于糖酵解，始终指明”净增益 2 ATP”。

9. Summary Table — The Complete Picture | 总结表 — 完整图像

Stage 阶段	Location 位置	O₂ Needed? 需要O₂？	Input 输入	Output 输出	ATP (net) ATP（净）
Glycolysis 糖酵解	Cytoplasm 细胞质	No 否	Glucose	2 Pyruvate, 2 NADH	2
Link Reaction 连接反应	Mitochondrial matrix 线粒体基质	Indirectly 间接	2 Pyruvate	2 Acetyl-CoA, 2 CO₂, 2 NADH	0
Krebs Cycle 克雷布斯循环	Mitochondrial matrix 线粒体基质	Indirectly 间接	2 Acetyl-CoA	4 CO₂, 6 NADH, 2 FADH₂, 2 ATP	2
Oxidative Phosphorylation 氧化磷酸化	Inner mitochondrial membrane 线粒体内膜	Yes 是	NADH, FADH₂, O₂	H₂O, NAD⁺, FAD	~26–28

10. Practice Questions | 练习题

Test your understanding with these exam-style questions:

用以下考试风格的题目测试你的理解：

1. Explain why the theoretical yield of ATP from aerobic respiration is rarely achieved in living cells. (4 marks)
   解释为什么有氧呼吸的理论 ATP 产量在活细胞中很少达到。（4分）
2. Describe the role of NAD in cellular respiration and explain why its regeneration is essential. (6 marks)
   描述 NAD 在细胞呼吸中的作用，并解释为什么其再生至关重要。（6分）
3. Compare and contrast aerobic and anaerobic respiration in mammalian cells. (5 marks)
   比较和对比哺乳动物细胞中的有氧呼吸和无氧呼吸。（5分）
4. Explain how the structure of mitochondria is adapted for oxidative phosphorylation. (4 marks)
   解释线粒体的结构如何适应氧化磷酸化。（4分）

Answer Hints | 答案提示:

Q1: Some protons leak back across the membrane without passing through ATP synthase; some ATP is used to transport pyruvate and NADH into mitochondria; the theoretical maximum assumes 100% efficiency which doesn’t occur in reality.

Q1 提示： 部分质子泄漏穿过膜而不经过 ATP 合酶；部分 ATP 用于将丙酮酸和 NADH 运入线粒体；理论最大值假设 100% 效率，实际并不存在。

Q2: NAD acts as a hydrogen carrier — it accepts hydrogen atoms (is reduced to NADH) during glycolysis, the Link Reaction, and the Krebs cycle. It then carries these to the ETC where it’s oxidised back to NAD⁺. Without regeneration, glycolysis and the Krebs cycle would stop due to lack of NAD⁺.

Q2 提示： NAD 作为氢载体——在糖酵解、连接反应和克雷布斯循环中接受氢原子（被还原为 NADH）。然后将其携带至 ETC，在此被氧化回 NAD⁺。没有再生，糖酵解和克雷布斯循环将因缺乏 NAD⁺ 而停止。

Q3: Aerobic: uses O₂, produces ~30-32 ATP per glucose, CO₂ and H₂O as waste, occurs in cytoplasm + mitochondria. Anaerobic: no O₂, produces only 2 ATP per glucose (from glycolysis only), lactate as waste in animals (no CO₂), occurs only in cytoplasm.

Q3 提示： 有氧：使用 O₂，每分子葡萄糖产生约 30-32 ATP，CO₂ 和 H₂O 为废物，发生在细胞质+线粒体中。无氧：不使用 O₂，每分子葡萄糖仅产生 2 ATP（仅来自糖酵解），动物中乳酸为废物（无 CO₂），仅发生在细胞质中。

Q4: Cristae (folded inner membrane) provide a large surface area for ETC proteins and ATP synthase; the intermembrane space is small so proton concentration builds up quickly; the matrix contains enzymes for the Link Reaction and Krebs cycle.

Q4 提示： 嵴（折叠的内膜）为 ETC 蛋白和 ATP 合酶提供了大表面积；膜间空间狭小，因此质子浓度迅速积累；基质含有连接反应和克雷布斯循环所需的酶。

Final Thoughts | 最后的话

Cellular respiration is a topic that rewards systematic study. Rather than memorising isolated facts, focus on understanding the flow — how the product of one stage becomes the substrate for the next. Draw the pathways repeatedly from memory. Trace the carbon atoms from glucose all the way to CO₂. And most importantly, practise explaining the process in your own words — this is the skill that will earn you top marks in the exam.

细胞呼吸是一个值得系统学习的主题。与其记忆孤立的事实，不如专注于理解流程——一个阶段的产物如何成为下一个阶段的底物。反复凭记忆画出通路。追踪碳原子从葡萄糖一直到 CO₂ 的路径。最重要的是，练习用自己的语言解释这个过程——这是让你在考试中获得高分的关键技能。

Good luck with your studies! 祝学习顺利！🎓