📚 Respiration: Key Points and Exam Tips for IB and CIE Biology | IB与CIE生物：呼吸作用考点精讲
Respiration is a fundamental metabolic process that releases energy from organic molecules to fuel cellular activities. For IB and CIE Biology examinations, understanding both aerobic and anaerobic pathways, their locations within the cell, the role of coenzymes, and the principles of chemiosmosis is essential. This article provides a comprehensive, syllabus-focused review of respiration, equipping you with the precise knowledge and exam strategies needed to score top marks.
呼吸作用是释放有机分子中的能量以驱动细胞活动的基本代谢过程。在IB和CIE生物考试中，理解需氧和厌氧途径、它们在细胞内的位置、辅酶的作用以及化学渗透原理至关重要。本文提供了一份紧扣大纲的呼吸作用全面复习，让你掌握取得高分所需的精准知识和考试策略。
1. Overview of Respiration | 呼吸作用总览
Respiration is the enzyme-controlled release of chemical potential energy from organic compounds, typically glucose, to produce ATP. It occurs in all living cells and can be aerobic (requiring oxygen) or anaerobic (occurring in the absence of oxygen). The overall equation for aerobic respiration is: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy (as ATP). However, this single equation masks a complex series of metabolic pathways: glycolysis, the link reaction, the Krebs cycle, and oxidative phosphorylation.
呼吸作用是酶控的从有机化合物（通常是葡萄糖）中释放化学势能以产生ATP的过程。它发生在所有活细胞中，可以是需氧的（需要氧气）或厌氧的（在无氧条件下发生）。需氧呼吸的总方程式为：C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + 能量（以ATP形式）。然而，这个简单方程式掩盖了一系列复杂的代谢途径：糖酵解、链接反应、克雷伯氏循环和氧化磷酸化。
In eukaryotic cells, respiration is compartmentalised: glycolysis takes place in the cytoplasm, while the link reaction, Krebs cycle, and oxidative phosphorylation occur in the mitochondria. Prokaryotes carry out respiration across the cell membrane and cytoplasm. Both IB and CIE syllabi expect you to be able to identify where each stage occurs and explain why this compartmentalisation is vital for efficiency.
在真核细胞中，呼吸作用是分区进行的：糖酵解发生在细胞质中，而链接反应、克雷伯氏循环和氧化磷酸化在线粒体中进行。原核生物则跨细胞膜和细胞质进行呼吸作用。IB和CIE大纲都要求你能够识别每个阶段发生的位置，并解释为什么这种分区对效率至关重要。
2. Glycolysis: The Universal First Step | 糖酵解：通用的第一步
Glycolysis is the breakdown of one molecule of glucose (a 6‑carbon sugar) into two molecules of pyruvate (each 3‑carbon). It occurs in the cytoplasm, does not require oxygen, and is found in nearly all organisms. The process consumes 2 ATP in the energy investment phase but produces 4 ATP by substrate‑level phosphorylation, yielding a net gain of 2 ATP per glucose. Additionally, 2 molecules of NAD⁺ are reduced to NADH + H⁺, which will be used later in the electron transport chain.
糖酵解是将一分子葡萄糖（六碳糖）分解为两分子丙酮酸（各为三碳）的过程。它发生在细胞质中，不需要氧气，几乎存在于所有生物体中。该过程在能量投入阶段消耗2个ATP，但通过底物水平磷酸化产生4个ATP，每分子葡萄糖净得2个ATP。此外，2分子NAD⁺被还原为NADH + H⁺，这将在后续的电子传递链中被使用。
The key steps to remember for the exam are the phosphorylation of glucose (using ATP) to make it more reactive, lysis of the hexose bisphosphate into two triose phosphates, and the subsequent oxidation and dephosphorylation to form pyruvate. In IB exams, you may be asked to outline glycolysis in terms of phosphorylation, lysis, oxidation, and ATP formation; CIE expects you to recall the involvement of NAD and the net yield.
考试中需要记住的关键步骤是：葡萄糖的磷酸化（消耗ATP）使其更活泼，己糖二磷酸裂解为两分子丙糖磷酸，随后经氧化和去磷酸化形成丙酮酸。在IB考试中，你可能被要求从磷酸化、裂解、氧化和ATP形成的角度概述糖酵解；CIE则期望你记住NAD的参与和净产量。
3. The Link Reaction: Connecting Glycolysis to the Krebs Cycle | 链接反应：连接糖酵解与克雷伯氏循环
Once pyruvate enters the mitochondrial matrix, it undergoes the link reaction (also called pyruvate decarboxylation). Each pyruvate molecule is decarboxylated (CO₂ is removed) and oxidised by NAD⁺ to form an acetyl group, which immediately combines with coenzyme A to produce acetyl‑CoA. For one glucose, two pyruvates yield two acetyl‑CoA, two CO₂, and two reduced NAD (NADH + H⁺). No ATP is produced directly in this step.
一旦丙酮酸进入线粒体基质，就会发生链接反应（也称丙酮酸脱羧）。每分子丙酮酸被脱羧（脱去CO₂）并被NAD⁺氧化生成乙酰基，乙酰基立即与辅酶A结合形成乙酰辅酶A。对于一分子葡萄糖，两分子丙酮酸产生两分子乙酰辅酶A、两分子CO₂和两分子还原型NAD（NADH + H⁺）。此步骤不直接产生ATP。
This reaction is irreversible and crucial because it commits the carbon atoms from glucose to complete oxidation. Both IB and CIE require you to name the location (mitochondrial matrix) and the products per glucose. Be precise: decarboxylation and oxidation are the key processes, and NAD is the hydrogen acceptor.
该反应不可逆且至关重要，因为它使来自葡萄糖的碳原子进入彻底氧化阶段。IB和CIE都要求你指明发生位置（线粒体基质）和每分子葡萄糖的产物。注意细节：脱羧和氧化是关键过程，NAD是氢受体。
4. The Krebs Cycle (Citric Acid Cycle) | 克雷伯氏循环（柠檬酸循环）
The acetyl‑CoA (2‑carbon) enters the Krebs cycle by combining with oxaloacetate (4‑carbon) to form citrate (6‑carbon). Through a series of enzyme‑controlled reactions, citrate is gradually decarboxylated and oxidised, regenerating oxaloacetate. For each turn of the cycle, two CO₂ molecules are released, one ATP (or GTP) is produced by substrate‑level phosphorylation, and three NADH + H⁺ and one FADH₂ are generated. Since two acetyl‑CoA are produced per glucose, the cycle turns twice, doubling these products.
乙酰辅酶A（2碳）进入克雷伯氏循环，与草酰乙酸（4碳）结合生成柠檬酸（6碳）。通过一系列酶控反应，柠檬酸逐步脱羧氧化，重新生成草酰乙酸。循环每转一圈，释放两分子CO₂，通过底物水平磷酸化产生一分子ATP（或GTP），并生成三分子NADH + H⁺和一分子FADH₂。由于每分子葡萄糖产生两分子乙酰辅酶A，循环需转两圈，产物相应翻倍。
Remember that the Krebs cycle occurs in the mitochondrial matrix. The electron carriers NADH and FADH₂ are vital—they store energy in the form of high‑energy electrons that will be transferred to the electron transport chain. Exam questions frequently ask for the numbers of reduced coenzymes produced per glucose: 6 NADH, 2 FADH₂, and 2 ATP. You must also be able to name the 4‑carbon starter molecule, oxaloacetate.
请记住克雷伯氏循环发生在线粒体基质中。电子载体NADH和FADH₂至关重要——它们以高能电子的形式储存能量，后续将传递给电子传递链。考题经常询问每分子葡萄糖产生的还原型辅酶数量：6个NADH、2个FADH₂和2个ATP。你还需能说出四碳起始分子草酰乙酸的名称。
5. Oxidative Phosphorylation: Electron Transport Chain and Chemiosmosis | 氧化磷酸化：电子传递链与化学渗透
Oxidative phosphorylation is the final stage of aerobic respiration, taking place on the inner mitochondrial membrane (cristae). It comprises the electron transport chain (ETC) and chemiosmosis. Reduced NAD and FADH₂ donate electrons to the ETC, a series of carrier proteins embedded in the cristae. As electrons pass from one carrier to the next, energy is released to pump protons (H⁺) from the matrix into the intermembrane space, creating a proton gradient (an electrochemical gradient).
氧化磷酸化是需氧呼吸的最后阶段，发生在线粒体内膜（嵴）上。它包括电子传递链（ETC）和化学渗透。还原型NAD和FADH₂将电子提供给ETC，即嵌入嵴中的一系列载体蛋白。电子从一个载体传递到下一个时释放能量，用于将质子（H⁺）从基质泵入膜间隙，形成质子梯度（电化学梯度）。
Oxygen acts as the final electron acceptor, combining with electrons and H⁺ to form water: ½O₂ + 2e⁻ + 2H⁺ → H₂O. This step is vital; without oxygen, the ETC backs up and respiration halts. The proton gradient powers ATP synthase, a channel protein that allows H⁺ to flow back into the matrix. The energy of this flow drives the phosphorylation of ADP + Pᵢ to form ATP. This coupling of electron transport to ATP synthesis is chemiosmosis.
氧气作为最终电子受体，与电子和H⁺结合生成水：½O₂ + 2e⁻ + 2H⁺ → H₂O。此步骤至关重要；没有氧气，ETC会阻塞，呼吸作用停止。质子梯度驱动ATP合酶，这是一种通道蛋白，允许H⁺流回基质。这股流动的能量驱动ADP + Pᵢ磷酸化生成ATP。这种电子传递与ATP合成的偶联即为化学渗透。
6. ATP Yield and Theoretical Efficiency | ATP产量与理论效率
The total ATP yield from one glucose molecule varies slightly between textbooks and organisms due to the shuttle systems for cytoplasmic NADH. In the ideal scenario, the yields are: glycolysis – 2 ATP and 2 NADH; link reaction – 2 NADH; Krebs cycle – 2 ATP, 6 NADH, and 2 FADH₂. Each NADH is theoretically used to synthesise about 2.5 ATP in the ETC, and each FADH₂ about 1.5 ATP. Summing these gives a total of approximately 30–32 ATP per glucose in eukaryotes. Prokaryotes can yield up to 38 ATP because no energy is spent shuttling NADH into mitochondria.
由于细胞质NADH的穿梭系统不同，一分子葡萄糖的总ATP产量在不同教科书和生物体之间略有差异。理想情况下，产量为：糖酵解——2 ATP和2 NADH；链接反应——2 NADH；克雷伯氏循环——2 ATP、6 NADH和2 FADH₂。理论上每个NADH在ETC中可合成约2.5个ATP，每个FADH₂约1.5个ATP。求和后真核生物每分子葡萄糖约产生30–32个ATP。原核生物最多可产生38个ATP，因为无需消耗能量将NADH转运入线粒体。
For CIE exams, it is common to state the theoretical maximum as 38 ATP, but you should follow your specification. IB typically presents 36 ATP as a standard value or expects you to explain the variation. In both boards, you must show that the vast majority of ATP is produced during oxidative phosphorylation, highlighting the importance of oxygen.
CIE考试中通常陈述理论最大值为38 ATP，但应遵循你所学的大纲。IB一般以36 ATP为标准值，或期望你解释差异的原因。在两个考试局中，你都必须表明绝大多数ATP是在氧化磷酸化过程中产生的，这凸显了氧气的重要性。
7. Anaerobic Respiration: Fermentation Pathways | 厌氧呼吸：发酵途径
When oxygen is not available, cells can regenerate NAD⁺ through fermentation, allowing glycolysis to continue producing a small amount of ATP. There are two main types: lactate fermentation (in animal muscles and some bacteria) and alcoholic fermentation (in yeast and some plants). In lactate fermentation, pyruvate is directly reduced by NADH, forming lactate and NAD⁺. In alcoholic fermentation, pyruvate is first decarboxylated to ethanal, releasing CO₂, and then reduced by NADH to ethanol, regenerating NAD⁺.
当氧气不足时，细胞可通过发酵再生NAD⁺，使糖酵解得以继续产生少量ATP。主要有两种类型：乳酸发酵（动物肌肉和一些细菌中）和酒精发酵（酵母和一些植物中）。在乳酸发酵中，丙酮酸直接被NADH还原，生成乳酸和NAD⁺。在酒精发酵中，丙酮酸首先脱羧生成乙醛，释放CO₂，然后被NADH还原为乙醇，再生NAD⁺。
Note that the sole purpose of fermentation is to oxidise NADH back to NAD⁺ so that glycolysis can keep running. No further ATP is made beyond the 2 net ATP from glycolysis. In IB, you may need to discuss the production of biotechnological products like bioethanol; CIE often links fermentation to yeast respiration experiments and the use of anaerobic respiration in bread‑making and brewing.
请注意，发酵的唯一目的是将NADH氧化回NAD⁺，以便糖酵解能继续进行。除糖酵解净产的2个ATP外，不再生成更多ATP。在IB中，你可能需要讨论生物技术产品如生物乙醇的生产；CIE通常将发酵与酵母呼吸实验以及面包制作和酿造中的厌氧呼吸联系起来。
8. Mitochondrial Structure Related to Function | 线粒体结构与其功能的关系
Mitochondria are double‑membraned organelles with a smooth outer membrane and a highly folded inner membrane (cristae). The cristae greatly increase the surface area for the electron transport chain and ATP synthase. The intermembrane space is small, allowing protons to accumulate quickly and build a steep gradient. The matrix contains enzymes for the link reaction and the Krebs cycle, as well as mitochondrial DNA and ribosomes for synthesising some of the respiratory proteins.
线粒体是双膜细胞器，外膜平滑，内膜高度折叠形成嵴。嵴极大地增加了电子传递链和ATP合酶的表面积。膜间隙狭小，使质子能快速积累并建立陡峭的浓度梯度。基质含有链接反应和克雷伯氏循环所需的酶，以及线粒体DNA和核糖体，用于合成部分呼吸蛋白。
Both IB and CIE require you to relate structure to function: the compartmentalisation keeps enzymes and substrates concentrated, the pH difference across the inner membrane drives ATP synthesis, and the high surface area of the cristae accommodates many ETC complexes. Diagrams often appear in exams, and you should be able to annotate them accurately.
IB与CIE均要求你将结构与功能联系起来：区域化使酶和底物保持浓缩，跨内膜的pH差驱动ATP合成，嵴的大表面积可容纳大量ETC复合物。考试中常出现结构图，你应能准确标注。
9. Respiratory Quotient (RQ) | 呼吸商
The respiratory quotient (RQ) is a ratio that provides information about the substrate being respired and whether respiration is aerobic or anaerobic. It is defined as the volume of CO₂ produced divided by the volume of O₂ consumed: RQ = CO₂ produced / O₂ consumed. For carbohydrates, RQ = 1.0 (e.g., C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O gives 6/6 = 1.0). For lipids, RQ is about 0.7 because lipids are more reduced and require more oxygen for oxidation. Proteins yield RQ values around 0.9.
呼吸商（RQ）是一个比值，提供有关被呼吸废物类型以及呼吸是需氧还是厌氧的信息。它定义为产生的CO₂体积与消耗的O₂体积之比：RQ = 产生的CO₂ / 消耗的O₂。对于碳水化合物，RQ = 1.0（例如，C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O，6/6 = 1.0）。脂质的RQ约为0.7，因为脂质还原度更高，氧化时需要更多氧气。蛋白质的RQ值约为0.9。
If anaerobic respiration occurs alongside aerobic respiration, the RQ value can rise above 1.0 because CO₂ is produced without a corresponding consumption of O₂ (as in alcoholic fermentation). CIE particularly likes to include RQ calculations and interpretations in practical assessments and data‑response questions. You must be comfortable with the formula and able to infer the substrate or metabolic state.
如果厌氧呼吸与需氧呼吸同时发生，RQ值可升至1.0以上，因为产生CO₂却没有相应地消耗O₂（如酒精发酵中）。CIE特别喜欢在实验评估和数据分析题中包含RQ计算和解读。你必须熟练掌握公式，并能推断底物或代谢状态。
10. Measuring Respiration Rate: The Respirometer | 测量呼吸速率：呼吸计
A respirometer is a commonly used apparatus to measure the rate of aerobic respiration by tracking oxygen consumption. The test organism (e.g., germinating seeds or small invertebrates) is sealed in a chamber with a carbon dioxide absorbent (such as KOH or soda lime). As the organism respires, it consumes O₂ and releases CO₂, but the CO₂ is absorbed, causing a decrease in pressure. This pressure change draws a coloured liquid along a manometer tube, and the volume of O₂ consumed per unit time can be calculated.
呼吸计是一种常用装置，通过跟踪氧气消耗来测量需氧呼吸速率。实验生物（如萌发种子或小型无脊椎动物）被密封在一个含二氧化碳吸收剂（如KOH或碱石灰）的容器中。当生物呼吸时，消耗O₂并释放CO₂，但CO₂被吸收，导致压力下降。这一压力变化使有色液体沿压力计管移动，单位时间内消耗的O₂体积即可计算出来。
Key precautions include: maintaining a constant temperature (using a water bath), ensuring an airtight seal, allowing the organism to acclimatise, and repeating readings to ensure reliability. In both IB and CIE, you may be asked to design or evaluate respirometer experiments, explain why KOH is used, or predict the effect of changing temperature on respiration rate.
主要注意事项包括：保持恒温（使用水浴），确保密封性，让生物适应环境，并重复读数以确保可靠性。在IB和CIE中，你可能被要求设计或评价呼吸计实验，解释为何使用KOH，或预测温度变化对呼吸速率的影响。
11. Comparison of Aerobic and Anaerobic Respiration | 需氧与厌氧呼吸的比较
Aerobic respiration yields a large amount of ATP (30–32 per glucose) through the complete oxidation of glucose, with water and CO₂ as final products. It depends on oxygen as the final electron acceptor and involves the mitochondria in eukaryotes. Anaerobic respiration produces only 2 ATP per glucose via glycolysis, and its end products are either lactate or ethanol and CO₂. It occurs entirely in the cytoplasm and does not use an electron transport chain.
需氧呼吸通过对葡萄糖的彻底氧化产生大量ATP（每分子葡萄糖30–32个），终产物为水和CO₂。它依赖氧气作为最终电子受体，并在真核生物的线粒体中进行。厌氧呼吸通过糖酵解仅产生2个ATP，终产物是乳酸或乙醇与CO₂。它完全在细胞质中进行，不使用电子传递链。
A useful way to structure a comparison question is to create a table covering location, oxygen requirement, ATP yield, final electron acceptor, products, and stages involved. Both syllabi will often ask you to evaluate the advantages and disadvantages: aerobic respiration is more efficient but slower to respond, while anaerobic respiration supplies ATP quickly but causes a build‑up of waste products, leading to muscle fatigue or ethanol toxicity.
构建比较题答案时，一个实用方法是制作表格，涵盖位置、氧气需求、ATP产量、最终电子受体、产物和涉及的阶段。两个大纲常要求你评价优缺点：需氧呼吸效率更高但响应较慢，而厌氧呼吸供能快但导致废物积累，引起肌肉疲劳或乙醇毒性。
12. Exam Tips and Common Pitfalls | 考试技巧与常见易错点
(1) Be specific with terms: distinguish between substrate‑level phosphorylation (direct formation of ATP in glycolysis and Krebs cycle) and oxidative phosphorylation (ATP synthesis using the proton gradient). (2) Always state the location of each stage; students often lose marks by leaving out ‘mitochondrial matrix’ or ‘inner membrane’. (3) When describing chemiosmosis, use the exact sequence: electron transport → proton pumping → proton gradient → H⁺ flow through ATP synthase → ATP production.
(1) 术语要精准：区分底物水平磷酸化（糖酵解和克雷伯氏循环中直接生成ATP）和氧化磷酸化（利用质子梯度合成ATP）。(2) 始终说明每个阶段的位置；学生常因漏写“线粒体基质”或“内膜”而失分。(3) 描述化学渗透时，使用准确顺序：电子传递→质子泵出→质子梯度→H⁺流经ATP合酶→ATP生成。
(4) In equations, always balance and indicate enzymes/coenzymes where relevant. (5) For respirometer questions, explicitly state that CO₂ absorption prevents interference and that pressure drop is due to O₂ consumption. (6) Do not confuse NAD⁺ regeneration in fermentation with ATP production; fermentation only regenerates NAD⁺ to keep glycolysis going. (7) Practice drawing the cristae, ETC, and ATP synthase for labelling tasks. By focusing on these points and using the bilingual explanations above, you will be well prepared for any IB or CIE respiration question.
(4) 写方程式时，始终配平并在相关处标明酶/辅酶。(5) 呼吸计题中，明确说明吸收CO₂是为了防止干扰，压力下降是由于O₂消耗。(6) 不要将发酵中NAD⁺的再生与ATP生成混淆；发酵仅再生NAD⁺以维持糖酵解。(7) 练习绘制嵴、ETC和ATP合酶以便应对标注题。专注于以上要点，并借助上文双语讲解，你将能轻松应对任何IB或CIE呼吸作用考题。
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