📚 5.7 Respiration Exam Practice | 5.7 呼吸作用真题精练
Respiration is a core A‑Level Biology topic that frequently appears in both structured and data‑based exam questions. This article consolidates key facts, common question types and typical pitfalls, giving you a thorough revision workout. Work through each section, test your understanding and be ready to tackle any respiration question with confidence.
呼吸作用是A‑Level生物的核心考点，常以结构化问答题和数据分析题出现。本文梳理了关键知识点、常见题型和易错点，为你提供一次全面的复习训练。逐节练习、自我检测，你就能满怀信心地应对任何呼吸作用题目。
1. Overview of Aerobic Respiration | 有氧呼吸概述
Aerobic respiration is the complete breakdown of glucose to carbon dioxide and water in the presence of oxygen. The overall equation is:
有氧呼吸是在有氧条件下将葡萄糖彻底分解为二氧化碳和水的过程。总反应式为：
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy (ATP + heat)
The process involves four main stages: glycolysis in the cytoplasm, the link reaction, the Krebs cycle and oxidative phosphorylation—the latter three all occurring inside the mitochondria. The theoretical maximum ATP yield from one glucose molecule is about 30–32 ATP, though actual values vary slightly depending on the shuttle system used.
整个过程包含四个主要阶段：细胞质中的糖酵解、线粒体基质中的连接反应和克雷布斯循环，以及发生在线粒体内膜的氧化磷酸化。一分子葡萄糖的理论最大ATP产量约为30–32个，具体数值因穿梭系统不同而略有差异。
Tip: When labelling a diagram, be precise about mitochondrial structures. The matrix is the site of the link reaction and Krebs cycle, while the cristae (inner membrane) contain the electron transport chain and ATP synthase.
小贴士：贴标签图题时务必精确标示线粒体结构。基质是连接反应和克雷布斯循环的场所，嵴（内膜）上则分布着电子传递链和ATP合酶。
2. Glycolysis: The First Stage | 糖酵解：第一阶段
Glycolysis takes place in the cytoplasm and does not require oxygen. Glucose (6C) is first phosphorylated using 2 ATP to form hexose bisphosphate. This makes glucose more reactive and prevents it from diffusing out of the cell. The 6‑carbon sugar is then split into two molecules of triose phosphate (3C). Triose phosphate is oxidised to pyruvate, during which 2 molecules of reduced NAD (NADH) are formed and 4 ATP are produced by substrate‑level phosphorylation. The net gain is thus 2 ATP per glucose.
糖酵解在细胞质中进行，不需要氧气。葡萄糖（6C）首先被2分子ATP磷酸化，形成己糖二磷酸。这使葡萄糖更具反应活性，也阻止其扩散出细胞。随后六碳糖裂解为两分子磷酸丙糖（3C）。磷酸丙糖被氧化为丙酮酸，在此过程中生成2分子还原型NAD（NADH），并通过底物水平磷酸化产生4分子ATP。因此每分子葡萄糖的净得ATP为2个。
An exam favourite: “Explain why glycolysis is described as a universal stage.” Answer: It occurs in the cytoplasm of almost all living organisms, does not need oxygen, and is therefore common to both aerobic and anaerobic respiration.
真题常问：“解释为什么糖酵解被称为普适阶段？” 因为它几乎发生在所有生物体的细胞质中，不需要氧气，因此无论有氧还是无氧呼吸都离不开它。
Also note: The oxidation of triose phosphate produces 2 NADH per glucose, which must be re‑oxidised back to NAD⁺ to allow glycolysis to continue, especially under anaerobic conditions.
此外注意：磷酸丙糖的氧化每分子葡萄糖产生2分子NADH，这些NADH必须被重新氧化为NAD⁺，才能维持糖酵解的继续进行，尤其在无氧条件下更为关键。
3. Link Reaction: The Bridging Step | 连接反应：承上启下
Pyruvate is actively transported into the mitochondrial matrix, where it undergoes oxidative decarboxylation. Pyruvate (3C) is decarboxylated: one CO₂ molecule is removed. Simultaneously it is oxidised, transferring hydrogen atoms to NAD⁺ to form reduced NAD. The remaining 2‑carbon acetyl group combines with coenzyme A to produce acetyl coenzyme A (acetyl CoA). For each glucose, this occurs twice, yielding 2 acetyl CoA, 2 CO₂ and 2 reduced NAD.
丙酮酸被主动转运进入线粒体基质，在那里发生氧化脱羧。丙酮酸（3C）脱羧并释放一分子CO₂；同时被氧化，将氢原子传递给NAD⁺生成还原型NAD。剩余的2碳乙酰基与辅酶A结合，形成乙酰辅酶A。每分子葡萄糖该过程进行两次，总共产生2个乙酰辅酶A、2个CO₂和2个还原型NAD。
Watch out: The link reaction does not produce ATP directly, but the NADH formed feeds into oxidative phosphorylation, driving considerable ATP synthesis. Students often mistakenly count zero ATP here—correct, but the potential energy in NADH is later converted.
注意：连接反应本身不直接产生ATP，但生成的NADH会进入氧化磷酸化，驱动后续大量ATP合成。同学们常误以为此阶段产能为零——确实不直接产ATP，但NADH中储存的势能会被后续转化。
4. Krebs Cycle: The Hub of Oxidation | 克雷布斯循环：氧化枢纽
The Krebs cycle (citric acid cycle) occurs in the mitochondrial matrix. Acetyl CoA (2C) combines with oxaloacetate (4C) to form citrate (6C). Through a series of enzyme‑controlled steps, citrate is decarboxylated and dehydrogenated, regenerating oxaloacetate. For each turn of the cycle, one acetyl CoA yields 2 CO₂, 3 reduced NAD, 1 reduced FAD and 1 ATP via substrate‑level phosphorylation. Since each glucose produces 2 acetyl CoA, the Krebs cycle turns twice, doubling these products.
克雷布斯循环（柠檬酸循环）发生在线粒体基质中。乙酰辅酶A（2C）与草酰乙酸（4C）结合形成柠檬酸（6C）。经过一系列酶促反应，柠檬酸逐步脱羧、脱氢，最终再生草酰乙酸。每一次循环，一个乙酰辅酶A产生2 CO₂、3个还原型NAD、1个还原型FAD和通过底物水平磷酸化得到的1个ATP。每分子葡萄糖生成2个乙酰辅酶A，因此循环运转两次，上述产物翻倍。
Common exam challenge: “Malonate inhibits succinate dehydrogenase. Predict and explain its effect on ATP production.” Malonate competes with succinate for the active site, reducing the production of FADH₂ and therefore decreasing the flow of electrons into oxidative phosphorylation, so fewer ATP are made.
真题挑战：“丙二酸抑制琥珀酸脱氢酶。请预测并解释其对ATP产量的影响。”丙二酸与琥珀酸竞争酶的活性位点，减少了FADH₂的生成，从而降低电子流入氧化磷酸化的速率，ATP产量随之下降。
Key to remember: The Krebs cycle intermediates such as oxaloacetate are constantly recycled. It also provides precursors for biosynthesis, linking respiration to metabolism.
关键在于：草酰乙酸等循环中间体被不断再生。克雷布斯循环还能为生物合成提供前体，将呼吸作用与代谢网络紧密连接。
5. Oxidative Phosphorylation & Chemiosmosis | 氧化磷酸化与化学渗透
Oxidative phosphorylation is the final stage, occurring across the inner mitochondrial membrane (cristae). Reduced NAD and reduced FAD from earlier stages donate electrons to the electron transport chain (ETC). As electrons are passed along a series of carriers at progressively lower energy levels, the energy released is used to pump protons (H⁺) from the matrix into the intermembrane space. This creates a proton gradient—a store of potential energy. Protons flow back through ATP synthase (chemiosmosis), driving the phosphorylation of ADP to ATP. Oxygen acts as the final electron acceptor, combining with electrons and protons to form water. Without oxygen, the ETC stops and no ATP can be made this way.
氧化磷酸化是最后阶段，发生于线粒体内膜（嵴）上。前几个阶段产生的还原型NAD和还原型FAD将电子传递给电子传递链。电子沿一系列载体传递时能量逐渐降低，释放的能量将质子（H⁺）从基质泵入膜间隙，形成质子梯度——即势能的储存。质子通过ATP合酶回流（化学渗透），驱动ADP磷酸化为ATP。氧气作为最终电子受体，与电子和质子结合生成水。若没有氧气，电子传递链中止，此途径无法合成ATP。
Many questions focus on the number of ATP produced per reduced NAD (∼2.5) and per reduced FAD (∼1.5). Explain why FADH₂ yields fewer ATP: FADH₂ donates electrons further down the chain, so fewer protons are pumped across the membrane.
很多题目都会问及每分子还原型NAD（≈2.5）和还原型FAD（≈1.5）的ATP产量。解释FADH₂产能较少的原因：FADH₂在电子传递链较后的位置提供电子，因而泵出膜的质子较少。
Also essential: Understand the action of uncouplers such as DNP. DNP makes the inner membrane permeable to H⁺, collapsing the proton gradient. Energy is released as heat instead of ATP synthesis.
同样重要：理解解偶联剂如DNP的作用。DNP使内膜对H⁺通透，消散质子梯度，能量以热能形式释放，无法合成ATP。
6. Anaerobic Respiration | 无氧呼吸
When oxygen is insufficient, cells rely on glycolysis to produce ATP. The key problem is that NAD⁺ must be regenerated from reduced NAD so that glycolysis can continue. In animals, pyruvate is reduced by NADH to lactate, catalysed by lactate dehydrogenase, producing no CO₂. In plants and yeast, pyruvate is first decarboxylated to ethanal, releasing CO₂, then ethanal is reduced to ethanol by NADH. The sole purpose is to regenerate NAD⁺; the ATP yield is just 2 per glucose from glycolysis.
当氧气不足时，细胞依靠糖酵解产生ATP。关键问题在于必须将还原型NAD重新氧化为NAD⁺，否则糖酵解无法持续。在动物体内，丙酮酸被NADH还原为乳酸，由乳酸脱氢酶催化，不释放CO₂。在植物和酵母中，丙酮酸先脱羧生成乙醛，释放CO₂，然后乙醛被NADH还原为乙醇。两种方式的唯一目的都是再生NAD⁺；每分子葡萄糖只从糖酵解获得2个ATP。
Compare energy yields: aerobic respiration can produce >30 ATP per glucose; anaerobic respiration produces only 2 ATP. The remaining energy stays in lactate or ethanol, which are relatively large and can be excreted.
能量产量对比：有氧呼吸每分子葡萄糖可合成30多个ATP；无氧呼吸仅产生2个ATP。大量能量保留在乳酸或乙醇中，这些分子体积相对较大，可被排出体外。
Exam alert: Students often forget that in anaerobic respiration, glycolysis is the only ATP‑producing stage. Also, note that CO₂ is released only in the yeast/plant pathway, not in animal lactate production.
考试预警：学生常忘记在无氧呼吸中，糖酵解是唯一产生ATP的阶段。此外，只有酵母/植物途径释放CO₂，动物乳酸发酵不产生CO₂。
7. Respiratory Quotient (RQ) | 呼吸商
The respiratory quotient (RQ) is the ratio of carbon dioxide produced to oxygen consumed over a given period:
呼吸商（RQ）是一定时间内产生的二氧化碳与消耗的氧气的体积比（或摩尔比）：
RQ = CO₂ produced ÷ O₂ consumed
RQ values vary with the substrate being respired: carbohydrate gives an RQ of approximately 1.0 (equal volumes of CO₂ and O₂); lipid yields around 0.7 (needs more O₂ per CO₂); protein about 0.9. In respirometer experiments, RQ can indicate which fuel is being used. If an organism shifts from using carbohydrate to lipids, its RQ drops.
RQ值因呼吸底物而异：碳水化合物约为1.0（CO₂与O₂等体积）；脂类约为0.7（每分子CO₂需要更多O₂）；蛋白质约为0.9。在呼吸计实验中，RQ可用来判断被消耗的燃料类型。生物体若从消耗碳水化合物转向脂质，其RQ会下降。
A typical data question: “A small animal gave a decrease in gas volume of 0.5 cm³ when CO₂ was absorbed, and an O₂ uptake of 1.2 cm³ in a control. Calculate RQ and suggest the main substrate.” Here, CO₂ produced = 1.2 − 0.5 = 0.7 cm³? Wait—if the manometer showed a decrease of 0.5 cm³ after absorbing CO₂, that means CO₂ produced was 0.5, O₂ uptake 1.2, RQ = 0.5/1.2 ≈ 0.42. This might indicate a mixed substrate or an error—always check the logic. (For a true 0.42, it is below the range of common substrates; perhaps the animal was respiring a substance with very low RQ or the data represent some artefact.)
典型数据分析题：“一个小动物呼吸计中，吸收CO₂后气体总体积减少了0.5 cm³，对照组测得O₂消耗量为1.2 cm³。求RQ并推测主要底物。”这里CO₂产生量 = 0.5 cm³，O₂消耗量 = 1.2 cm³，RQ = 0.5/1.2 ≈ 0.42。这样低的RQ已偏离常规底物范围，可能暗示实验误差或极端脂类——需要仔细检查逻辑。实际考卷会给出合理数值。
Be prepared to calculate RQ from given volumes and to interpret values in terms of substrate usage. Remember that RQ can also exceed 1.0 when anaerobic respiration contributes (producing CO₂ without O₂ intake).
做题时要能从体积数据计算RQ，并根据数值推断底物类型。切记若有缺氧呼吸参与，RQ可大于1.0，因为部分CO₂是在无氧消耗的情况下产生的。
8. Common Exam Mistakes & Tips | 常见考试错误与技巧
Mistake 1: Confusing NAD with NADH. NAD⁺ is the oxidised form; NADH is the reduced form. In equations, specify which one is being produced or used. Mark schemes are strict about the notation.
错误1：混淆NAD与NADH。NAD⁺为氧化型，NADH为还原型。作答时务必指明产生或消耗的是哪一种。评分标准对标示要求严格。
Mistake 2: Stating that oxygen is used in the Krebs cycle. Oxygen is only used at the very end of the electron transport chain as the final electron acceptor.
错误2：声称克雷布斯循环用到了氧气。氧分子仅在电子传递链最末端作为最终电子受体被使用。
Mistake 3: Forgetting that glycolysis occurs in the cytoplasm, not in the mitochondrion. Many students write “in the mitochondria” for all stages.
错误3：忘记糖酵解发生在细胞质而非线粒体。不少同学将所有阶段都写在线粒体中。
Mistake 4: Mixing up substrate‑level phosphorylation and oxidative phosphorylation. Substrate‑level phosphorylation happens in glycolysis and the Krebs cycle when a phosphate group is transferred directly from a phosphorylated intermediate to ADP. Oxidative phosphorylation uses the proton gradient and ATP synthase.
错误4：混淆底物水平磷酸化与氧化磷酸化。底物水平磷酸化发生于糖酵解及克雷布斯循环中，由磷酸化中间体直接将磷酸基转移给ADP；氧化磷酸化则依赖质子梯度和ATP合酶。
Mistake 5: Getting ATP totals wrong. Be systematic: glycolysis net 2 ATP + 2 NADH (≈5 ATP after oxidative phosphorylation), link reaction 2 NADH (≈5), Krebs cycle 2 ATP + 6 NADH (≈15) + 2 FADH₂ (≈3). Sum = ~30 ATP. Count reduced coenzymes first, then convert to ATP.
错误5：ATP总数算不准。系统计算：糖酵解净得2 ATP + 2 NADH（≈5 ATP），连接反应2 NADH（≈5），克雷布斯循环2 ATP + 6 NADH（≈15）+ 2 FADH₂（≈3），总和约30 ATP。先数清还原性辅酶，再换算为ATP。
Mistake 6: In anaerobic respiration, forgetting to state that NAD⁺ is regenerated for glycolysis, or wrongly claiming that oxygen is still needed. Practise describing the pathway clearly.
错误6：在无氧呼吸中忘了说明NAD⁺被再生以维持糖酵解，或错误声称仍需要氧气。务必能清晰描述整个途径。
Tip: When analysing respirometer data, always account for CO₂ absorption when calculating O₂ uptake. Understanding the role of soda lime or potassium hydroxide is crucial.
技巧：分析呼吸计数据时，计算氧耗量一定要扣除被碱石灰或KOH溶液吸收的CO₂。理解这一步骤至关重要。
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