PART I · SHORT RESPONSE · SOLUTIONS第一部分 · 短答题 · 详解AP-style MCQ + ON/BC short answer · 25 marksAP 风格选择题 + 安/卑省考短答 · 共 25 分
Section A · Worked SolutionsA 部分 · 详细解答
Q1EASY易🇺🇸 US美AP-style MCQAP 风格选择题§1 ATP and EnergyATP 与能量 · HS-LS1-7[3 marks][3 分]
Which of the following correctly describes what is released when one molecule of ATP is hydrolyzed inside a cell?下列哪项正确描述了细胞内一个 ATP 分子水解时释放的物质？
Answer:答案：(C) ADP, one inorganic phosphate ($\text{P}_i$), and energy
Hydrolysis of ATPATP 的水解A1·A1·A1
When one ATP molecule is hydrolyzed, a water molecule breaks the terminal phosphate bond. The products are ADP (adenosine diphosphate), one inorganic phosphate group ($\text{P}_i$), and released free energy that can drive cellular work. The reaction is:当一个 ATP 分子水解时，水分子断裂末端磷酸键，产物为 ADP（二磷酸腺苷）、一个无机磷酸基团（$\text{P}_i$）以及可驱动细胞工作的自由能。反应为：
$$ \text{ATP} + \text{H}_2\text{O} \;\longrightarrow\; \text{ADP} + \text{P}_i + \text{energy} $$
Why the distractors fail.干扰项分析。
(A)AMP and two phosphates would result from removing two phosphate groups (two hydrolysis steps), not one.AMP 和两个磷酸是移除两个磷酸基团（两步水解）的结果，而非一步。
(B)Glucose and oxygen are reactants in cellular respiration, not products of ATP hydrolysis.葡萄糖和氧气是细胞呼吸的反应物，并非 ATP 水解的产物。
(D)Carbon dioxide and water are products of aerobic respiration, not ATP hydrolysis.二氧化碳和水是有氧呼吸的产物，并非 ATP 水解的产物。
ATP is the cell's energy currency, not its energy storage molecule.ATP 是细胞的能量货币，而非能量储存分子。Each hydrolysis releases approximately 30.5 kJ/mol under standard conditions (more under cellular conditions). The energy released is immediately used for mechanical work (muscle contraction), active transport, or biosynthesis. The cell regenerates ATP from ADP and $\text{P}_i$ using aerobic respiration (primarily) or fermentation, keeping the ATP pool constantly cycling. Never conflate ATP hydrolysis with combustion of glucose: they are completely different reactions at different scales.在标准条件下，每次水解约释放 30.5 kJ/mol（细胞内条件下更多）。释放的能量立即用于机械做功（肌肉收缩）、主动运输或生物合成。细胞通过有氧呼吸（主要途径）或发酵将 ADP 和 $\text{P}_i$ 重新合成 ATP，使 ATP 池持续循环。切勿将 ATP 水解与葡萄糖燃烧混淆：二者是完全不同的反应，尺度也不同。
An enzyme is placed in a solution at 80 °C. After 10 minutes, its reaction rate drops to nearly zero even though substrate is still present. What best explains this observation?将一种酶置于 80 °C 的溶液中。10 分钟后，即使底物仍然存在，其反应速率也降至接近零。对此最合理的解释是什么？
Answer:答案：(B) The high temperature has denatured the enzyme, destroying its active site高温使酶变性，破坏了其活性位点
Enzyme denaturation at high temperature高温导致酶变性A1·A1·A1
Enzymes are proteins whose function depends on a precise 3D shape maintained by hydrogen bonds, ionic interactions, and other noncovalent forces. At 80 °C (far above physiological temperature for most organisms), these bonds break and the enzyme's tertiary structure unfolds irreversibly. The active site loses its complementary shape to the substrate, so no enzyme-substrate complexes can form and the reaction rate falls to zero. Substrate is still present, ruling out substrate depletion as the cause.酶是蛋白质，其功能依赖于由氢键、离子相互作用等非共价力维持的精确三维结构。在 80 °C（远高于大多数生物的生理温度）下，这些键断裂，酶的三级结构不可逆地展开。活性位点失去与底物互补的形状，无法形成酶-底物复合物，反应速率降至零。底物仍然存在，排除了底物耗尽的可能。
Why the distractors fail.干扰项分析。
(A)Competitive inhibition by a product would still allow increased substrate to overcome it; here more substrate makes no difference because the enzyme's shape is destroyed.产物的竞争性抑制可通过增加底物浓度克服；但此处增加底物无效，因为酶的形状已被破坏。
(C)Exceeding $V_{max}$ means all active sites are saturated, giving maximum rate, not zero rate.超过 $V_{max}$ 意味着所有活性位点被饱和，速率达到最大值，而非零。
(D)The question states the enzyme is in a temperature-controlled solution, not an acidic one; pH is not mentioned as changing.题目指出酶置于恒温溶液中，并未提到 pH 改变。
Enzyme activity is lost permanently on denaturation; reducing the temperature does not restore a denatured enzyme.酶变性后活性永久丧失；降温无法恢复变性酶的活性。This distinguishes denaturation from simple inhibition, which is often reversible. Each enzyme has an optimal temperature range (typically 35-40 °C for human enzymes); above that range, denaturation outpaces any increase in reaction rate from higher thermal energy. Thermophilic bacteria have evolved enzymes with extra stabilizing bonds that remain functional at 70-80 °C, but the question's enzyme clearly does not. Note that denaturation unfolds secondary, tertiary, and sometimes quaternary structure, but does not break covalent peptide bonds.这将变性与通常可逆的单纯抑制区分开来。每种酶都有最适温度范围（人体酶通常为 35-40 °C）；超过该范围，变性速率超过热能增加带来的速率提升。嗜热菌已进化出具有额外稳定键的酶，可在 70-80 °C 下正常工作，但题目中的酶显然不是。注意变性会破坏二级、三级乃至四级结构，但不断裂共价肽键。
Aerobic cellular respiration in a eukaryotic cell occurs in three stages. (a) Name the three stages in order and state the location of each. (b) Write the balanced overall summary equation. (c) State whether oxygen is a reactant or product and identify the stage at which it is consumed.真核细胞中有氧细胞呼吸分三个阶段进行。(a) 按顺序命名三个阶段并说明各阶段位置。(b) 写出平衡总方程式。(c) 说明氧气是反应物还是产物，并确定消耗氧气的阶段。
Answer:答案：(a)Glycolysis (cytoplasm), Krebs cycle (mitochondrial matrix), electron transport chain (inner mitochondrial membrane)糖酵解（细胞质）、克雷布斯循环（线粒体基质）、电子传递链（线粒体内膜） · (b) $\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \to 6\text{CO}_2 + 6\text{H}_2\text{O}$ · (c)reactant; electron transport chain反应物；电子传递链
(a) Three stages of aerobic respiration and their locations有氧呼吸三个阶段及其位置A1·A1·A1
Glycolysis occurs in the cytoplasm (cytosol). Glucose (6C) is split into two pyruvate (3C) molecules, producing a net of 2 ATP and 2 NADH.糖酵解发生在细胞质（细胞液）中。葡萄糖（6C）被分解为两个丙酮酸（3C），净产生 2 个 ATP 和 2 个 NADH。
Krebs cycle (citric acid cycle) occurs in the mitochondrial matrix. Pyruvate is converted to acetyl-CoA and fed into the cycle, producing NADH, FADH$_2$, ATP, and $\text{CO}_2$.克雷布斯循环（柠檬酸循环）发生在线粒体基质中。丙酮酸转变为乙酰辅酶A并进入循环，产生 NADH、FADH$_2$、ATP 和 $\text{CO}_2$。
Electron transport chain (ETC) and chemiosmosis occur at the inner mitochondrial membrane. NADH and FADH$_2$ donate electrons, a proton gradient drives ATP synthase, producing most of the ATP (~26-28 ATP).电子传递链（ETC）与化学渗透发生在线粒体内膜上。NADH 和 FADH$_2$ 提供电子，质子梯度驱动 ATP 合酶，产生绝大部分 ATP（约 26-28 个）。
(b) Balanced summary equation for aerobic respiration有氧呼吸平衡总方程式A1
$$ \text{C}_6\text{H}_{12}\text{O}_6 + 6\,\text{O}_2 \;\longrightarrow\; 6\,\text{CO}_2 + 6\,\text{H}_2\text{O} + \text{ATP (energy)} $$
All atoms balance: 6C, 12H, 18O on each side.原子守恒：两边各有 6C、12H、18O。
(c) Role and stage of oxygen consumption氧气的作用及消耗阶段A1
Oxygen is a reactant. It is consumed at the very end of the electron transport chain, where it serves as the terminal electron acceptor, combining with electrons and protons to form water: $\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \to \text{H}_2\text{O}$. This step, part of the ETC, is what makes the process "aerobic."氧气是反应物。它在电子传递链的末端被消耗，充当最终电子受体，与电子和质子结合生成水：$\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \to \text{H}_2\text{O}$。这一步骤是该过程被称为"有氧"的原因。
The three-stage mental map: cytoplasm to matrix to inner membrane.三阶段记忆框架：细胞质到基质到内膜。A helpful spatial anchor: glycolysis requires no organelle at all (it predates mitochondria evolutionarily); the Krebs cycle needs the aqueous interior (matrix) for its enzyme-driven reactions; the ETC needs the membrane itself as a scaffold for the protein complexes and as a barrier for the proton gradient. Oxygen is only involved at the very last step of the ETC and is not consumed during glycolysis or the Krebs cycle. This is why cyanide (which blocks the ETC) kills aerobic respiration even though glycolysis and the Krebs cycle are unaffected.一个有用的空间锚点：糖酵解完全不需要细胞器（从进化角度早于线粒体出现）；克雷布斯循环需要水性内环境（基质）进行酶促反应；ETC 需要膜本身作为蛋白质复合体的支架和质子梯度的屏障。氧气仅在 ETC 的最后一步参与，在糖酵解或克雷布斯循环中不被消耗。这就是为何氰化物（阻断 ETC）能杀死有氧呼吸，即使糖酵解和克雷布斯循环不受影响。
Q4MEDIUM中🇨🇦 BC卑BC Provincial-style卑诗省考风格§4 Anaerobic Respiration and Fermentation无氧呼吸与发酵 · Life Sciences 11[6 marks][6 分]
A student bakes bread using yeast. She notices that the dough rises before baking but the loaf does not taste alcoholic after baking.一名学生用酵母烤面包。她注意到面团在烘焙前会膨胀，但烤好的面包没有酒精味道。
Answer:答案：(a)alcoholic (ethanol) fermentation; ethanol and carbon dioxide乙醇发酵；乙醇和二氧化碳 · (b)$\text{CO}_2$ gas expands pockets in the dough$\text{CO}_2$ 气体使面团气泡膨胀 · (c)ethanol evaporates during baking乙醇在烘焙中挥发 · (d)fermentation: 2 ATP; aerobic: ~30-32 ATP; aerobic yields far more because electrons in NADH/FADH$_2$ are passed through the ETC to drive chemiosmosis发酵：2 个 ATP；有氧：约 30-32 个；有氧产量远多，因为 NADH/FADH$_2$ 中的电子经 ETC 驱动化学渗透
(a) Type of fermentation and its products发酵类型及产物A1·A1
Yeast performs alcoholic (ethanol) fermentation under anaerobic conditions in dough. The two products (other than ATP) are ethanol ($\text{C}_2\text{H}_5\text{OH}$) and carbon dioxide ($\text{CO}_2$). The summary equation is:酵母在面团的无氧条件下进行乙醇（酒精）发酵。除 ATP 外的两种产物为乙醇（$\text{C}_2\text{H}_5\text{OH}$）和二氧化碳（$\text{CO}_2$）。总方程式为：
$$ \text{glucose} \;\longrightarrow\; 2\,\text{ethanol} + 2\,\text{CO}_2 + 2\,\text{ATP} $$
(b) Why the dough rises面团膨胀的原因A1
Yeast cells release carbon dioxide gas as a byproduct of fermentation. The gas forms bubbles throughout the dough. The gluten network traps these gas pockets, causing the dough to expand and rise before baking.酵母细胞在发酵过程中释放二氧化碳气体。气体在面团中形成气泡，被面筋网络捕获，导致面团在烘焙前膨胀。
(c) Why the finished loaf is not alcoholic成品面包无酒精味的原因A1
Ethanol has a boiling point of 78 °C. During baking (typically 180-220 °C), the ethanol evaporates (volatilizes) and escapes from the bread, leaving no alcohol in the finished product.乙醇的沸点为 78 °C。在烘焙过程中（通常为 180-220 °C），乙醇挥发并从面包中散逸，成品中不留任何酒精。
(d) ATP yield comparison and reason for the differenceATP 产量比较及差异原因A1·A1
Ethanol fermentation yields only 2 ATP per glucose (from glycolysis alone). Aerobic respiration yields approximately 30-32 ATP per glucose. The dramatic difference arises because fermentation cannot use the electron carriers NADH produced during glycolysis to generate more ATP. Instead, those electrons are transferred to acetaldehyde to regenerate $\text{NAD}^+$, wasting the energy they carry. Aerobic respiration feeds NADH and FADH$_2$ into the electron transport chain, where the energy in each electron is used to pump protons and drive chemiosmosis (ATP synthase), generating the bulk of the ATP.乙醇发酵每分子葡萄糖仅产生 2 个 ATP（仅来自糖酵解）。有氧呼吸每分子葡萄糖约产生 30-32 个 ATP。巨大差异的原因在于：发酵无法利用糖酵解中产生的电子载体 NADH 来产生更多 ATP，而是将这些电子转移给乙醛以再生 $\text{NAD}^+$，白白浪费了其携带的能量。有氧呼吸将 NADH 和 FADH$_2$ 送入电子传递链，每个电子携带的能量用于泵送质子并驱动化学渗透（ATP 合酶），从而产生绝大部分 ATP。
Fermentation's purpose is $\text{NAD}^+$ regeneration, not ATP production.发酵的目的是再生 $\text{NAD}^+$，而非产生 ATP。Fermentation exists solely to recycle NADH back to $\text{NAD}^+$ so glycolysis can continue. Without $\text{NAD}^+$, glycolysis halts because it needs $\text{NAD}^+$ as an electron acceptor. The 2 ATP from glycolysis keep the cell barely alive under anaerobic conditions. This is also why lactic acid fermentation in muscle cells produces burning sensations: accumulating lactate (the converted form of pyruvate) lowers pH, interfering with muscle contraction enzymes.发酵的唯一目的是将 NADH 再生为 $\text{NAD}^+$，使糖酵解得以继续。没有 $\text{NAD}^+$，糖酵解就会停止，因为它需要 $\text{NAD}^+$ 作为电子受体。糖酵解产生的 2 个 ATP 使细胞在无氧条件下勉强维持生命。这也解释了为何肌肉细胞中的乳酸发酵会产生灼热感：积累的乳酸（丙酮酸的转化形式）降低 pH，干扰肌肉收缩酶的功能。
Which of the following correctly identifies the source of the oxygen ($O_2$) released during photosynthesis?下列哪项正确指出了光合作用中释放的氧气（$O_2$）的来源？
Answer:答案：(B) Water molecules ($H_2O$) split during the light reactions (photolysis)光反应（光解）期间分解的水分子（$H_2O$）
Photolysis of water as the source of $O_2$水的光解是 $O_2$ 的来源A1·A1·A1
During the light reactions, light energy absorbed by Photosystem II drives the splitting of water molecules (photolysis) at the oxygen-evolving complex: $2\text{H}_2\text{O} \to 4\text{H}^+ + 4e^- + \text{O}_2$. The oxygen atoms come entirely from water, not from $\text{CO}_2$. This was confirmed experimentally using isotope-labelled water ($\text{H}_2^{18}\text{O}$): the $^{18}\text{O}$ label appeared in the released $\text{O}_2$, not in $\text{CO}_2$.光反应中，光系统 II 吸收光能，在放氧复合物处驱动水分子的光解：$2\text{H}_2\text{O} \to 4\text{H}^+ + 4e^- + \text{O}_2$。氧原子完全来自水，而非 $\text{CO}_2$。这通过同位素标记水（$\text{H}_2^{18}\text{O}$）实验得到确认：$^{18}\text{O}$ 标记出现在释放的 $\text{O}_2$ 中，而非 $\text{CO}_2$ 中。
Why the distractors fail.干扰项分析。
(A)$\text{CO}_2$ enters the leaf and its carbon is fixed into glucose in the Calvin cycle; no $\text{O}_2$ is released from $\text{CO}_2$.$\text{CO}_2$ 进入叶片，其碳在卡尔文循环中被固定为葡萄糖；$\text{CO}_2$ 中不释放 $\text{O}_2$。
(C)Glucose is a product of photosynthesis, not a reactant that is broken down in the chloroplast.葡萄糖是光合作用的产物，而非在叶绿体中被分解的反应物。
(D)ATP hydrolysis releases a phosphate group and energy, not oxygen gas.ATP 水解释放磷酸基团和能量，而非氧气。
$O_2$ from photosynthesis comes from $H_2O$, not $CO_2$ -- confirmed by isotope tracing.光合作用产生的 $O_2$ 来自 $H_2O$，而非 $CO_2$，已由同位素示踪实验证实。This is one of biology's most counterintuitive facts. Students naturally assume that if $\text{CO}_2$ goes in and $\text{O}_2$ comes out, the oxygen must come from $\text{CO}_2$. The isotope experiment (van Niel, 1930s; confirmed with heavy oxygen later) definitively proved otherwise. Photolysis generates the protons and electrons needed to power the ETC of the light reactions, regenerating ATP and NADPH. Oxygen is simply the byproduct of stripping hydrogen (electrons + protons) from water. Without photolysis, the entire electron flow of the light reactions collapses.这是生物学中最违反直觉的事实之一。学生自然会认为 $\text{CO}_2$ 进入、$\text{O}_2$ 释放，氧气应来自 $\text{CO}_2$。同位素实验（van Niel，1930年代；后经重氧实验确认）最终证明并非如此。光解产生的质子和电子为光反应的电子传递链提供动力，再生 ATP 和 NADPH。氧气只是从水中剥离氢（电子和质子）后的副产品。没有光解，整个光反应的电子流动就会崩溃。
Q6MEDIUM中🇺🇸 US美AP-feeder FRQAP 衔接简答题§1 + §3 ATP role and aerobic respiration overviewATP 作用与有氧呼吸概述 · HS-LS1-7[8 marks][8 分]
A biology student claims: "Cells store large amounts of ATP the way a battery stores charge, drawing on that reserve whenever they need energy." Evaluate this claim and explain the true role of ATP in cell metabolism.一名生物学生声称："细胞像电池储存电荷一样大量储存 ATP，并在需要能量时从该储备中提取。"评价这一说法，并解释 ATP 在细胞代谢中的真实作用。
Answer summary:答案摘要：(a)Incorrect; ATP half-life in a cell is only seconds错误；ATP 在细胞内的半衰期仅为数秒 · (b)e.g. active transport and protein synthesis; aerobic cellular respiration regenerates ATP如主动运输和蛋白质合成；有氧细胞呼吸再生 ATP · (c)glucose stores chemical energy long-term; its energy is ultimately transferred to ATP bonds葡萄糖长期储存化学能；其能量最终转移至 ATP 键中
(a) Is the student's claim correct? Justify with ATP half-life学生的说法是否正确？结合 ATP 半衰期说明理由A1·A1
The claim is incorrect. Cells do not maintain a large reserve of ATP. The half-life of ATP in a typical cell is only about 1-2 seconds; an ATP molecule is hydrolyzed and regenerated approximately 500 times per day per cell. The cell keeps a very small, constantly cycling pool of ATP (roughly 250 g in the human body total), not a stockpile. This contrasts sharply with a battery, which stores charge passively for long periods. ATP is better described as a rapidly cycling energy carrier, not a reservoir.该说法错误。细胞并不维持大量 ATP 储备。典型细胞中 ATP 的半衰期仅约 1-2 秒；一个 ATP 分子每天被水解和再生约 500 次。细胞维持的是一个极小的、持续循环的 ATP 池（人体总量约 250 g），而非储存库。这与电池被动长期储存电荷形成鲜明对比。ATP 更准确地被描述为快速循环的能量载体，而非储能库。
(b) Two processes that consume ATP; process that regenerates ATP两种消耗 ATP 的过程；再生 ATP 的过程A1·A1·A1
Two cellular processes that consume ATP include:两种消耗 ATP 的细胞过程包括：
Active transport: membrane pumps (e.g., the sodium-potassium pump, Na$^+$/K$^+$-ATPase) use ATP hydrolysis to move ions against their concentration gradients.主动运输：膜泵（如钠钾泵 Na$^+$/K$^+$-ATPase）利用 ATP 水解逆浓度梯度运输离子。
Protein synthesis (biosynthesis): ribosomes use ATP (and GTP) to form peptide bonds and translocate along mRNA.蛋白质合成（生物合成）：核糖体利用 ATP（和 GTP）形成肽键并在 mRNA 上移位。
(Any two of: muscle contraction, active transport, biosynthesis, cell signaling, DNA replication, etc., are acceptable.)（以下任意两项均可接受：肌肉收缩、主动运输、生物合成、细胞信号传导、DNA 复制等。） Under aerobic conditions, ADP is phosphorylated back to ATP by aerobic cellular respiration, primarily through chemiosmosis at the electron transport chain in the mitochondria.在有氧条件下，ADP 通过有氧细胞呼吸重新磷酸化为 ATP，主要通过线粒体电子传递链上的化学渗透实现。
(c) Why glucose, not ATP, is the long-term energy storage molecule为何葡萄糖而非 ATP 是长期能量储存分子A1·A1·A1
Glucose (and its storage polymer, glycogen, or fats) is used for long-term energy storage for several reasons. First, glucose is chemically stable and does not spontaneously hydrolyze under cellular conditions, whereas ATP would release its energy prematurely if stockpiled. Second, glucose is energy-dense: one glucose molecule yields approximately 30-32 ATP through aerobic respiration. Third, glucose molecules are compact and can be polymerized into glycogen (in animals) or starch (in plants) for efficient storage. The chemical energy in glucose (stored in its C-H and C-C bonds) is ultimately transferred to the phosphoanhydride bonds of ATP during the three stages of aerobic respiration, where it becomes available for immediate use in cellular work.葡萄糖（及其储存聚合物糖原，或脂肪）用于长期能量储存，原因如下。第一，葡萄糖在细胞条件下化学稳定，不会自发水解，而大量储存的 ATP 会过早释放能量。第二，葡萄糖能量密度高：一个葡萄糖分子通过有氧呼吸可产生约 30-32 个 ATP。第三，葡萄糖分子紧凑，可聚合为糖原（动物体内）或淀粉（植物体内）高效储存。葡萄糖中的化学能（储存在其 C-H 和 C-C 键中）在有氧呼吸的三个阶段中最终转移到 ATP 的磷酸酐键中，从而可立即用于细胞做功。
ATP is the cell's energy currency; glucose is its savings account.ATP 是细胞的能量货币；葡萄糖是细胞的储蓄账户。The battery analogy fails because batteries store charge statically, but ATP is dynamic: it is continuously made and broken down. A more accurate analogy is that glucose is the banknote (stored value) and ATP is the coin used for each small transaction. Every time a cell needs to do work, it "spends" ATP; every time it processes glucose through respiration, it "earns" more ATP. The body regulates this balance tightly through hormones like insulin and glucagon, ensuring ATP availability matches cellular demand moment by moment.电池类比不恰当，因为电池静态储存电荷，而 ATP 是动态的：它被持续合成和分解。更准确的类比是：葡萄糖是纸币（储存的价值），ATP 是用于每笔小交易的硬币。每次细胞需要做功，就"花掉"ATP；每次细胞通过呼吸处理葡萄糖，就"获得"更多 ATP。身体通过胰岛素和胰高血糖素等激素严格调节这一平衡，确保 ATP 供应随时匹配细胞需求。
During an 800-metre race, an athlete's muscle cells initially use aerobic respiration but switch to lactic acid fermentation as the race intensifies.在一场 800 米比赛中，运动员的肌肉细胞起初进行有氧呼吸，但随着比赛强度增大，转向乳酸发酵。
Answer summary:答案摘要：(a)oxygen supply to muscles falls below demand; aerobic respiration requires $O_2$肌肉供氧量低于需求；有氧呼吸需要 $O_2$ · (b)fermentation 2 ATP vs. aerobic ~30-32 ATP; ETC oxidizes NADH/FADH$_2$ but fermentation cannot发酵 2 个 ATP 对比有氧约 30-32 个；ETC 氧化 NADH/FADH$_2$ 而发酵不能 · (c)$\text{NAD}^+$ regeneration allows glycolysis to continue accepting electrons; without it glycolysis stops再生 $\text{NAD}^+$ 使糖酵解得以继续接受电子；否则糖酵解停止
(a) Why muscle cells switch from aerobic respiration to lactic acid fermentation肌肉细胞从有氧呼吸转向乳酸发酵的原因A1·A1
During intense exercise, muscles contract rapidly and demand ATP faster than the cardiovascular system can supply oxygen. When the rate of $\text{O}_2$ delivery falls below the rate of $\text{O}_2$ consumption, the cells enter an anaerobic state. Aerobic respiration requires oxygen as the terminal electron acceptor in the electron transport chain; without sufficient $\text{O}_2$, the ETC stalls, NADH builds up, and the cell switches to lactic acid fermentation to regenerate $\text{NAD}^+$ and keep ATP production going.在剧烈运动中，肌肉快速收缩，对 ATP 的需求超过心血管系统供应氧气的速率。当 $\text{O}_2$ 输送速率低于消耗速率时，细胞进入无氧状态。有氧呼吸需要氧气作为电子传递链的最终电子受体；缺乏足够 $\text{O}_2$ 时，ETC 停滞，NADH 积累，细胞转向乳酸发酵以再生 $\text{NAD}^+$ 并维持 ATP 生产。
(b) ATP yield comparison and metabolic reason for the differenceATP 产量比较及代谢原因A1·A1·A1
Lactic acid fermentation yields 2 ATP per glucose (glycolysis only). Aerobic respiration yields approximately 30-32 ATP per glucose. The metabolic reason for this difference is that lactic acid fermentation uses only glycolysis; the pyruvate produced is reduced to lactate to recycle $\text{NAD}^+$, but the electrons and chemical energy in NADH are discarded in lactate. Aerobic respiration additionally runs the Krebs cycle and the electron transport chain: NADH and FADH$_2$ donate high-energy electrons to protein complexes in the inner mitochondrial membrane, which pump protons to create a gradient. This proton gradient drives ATP synthase (chemiosmosis), producing the vast majority of the ATP (~26-28 from the ETC, versus only 4 from glycolysis + Krebs substrate-level phosphorylation).乳酸发酵每分子葡萄糖产生 2 个 ATP（仅糖酵解）。有氧呼吸每分子葡萄糖约产生 30-32 个 ATP。差异的代谢原因在于：乳酸发酵仅使用糖酵解，产生的丙酮酸被还原为乳酸以循环利用 $\text{NAD}^+$，但 NADH 中的电子和化学能被浪费在乳酸中。有氧呼吸还运行克雷布斯循环和电子传递链：NADH 和 FADH$_2$ 将高能电子传递给线粒体内膜上的蛋白质复合体，后者泵送质子形成梯度。该质子梯度驱动 ATP 合酶（化学渗透），产生绝大部分 ATP（ETC 约产生 26-28 个，而糖酵解和克雷布斯循环底物水平磷酸化仅产生 4 个）。
(c) Why $\text{NAD}^+$ regeneration is essential for glycolysis to continue为何 $\text{NAD}^+$ 再生对糖酵解持续运行至关重要A1·A1·A1
Glycolysis requires $\text{NAD}^+$ as an electron acceptor in one of its key oxidation steps (the oxidation of glyceraldehyde-3-phosphate by GAPDH). During this step, $\text{NAD}^+$ accepts 2 electrons and a proton to become NADH. If the cell is anaerobic and NADH cannot be oxidized by the ETC (because there is no oxygen to accept the electrons at the end), $\text{NAD}^+$ levels fall to zero and glycolysis halts. Lactic acid fermentation solves this by transferring the electrons from NADH to pyruvate, converting it to lactate (in animal cells) and regenerating $\text{NAD}^+$. This restored $\text{NAD}^+$ allows glycolysis to continue, keeping the cell alive with the 2 ATP per glucose it still produces.糖酵解需要 $\text{NAD}^+$ 作为其关键氧化步骤（GAPDH 对甘油醛-3-磷酸的氧化）的电子受体。在此步骤中，$\text{NAD}^+$ 接受 2 个电子和一个质子变为 NADH。若细胞处于无氧状态，NADH 无法被 ETC 氧化（因为末端无氧接受电子），$\text{NAD}^+$ 水平降至零，糖酵解停止。乳酸发酵通过将 NADH 中的电子转移至丙酮酸，将其转化为乳酸（动物细胞中）并再生 $\text{NAD}^+$ 来解决这一问题。恢复的 $\text{NAD}^+$ 使糖酵解得以继续，使细胞依靠每分子葡萄糖仍能产生的 2 个 ATP 维持生命。
Glycolysis is the ancient, oxygen-independent core of cellular energy metabolism.糖酵解是细胞能量代谢古老的、不依赖氧气的核心途径。Glycolysis evolved before oxygen existed in Earth's atmosphere, so it makes no use of oxygen and functions under any conditions. Fermentation is simply glycolysis's "exhaust pipe" for disposing of electrons when the aerobic pathway is unavailable. The athlete's muscle cells during the 800 m race are a living demonstration: for the first ~60 seconds, aerobic respiration dominates; as intensity peaks, lactic acid fermentation supplements it. The "burn" sensation is partly from $\text{H}^+$ ions (lowered pH), not lactate itself. After the race, the athlete repays the "oxygen debt" by breathing hard to oxidize accumulated lactate back through aerobic pathways.糖酵解在地球大气层中氧气出现之前就已进化，因此不使用氧气，在任何条件下都能运作。发酵只是糖酵解在无氧途径不可用时处理电子的"排气管"。800 米比赛中运动员的肌肉细胞是生动的示例：前约 60 秒有氧呼吸主导；随着强度达到峰值，乳酸发酵开始补充。灼烧感部分来自 $\text{H}^+$ 离子（pH 降低），而非乳酸本身。比赛结束后，运动员通过大口呼吸偿还"氧债"，将积累的乳酸氧化回有氧途径。
Q8HARD难🇨🇦 AB阿AB Diploma-style阿尔伯塔毕业考风格§5 + §6 Photosynthesis light reactions and Calvin cycle光合作用光反应与卡尔文循环 · Biology 20 Unit C GO1[8 marks][8 分]
A plant leaf is illuminated with bright white light and provided with $CO_2$ and water. (a) Write the balanced overall summary equation for photosynthesis, labelling what each stage uses or produces. (b) Explain the role of ATP and NADPH and where they are used in the chloroplast. (c) Identify the source of the oxygen released and explain why this surprises students.一片植物叶片被强白光照射，并提供 $CO_2$ 和水。(a) 写出光合作用平衡总方程式，标注各阶段使用或产生的物质。(b) 解释 ATP 和 NADPH 的作用及其在叶绿体中的使用位置。(c) 指出释放氧气的来源，并解释为何令学生惊讶。
Answer summary:答案摘要：(a) $6\text{CO}_2 + 6\text{H}_2\text{O} \to \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$ · (b)ATP and NADPH from light reactions power the Calvin cycle in the stroma光反应产生的 ATP 和 NADPH 在基质中驱动卡尔文循环 · (c)$O_2$ comes from $H_2O$ photolysis, not $CO_2$$O_2$ 来自水的光解，而非 $CO_2$
(a) Balanced summary equation for photosynthesis with stage labels光合作用平衡总方程式及阶段标注A1·A1·A1
$$ 6\,\text{CO}_2 + 6\,\text{H}_2\text{O} \;\xrightarrow{\text{light energy}}\; \text{C}_6\text{H}_{12}\text{O}_6 + 6\,\text{O}_2 $$
Light reactions (thylakoid membranes): use $\text{H}_2\text{O}$ and light energy; produce $\text{O}_2$, ATP, and NADPH. Calvin cycle (stroma): uses $\text{CO}_2$, ATP, and NADPH; produces glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) and regenerates ADP + $\text{P}_i$ and NADP$^+$.光反应（类囊体膜）：使用 $\text{H}_2\text{O}$ 和光能；产生 $\text{O}_2$、ATP 和 NADPH。卡尔文循环（基质）：使用 $\text{CO}_2$、ATP 和 NADPH；产生葡萄糖（$\text{C}_6\text{H}_{12}\text{O}_6$）并再生 ADP + $\text{P}_i$ 和 NADP$^+$。
(b) Role of ATP and NADPH; location within the chloroplastATP 和 NADPH 的作用；在叶绿体中的位置A1·A1·A1
ATP and NADPH are both produced in the thylakoid membranes (grana) during the light reactions. They then diffuse into the stroma where the Calvin cycle takes place. In the Calvin cycle:ATP 和 NADPH 均在光反应中由类囊体膜（基粒）产生，随后扩散到基质中（卡尔文循环发生的场所）。在卡尔文循环中：
ATP provides the phosphate group and energy needed to phosphorylate 3-carbon intermediates (converting 3-phosphoglycerate to glyceraldehyde-3-phosphate, G3P).ATP 提供磷酸基团和能量，用于磷酸化三碳中间体（将 3-磷酸甘油酸转化为 3-磷酸甘油醛，G3P）。
NADPH provides electrons (reducing power) to reduce the 3-carbon intermediates into G3P, the building block for glucose synthesis.NADPH 提供电子（还原力），将三碳中间体还原为 G3P，即合成葡萄糖的构件。
Without ATP and NADPH from the light reactions, the Calvin cycle cannot run; this is why photosynthesis stops in darkness (the Calvin cycle has no direct source of these molecules in the dark).没有光反应提供的 ATP 和 NADPH，卡尔文循环就无法运行；这就是为什么光合作用在黑暗中停止（卡尔文循环在黑暗中没有这些分子的直接来源）。
(c) Source of released oxygen and why it is surprising释放氧气的来源及令学生惊讶的原因A1·A1
The oxygen released by plants comes from water ($\text{H}_2\text{O}$), not from $\text{CO}_2$. During the light reactions, the oxygen-evolving complex in Photosystem II splits water molecules (photolysis): $2\text{H}_2\text{O} \to 4\text{H}^+ + 4e^- + \text{O}_2$. Students assume the oxygen must come from $\text{CO}_2$ because $\text{CO}_2$ contains oxygen and they observe that $\text{CO}_2$ goes in while $\text{O}_2$ comes out. However, the oxygen atoms in $\text{CO}_2$ are incorporated into glucose (C-O bonds), not released as gas. Isotope labelling experiments confirm this: when plants are given $^{18}\text{O}$-labelled water, the $^{18}\text{O}$ appears in released $\text{O}_2$.植物释放的氧气来自水（$\text{H}_2\text{O}$），而非 $\text{CO}_2$。光反应中，光系统 II 的放氧复合物分解水分子（光解）：$2\text{H}_2\text{O} \to 4\text{H}^+ + 4e^- + \text{O}_2$。学生们认为氧气应来自 $\text{CO}_2$，因为 $\text{CO}_2$ 含有氧，而他们观察到 $\text{CO}_2$ 进入、$\text{O}_2$ 释放。然而，$\text{CO}_2$ 中的氧原子被结合进葡萄糖（C-O 键），而非以气体形式释放。同位素标记实验证实了这一点：给植物提供 $^{18}\text{O}$ 标记的水，$^{18}\text{O}$ 出现在释放的 $\text{O}_2$ 中。
Photosynthesis runs the overall chemical equation in reverse compared to respiration, but uses completely different pathways.光合作用的总化学方程式与呼吸作用方向相反，但使用完全不同的途径。Respiration: $\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \to 6\text{CO}_2 + 6\text{H}_2\text{O}$. Photosynthesis: $6\text{CO}_2 + 6\text{H}_2\text{O} \to \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$. Despite the apparent reversal, the enzymes, organelles, electron carriers, and intermediates are completely different. Photosynthesis captures light energy and stores it in chemical bonds; respiration releases that stored energy as ATP. Together they form the global carbon cycle, with photosynthesis fixing atmospheric $\text{CO}_2$ into organic molecules and respiration returning it.呼吸作用：$\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \to 6\text{CO}_2 + 6\text{H}_2\text{O}$。光合作用：$6\text{CO}_2 + 6\text{H}_2\text{O} \to \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$。尽管表面上相反，但酶、细胞器、电子载体和中间体完全不同。光合作用捕获光能并将其储存在化学键中；呼吸作用将储存的能量以 ATP 形式释放。两者共同构成全球碳循环：光合作用将大气中的 $\text{CO}_2$ 固定为有机物，呼吸作用将其归还。
Chemiosmosis is the mechanism by which both aerobic respiration and photosynthesis produce most of their ATP. A student claims the two processes are essentially "mirror images" of each other. (a) Describe chemiosmosis in the mitochondrion. (b) Evaluate the "mirror image" claim by identifying two ways the processes are complementary and one key difference in energy flow.化学渗透是有氧呼吸和光合作用产生大部分 ATP 的机制。一名学生声称这两个过程本质上是彼此的"镜像"。(a) 描述线粒体中的化学渗透。(b) 评价"镜像"说法：指出两种互补方式和能量流动的一个关键差异。
Answer summary:答案摘要：(a)NADH/FADH$_2$ donate electrons to ETC; protons pumped to intermembrane space; flow back via ATP synthase; $O_2$ is terminal acceptorNADH/FADH$_2$ 向 ETC 提供电子；质子被泵入膜间隙；经 ATP 合酶流回；$O_2$ 是最终受体 · (b)complementary: reactants/products of one are reactants/products of the other; key difference: respiration releases stored energy as ATP, photosynthesis captures light energy into chemical bonds互补：一者的反应物/产物是另一者的产物/反应物；关键差异：呼吸以 ATP 形式释放储存能量，光合作用将光能固定为化学键
(a) Chemiosmosis in the mitochondrion during aerobic respiration有氧呼吸期间线粒体中的化学渗透A1·A1·A1·A1
The electron transport chain uses high-energy electrons donated by NADH and FADH$_2$ (produced in glycolysis and the Krebs cycle). These electron carriers pass electrons through a series of protein complexes (Complexes I-IV) embedded in the inner mitochondrial membrane. As electrons flow "downhill" in energy through these complexes, the energy released is used to actively pump protons ($\text{H}^+$) from the mitochondrial matrix into the intermembrane space, creating a steep electrochemical gradient (high $\text{H}^+$ concentration in the intermembrane space relative to the matrix). Protons then flow back down this gradient through ATP synthase (also called Complex V), a channel-enzyme embedded in the inner membrane. The flow of protons drives the rotation of ATP synthase's rotor, catalyzing the phosphorylation of ADP to ATP. At the end of the electron transport chain, electrons reduce oxygen ($\text{O}_2$), the terminal electron acceptor, combining with $\text{H}^+$ ions to form water ($\text{H}_2\text{O}$). Without $\text{O}_2$, the ETC stalls and chemiosmosis stops.电子传递链使用由NADH和FADH$_2$（在糖酵解和克雷布斯循环中产生）提供的高能电子。这些电子载体通过嵌入线粒体内膜的一系列蛋白质复合体（复合体 I-IV）传递电子。电子在这些复合体中"顺能量梯度"流动时，释放的能量被用于主动将质子（$\text{H}^+$）从线粒体基质泵入膜间隙，形成陡峭的电化学梯度（膜间隙中 $\text{H}^+$ 浓度远高于基质）。质子随后通过ATP 合酶（也称复合体 V，嵌入内膜的通道酶）沿梯度流回。质子流动驱动 ATP 合酶转子旋转，催化 ADP 磷酸化为 ATP。在电子传递链末端，电子还原氧气（$\text{O}_2$）（最终电子受体），与 $\text{H}^+$ 离子结合生成水（$\text{H}_2\text{O}$）。没有 $\text{O}_2$，ETC 停滞，化学渗透停止。
(b) Evaluating the "mirror image" claim: two complementary features, one key difference评价"镜像"说法：两种互补特征，一个关键差异A1·A1·A1·A1
Two ways the processes are genuinely complementary:两种真正互补的方式：
The reactants of one are the products of the other: aerobic respiration consumes $\text{O}_2$ and glucose and produces $\text{CO}_2$ and $\text{H}_2\text{O}$; photosynthesis consumes $\text{CO}_2$ and $\text{H}_2\text{O}$ and produces $\text{O}_2$ and glucose. Together they recycle carbon and oxygen atoms continuously.一者的反应物是另一者的产物：有氧呼吸消耗 $\text{O}_2$ 和葡萄糖，产生 $\text{CO}_2$ 和 $\text{H}_2\text{O}$；光合作用消耗 $\text{CO}_2$ 和 $\text{H}_2\text{O}$，产生 $\text{O}_2$ 和葡萄糖。两者共同持续循环碳原子和氧原子。
Both use chemiosmosis and ATP synthase to make ATP: a proton gradient across a membrane drives protons through ATP synthase to generate ATP (across the inner mitochondrial membrane in respiration; across the thylakoid membrane in photosynthesis).两者均使用化学渗透和 ATP 合酶合成 ATP：跨膜质子梯度驱动质子流经 ATP 合酶产生 ATP（呼吸中跨线粒体内膜；光合作用中跨类囊体膜）。
One key difference in energy flow: Aerobic respiration is an energy-releasing (catabolic) process: it breaks down glucose to release the chemical energy stored in its bonds, converting it to ATP (usable energy) plus heat. Photosynthesis is an energy-capturing (anabolic) process: it uses light energy from the sun to drive an energetically unfavorable reaction, building chemical energy into glucose bonds. In terms of entropy and free energy, respiration increases disorder (large molecule to small), while photosynthesis decreases local disorder (small molecules to a large, ordered glucose) by inputting solar energy. The "mirror image" claim is partially valid for the overall equation, but the two processes differ fundamentally in whether they release or capture free energy.能量流动的一个关键差异：有氧呼吸是放能（分解代谢）过程：它分解葡萄糖，释放其化学键中储存的能量，转化为 ATP（可用能量）加热量。光合作用是储能（合成代谢）过程：它利用太阳光能驱动热力学上不利的反应，将化学能储存到葡萄糖键中。从熵和自由能角度看，呼吸增加无序度（大分子变小），而光合作用通过输入太阳能降低局部无序度（小分子变为有序的大葡萄糖分子）。"镜像"说法就总方程式而言有一定道理，但两者在是否释放或捕获自由能方面存在根本差异。
The proton gradient is the cell's universal energy intermediate -- the same mechanism powers both photosynthesis and respiration.质子梯度是细胞通用的能量中间体，同一机制同时驱动光合作用和呼吸作用。Peter Mitchell's chemiosmotic theory (Nobel Prize 1978) revealed that the proton gradient, not ATP itself, is the primary energy currency at the membrane level. Both photosynthesis and respiration exploit this same elegant mechanism. The key evolutionary insight is that ATP synthase is so ancient that it is found in essentially all living cells (bacteria, archaea, eukaryotes), and the proton gradient mechanism was likely present in the earliest forms of life. The directionality of proton pumping differs (into the intermembrane space in mitochondria vs. into the thylakoid lumen in chloroplasts), but the rotary motor mechanism of ATP synthase is identical.彼得·米切尔的化学渗透理论（1978 年诺贝尔奖）揭示了质子梯度而非 ATP 本身才是膜层面的主要能量货币。光合作用和呼吸作用都利用了这同一优雅机制。关键的进化洞见是：ATP 合酶如此古老，几乎存在于所有生物（细菌、古菌、真核生物）中，质子梯度机制很可能存在于最早期的生命形式中。质子泵送方向不同（线粒体中泵入膜间隙，叶绿体中泵入类囊体腔），但 ATP 合酶的旋转马达机制完全相同。
Q10MEDIUM中🇨🇦 AB阿AB Diploma-style阿尔伯塔毕业考风格§2 + §3 Enzyme inhibition and metabolic disruption酶抑制与代谢干扰 · Biology 20 C GO2[8 marks][8 分]
Cyanide is a toxin that binds irreversibly to the final protein complex of the electron transport chain in the mitochondrion, preventing oxygen from accepting electrons. (a) Classify cyanide as competitive or non-competitive inhibitor and justify. (b) Explain what happens to the proton gradient and why ATP synthesis stops. (c) A cyanide-poisoned cell can still produce a small amount of ATP briefly. Identify the pathway and explain why it eventually also fails.氰化物是一种毒素，不可逆地与线粒体电子传递链最终蛋白复合体结合，阻止氧气接受电子。(a) 将氰化物分类为竞争性或非竞争性抑制剂并说明理由。(b) 解释质子梯度发生的变化及 ATP 合成为何停止。(c) 被氰化物毒害的细胞在短时间内仍能产生少量 ATP。确定该途径并解释为何最终也会失败。
Answer summary:答案摘要：(a)non-competitive (binds irreversibly to a different site; not displaced by substrate)非竞争性（不可逆结合于不同位点；底物无法置换） · (b)electrons stall; no proton pumping; gradient collapses; ATP synthase stops电子传递停滞；质子无法泵送；梯度消散；ATP 合酶停止 · (c)glycolysis + fermentation; fails when $\text{NAD}^+$ runs out or glucose is depleted糖酵解和发酵；当 $\text{NAD}^+$ 耗尽或葡萄糖耗尽时失败
(a) Classifying cyanide as a competitive or non-competitive inhibitor将氰化物分类为竞争性或非竞争性抑制剂A1·A1
Cyanide is a non-competitive inhibitor (more precisely, an irreversible inhibitor). Justification: cyanide binds to the iron (Fe) center of cytochrome c oxidase (Complex IV) at a site that is not the active site where oxygen binds as the normal substrate. Because it does not compete with oxygen for the same binding site, increasing oxygen concentration does not displace cyanide or restore enzyme activity. Additionally, the binding is irreversible, meaning it cannot be overcome at all. In classical enzyme kinetics, competitive inhibitors bind reversibly at the active site and can be outcompeted by substrate; cyanide does neither.氰化物是非竞争性抑制剂（更准确地说是不可逆抑制剂）。理由：氰化物结合到细胞色素 c 氧化酶（复合体 IV）的铁（Fe）中心，该位点并非氧气作为正常底物结合的活性位点。由于它不与氧气竞争同一结合位点，增加氧气浓度无法置换氰化物或恢复酶活性。此外，结合是不可逆的，意味着根本无法克服。在经典酶动力学中，竞争性抑制剂可逆地结合在活性位点，可被底物竞争置换；氰化物两者都不符合。
(b) What happens to the proton gradient; why ATP synthesis stops质子梯度的变化；ATP 合成停止的原因A1·A1·A1
When cyanide blocks Complex IV, electrons can no longer flow through the end of the ETC to reduce oxygen. Because electron flow through the earlier complexes (I, III) is what drives the pumping of protons from the matrix into the intermembrane space, that pumping stops. Protons already in the intermembrane space continue to flow back through ATP synthase briefly, but no new protons are pumped in to maintain the gradient. The proton gradient (electrochemical gradient) rapidly collapses. ATP synthase requires a proton gradient to function (proton flow through it drives the rotation of the rotor that synthesizes ATP). Without the gradient, ATP synthase halts and ATP synthesis from chemiosmosis ceases completely.氰化物阻断复合体 IV 后，电子无法继续流经 ETC 末端还原氧气。由于流过早期复合体（I、III）的电子流驱动质子从基质泵入膜间隙，泵送停止。膜间隙中已有的质子继续短暂流经 ATP 合酶，但没有新质子被泵入以维持梯度。质子梯度（电化学梯度）迅速消散。ATP 合酶需要质子梯度才能运作（质子流过它驱动合成 ATP 的转子旋转）。没有梯度，ATP 合酶停止，化学渗透产生的 ATP 合成完全停止。
(c) Pathway still producing ATP briefly; why it eventually fails短暂仍能产生 ATP 的途径；最终失败的原因A1·A1·A1
The pathway responsible is glycolysis (combined with lactic acid fermentation to regenerate $\text{NAD}^+$). Glycolysis occurs in the cytoplasm and does not require oxygen or the ETC; it produces 2 ATP per glucose via substrate-level phosphorylation. However, this ATP supply is temporary for two reasons. First, if fermentation does not adequately regenerate $\text{NAD}^+$, NADH accumulates, $\text{NAD}^+$ is depleted, and glycolysis halts (because it requires $\text{NAD}^+$ as an electron acceptor). Second, even if fermentation runs, glucose stores are limited and are consumed rapidly when it is the only source of ATP. Ultimately, the cell cannot survive on 2 ATP per glucose from glycolysis alone, and critical ion pumps, biosynthetic processes, and signaling pathways shut down, leading to cell death.负责此过程的途径是糖酵解（结合乳酸发酵再生 $\text{NAD}^+$）。糖酵解发生在细胞质中，不需要氧气或 ETC；通过底物水平磷酸化每分子葡萄糖产生 2 个 ATP。然而，这种 ATP 供应是短暂的，原因有二。第一，若发酵不能充分再生 $\text{NAD}^+$，NADH 积累，$\text{NAD}^+$ 耗尽，糖酵解停止（因为它需要 $\text{NAD}^+$ 作为电子受体）。第二，即使发酵运行，葡萄糖储量有限，且在它是唯一 ATP 来源时被迅速消耗。最终，细胞无法仅靠糖酵解每分子葡萄糖产生的 2 个 ATP 维持生存，关键离子泵、生物合成过程和信号通路关闭，导致细胞死亡。
Cyanide poisoning illustrates why the ETC is the critical bottleneck of aerobic respiration.氰化物中毒说明了为何 ETC 是有氧呼吸的关键瓶颈。This scenario ties together enzyme inhibition, the ETC, chemiosmosis, and the role of glycolysis as a backup pathway. It also reveals a pharmacological principle: drugs targeting the ETC can be lethal but also therapeutic (some antibiotics and cancer drugs work by disrupting mitochondrial function in bacteria or rapidly dividing cells). The reason cyanide acts so rapidly is that the human body generates 90% of its ATP through the ETC; blocking it causes immediate cellular energy failure in energy-hungry tissues like the brain and heart.这一情境将酶抑制、ETC、化学渗透和糖酵解作为备用途径联系在一起。它还揭示了一个药理学原理：靶向 ETC 的药物可以是致命的，也可以是治疗性的（某些抗生素和癌症药物通过破坏细菌或快速分裂细胞的线粒体功能发挥作用）。氰化物之所以如此迅速地起效，是因为人体 90% 的 ATP 通过 ETC 产生；阻断它会导致大脑和心脏等高耗能组织立即出现细胞能量衰竭。
A potted green plant is placed in a sealed transparent container. Sensors measure $O_2$ and $CO_2$ concentrations over 24 hours: 12 hours of bright light followed by 12 hours of complete darkness. (a) During the light period $O_2$ rises and $CO_2$ falls. Explain why. (b) During the dark period $O_2$ falls and $CO_2$ rises. Explain why. (c) A classmate claims plants only photosynthesize. Evaluate this claim using the dark-period data.一盆绿色植物被置于密封透明容器中。传感器测量 24 小时内 $O_2$ 和 $CO_2$ 浓度：12 小时强光照射，随后 12 小时完全黑暗。(a) 光照期间 $O_2$ 上升、$CO_2$ 下降。解释原因。(b) 黑暗期间 $O_2$ 下降、$CO_2$ 上升。解释原因。(c) 一位同学声称植物只进行光合作用。用黑暗期数据评价这一说法。
Answer summary:答案摘要：(a)photosynthesis rate exceeds respiration; net $O_2$ produced, net $CO_2$ consumed光合速率超过呼吸速率；净产生 $O_2$，净消耗 $CO_2$ · (b)only aerobic respiration in mitochondria; consumes $O_2$, produces $CO_2$仅线粒体中进行有氧呼吸；消耗 $O_2$，产生 $CO_2$ · (c)incorrect; dark-period data shows plant consuming $O_2$ and releasing $CO_2$ (aerobic respiration)错误；黑暗期数据表明植物消耗 $O_2$ 并释放 $CO_2$（有氧呼吸）
(a) Why $O_2$ rises and $CO_2$ falls during the 12-hour light period光照 12 小时期间 $O_2$ 上升、$CO_2$ 下降的原因A1·A1·A1·A1
During the light period, both photosynthesis and aerobic respiration occur simultaneously in the plant. Photosynthesis (in chloroplasts) consumes $\text{CO}_2$ and $\text{H}_2\text{O}$, and produces $\text{O}_2$ and glucose. Aerobic respiration (in mitochondria) consumes $\text{O}_2$ and glucose, and produces $\text{CO}_2$ and $\text{H}_2\text{O}$. Because the light is bright, the rate of photosynthesis greatly exceeds the rate of respiration: the plant fixes more $\text{CO}_2$ than it releases and produces more $\text{O}_2$ than it consumes. The net direction is therefore: $\text{O}_2$ accumulates in the sealed container (rises) and $\text{CO}_2$ is drawn down (falls). This net change reflects apparent (net) photosynthesis, which is total photosynthesis minus respiration.光照期间，植物同时进行光合作用和有氧呼吸。光合作用（在叶绿体中）消耗 $\text{CO}_2$ 和 $\text{H}_2\text{O}$，产生 $\text{O}_2$ 和葡萄糖。有氧呼吸（在线粒体中）消耗 $\text{O}_2$ 和葡萄糖，产生 $\text{CO}_2$ 和 $\text{H}_2\text{O}$。由于光照强烈，光合速率远超呼吸速率：植物固定的 $\text{CO}_2$ 多于释放的，产生的 $\text{O}_2$ 多于消耗的。因此净方向为：$\text{O}_2$ 在密封容器中积累（上升），$\text{CO}_2$ 被消耗（下降）。这种净变化反映的是表观（净）光合速率，即总光合速率减去呼吸速率。
(b) Why $O_2$ falls and $CO_2$ rises during the 12-hour dark period黑暗 12 小时期间 $O_2$ 下降、$CO_2$ 上升的原因A1·A1
In complete darkness, the light reactions of photosynthesis cannot occur (no light energy to split water or produce ATP/NADPH), so the plant performs no photosynthesis. However, the plant's cells continue to perform aerobic cellular respiration around the clock in the mitochondria. Aerobic respiration breaks down glucose using $\text{O}_2$ and releases $\text{CO}_2$ and $\text{H}_2\text{O}$. In the sealed container, this causes $\text{O}_2$ concentration to fall and $\text{CO}_2$ concentration to rise throughout the dark period.在完全黑暗中，光合作用的光反应无法进行（没有光能分解水或产生 ATP/NADPH），所以植物不进行光合作用。然而，植物细胞的线粒体中全天候持续进行有氧细胞呼吸。有氧呼吸分解葡萄糖，消耗 $\text{O}_2$ 并释放 $\text{CO}_2$ 和 $\text{H}_2\text{O}$。在密封容器中，这导致整个黑暗期 $\text{O}_2$ 浓度下降，$\text{CO}_2$ 浓度上升。
(c) Evaluating the claim that plants only photosynthesize评价"植物只进行光合作用"的说法A1·A1·A1
The classmate's claim is incorrect. The dark-period data provides clear evidence that plants also perform aerobic respiration. During the 12-hour dark period, the sensor data shows $\text{O}_2$ falling and $\text{CO}_2$ rising inside the sealed container. Because no photosynthesis can occur in darkness, the only explanation for consuming $\text{O}_2$ and releasing $\text{CO}_2$ is aerobic cellular respiration. Plants are like all other eukaryotes: they continuously respire 24 hours a day to produce ATP for active transport, protein synthesis, growth, and other cellular processes. Photosynthesis only adds on top of this during daylight hours when light is available.该同学的说法错误。黑暗期数据提供了植物也进行有氧呼吸的清晰证据。在 12 小时黑暗期间，传感器数据显示密封容器内 $\text{O}_2$ 下降、$\text{CO}_2$ 上升。由于黑暗中不能进行光合作用，消耗 $\text{O}_2$ 并释放 $\text{CO}_2$ 的唯一解释就是有氧细胞呼吸。植物和其他所有真核生物一样：每天 24 小时持续呼吸，以产生 ATP 用于主动运输、蛋白质合成、生长和其他细胞过程。光合作用只是在白天有光时叠加在此基础之上。
The light compensation point separates net $O_2$ production from net $O_2$ consumption in plants.光补偿点是植物净产 $O_2$ 与净消耗 $O_2$ 的分界线。This experiment perfectly demonstrates the concept of the light compensation point: the light intensity at which the rate of photosynthesis exactly equals the rate of respiration (so net gas exchange is zero). At bright light, photosynthesis dominates (net $O_2$ produced); in darkness, respiration dominates (net $O_2$ consumed). Between these extremes lies the compensation point. Farmers and ecologists use this concept to determine how much light crops need to grow (they must be above the compensation point long enough each day to accumulate net organic matter). This also explains why indoor plants with insufficient light fail to thrive: they respire more than they photosynthesize and slowly use up their stored glucose.该实验完美地展示了光补偿点的概念：光合速率恰好等于呼吸速率时的光照强度（净气体交换为零）。在强光下，光合作用主导（净产 $O_2$）；在黑暗中，呼吸作用主导（净消耗 $O_2$）。光补偿点介于两者之间。农民和生态学家利用这一概念来确定农作物需要多少光照才能生长（每天必须超过补偿点足够长时间以积累净有机物）。这也解释了为什么光照不足的室内植物无法茁壮成长：它们的呼吸多于光合，慢慢消耗掉储存的葡萄糖。
Scientists studying a wetland ecosystem observe that submerged aquatic plants produce oxygen bubbles during the day but none at night. Meanwhile, bacteria living in the oxygen-depleted sediment at the bottom continue to survive by producing ethanol.研究湿地生态系统的科学家观察到，沉水植物白天产生氧气气泡，夜间则没有。与此同时，生活在底部缺氧沉积物中的细菌通过产生乙醇继续存活。
Answer summary:答案摘要：(a) $6\text{CO}_2 + 6\text{H}_2\text{O} \to \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$; $O_2$ atoms come from $H_2O$ (photolysis); confirmed by $^{18}O$ isotope tracing$O_2$ 中的氧原子来自 $H_2O$（光解）；$^{18}O$ 同位素示踪实验确认 · (b)no light, so light reactions stop; no ATP/NADPH for Calvin cycle; Calvin cycle halts无光，光反应停止；卡尔文循环无 ATP/NADPH；卡尔文循环停止 · (c)starting molecule: glucose; glycolysis + ethanol fermentation generate 2 ATP without $O_2$; limitation: far less ATP than aerobic respiration起始分子：葡萄糖；糖酵解和乙醇发酵在无氧条件下产生 2 个 ATP；局限性：远少于有氧呼吸产生的 ATP
(a) Balanced summary equation for photosynthesis; origin of oxygen atoms in $O_2$ bubbles光合作用平衡总方程式；$O_2$ 气泡中氧原子的来源A1·A1·A1
$$ 6\,\text{CO}_2 + 6\,\text{H}_2\text{O} \;\xrightarrow{\text{light energy}}\; \text{C}_6\text{H}_{12}\text{O}_6 + 6\,\text{O}_2 $$
The oxygen atoms in the $\text{O}_2$ bubbles originated from water ($\text{H}_2\text{O}$), not from $\text{CO}_2$. During the light reactions in Photosystem II, water molecules are split by photolysis: $2\text{H}_2\text{O} \to 4\text{H}^+ + 4e^- + \text{O}_2$. The oxygen atoms in $\text{CO}_2$ are incorporated into the organic product (glucose) during the Calvin cycle, not released as $\text{O}_2$. The evidence confirming this is isotope tracing: when aquatic plants (or any photosynthetic organism) are given water enriched with heavy oxygen ($^{18}\text{O}$-labelled $\text{H}_2\text{O}$), the $^{18}\text{O}$ label appears in the released $\text{O}_2$ gas, not in any $\text{CO}_2$ produced. This experiment definitively showed that the cleavage of water, not $\text{CO}_2$, is the source of the bubbles.$\text{O}_2$ 气泡中的氧原子来自水（$\text{H}_2\text{O}$），而非 $\text{CO}_2$。光反应中，光系统 II 通过光解分解水分子：$2\text{H}_2\text{O} \to 4\text{H}^+ + 4e^- + \text{O}_2$。$\text{CO}_2$ 中的氧原子在卡尔文循环中被结合进有机产物（葡萄糖），而非以 $\text{O}_2$ 形式释放。确认这一点的证据是同位素示踪：当给沉水植物（或任何光合生物）提供富含重氧（$^{18}\text{O}$ 标记的 $\text{H}_2\text{O}$）的水时，$^{18}\text{O}$ 标记出现在释放的 $\text{O}_2$ 气体中，而非产生的 $\text{CO}_2$ 中。该实验明确证明了气泡的来源是水的分裂，而非 $\text{CO}_2$。
(b) Why aquatic plants produce no $O_2$ bubbles at night despite having chloroplasts为何沉水植物夜间没有 $O_2$ 气泡，即使它们有叶绿体A1·A1·A1
At night, the aquatic plants produce no oxygen bubbles because the light reactions stop. The light reactions require light energy to drive the photolysis of water (splitting $\text{H}_2\text{O}$ to release $\text{O}_2$) and to generate ATP and NADPH. Without light, Photosystem I and Photosystem II cannot absorb photons and cannot drive the electron transport chain of the chloroplast. As a consequence, ATP and NADPH are no longer produced by the light reactions. This has an immediate consequence for the Calvin cycle: the Calvin cycle in the stroma requires a continuous supply of ATP and NADPH to fix $\text{CO}_2$ into G3P and ultimately glucose. Without ATP and NADPH, the Calvin cycle halts. The plant still has chloroplasts, but possessing the organelle is insufficient; the organelle needs light energy to function. All that occurs in the darkness is aerobic respiration in the mitochondria, which consumes $\text{O}_2$ (producing no bubbles).夜间，沉水植物不产生氧气气泡，因为光反应停止。光反应需要光能驱动水的光解（分解 $\text{H}_2\text{O}$ 释放 $\text{O}_2$）并产生 ATP 和 NADPH。没有光，光系统 I 和光系统 II 无法吸收光子，无法驱动叶绿体的电子传递链。因此，光反应不再产生 ATP 和 NADPH。这对卡尔文循环有直接影响：基质中的卡尔文循环需要持续供应 ATP 和 NADPH，将 $\text{CO}_2$ 固定为 G3P 并最终合成葡萄糖。没有 ATP 和 NADPH，卡尔文循环停止。植物仍有叶绿体，但拥有该细胞器是不够的；细胞器需要光能才能运作。黑暗中发生的只是线粒体中的有氧呼吸，消耗 $\text{O}_2$（不产生气泡）。
(c) Starting molecule for ethanol fermentation; how it generates ATP without $O_2$; one limitation乙醇发酵的起始分子；如何在无氧条件下产生 ATP；一个局限性A1·A1
The starting molecule for ethanol fermentation is glucose ($\text{C}_6\text{H}_{12}\text{O}_6$). The bacteria generate ATP in two steps without requiring oxygen: (1) Glycolysis in the cytoplasm breaks glucose into two pyruvate molecules and produces 2 ATP and 2 NADH. (2) Ethanol fermentation then converts pyruvate to acetaldehyde and then to ethanol, regenerating $\text{NAD}^+$ from NADH. This allows glycolysis to continue running (since it needs $\text{NAD}^+$ as an electron acceptor). The net result is 2 ATP per glucose, all from substrate-level phosphorylation, with no need for oxygen. One significant limitation compared to aerobic respiration is that ethanol fermentation yields only 2 ATP per glucose instead of approximately 30-32 ATP. This far lower energy yield means the bacteria must consume far more glucose to meet the same ATP demands, and they grow and reproduce much more slowly than aerobic organisms would under the same conditions.乙醇发酵的起始分子是葡萄糖（$\text{C}_6\text{H}_{12}\text{O}_6$）。细菌在不需要氧气的情况下通过两步产生 ATP：(1) 细胞质中的糖酵解将葡萄糖分解为两个丙酮酸分子，产生 2 个 ATP 和 2 个 NADH。(2) 乙醇发酵将丙酮酸转化为乙醛，再转化为乙醇，将 NADH 再生为 $\text{NAD}^+$。这使糖酵解得以继续运行（因为它需要 $\text{NAD}^+$ 作为电子受体）。净结果是每分子葡萄糖产生 2 个 ATP，全部来自底物水平磷酸化，无需氧气。与有氧呼吸相比，一个显著的局限性是乙醇发酵每分子葡萄糖仅产生 2 个 ATP，而非约 30-32 个。这种远低的能量产出意味着细菌必须消耗更多葡萄糖才能满足相同的 ATP 需求，且在相同条件下，其生长和繁殖速度远慢于有氧生物。
The wetland scenario illustrates two distinct energy worlds coexisting within centimetres of each other.湿地情境展示了在相距数厘米内共存的两个截然不同的能量世界。The surface water supports photosynthetic plants that harness solar energy and produce $\text{O}_2$; just below in the anoxic sediment, anaerobic bacteria survive on fermentation. This "oxic-anoxic interface" is a fundamental feature of many ecosystems (wetlands, lake bottoms, ocean floor). The bacteria in the sediment illustrate that life does not require oxygen per se; it requires a source of free energy and a way to extract it. Fermentation is ancient (preceding the rise of atmospheric oxygen ~2.4 billion years ago) and remains essential for organisms in oxygen-depleted environments today. The low ATP yield of fermentation explains why anaerobic zones support far less biodiversity and biomass than aerobic zones.水面支持光合植物，它们利用太阳能产生 $\text{O}_2$；就在数厘米以下的缺氧沉积物中，厌氧细菌靠发酵生存。这个"有氧-无氧界面"是许多生态系统（湿地、湖底、海底）的基本特征。沉积物中的细菌说明生命本身并不需要氧气；它需要自由能的来源和提取方式。发酵是古老的（早于约 24 亿年前大气氧的升高），至今对缺氧环境中的生物仍至关重要。发酵低效的 ATP 产量解释了为何厌氧区域的生物多样性和生物量远低于有氧区域。