High School Biology

Cellular Energetics细胞能量学

Every cell runs on ATP — the universal energy currency. This guide traces how cells make and spend that energy: ATP structure and the role of adenosine triphosphate, enzymes and activation energy, aerobic cellular respiration through glycolysis, the Krebs cycle, and the electron transport chain, anaerobic respiration and fermentation, photosynthesis through the light-dependent reactions and the Calvin cycle, and finally how respiration and photosynthesis are mirror processes that together drive the global carbon cycle. Worked examples and quiz questions use the summary equations and real biochemical contexts throughout.每个细胞都依赖 ATP——通用能量货币——运转。本指南追踪细胞如何制造和消耗能量：ATP 结构与三磷酸腺苷的作用、酶与活化能、通过糖酵解（glycolysis，糖酵解）、克雷布斯循环（Krebs cycle，克雷布斯循环）与电子传递链（electron transport chain，电子传递链）进行的有氧细胞呼吸（cellular respiration，细胞呼吸）、无氧呼吸与发酵（fermentation，发酵），以及通过光反应与卡尔文循环（Calvin cycle，卡尔文循环）进行的光合作用（photosynthesis，光合作用），最后比较呼吸与光合作用如何共同驱动全球碳循环。所有例题与测验均基于总结方程式与真实生化情境。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB SBI4U pathway detail marked HonorsSBI4U 代谢途径详解标为荣誉级
Read me first先读这里

How to use this guide如何使用本指南

Cellular energetics is the unit where the four curricula diverge most sharply. All four agree on the big picture: cells need ATP, plants make sugar by photosynthesis, and organisms release energy from sugar by cellular respiration. But they differ enormously on biochemical depth. US NGSS HS-LS1-5 and HS-LS1-7 explicitly exclude the biochemical steps (Assessment Boundaries state "does not include details of the biochemical steps"). BC Life Sciences 11 sits at the same conceptual level. Alberta Biology 20 Unit C names glycolysis, Krebs cycle, NADH/FADH, and chemiosmosis but repeatedly notes "detailed knowledge of metabolic intermediates is not required." Ontario SBI4U Strand C is the deepest: students must know glycolysis, Krebs, ETC, oxidative phosphorylation, Calvin cycle, and the named coenzymes. The table below locates each section in your curriculum; each row cites the source document it was verified against.细胞能量学是四套大纲分歧最大的单元。四者在大方向上一致：细胞需要 ATP，植物通过光合作用制造糖，生物体通过细胞呼吸从糖中释放能量。但在生化深度上分歧极大。US NGSS HS-LS1-5 与 HS-LS1-7 明确排除生化步骤（评估边界均注明"不含生化步骤细节"）。BC Life Sciences 11 处于同样的概念层面。阿尔伯塔 Biology 20 Unit C 命名了糖酵解、克雷布斯循环、NADH/FADH 与化学渗透，但反复注明"不要求对代谢中间体的详细了解"。安大略 SBI4U C 单元最深：学生须掌握糖酵解、克雷布斯循环、电子传递链、氧化磷酸化、卡尔文循环及命名辅酶。下表定位各节在你大纲中的位置；每行均注明所依据的来源文件。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Life Sciences美国 NGSS 生命科学 §1 (ATP role), §3 (respiration concept), §5 (photosynthesis concept), §7 (cycling of matter) — the overall reaction, organelle location, and energy-transfer model are the expectation§1（ATP 作用）、§3（细胞呼吸概念）、§5（光合作用概念）、§7（物质循环）——总反应、细胞器位置与能量转移模型为预期要求 Detailed biochemical pathways (going-deeper boxes): NGSS Assessment Boundaries explicitly exclude these steps — read for enrichment only详细生化途径（深入内容框）：NGSS 评估边界明确排除这些步骤——仅作拓展阅读 NGSS HS Life Science — HS-LS1-5, HS-LS1-7, HS-LS2-3 PEs and Assessment Boundaries— HS-LS1-5、HS-LS1-7、HS-LS2-3 表现期望及评估边界
🇨🇦 ON Grade 12 — SBI4U Honors安大略 12 年级 — SBI4U 荣誉 All 7 sections in full. SBI4U C2.1 requires named terminology (glycolysis, Krebs cycle, ETC, ATP synthase, chemiosmosis, Calvin cycle); C3.1 and C3.2 require explanation of chemical changes and energy conversions for both processes全部 7 节完整学习。SBI4U C2.1 要求命名术语（糖酵解、克雷布斯循环、ETC、ATP 合酶、化学渗透、卡尔文循环）；C3.1 与 C3.2 要求解释两个过程的化学变化与能量转化 Nothing — this is the primary curriculum for this unit at the deepest level无 — 本单元在此大纲中的深度最高 Ontario SBI3U/4U Biology — SBI4U Strand C C2.1, C3.1, C3.2— SBI4U C 单元 C2.1、C3.1、C3.2
🇨🇦 BC Life Sciences 11BC Life Sciences 11 §1, §3 cram-cheat, §5 cram-cheat, §7 concept: the Content bullet names the overall reactions ("glucose broken down yields ATP and CO2"; "consumes CO2 and water, produces O2 and sugars") but no biochemical steps§1、§3 速记框、§5 速记框、§7 概念：内容条目仅命名总反应（"葡萄糖分解产生 ATP 和 CO2"；"消耗 CO2 和水，产生 O2 和糖"），无生化步骤 All going-deeper boxes (§2 enzyme kinetics, §3 Krebs/ETC detail, §5 photosystem detail, §6 Calvin intermediate)所有深入内容框（§2 酶动力学、§3 克雷布斯/ETC 细节、§5 光系统细节、§6 卡尔文中间体） BC Life Sciences 11 / Anatomy 12 — Life Sciences 11 Content bullet "energy transformations in cells"— Life Sciences 11 内容条目"细胞中的能量转化"
🇨🇦 AB Biology 20阿尔伯塔 Biology 20 §1–§7 all relevant. Biology 20 Unit C names glycolysis, Krebs, NADH/FADH, chemiosmosis, light-dependent/light-independent reactions — but repeatedly notes "detailed knowledge of metabolic intermediates is not required." Aerobic vs anaerobic vs fermentation distinction (§4) is explicitly required (20–C2.3k)§1–§7 全部相关。Biology 20 Unit C 命名了糖酵解、克雷布斯循环、NADH/FADH、化学渗透、光依赖/光独立反应——但反复注明"不要求对代谢中间体的详细了解"。有氧/无氧/发酵之分（§4）明确要求（20–C2.3k） Intermediate names (pyruvate, acetyl-CoA, G3P, NADPH specific roles): Alberta notes these as not required. Read for context only.中间体名称（丙酮酸、乙酰辅酶 A、G3P、NADPH 具体角色）：阿尔伯塔注明这些不作要求。仅作上下文阅读。 Alberta Biology 20/30 — Biology 20 Unit C GO1 (20–C1.1k, 20–C1.2k) and GO2 (20–C2.1k–20–C2.4k)— Biology 20 Unit C GO1（20–C1.1k、20–C1.2k）与 GO2（20–C2.1k–20–C2.4k）

Once you have located your row, use the two cards below for the approach that fits your timeline.找到所在行后，用下面两张卡片选择适合你时间安排的方式。

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If you are cramming the night before如果你在临阵磨枪

Memorize both summary equations ($C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O$ and the reverse for photosynthesis), know that respiration happens in mitochondria and photosynthesis in chloroplasts, know lactic acid vs ethanol fermentation, and be able to explain why ATP is the energy currency. Read every cram-cheat box. Skip the going-deeper boxes on Krebs intermediates and photosystem protein complexes.记住两个总方程式（$C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O$ 及其光合作用的逆式），了解细胞呼吸发生在线粒体（mitochondria，线粒体）、光合作用发生在叶绿体（chloroplast，叶绿体），了解乳酸与乙醇发酵之分，并能解释 ATP 为何是能量货币。读每个速记框，跳过克雷布斯中间体与光系统蛋白复合体的深入内容。

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If you are going for the top mark如果你目标顶分

For SBI4U: trace every named intermediate and coenzyme. Distinguish glycolysis (cytoplasm), link reaction (mitochondrial matrix), Krebs cycle (matrix), and ETC (inner mitochondrial membrane). Know that chemiosmosis drives ATP synthase via a proton gradient. For photosynthesis: distinguish Photosystem II (splits water, releases O2, excites electrons) from Photosystem I (reduces NADP+ to NADPH); trace carbon fixation in the Calvin cycle through 3 turns to produce one G3P. Explain why the net ATP yield from aerobic respiration (~30–32 ATP) is so much higher than from fermentation (2 ATP).SBI4U 轨道：追踪每个命名中间体与辅酶。区分糖酵解（细胞质）、连接反应（线粒体基质）、克雷布斯循环（基质）与电子传递链（线粒体内膜）。了解化学渗透通过质子梯度驱动 ATP 合酶。光合作用：区分光系统 II（分解水、释放 O2、激发电子）与光系统 I（将 NADP+ 还原为 NADPH）；追踪卡尔文循环碳固定经 3 轮产生一个 G3P 的过程。解释为何有氧呼吸的净 ATP 产量（约 30–32 个 ATP）远高于发酵（2 个 ATP）。

Honors flag.荣誉级标记。 The going-deeper boxes in §2 (enzyme kinetics), §3 (Krebs cycle intermediates, ETC complexes), §5 (Photosystem I vs II, Z-scheme), and §6 (Calvin cycle C3 intermediates) carry the Honors SBI4U chip, because Ontario SBI4U C2.1/C3.1/C3.2 expects this depth while NGSS and BC Life Sciences 11 explicitly exclude it. Alberta Biology 20 Unit C names the steps in general terms but does not require intermediates. If your curriculum is NGSS or BC Life Sciences 11, read the going-deeper sections for conceptual enrichment but do not memorize the named lists.§2（酶动力学）、§3（克雷布斯循环中间体、ETC 复合体）、§5（光系统 I 与 II、Z 方案）和 §6（卡尔文循环 C3 中间体）的深入内容框标注 荣誉 SBI4U，因为安大略 SBI4U C2.1/C3.1/C3.2 要求这一深度，而 NGSS 与 BC Life Sciences 11 明确排除。阿尔伯塔 Biology 20 Unit C 以概括方式命名步骤但不要求中间体。若你的大纲是 NGSS 或 BC Life Sciences 11，阅读深入内容以获得概念拓展，但无需背诵命名列表。
Section 1 · ATP and energy in cells第 1 节 · ATP 与细胞能量

ATP and Energy in CellsATP 与细胞能量

ATP: adenosine triphosphate, the universal energy currency.ATP：三磷酸腺苷，通用能量货币。
  • Structure: adenine base + ribose sugar + three phosphate groups. The bond between the second and third phosphate stores chemical potential energy.结构：腺嘌呤碱基 + 核糖 + 三个磷酸基团。第二与第三磷酸基团之间的键储存化学势能。
  • Hydrolysis releases energy: $\text{ATP} + \text{H}_2\text{O} \to \text{ADP} + \text{P}_i + \text{energy}$. The cell uses this energy for active transport, muscle contraction, biosynthesis, and other work.水解释放能量：$\text{ATP} + \text{H}_2\text{O} \to \text{ADP} + \text{P}_i + \text{energy}$。细胞用这些能量进行主动运输、肌肉收缩、生物合成和其他工作。
  • Regeneration: ADP is recharged back to ATP by cellular respiration (in mitochondria) and by photosynthesis (in chloroplasts in plant cells). ATP is recycled — it is not made once and stored.再生：ADP 通过细胞呼吸（在线粒体中）和光合作用（在植物细胞叶绿体中）重新充能为 ATP。ATP 是循环利用的——不是一次性制造后储存。
  • Energy scale: a typical human cell uses and recycles roughly its own weight in ATP every day. Muscle cells at maximum effort can cycle ATP in under a second.能量规模：典型人体细胞每天消耗和循环的 ATP 量大约相当于自身重量。肌肉细胞在最大用力时可在不到一秒内循环 ATP。
ATP hydrolysis and the overall energy logicATP 水解与整体能量逻辑
$$\text{ATP} + \text{H}_2\text{O} \;\longrightarrow\; \text{ADP} + \text{P}_i \quad (\Delta G \approx -30.5\ \text{kJ mol}^{-1})$$

The cell is an energy transformer: food (chemical energy) → ATP → mechanical, electrical, or chemical work. Photosynthesis runs this in reverse: light energy → ATP → glucose (chemical storage).细胞是能量转换器：食物（化学能）→ ATP → 机械、电或化学做功。光合作用逆向运行：光能 → ATP → 葡萄糖（化学储存）。

Worked Example 1.例题 1。 A student claims that cells store large amounts of ATP for future use the way a battery stores charge. Evaluate this claim.一名学生声称细胞像电池储电一样储存大量 ATP 备用。请评价这一说法。

Evaluation: incorrect. ATP is not stored in bulk — it is synthesized on demand and immediately consumed. The energy is stored long-term in glucose (and fats). ATP has a half-life of seconds to minutes in a cell. The correct analogy is a rechargeable battery that is constantly being charged and discharged, not a reservoir.评价：不正确。ATP 不会大量储存——它按需合成并立即消耗。能量以葡萄糖（和脂肪）形式长期储存。ATP 在细胞中的半衰期为秒到分钟。正确的类比是一块不断被充电和放电的可充电电池，而非储能库。

What is released when one ATP molecule is hydrolyzed inside a cell?细胞内一个 ATP 分子水解时释放什么？
§1 · Q1
AMP and two phosphate groupsAMP 和两个磷酸基团
Glucose and oxygen葡萄糖和氧气
ADP, inorganic phosphate ($\text{P}_i$), and energyADP、无机磷酸（$\text{P}_i$）和能量
Carbon dioxide and water二氧化碳和水
ATP hydrolysis: ATP + H₂O → ADP + Pi + energy (~30.5 kJ mol⁻¹). The released energy drives cellular work; ADP and Pi are recycled by respiration or photosynthesis back to ATP.ATP 水解：ATP + H₂O → ADP + Pi + 能量（约 30.5 kJ mol⁻¹）。释放的能量驱动细胞做功；ADP 和 Pi 经呼吸或光合作用循环回 ATP。
ATP loses one phosphate group when hydrolyzed: ATP → ADP + Pi + energy. CO₂ and water are products of cellular respiration, not ATP hydrolysis.ATP 水解失去一个磷酸基团：ATP → ADP + Pi + 能量。CO₂ 和水是细胞呼吸的产物，不是 ATP 水解的产物。
Which organelle is the primary site of ATP production in aerobic eukaryotic cells?在有氧真核细胞中，哪种细胞器是产生 ATP 的主要场所？
§1 · Q2
Mitochondrion线粒体
Nucleus细胞核
Ribosome核糖体
Lysosome溶酶体
Mitochondria are the site of aerobic cellular respiration and produce the vast majority of ATP (via the electron transport chain and chemiosmosis). A small amount is made in glycolysis in the cytoplasm.线粒体（mitochondria，线粒体）是有氧细胞呼吸（cellular respiration，细胞呼吸）的场所，通过电子传递链与化学渗透产生绝大多数 ATP。少量 ATP 在细胞质中的糖酵解产生。
The mitochondrion is the "powerhouse of the cell." The nucleus stores DNA; ribosomes make proteins; lysosomes digest material. ATP production requires the mitochondrial inner membrane.线粒体是细胞的"能量工厂"。细胞核储存 DNA；核糖体合成蛋白质；溶酶体消化物质。ATP 生产需要线粒体内膜。
Section 2 · Enzymes in metabolism第 2 节 · 代谢中的酶

Enzymes in Metabolism代谢中的酶

Enzymes are biological catalysts — they lower activation energy without being consumed.酶是生物催化剂——它们降低活化能，但自身不被消耗。
  • Activation energy ($E_a$):活化能（$E_a$）： the minimum energy needed to start a chemical reaction. Enzymes lower $E_a$, so reactions proceed faster at body temperature than they would spontaneously.启动化学反应所需的最低能量。酶降低 $E_a$，使反应在体温下比自发进行快得多。
  • Active site and lock-and-key / induced-fit:活性位点与锁钥/诱导契合模型： the substrate binds the enzyme's active site. Lock-and-key: rigid complementary shapes. Induced-fit: the enzyme changes shape slightly when the substrate binds, tightening the interaction.底物与酶的活性位点结合。锁钥模型：形状互补且固定。诱导契合模型：底物结合时酶形状略有改变，加强互作。
  • Factors affecting enzyme activity:影响酶活性的因素：
    • Temperature: increasing temperature raises reaction rate until the optimum; above the optimum the enzyme denatures (bonds break, active site distorted).温度：升温提高反应速率，直至最适温度；超过最适温度，酶变性（键断裂，活性位点扭曲）。
    • pH: each enzyme has an optimum pH (e.g. pepsin at pH 2, salivary amylase at pH 7). Outside the optimum, ionization of the active site changes and activity drops.pH：每种酶有最适 pH（如胃蛋白酶 pH 2，唾液淀粉酶 pH 7）。偏离最适值，活性位点离子化状态改变，活性下降。
    • Substrate concentration: rate increases with substrate concentration until all enzyme active sites are saturated (maximum rate, $V_{max}$).底物浓度：速率随底物浓度增加而升高，直至所有酶活性位点饱和（最大速率，$V_{max}$）。
    • Inhibitors: competitive inhibitors block the active site (reversible if substrate out-competes); non-competitive inhibitors bind elsewhere and change the active site shape.抑制剂：竞争性抑制剂阻断活性位点（若底物能竞争胜出则可逆）；非竞争性抑制剂结合在其他位置，改变活性位点形状。
  • Metabolic pathways:代谢途径： both cellular respiration and photosynthesis are chains of enzyme-catalyzed reactions. Each step produces an intermediate that feeds the next reaction. Blocking one enzyme can halt the entire pathway.细胞呼吸与光合作用均为酶催化反应链。每步产生中间体，为下一步反应提供原料。阻断一个酶可使整个途径停止。
Worked Example 2 · Enzyme and temperature例题 2 · 酶与温度

An experiment measures the rate of an enzyme-catalyzed reaction at 10 °C, 25 °C, 37 °C, and 75 °C. At 75 °C the rate drops to near zero. Explain why, using the terms active site, denaturation, and enzyme–substrate complex.实验测量酶催化反应在 10 °C、25 °C、37 °C 和 75 °C 下的速率。75 °C 时速率接近零。请用"活性位点""变性"和"酶-底物复合物"解释原因。

At 10–37 °C, increasing temperature provides more kinetic energy, so substrate molecules collide with the active site more frequently, forming more enzyme–substrate complexes per second. At 37 °C the rate is near the optimum for most human enzymes. At 75 °C, the high temperature breaks the hydrogen bonds and other non-covalent interactions that maintain the enzyme's 3-D shape. The active site loses its specific geometry; substrate can no longer fit. This is denaturation — an irreversible change in protein structure. The reaction rate drops to near zero because no functional enzyme remains.在 10–37 °C，温度升高提供更多动能，底物分子更频繁地与活性位点碰撞，每秒形成更多酶-底物复合物。37 °C 时速率接近大多数人体酶的最适值。在 75 °C，高温断裂维持酶三维结构的氢键等非共价相互作用。活性位点失去特定几何形状；底物无法再与之结合。这是变性——蛋白质结构的不可逆变化。由于没有功能性酶存在，反应速率降至接近零。

How do enzymes speed up biochemical reactions?酶如何加速生化反应？
§2 · Q1
They supply energy directly to the reaction直接向反应提供能量
They lower the activation energy required to start the reaction降低启动反应所需的活化能
They increase the temperature inside the cell升高细胞内的温度
They are consumed in the reaction and regenerated separately在反应中被消耗，然后单独再生
Enzymes are catalysts: they lower the activation energy (Ea) so the reaction can proceed quickly at body temperature. They are not consumed — the same enzyme molecule can catalyze the same reaction millions of times.酶是催化剂：它们降低活化能（Ea），使反应在体温下能快速进行。酶不被消耗——同一酶分子可催化同一反应数百万次。
Enzymes do not add energy or heat; they lower activation energy so the reaction can proceed faster without extra energy input. Enzymes are also not consumed.酶不添加能量或热量；它们降低活化能，使反应无需额外能量输入即可更快进行。酶也不会被消耗。
A competitive inhibitor of an enzyme resembles the natural substrate. What does it do?某酶的竞争性抑制剂与天然底物相似。它的作用是什么？
§2 · Q2
It permanently destroys the active site永久破坏活性位点
It binds to the allosteric site and changes the enzyme's shape结合变构位点并改变酶的形状
It raises the activation energy升高活化能
It competes with the substrate for the active site, reducing the rate of product formation与底物竞争活性位点，降低产物形成速率
A competitive inhibitor occupies the active site, blocking the natural substrate. The inhibition is reversible — increasing substrate concentration can out-compete the inhibitor and restore activity. Many drugs work this way (e.g., statins block an enzyme in cholesterol synthesis).竞争性抑制剂占据活性位点，阻止天然底物结合。该抑制是可逆的——增加底物浓度可竞争胜过抑制剂并恢复活性。许多药物就是这样起作用的（如他汀类药物阻断胆固醇合成中的一个酶）。
A competitive inhibitor binds the active site (reversibly). An allosteric/non-competitive inhibitor binds elsewhere. Inhibitors do not raise activation energy — they reduce the rate of productive substrate binding.竞争性抑制剂与活性位点结合（可逆）。变构/非竞争性抑制剂结合在其他位置。抑制剂不升高活化能——它们降低有效底物结合的速率。
Going deeper — enzyme kinetics: Vmax, Km, and allosteric regulation (SBI4U Biochemistry)深入 — 酶动力学：Vmax、Km 与变构调节（SBI4U 生物化学）

The Michaelis-Menten model describes how reaction rate ($v$) increases with substrate concentration ($[S]$) toward a maximum rate $V_{max}$. The Michaelis constant $K_m$ is the substrate concentration at half-$V_{max}$: a low $K_m$ means high affinity (the enzyme works well even at low substrate). In allosteric regulation, a molecule binds a site other than the active site (the allosteric site), changing the enzyme's 3-D shape and either activating or inhibiting it. This is how metabolic pathways self-regulate: the end-product often acts as an allosteric inhibitor of an early enzyme in the same pathway (feedback inhibition), preventing overproduction.米氏模型描述反应速率（$v$）如何随底物浓度（$[S]$）增加趋向最大速率 $V_{max}$。米氏常数 $K_m$ 是半最大速率时的底物浓度：$K_m$ 低意味着亲和力高（即使底物浓度低，酶也能高效工作）。在变构调节中，分子结合活性位点以外的位置（变构位点），改变酶的三维结构，从而激活或抑制酶。代谢途径就是这样自我调节的：终产物通常充当同一途径早期酶的变构抑制剂（反馈抑制），防止过度生产。

Section 3 · Cellular respiration overview第 3 节 · 细胞呼吸概述

Cellular Respiration: Glycolysis, Krebs Cycle, and the Electron Transport Chain细胞呼吸：糖酵解、克雷布斯循环与电子传递链

Summary equation (aerobic respiration) — memorize this.总方程式（有氧呼吸）— 记住它。 $$C_6H_{12}O_6 + 6O_2 \;\longrightarrow\; 6CO_2 + 6H_2O + \text{ATP (energy)}$$
  • Reactants:反应物： glucose ($C_6H_{12}O_6$) and oxygen ($O_2$). Reactants enter via food digestion and breathing.葡萄糖（$C_6H_{12}O_6$）和氧气（$O_2$）。反应物通过食物消化和呼吸进入。
  • Products:产物： carbon dioxide ($CO_2$) exhaled, water ($H_2O$) released, and ATP produced (~30–32 ATP per glucose in aerobic conditions).二氧化碳（$CO_2$）呼出，水（$H_2O$）释放，产生 ATP（有氧条件下每分子葡萄糖约 30–32 个 ATP）。
  • Three stages (NGSS: know the overall; AB: know the named stages; SBI4U: know each stage in detail):三个阶段（NGSS：了解总体；AB：了解命名阶段；SBI4U：了解每个阶段的详细内容）：
    1. Glycolysis (cytoplasm): glucose (6C) split into 2 pyruvate (3C each). Net gain: 2 ATP, 2 NADH. No oxygen required.糖酵解（glycolysis，糖酵解）（细胞质）：葡萄糖（6C）分解为 2 个丙酮酸（各 3C）。净产出：2 个 ATP、2 个 NADH。无需氧气。
    2. Krebs cycle (mitochondrial matrix): pyruvate converted to acetyl-CoA, then fed into the cycle. Each turn: 1 ATP, 3 NADH, 1 FADH₂; CO₂ released. Two turns per glucose. Honors SBI4U克雷布斯循环（Krebs cycle，克雷布斯循环）（线粒体基质）：丙酮酸转化为乙酰辅酶 A，再进入循环。每轮：1 个 ATP、3 个 NADH、1 个 FADH₂；释放 CO₂。每分子葡萄糖进行两轮。荣誉 SBI4U
    3. Electron Transport Chain (ETC) (inner mitochondrial membrane): NADH and FADH₂ donate electrons; electrons move through protein complexes; protons are pumped across the membrane; ATP synthase uses the proton gradient (chemiosmosis) to make ATP. Final electron acceptor: O₂ → H₂O. Produces ~26–28 ATP. Honors SBI4U电子传递链（electron transport chain，电子传递链，ETC）（线粒体内膜）：NADH 与 FADH₂ 提供电子；电子沿蛋白质复合体传递；质子被泵过膜；ATP 合酶利用质子梯度（化学渗透）合成 ATP。最终电子受体：O₂ → H₂O。产生约 26–28 个 ATP。荣誉 SBI4U
Worked Example 3 · Location of each stage例题 3 · 各阶段的位置

Identify the location of each stage of aerobic respiration and explain why that location makes sense structurally.确定有氧呼吸各阶段的位置，并解释为何在结构上合理。

Glycolysis: cytoplasm. It evolved in ancient anaerobic cells before mitochondria existed. It requires no membrane and produces 2 ATP quickly from glucose.糖酵解（glycolysis，糖酵解）：细胞质。它在线粒体存在之前的古代无氧细胞中演化形成。不需要膜结构，可从葡萄糖快速产生 2 个 ATP。

Krebs cycle: mitochondrial matrix. Pyruvate is transported into the mitochondrion, converted to acetyl-CoA (releasing CO₂), and fed into the cycle. The matrix contains all the enzymes needed.克雷布斯循环（Krebs cycle，克雷布斯循环）：线粒体基质。丙酮酸被转运到线粒体内，转化为乙酰辅酶 A（释放 CO₂），再进入循环。基质含有所需的全部酶。

ETC: inner mitochondrial membrane. The cristae (folds) dramatically increase the surface area of the inner membrane, maximizing the density of ATP synthase proteins. Protons are pumped into the intermembrane space and flow back through ATP synthase — chemiosmosis.电子传递链（electron transport chain，电子传递链）：线粒体内膜。嵴（折叠）大大增加内膜的表面积，使 ATP 合酶蛋白的密度最大化。质子被泵入膜间隙，再通过 ATP 合酶流回——化学渗透。

In which location does glycolysis occur?糖酵解发生在哪里？
§3 · Q1
Cytoplasm细胞质
Mitochondrial matrix线粒体基质
Inner mitochondrial membrane线粒体内膜
Chloroplast stroma叶绿体基质
Glycolysis occurs in the cytoplasm of all cells (prokaryotic and eukaryotic). It requires no organelle and proceeds without oxygen, splitting one glucose into two pyruvate molecules with a net gain of 2 ATP.糖酵解（glycolysis，糖酵解）发生在所有细胞（原核和真核）的细胞质中。不需要细胞器，在无氧条件下进行，将一个葡萄糖分解为两个丙酮酸分子，净产出 2 个 ATP。
Glycolysis is in the cytoplasm. The Krebs cycle is in the mitochondrial matrix; the ETC is on the inner mitochondrial membrane; the Calvin cycle (not respiration) is in the chloroplast stroma.糖酵解在细胞质中。克雷布斯循环在线粒体基质；ETC 在线粒体内膜；卡尔文循环（不是呼吸作用）在叶绿体基质。
What is the final electron acceptor in the electron transport chain of aerobic respiration?有氧呼吸电子传递链中的最终电子受体是什么？
§3 · Q2
Carbon dioxide ($CO_2$)二氧化碳（$CO_2$）
NADHNADH
Oxygen ($O_2$)氧气（$O_2$）
Glucose ($C_6H_{12}O_6$)葡萄糖（$C_6H_{12}O_6$）
Oxygen is the final electron acceptor in aerobic respiration. It accepts electrons from the ETC and combines with protons to form water (H₂O). This is why aerobic respiration requires oxygen and why the process is called "aerobic."氧气（$O_2$）是有氧呼吸中的最终电子受体。它接受来自电子传递链（electron transport chain，电子传递链）的电子，并与质子结合形成水（H₂O）。这就是为什么有氧呼吸需要氧气，以及该过程被称为"有氧"的原因。
O₂ is the final electron acceptor in aerobic respiration — it becomes H₂O. CO₂ is a waste product of decarboxylation in the Krebs cycle; NADH is an electron carrier, not acceptor; glucose is the starting substrate.O₂ 是有氧呼吸中的最终电子受体——它变成 H₂O。CO₂ 是克雷布斯循环脱羧反应的废物；NADH 是电子载体而非受体；葡萄糖是初始底物。
Going deeper — chemiosmosis and the proton gradient (SBI4U C2.1, C3.1)深入 — 化学渗透与质子梯度（SBI4U C2.1、C3.1）

As electrons pass through the four protein complexes of the ETC (Complexes I–IV), protons (H⁺) are actively pumped from the mitochondrial matrix into the intermembrane space, building a proton gradient (also called a proton-motive force). This gradient represents stored potential energy. Protons flow back into the matrix through ATP synthase (Complex V) — a molecular turbine that uses the flow to rotate and phosphorylate ADP + Pi to ATP. This coupling of electron flow to ATP synthesis through a proton gradient is chemiosmosis. Peter Mitchell won the Nobel Prize in 1978 for discovering this mechanism. The SBI4U expectation is to name oxidative phosphorylation as the overall process, and chemiosmosis as the mechanism within it.电子通过电子传递链（electron transport chain，电子传递链）的四个蛋白复合体（复合体 I–IV）时，质子（H⁺）从线粒体基质主动泵入膜间隙，建立质子梯度（也称质子驱动力）。该梯度代表储存的势能。质子通过 ATP 合酶（复合体 V）流回基质——这是一个分子涡轮，利用质子流旋转，将 ADP + Pi 磷酸化为 ATP。通过质子梯度将电子流与 ATP 合成耦合的过程称为化学渗透。彼得·米切尔因发现这一机制于 1978 年获得诺贝尔奖。SBI4U 的要求是将氧化磷酸化命名为整体过程，将化学渗透命名为其中的机制。

Section 4 · Anaerobic respiration and fermentation第 4 节 · 无氧呼吸与发酵

Anaerobic Respiration and Fermentation无氧呼吸与发酵

When oxygen is absent, cells still need ATP — they use fermentation.当氧气缺乏时，细胞仍需 ATP——它们使用发酵（fermentation，发酵）。
  • Starting point:起点： glycolysis still runs (in cytoplasm), producing 2 pyruvate and 2 ATP. The problem: glycolysis also produces NADH, which must be oxidized back to NAD⁺ for glycolysis to continue. Without O₂, the ETC cannot do this — fermentation steps in.糖酵解（glycolysis，糖酵解）仍在进行（细胞质中），产生 2 个丙酮酸和 2 个 ATP。问题是：糖酵解还产生 NADH，必须将其氧化回 NAD⁺ 才能使糖酵解继续。没有 O₂，电子传递链（electron transport chain，电子传递链）无法做到这一点——发酵介入。
  • Lactic acid fermentation (animals, bacteria):乳酸发酵（动物、细菌）： $$\text{Pyruvate} + \text{NADH} \;\longrightarrow\; \text{Lactate} + \text{NAD}^+$$ Example: muscle cells during intense exercise. Lactate accumulates, causing the burning sensation. Removed by the liver later. Used to make yogurt and cheese.例：剧烈运动时的肌肉细胞。乳酸积累，产生灼烧感。之后由肝脏清除。也用于制作酸奶和奶酪。
  • Ethanol fermentation (yeast, some bacteria):乙醇发酵（酵母菌、部分细菌）： $$\text{Pyruvate} \;\longrightarrow\; \text{Acetaldehyde} + CO_2 \;\longrightarrow\; \text{Ethanol} + \text{NAD}^+$$ Example: yeast in bread dough (CO₂ causes rising; ethanol evaporates during baking) and in brewing beer/wine. Only 2 ATP per glucose vs ~30–32 in aerobic respiration.例：面团中的酵母（CO₂ 使面团发酵；乙醇在烘烤时蒸发），以及酿造啤酒/葡萄酒。每分子葡萄糖仅产生 2 个 ATP，而有氧呼吸约产生 30–32 个。
  • Key comparison:关键对比： Aerobic: ~30–32 ATP/glucose, requires O₂, complete oxidation. Anaerobic: 2 ATP/glucose, no O₂ needed, partial oxidation only.有氧：约 30–32 个 ATP/葡萄糖，需要 O₂，完全氧化。无氧：2 个 ATP/葡萄糖，不需要 O₂，仅部分氧化。
Why fermentation matters for survival.发酵对生存的意义。

Fermentation allows cells to survive brief periods of oxygen deprivation. Sprinters rely on lactic acid fermentation when oxygen delivery to muscles cannot keep up with ATP demand. Deep-sea anaerobic bacteria use fermentation as their only ATP source. Comparing aerobic and anaerobic conditions is explicitly required by Alberta Biology 20 Unit C GO2 (20–C2.3k: "distinguish, in general terms, between aerobic and anaerobic respiration and fermentation in plants, animals and yeast") and by NGSS HS-LS2-3 (cycling of matter in aerobic and anaerobic conditions).发酵使细胞能够在短暂缺氧期间存活。短跑运动员在肌肉供氧无法满足 ATP 需求时依赖乳酸发酵（fermentation，发酵）。深海无氧细菌将发酵作为其唯一的 ATP 来源。阿尔伯塔 Biology 20 Unit C GO2（20–C2.3k："概括描述植物、动物和酵母中有氧和无氧呼吸及发酵的区别"）和 NGSS HS-LS2-3（有氧和无氧条件下物质循环）明确要求比较有氧与无氧条件。

Which statement correctly compares aerobic respiration and lactic acid fermentation?下列哪项正确比较了有氧呼吸与乳酸发酵？
§4 · Q1
Both produce the same amount of ATP per glucose两者每分子葡萄糖产生相同数量的 ATP
Aerobic respiration requires oxygen; lactic acid fermentation does not, and produces far less ATP有氧呼吸需要氧气；乳酸发酵不需要，且产生的 ATP 少得多
Lactic acid fermentation occurs in the mitochondria; aerobic respiration occurs in the cytoplasm乳酸发酵发生在线粒体；有氧呼吸发生在细胞质
Both processes produce CO₂ as a waste product两个过程都产生 CO₂ 作为废物
Aerobic respiration requires O₂ and produces ~30–32 ATP per glucose. Lactic acid fermentation requires no O₂ and produces only 2 ATP per glucose (from glycolysis). Lactic acid fermentation does not produce CO₂ (ethanol fermentation does).有氧呼吸需要 O₂，每分子葡萄糖产生约 30–32 个 ATP。乳酸发酵（fermentation，发酵）不需要 O₂，每分子葡萄糖仅产生 2 个 ATP（来自糖酵解）。乳酸发酵不产生 CO₂（乙醇发酵才产生）。
Aerobic respiration: O₂ required, ~30–32 ATP. Lactic acid fermentation: no O₂, only 2 ATP. Both start with glycolysis in the cytoplasm; the Krebs/ETC steps only occur aerobically. Lactic acid fermentation does not release CO₂.有氧呼吸：需要 O₂，约 30–32 个 ATP。乳酸发酵：不需要 O₂，仅 2 个 ATP。两者都从细胞质中的糖酵解开始；克雷布斯/ETC 步骤仅在有氧条件下发生。乳酸发酵不释放 CO₂。
Why does yeast produce CO₂ during ethanol fermentation?为什么酵母菌在乙醇发酵过程中产生 CO₂？
§4 · Q2
Because the ETC releases CO₂ as electrons pass through因为电子通过时电子传递链释放 CO₂
Because glucose is directly converted to CO₂ without any intermediate steps因为葡萄糖没有任何中间步骤直接转化为 CO₂
CO₂ is produced during lactic acid fermentation, not ethanol fermentationCO₂ 在乳酸发酵中产生，而非乙醇发酵
Pyruvate is decarboxylated to acetaldehyde, releasing CO₂ before ethanol is formed丙酮酸脱羧为乙醛，在乙醇形成之前释放 CO₂
In ethanol fermentation: pyruvate → acetaldehyde + CO₂ (decarboxylation step), then acetaldehyde + NADH → ethanol + NAD⁺. The CO₂ released in bread-making causes the dough to rise. Lactic acid fermentation does NOT release CO₂.乙醇发酵（fermentation，发酵）中：丙酮酸 → 乙醛 + CO₂（脱羧步骤），然后乙醛 + NADH → 乙醇 + NAD⁺。面包制作中释放的 CO₂ 使面团发酵。乳酸发酵不释放 CO₂。
Ethanol fermentation decarboxylates pyruvate to acetaldehyde (releasing CO₂), then reduces it to ethanol. Lactic acid fermentation does not produce CO₂ — pyruvate is directly converted to lactate.乙醇发酵将丙酮酸脱羧为乙醛（释放 CO₂），然后还原为乙醇。乳酸发酵不产生 CO₂——丙酮酸直接转化为乳酸。
Section 5 · Photosynthesis — light-dependent reactions第 5 节 · 光合作用——光反应

Photosynthesis: Light-Dependent Reactions光合作用（photosynthesis，光合作用）：光依赖反应

Summary equation for photosynthesis (reverse of respiration).光合作用总方程式（呼吸作用的逆过程）。 $$6CO_2 + 6H_2O + \text{light energy} \;\longrightarrow\; C_6H_{12}O_6 + 6O_2$$
  • Location:位置： chloroplasts in plant and algae cells. The light reactions occur in the thylakoid membranes (stacked discs); the Calvin cycle occurs in the stroma (fluid surrounding the thylakoids).植物和藻类细胞的叶绿体（chloroplast，叶绿体）中。光反应发生在类囊体膜（叠状圆盘）上；卡尔文循环（Calvin cycle，卡尔文循环）发生在基质（类囊体周围的液体）中。
  • Pigments:色素： chlorophyll a and b absorb mainly red and blue light; they reflect green, which is why plants look green. Carotenoids absorb additional wavelengths, extending the light spectrum captured.叶绿素 a 和 b 主要吸收红光和蓝光；反射绿光，这就是植物呈绿色的原因。类胡萝卜素吸收额外波长，扩大了捕获的光谱范围。
  • Light reactions (thylakoid membrane):光反应（类囊体膜）：
    1. Light is absorbed by Photosystem II (PSII). Water is split (photolysis): $2H_2O \to 4H^+ + 4e^- + O_2$. The O₂ is released as a byproduct.光被光系统 II（PSII）吸收。水分解（光解）：$2H_2O \to 4H^+ + 4e^- + O_2$。O₂ 作为副产品释放。
    2. Energized electrons pass through the electron transport chain to Photosystem I (PSI), pumping H⁺ across the thylakoid membrane. ATP is made by chemiosmosis (ATP synthase). Honors SBI4U激发电子通过电子传递链传递到光系统 I（PSI），将 H⁺ 泵过类囊体膜。通过化学渗透（ATP 合酶）合成 ATP。荣誉 SBI4U
    3. At PSI, electrons are re-energized by light and used to reduce NADP⁺ to NADPH.在 PSI 处，电子被光重新激发，用于将 NADP⁺ 还原为 NADPH。
  • Outputs of the light reactions:光反应的产物： ATP, NADPH (both used in the Calvin cycle), and O₂ (released into the atmosphere).ATP、NADPH（均用于卡尔文循环）和 O₂（释放到大气中）。
Worked Example 4 · What happens to O₂ produced in photosynthesis?例题 4 · 光合作用产生的 O₂ 去哪里了？

A student says: "Plants produce O₂ in the Calvin cycle by combining carbon and oxygen." Identify the error and give the correct account.一名学生说："植物通过将碳和氧结合，在卡尔文循环中产生 O₂。"找出错误并给出正确说明。

Error: O₂ is produced in the light reactions, not the Calvin cycle, and it comes from the splitting of water (photolysis), not from CO₂. Correct account: in the light-dependent reactions at Photosystem II, water molecules are split: $2H_2O \to 4H^+ + 4e^- + O_2$. The oxygen atoms from water are released as O₂ gas into the atmosphere. The Calvin cycle uses CO₂ and the NADPH and ATP from the light reactions to build glucose, but produces no O₂.错误：O₂ 产生于光反应，而非卡尔文循环（Calvin cycle，卡尔文循环），且来源于水的分解（光解），而非 CO₂。正确说明：在光系统 II 的光依赖反应中，水分子被分解：$2H_2O \to 4H^+ + 4e^- + O_2$。水中的氧原子以 O₂ 气体形式释放到大气中。卡尔文循环利用 CO₂ 以及光反应产生的 NADPH 和 ATP 来构建葡萄糖，但不产生 O₂。

Where does the oxygen (O₂) released by plants during photosynthesis come from?植物在光合作用过程中释放的氧气（O₂）来自哪里？
§5 · Q1
From carbon dioxide ($CO_2$) during the Calvin cycle来自卡尔文循环中的二氧化碳（$CO_2$）
From glucose during cellular respiration来自细胞呼吸中的葡萄糖
From the splitting of water (photolysis) in the light reactions来自光反应中水的分解（光解）
From NADPH reacting with atmospheric nitrogen来自 NADPH 与大气氮气的反应
O₂ comes from photolysis of water at Photosystem II: 2H₂O → 4H⁺ + 4e⁻ + O₂. This was confirmed by isotope labelling experiments (¹⁸O tracer). The O₂ is a byproduct of extracting the hydrogen atoms from water to pass to NADP⁺.O₂ 来自光系统 II 处水的光解：2H₂O → 4H⁺ + 4e⁻ + O₂。这已被同位素标记实验（¹⁸O 示踪剂）证实。O₂ 是从水中提取氢原子传递给 NADP⁺ 的副产品。
The O₂ released during photosynthesis comes from water molecules split in the light reactions at PSII, not from CO₂ or glucose. This was a famous early biology misconception corrected by isotope tracer experiments.光合作用（photosynthesis，光合作用）释放的 O₂ 来自光反应中在光系统 II 处分解的水分子，而非 CO₂ 或葡萄糖。这是一个著名的早期生物学误解，已被同位素示踪实验纠正。
What are the two main products of the light-dependent reactions that are used in the Calvin cycle?光依赖反应中用于卡尔文循环的两种主要产物是什么？
§5 · Q2
ATP and NADPHATP 和 NADPH
Glucose and oxygen葡萄糖和氧气
CO₂ and waterCO₂ 和水
NADH and FADH₂NADH 和 FADH₂
The light reactions produce ATP (via chemiosmosis) and NADPH (reduction of NADP⁺ at PSI). Both are "energy currencies" consumed in the Calvin cycle to fix CO₂ into glucose. O₂ is also a product but is released, not used in the Calvin cycle.光反应产生 ATP（通过化学渗透）和 NADPH（在光系统 I 处 NADP⁺ 的还原）。两者都是在卡尔文循环（Calvin cycle，卡尔文循环）中消耗的"能量货币"，用于将 CO₂ 固定成葡萄糖。O₂ 也是产物，但被释放，不用于卡尔文循环。
The light reactions produce ATP and NADPH for the Calvin cycle, and O₂ as a byproduct. NADH and FADH₂ are carriers in cellular respiration, not photosynthesis. Glucose is the product of the Calvin cycle, not the light reactions.光反应产生 ATP 和 NADPH 供卡尔文循环使用，O₂ 作为副产品。NADH 和 FADH₂ 是细胞呼吸（cellular respiration，细胞呼吸）中的载体，而非光合作用中的。葡萄糖是卡尔文循环的产物，而非光反应的产物。
Going deeper — the Z-scheme: Photosystem II to Photosystem I in detail (SBI4U C2.1)深入 — Z 方案：光系统 II 到光系统 I 的详细流程（SBI4U C2.1）

The "Z-scheme" describes the energy level of electrons as they move through the light reactions. Electrons start at low energy, are boosted to high energy by light at PSII (P680 reaction centre), drop in energy as they flow through the plastoquinone/cytochrome b₆f complex/plastocyanin chain (pumping H⁺ — this is the photosynthetic ETC, producing ATP by chemiosmosis), then are boosted again at PSI (P700 reaction centre), and finally reduce NADP⁺ to NADPH via ferredoxin. The diagram looks like a "Z" when electron energy is plotted vs position in the chain. Key terms SBI4U C2.1 lists: photolysis, cyclic and noncyclic phosphorylation (noncyclic = both PSI and PSII active, O₂ released; cyclic = PSI only, no O₂, no NADPH, only ATP)."Z 方案"描述电子在光反应中移动时的能级。电子从低能量开始，在光系统 II（P680 反应中心）被光提升到高能量，然后随着通过质体醌/细胞色素 b₆f 复合体/质体蓝素链（泵送 H⁺——这是光合作用中的电子传递链，通过化学渗透产生 ATP）而下降，再在光系统 I（P700 反应中心）被再次提升，最后通过铁氧还蛋白将 NADP⁺ 还原为 NADPH。当以电子位置为横轴、电子能量为纵轴绘图时，图形看起来像"Z"字形。SBI4U C2.1 列出的关键术语：光解、循环和非循环磷酸化（非循环 = 光系统 I 和 II 均激活，释放 O₂；循环 = 仅光系统 I，不释放 O₂，不产生 NADPH，仅产生 ATP）。

Section 6 · Photosynthesis — Calvin cycle第 6 节 · 光合作用——卡尔文循环

Photosynthesis: The Calvin Cycle (Light-Independent Reactions)光合作用：卡尔文循环（Calvin cycle，卡尔文循环）（光独立反应）

The Calvin cycle fixes CO₂ into glucose using ATP and NADPH from the light reactions.卡尔文循环利用光反应产生的 ATP 和 NADPH 将 CO₂ 固定成葡萄糖。
  • Location:位置： stroma of the chloroplast. No light is required directly (the cycle uses ATP and NADPH made in the light reactions, so it can run in the dark as long as those are available).叶绿体基质。不直接需要光（该循环使用光反应制造的 ATP 和 NADPH，因此只要这些可用，即使在黑暗中也能运行）。
  • Carbon fixation:碳固定： CO₂ from the air is attached to a 5-carbon molecule called RuBP (ribulose-1,5-bisphosphate) by the enzyme RuBisCO. This produces an unstable 6C compound that immediately splits into two 3-carbon molecules (3-PGA). Honors SBI4U空气中的 CO₂ 由酶 RuBisCO 连接到一个 5 碳分子 RuBP（1,5-二磷酸核酮糖）上。产生一个不稳定的 6C 化合物，立即分裂为两个 3 碳分子（3-磷酸甘油酸，3-PGA）。荣誉 SBI4U
  • Reduction:还原： 3-PGA is reduced using ATP and NADPH to form G3P (glyceraldehyde-3-phosphate), the 3-carbon sugar that is the first stable product. Some G3P exits to make glucose; the rest is used to regenerate RuBP. Honors SBI4U3-PGA 使用 ATP 和 NADPH 还原，形成 G3P（3-磷酸甘油醛），即第一个稳定的 3 碳糖产物。部分 G3P 离开循环用于制造葡萄糖；其余用于再生 RuBP。荣誉 SBI4U
  • Regeneration:再生： RuBP is regenerated from G3P using ATP, ready to accept another CO₂. 3 turns of the cycle fix 3 CO₂ to produce 1 net G3P (equivalent to half a glucose); 6 turns fix 6 CO₂ to produce 1 glucose.RuBP 使用 ATP 从 G3P 再生，准备接受下一个 CO₂。循环转 3 圈固定 3 个 CO₂ 净产生 1 个 G3P（相当于半个葡萄糖）；转 6 圈固定 6 个 CO₂ 产生 1 个葡萄糖。
  • Conceptual-level summary (NGSS / BC / AB):概念层面摘要（NGSS / BC / AB）： The Calvin cycle takes CO₂ + ATP + NADPH → sugar (glucose). It is the "building" stage of photosynthesis; the light reactions supply the energy.卡尔文循环：CO₂ + ATP + NADPH → 糖（葡萄糖）。它是光合作用的"构建"阶段；光反应供应能量。
Worked Example 5 · Inputs and outputs of the Calvin cycle例题 5 · 卡尔文循环的输入与输出

To produce one molecule of glucose ($C_6H_{12}O_6$), how many turns of the Calvin cycle are required, and what are the total inputs?生产一分子葡萄糖（$C_6H_{12}O_6$），需要卡尔文循环转多少圈，总输入是什么？

Each turn of the cycle fixes 1 CO₂. Glucose has 6 carbons, so 6 turns are needed. Total inputs for 6 turns: 6 CO₂, 18 ATP, and 12 NADPH (both supplied by the light reactions). Total output: 1 glucose + 6 H₂O regenerated. The glucose stores the chemical energy originally captured from sunlight.循环每转一圈固定 1 个 CO₂。葡萄糖含 6 个碳，因此需要 6 圈。6 圈的总输入：6 个 CO₂、18 个 ATP 和 12 个 NADPH（均由光反应供应）。总输出：1 个葡萄糖 + 再生 6 个 H₂O。葡萄糖储存了最初从阳光中捕获的化学能。

Where does the Calvin cycle take place within the chloroplast?卡尔文循环在叶绿体的哪个部位发生？
§6 · Q1
Thylakoid membrane类囊体膜
Stroma基质
Outer chloroplast membrane叶绿体外膜
Mitochondrial matrix线粒体基质
The Calvin cycle occurs in the stroma (the fluid-filled space surrounding the thylakoids) of the chloroplast. The light reactions occur in the thylakoid membranes. This spatial separation allows both processes to run simultaneously.卡尔文循环（Calvin cycle，卡尔文循环）发生在叶绿体（chloroplast，叶绿体）的基质（围绕类囊体的液体空间）中。光反应发生在类囊体膜上。这种空间分离使两个过程能够同时进行。
Stroma = Calvin cycle; thylakoid membrane = light reactions. The mitochondrial matrix is where the Krebs cycle of cellular respiration occurs, not photosynthesis.基质 = 卡尔文循环；类囊体膜 = 光反应。线粒体基质是细胞呼吸中克雷布斯循环（Krebs cycle，克雷布斯循环）发生的地方，不是光合作用。
Which raw material does the Calvin cycle use from outside the chloroplast?卡尔文循环使用来自叶绿体外部的哪种原料？
§6 · Q2
Water ($H_2O$)水（$H_2O$）
Oxygen ($O_2$)氧气（$O_2$）
NADPHNADPH
Carbon dioxide ($CO_2$)二氧化碳（$CO_2$）
The Calvin cycle fixes CO₂ from the atmosphere. ATP and NADPH come from inside the chloroplast (from the light reactions in the thylakoids). Water is split in the light reactions, not the Calvin cycle. Oxygen is a byproduct of water splitting, not an input to the Calvin cycle.卡尔文循环固定来自大气的 CO₂。ATP 和 NADPH 来自叶绿体内部（来自类囊体中的光反应）。水在光反应中分解，不是在卡尔文循环中。氧气是水分解的副产品，不是卡尔文循环的输入。
CO₂ from the air is the external raw material for the Calvin cycle. ATP and NADPH are made internally (light reactions). Water is used in the light reactions (at PSII). O₂ is released, not used.来自空气的 CO₂ 是卡尔文循环（Calvin cycle，卡尔文循环）的外部原料。ATP 和 NADPH 在内部产生（光反应）。水在光反应中使用（在光系统 II 处）。O₂ 被释放，而非被使用。
Going deeper — RuBisCO and C3 intermediates (SBI4U C2.1, C3.2)深入 — RuBisCO 与 C3 中间体（SBI4U C2.1、C3.2）

RuBisCO (ribulose-1,5-bisphosphate carboxylase/oxygenase) is Earth's most abundant enzyme and the entry point for most carbon into the biosphere. It catalyses the fixation of CO₂ to RuBP (5C) → unstable 6C → two molecules of 3-phosphoglycerate (3-PGA, a C3 compound). 3-PGA is phosphorylated by ATP to 1,3-bisphosphoglycerate, then reduced by NADPH to glyceraldehyde-3-phosphate (G3P, also a C3 compound). G3P is the first stable organic product. 5 out of every 6 G3P molecules are used to regenerate RuBP (using ATP); 1 out of 6 leaves to form glucose (and other organics). The cycle is named for Melvin Calvin, Andrew Benson, and James Bassham, who worked out its steps at Berkeley in the 1950s using ¹⁴C tracer experiments.RuBisCO（1,5-二磷酸核酮糖羧化酶/加氧酶）是地球上含量最丰富的酶，也是大多数碳进入生物圈的入口。它催化 CO₂ 固定到 RuBP（5C）→ 不稳定的 6C → 两分子 3-磷酸甘油酸（3-PGA，C3 化合物）。3-PGA 被 ATP 磷酸化为 1,3-二磷酸甘油酸，然后被 NADPH 还原为 3-磷酸甘油醛（G3P，也是 C3 化合物）。G3P 是第一个稳定的有机产物。每 6 个 G3P 分子中有 5 个用于再生 RuBP（使用 ATP）；1 个离开循环形成葡萄糖（和其他有机物）。该循环以梅尔文·卡尔文、安德鲁·本森和詹姆斯·巴沙姆命名，他们在 20 世纪 50 年代使用 ¹⁴C 示踪实验在伯克利确定了各步骤。

Section 7 · Comparing respiration and photosynthesis第 7 节 · 呼吸与光合作用的比较

Comparing and Connecting Respiration and Photosynthesis比较与联系细胞呼吸与光合作用

The two processes are chemical opposites that together cycle matter between the biosphere and the atmosphere.这两个过程互为化学逆过程，共同使物质在生物圈与大气之间循环。
Feature特征 Cellular Respiration细胞呼吸 Photosynthesis光合作用
Equation方程式$C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O$$6CO_2 + 6H_2O \to C_6H_{12}O_6 + 6O_2$
Energy change能量变化Releases energy (exothermic) → makes ATP释放能量（放热）→ 合成 ATPStores energy (endothermic) → uses light储存能量（吸热）→ 利用光能
Organelle细胞器Mitochondria (+ cytoplasm)线粒体（+ 细胞质）Chloroplasts叶绿体
Gas used消耗气体O₂ consumed消耗 O₂CO₂ consumed消耗 CO₂
Gas produced产生气体CO₂ produced产生 CO₂O₂ produced产生 O₂
Who does it谁进行All living cells所有活细胞Plants, algae, cyanobacteria植物、藻类、蓝藻
The global carbon cycle connection (NGSS HS-LS2-3).全球碳循环的联系（NGSS HS-LS2-3）。

Photosynthesis removes CO₂ from the atmosphere and fixes it into glucose; cellular respiration breaks glucose back down and releases CO₂. Together they cycle carbon between the biosphere (living organisms) and the atmosphere. In a balanced ecosystem, the rates are roughly equal. Human combustion of fossil fuels adds extra CO₂ that photosynthesis cannot fully absorb — driving the greenhouse effect. Alberta Biology 20 Unit A GO3.2 states: "explain how the equilibrium between gas exchanges in photosynthesis and cellular respiration influences atmospheric composition."光合作用从大气中去除 CO₂ 并将其固定成葡萄糖；细胞呼吸将葡萄糖分解并释放 CO₂。它们共同使碳在生物圈（活生物体）与大气之间循环。在平衡的生态系统中，两者的速率大致相等。人类燃烧化石燃料会添加额外的 CO₂，光合作用无法完全吸收——这推动了温室效应。阿尔伯塔 Biology 20 Unit A GO3.2 指出："解释光合作用和细胞呼吸中气体交换的平衡如何影响大气成分。"

Which statement correctly describes the relationship between cellular respiration and photosynthesis?下列哪项正确描述了细胞呼吸与光合作用之间的关系？
§7 · Q1
Both processes produce CO₂ and consume O₂两个过程都产生 CO₂ 并消耗 O₂
Photosynthesis occurs in all cells; respiration only in plant cells光合作用在所有细胞中发生；呼吸作用只在植物细胞中发生
Both processes occur only in the mitochondria两个过程都只在线粒体中发生
They are chemical opposites: photosynthesis stores energy in glucose using light; respiration releases that energy by breaking down glucose它们是化学逆过程：光合作用利用光能将能量储存在葡萄糖中；呼吸作用通过分解葡萄糖释放该能量
Photosynthesis (6CO₂ + 6H₂O + light → C₆H₁₂O₆ + 6O₂) and cellular respiration (C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + ATP) are chemical reverses of each other. Together they cycle carbon and oxygen between organisms and the atmosphere, forming the global carbon cycle.光合作用（photosynthesis，光合作用）（6CO₂ + 6H₂O + 光能 → C₆H₁₂O₆ + 6O₂）与细胞呼吸（cellular respiration，细胞呼吸）（C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + ATP）互为化学逆过程。它们共同使碳和氧在生物体与大气之间循环，形成全球碳循环。
Respiration (in all cells) consumes O₂ and produces CO₂; photosynthesis (in plants/algae) consumes CO₂ and produces O₂. Photosynthesis occurs in chloroplasts; respiration in mitochondria and cytoplasm. The two are chemical inverses.呼吸作用（在所有细胞中）消耗 O₂ 并产生 CO₂；光合作用（在植物/藻类中）消耗 CO₂ 并产生 O₂。光合作用在叶绿体（chloroplast，叶绿体）中发生；呼吸作用在线粒体（mitochondria，线粒体）和细胞质中发生。两者互为化学逆过程。
A plant is kept in the dark for 24 hours. Which process continues and which stops?一株植物在黑暗中放置 24 小时。哪个过程继续，哪个停止？
§7 · Q2
Both respiration and photosynthesis stop in the dark在黑暗中，呼吸作用和光合作用都停止
Cellular respiration continues; photosynthesis stops because there is no light细胞呼吸继续；光合作用停止，因为没有光
Photosynthesis continues using stored energy; respiration stops光合作用利用储存的能量继续进行；呼吸作用停止
Both continue unchanged because plants do not need light for either process两者都不变地继续，因为植物两个过程都不需要光
All living cells, including plant cells, carry out cellular respiration continuously (day and night) to produce ATP. Photosynthesis requires light energy for the light-dependent reactions; without light it stops. In the dark, a plant is a net consumer of O₂ and a net producer of CO₂ — this is why CO₂ levels rise and O₂ falls in a sealed chamber with a plant in the dark.所有活细胞，包括植物细胞，持续（昼夜）进行细胞呼吸（cellular respiration，细胞呼吸）以产生 ATP。光合作用（photosynthesis，光合作用）需要光能进行光依赖反应；没有光则停止。在黑暗中，植物净消耗 O₂ 并净产生 CO₂——这就是为什么在密封容器中有植物且处于黑暗时，CO₂ 浓度升高而 O₂ 下降。
Respiration is continuous in all living cells regardless of light. Photosynthesis requires light for its light-dependent stage (the Calvin cycle alone cannot run without the ATP and NADPH the light reactions supply).呼吸作用在所有活细胞中持续进行，无论是否有光。光合作用的光依赖阶段需要光（仅卡尔文循环在没有光反应提供的 ATP 和 NADPH 的情况下无法运行）。
Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Equation and location questions方程式与位置题
  • Always give the location when naming a stage.说明阶段时始终给出位置。 Not just "glycolysis" but "glycolysis in the cytoplasm." Not just "Krebs cycle" but "Krebs cycle in the mitochondrial matrix." Location is almost always part of the mark scheme.不只写"糖酵解"，要写"细胞质中的糖酵解（glycolysis，糖酵解）"。不只写"克雷布斯循环"，要写"线粒体基质中的克雷布斯循环（Krebs cycle，克雷布斯循环）"。位置几乎总是评分标准的一部分。
  • Know both summary equations cold.熟记两个总方程式。 Respiration: $C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O$. Photosynthesis: $6CO_2 + 6H_2O \to C_6H_{12}O_6 + 6O_2$. They are exact reverses — a common trap is mixing the reactants and products.呼吸：$C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O$。光合作用：$6CO_2 + 6H_2O \to C_6H_{12}O_6 + 6O_2$。它们是完全的逆过程——常见陷阱是混淆反应物和产物。
Source-of-O₂ and source-of-CO₂ questionsO₂ 来源与 CO₂ 来源题
  • O₂ in photosynthesis comes from water, not CO₂.光合作用中的 O₂ 来自水，而非 CO₂。 Photolysis at PSII: $2H_2O \to 4H^+ + 4e^- + O_2$. This was confirmed by ¹⁸O tracer experiments. The O in CO₂ goes into glucose.光系统 II 处光解：$2H_2O \to 4H^+ + 4e^- + O_2$。已由 ¹⁸O 示踪实验证实。CO₂ 中的氧进入葡萄糖。
  • CO₂ in respiration is released in the Krebs cycle (decarboxylation), not the ETC.呼吸作用中 CO₂ 在克雷布斯循环（脱羧反应）中释放，而非在电子传递链中。 The ETC's job is to use the electrons from NADH/FADH₂ to make ATP, not to release CO₂.电子传递链（electron transport chain，电子传递链）的作用是利用 NADH/FADH₂ 的电子合成 ATP，而非释放 CO₂。
Fermentation questions发酵题
  • Lactic acid fermentation does NOT release CO₂; ethanol fermentation does.乳酸发酵不释放 CO₂；乙醇发酵释放 CO₂。 A common error: stating that all fermentation produces CO₂. Only the decarboxylation of pyruvate to acetaldehyde (in ethanol fermentation) does.常见错误：认为所有发酵（fermentation，发酵）都产生 CO₂。只有丙酮酸脱羧为乙醛（乙醇发酵中）才会产生。
  • The purpose of fermentation is to regenerate NAD⁺ so glycolysis can continue.发酵的目的是再生 NAD⁺，使糖酵解能够继续。 Fermentation itself produces no additional ATP beyond the 2 from glycolysis. The ATP advantage of aerobic respiration (~30–32) comes entirely from chemiosmosis in the ETC.发酵本身在糖酵解产生的 2 个 ATP 之外不额外产生 ATP。有氧呼吸的 ATP 优势（约 30–32 个）完全来自电子传递链中的化学渗透。
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What is ATP and why is it the cell's energy currency?ATP 是什么？为何是细胞的能量货币？
Adenosine triphosphate. Hydrolysis of the third phosphate bond releases ~30.5 kJ mol⁻¹. Every energy-requiring process in the cell uses ATP. Recycled constantly — not stored.三磷酸腺苷。第三磷酸键的水解释放约 30.5 kJ mol⁻¹。细胞中每个需要能量的过程都使用 ATP。不断循环利用——不储存。
Overall equation for aerobic respiration?有氧呼吸总方程式？
$$C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O + \text{ATP}$$ ~30–32 ATP per glucose. Occurs in mitochondria + cytoplasm.每分子葡萄糖产生约 30–32 个 ATP。在线粒体 + 细胞质中进行。
Three stages of aerobic respiration — location of each?有氧呼吸的三个阶段——各阶段位置？
1. Glycolysis → cytoplasm. 2. Krebs cycle → mitochondrial matrix. 3. Electron transport chain → inner mitochondrial membrane.1. 糖酵解（glycolysis，糖酵解）→ 细胞质。2. 克雷布斯循环（Krebs cycle，克雷布斯循环）→ 线粒体基质。3. 电子传递链（electron transport chain，电子传递链）→ 线粒体内膜。
What is chemiosmosis?什么是化学渗透？
The use of a proton (H⁺) gradient across a membrane to drive ATP synthesis by ATP synthase. Occurs in both the mitochondrial inner membrane (respiration) and the thylakoid membrane (photosynthesis).利用跨膜质子（H⁺）梯度驱动 ATP 合酶合成 ATP 的过程。发生在线粒体内膜（呼吸）和类囊体膜（光合作用）上。
Lactic acid fermentation vs ethanol fermentation — key differences?乳酸发酵与乙醇发酵——关键区别？
Lactic acid: pyruvate → lactate (no CO₂). Animals & bacteria. Ethanol: pyruvate → acetaldehyde + CO₂ → ethanol. Yeast. Both yield only 2 ATP per glucose.乳酸：丙酮酸 → 乳酸（无 CO₂）。动物和细菌。乙醇：丙酮酸 → 乙醛 + CO₂ → 乙醇。酵母。两者每分子葡萄糖均只产生 2 个 ATP。
Overall equation for photosynthesis?光合作用总方程式？
$$6CO_2 + 6H_2O + \text{light} \to C_6H_{12}O_6 + 6O_2$$ Occurs in chloroplasts. Light reactions in thylakoids; Calvin cycle in stroma.在叶绿体（chloroplast，叶绿体）中进行。光反应在类囊体；卡尔文循环（Calvin cycle，卡尔文循环）在基质。
Where does O₂ come from in photosynthesis?光合作用中 O₂ 来自哪里？
From photolysis of water at Photosystem II: 2H₂O → 4H⁺ + 4e⁻ + O₂. NOT from CO₂. Confirmed by ¹⁸O tracer experiments.来自光系统 II 处水的光解：2H₂O → 4H⁺ + 4e⁻ + O₂。不来自 CO₂。已由 ¹⁸O 示踪实验证实。
Outputs of the light reactions used by the Calvin cycle?光反应中供卡尔文循环使用的产物？
ATP and NADPH. (O₂ is also produced but released, not used in the Calvin cycle.)ATP 和 NADPH。（O₂ 也是产物，但被释放而非用于卡尔文循环。）
What does the Calvin cycle do in one sentence?用一句话概括卡尔文循环的功能？
Uses ATP + NADPH (from light reactions) to fix CO₂ into G3P, which is used to make glucose. Location: chloroplast stroma.使用 ATP + NADPH（来自光反应）将 CO₂ 固定成 G3P，G3P 用于合成葡萄糖。位置：叶绿体基质。
How do enzymes speed up reactions?酶如何加速反应？
They lower the activation energy (Ea) by binding the substrate at the active site, forming an enzyme-substrate complex. They are not consumed. Rate depends on temperature, pH, and substrate concentration.通过在活性位点与底物结合形成酶-底物复合物来降低活化能（Ea）。酶不被消耗。速率取决于温度、pH 和底物浓度。
Respiration vs photosynthesis: which organelle and which gas consumed/produced?呼吸与光合作用：各在哪种细胞器中进行？消耗/产生哪种气体？
Respiration: mitochondria; consumes O₂, produces CO₂. Photosynthesis: chloroplasts; consumes CO₂, produces O₂. Mirror processes.呼吸：线粒体（mitochondria，线粒体）；消耗 O₂，产生 CO₂。光合作用：叶绿体（chloroplast，叶绿体）；消耗 CO₂，产生 O₂。互为镜像过程。
Why does aerobic respiration produce so much more ATP than fermentation?为何有氧呼吸比发酵产生更多 ATP？
Fermentation: only glycolysis → 2 ATP. Aerobic: glycolysis + Krebs + ETC/chemiosmosis → ~30–32 ATP. The ETC uses NADH/FADH₂ to drive proton pumps, making ~26–28 extra ATP via chemiosmosis.发酵（fermentation，发酵）：仅糖酵解 → 2 个 ATP。有氧：糖酵解 + 克雷布斯 + ETC/化学渗透 → 约 30–32 个 ATP。电子传递链利用 NADH/FADH₂ 驱动质子泵，通过化学渗透额外产生约 26–28 个 ATP。
Final electron acceptor in aerobic respiration?有氧呼吸中的最终电子受体？
Oxygen ($O_2$). It accepts electrons from the ETC and combines with H⁺ to form water. Without O₂ the ETC stalls and aerobic respiration stops.氧气（$O_2$）。它接受来自电子传递链（electron transport chain，电子传递链）的电子，与 H⁺ 结合形成水。没有 O₂，电子传递链停滞，有氧呼吸停止。
Connect respiration and photosynthesis to the global carbon cycle.将呼吸作用与光合作用与全球碳循环联系起来。
Photosynthesis fixes atmospheric CO₂ into glucose; respiration releases CO₂ back. Together they cycle carbon between organisms and the atmosphere. In balance, rates are equal. Human fossil-fuel burning adds extra CO₂, disrupting the equilibrium.光合作用（photosynthesis，光合作用）将大气 CO₂ 固定成葡萄糖；细胞呼吸（cellular respiration，细胞呼吸）将 CO₂ 释放回去。它们共同使碳在生物体与大气之间循环。平衡状态下，两者速率相等。人类燃烧化石燃料增加额外 CO₂，破坏平衡。
Mixed Practice综合练习

Practice Quiz综合测验

During a sprint, a runner's leg muscles temporarily run out of oxygen. What happens to ATP production?短跑时，跑步者的腿部肌肉暂时缺氧。ATP 产生会怎样？
Q1
ATP production stops completely because respiration requires O₂ATP 产生完全停止，因为呼吸需要 O₂
Photosynthesis compensates and produces ATP for the muscles光合作用补偿并为肌肉产生 ATP
The Krebs cycle runs faster to compensate for no ETC克雷布斯循环加速运行以补偿缺少 ETC
Glycolysis continues with lactic acid fermentation regenerating NAD⁺, producing 2 ATP per glucose糖酵解继续，乳酸发酵再生 NAD⁺，每分子葡萄糖产生 2 个 ATP
Without O₂, the ETC and Krebs cycle shut down, but glycolysis (cytoplasm) can still run. Lactic acid fermentation regenerates NAD⁺ (from NADH) so glycolysis can continue. ATP output drops dramatically from ~30 to just 2 per glucose, explaining why sprinting cannot be sustained.没有 O₂，电子传递链（electron transport chain，电子传递链）和克雷布斯循环（Krebs cycle，克雷布斯循环）关闭，但糖酵解（细胞质）仍可运行。乳酸发酵（fermentation，发酵）再生 NAD⁺（由 NADH），使糖酵解能够继续。ATP 产量从约 30 个骤降至每分子葡萄糖仅 2 个，这解释了为何短跑不能持续。
ATP production does not stop — glycolysis is anaerobic and continues with fermentation regenerating NAD⁺. The Krebs cycle does not run without O₂. Muscle cells do not photosynthesize.ATP 产生不会停止——糖酵解（glycolysis，糖酵解）是无氧的，随着发酵再生 NAD⁺ 而继续。没有 O₂ 克雷布斯循环无法运行。肌肉细胞不进行光合作用。
A plant is placed in a sealed glass jar in the dark for 12 hours. What happens to the CO₂ and O₂ concentrations inside the jar?将一株植物放入密封玻璃罐，在黑暗中放置 12 小时。罐内 CO₂ 和 O₂ 浓度如何变化？
Q2
CO₂ decreases; O₂ increases (photosynthesis dominates)CO₂ 减少；O₂ 增加（光合作用占主导）
No change because the plant is dormant in the dark无变化，因为植物在黑暗中处于休眠状态
CO₂ increases; O₂ decreases (cellular respiration continues; no photosynthesis)CO₂ 增加；O₂ 减少（细胞呼吸继续；无光合作用）
Both CO₂ and O₂ decrease because the plant absorbs bothCO₂ 和 O₂ 均减少，因为植物同时吸收两者
In the dark, photosynthesis cannot occur (no light for the light reactions). Cellular respiration continues 24/7 in all living cells, consuming O₂ and releasing CO₂. The plant is a net consumer of O₂ and a net producer of CO₂ in the dark.在黑暗中，光合作用（photosynthesis，光合作用）无法进行（光反应无光）。细胞呼吸（cellular respiration，细胞呼吸）在所有活细胞中全天候进行，消耗 O₂ 并释放 CO₂。黑暗中，植物净消耗 O₂ 并净产生 CO₂。
Without light, photosynthesis stops. Respiration continues, consuming O₂ and releasing CO₂. Plants are never dormant metabolically — they respire continuously.没有光，光合作用停止。呼吸作用继续，消耗 O₂ 并释放 CO₂。植物在代谢上从不休眠——它们持续呼吸。
An enzyme-catalyzed reaction is running at maximum rate ($V_{max}$). Adding more substrate does not increase the rate. Adding more enzyme does. What explains this?一个酶催化反应正在以最大速率（$V_{max}$）运行。增加更多底物不能提高速率，但增加更多酶可以。如何解释？
Q3
At $V_{max}$, all enzyme active sites are saturated with substrate; more enzyme molecules provide more active sites在 $V_{max}$ 时，所有酶的活性位点均被底物饱和；更多的酶分子提供更多活性位点
The enzyme is denatured and more enzyme compensates for the lost activity酶已变性，更多的酶补偿了失去的活性
Adding substrate increases temperature and inhibits the enzyme; more enzyme restores activity增加底物会升高温度并抑制酶；更多的酶恢复活性
At $V_{max}$, the substrate is all used up; more enzyme breaks down the substrate faster在 $V_{max}$ 时，底物全部耗尽；更多酶更快分解底物
$V_{max}$ is reached when every enzyme molecule's active site is occupied (saturated). Adding substrate cannot increase the rate because there are no free active sites. Adding enzyme creates more active sites, so more substrate can be processed simultaneously, raising the rate (and the new $V_{max}$).当每个酶分子的活性位点都被占据（饱和）时，达到 $V_{max}$。增加底物无法提高速率，因为没有空闲的活性位点。增加酶会创造更多活性位点，使更多底物能够同时被处理，提高速率（以及新的 $V_{max}$）。
At $V_{max}$, active sites are saturated — not denatured. Adding enzyme is effective because it increases the total number of active sites available to bind substrate.在 $V_{max}$ 时，活性位点是饱和的——不是变性的。增加酶是有效的，因为它增加了可与底物结合的活性位点总数。
Which correctly describes what happens to NADPH after it is produced in the light reactions?下列哪项正确描述了光反应产生的 NADPH 的去向？
Q4
It is used as the final electron acceptor in the electron transport chain of the chloroplast它用作叶绿体电子传递链中的最终电子受体
It donates its electrons/hydrogen to reduce 3-PGA in the Calvin cycle, becoming NADP⁺它将电子/氢捐给卡尔文循环中的 3-PGA 进行还原，自身变为 NADP⁺
It is pumped across the thylakoid membrane to generate ATP它被泵过类囊体膜以产生 ATP
It is exported to the mitochondria to drive aerobic respiration它被输出到线粒体驱动有氧呼吸
NADPH (produced at Photosystem I) carries a hydride ion (H⁻, i.e. 2e⁻ + H⁺) to the Calvin cycle in the stroma. It donates those electrons to reduce 3-phosphoglycerate (3-PGA) to G3P, becoming NADP⁺ which cycles back to the light reactions.NADPH（在光系统 I 处产生）携带氢负离子（H⁻，即 2e⁻ + H⁺）到基质中的卡尔文循环（Calvin cycle，卡尔文循环）。它将这些电子捐给 3-磷酸甘油酸（3-PGA）以将其还原为 G3P，自身变为 NADP⁺，循环回光反应。
NADPH is used in the Calvin cycle (stroma) to reduce 3-PGA to G3P. It is not a proton pump, not the final electron acceptor, and does not travel to the mitochondria.NADPH 在卡尔文循环（基质）中用于将 3-PGA 还原为 G3P。它不是质子泵，不是最终电子受体，也不会前往线粒体（mitochondria，线粒体）。
Alberta Biology 20 Unit C states that "detailed knowledge of metabolic intermediates is not required." Which of the following IS still required for Alberta students?阿尔伯塔 Biology 20 Unit C 指出"不要求对代谢中间体的详细了解"。对阿尔伯塔学生来说，以下哪项仍然是要求的？
Q5
The exact number of NADH molecules produced per turn of the Krebs cycle每轮克雷布斯循环产生的 NADH 分子数
The names of all 8 Krebs cycle intermediates克雷布斯循环全部 8 种中间体的名称
That cellular respiration occurs in two phases (glycolysis and then the Krebs/ETC in the mitochondrion) and that ATP is the energy product细胞呼吸分两个阶段进行（糖酵解，然后在线粒体中进行克雷布斯/ETC），ATP 是能量产物
The specific roles of cytochrome c and ubiquinone in the electron transport chain细胞色素 c 和泛醌在电子传递链中的具体作用
Alberta Biology 20 Unit C GO2 explicitly requires students to explain "in general terms" how glucose is oxidized during glycolysis and the Krebs cycle to produce NADH/FADH, and how chemiosmosis converts reducing power to ATP. Named intermediates (exact NADH counts, specific ETC proteins) are excluded by the "no detailed intermediates" caveat.阿尔伯塔 Biology 20 Unit C GO2 明确要求学生"概括"解释葡萄糖如何在糖酵解（glycolysis，糖酵解）和克雷布斯循环（Krebs cycle，克雷布斯循环）中被氧化产生 NADH/FADH，以及化学渗透如何将还原力转化为 ATP。命名中间体（精确 NADH 数量、特定 ETC 蛋白）被"不要求详细中间体"的注释排除。
Alberta requires the general framework: glycolysis in cytoplasm, Krebs/ETC in mitochondrion, chemiosmosis produces ATP. Named intermediates, exact coenzyme counts, and specific ETC protein names are all excluded by the Alberta "no detailed metabolic intermediates" note (20–C2.1k, 20–C2.2k).阿尔伯塔要求一般框架：糖酵解在细胞质，克雷布斯/ETC 在线粒体（mitochondria，线粒体），化学渗透产生 ATP。命名中间体、精确辅酶数量和特定 ETC 蛋白名称均被阿尔伯塔"不要求详细代谢中间体"注释（20–C2.1k、20–C2.2k）排除。
Self-check自查

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

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Next Steps后续衔接

What This Feeds Into本单元的去向

Cellular energetics is the metabolic engine that powers every other topic in biology. ATP made here fuels muscle contraction (Human A&P, Unit 10), DNA replication and cell division (Unit 4), and active transport across membranes (Unit 1). The carbon cycle connection (NGSS HS-LS2-3) feeds directly into Ecology and Ecosystems (Unit 9). The enzyme concepts introduced in this unit reappear in Biochemistry (Unit 2) and Molecular Genetics (Unit 6).细胞能量学是为生物学所有其他主题提供动力的代谢引擎。这里制造的 ATP 驱动肌肉收缩（人体解剖与生理，第 10 单元）、DNA 复制和细胞分裂（第 4 单元），以及跨膜主动运输（第 1 单元）。碳循环联系（NGSS HS-LS2-3）直接输入生态学与生态系统（第 9 单元）。本单元介绍的酶概念将在《生物化学》（第 2 单元）和《分子遗传学》（第 6 单元）中再次出现。

Within High School Biology.在 HS Biology 内部。

The Biochemistry guide (Unit 2) deepens enzyme kinetics and the organic molecules (glucose, fatty acids) that feed into respiration. Cell Division (Unit 4) requires ATP from respiration to power the chromosome-segregation machinery. Molecular Genetics (Unit 6) builds on the enzyme-as-catalyst concept for DNA polymerase and RNA polymerase. Ecology and Ecosystems (Unit 9) uses the photosynthesis/respiration balance to explain the global carbon cycle and ecosystem productivity. Human Anatomy and Physiology (Unit 10) applies mitochondrial density (in muscle cells, cardiac cells) as a direct application of the energetics concepts here.《生物化学》（第 2 单元）深化酶动力学和输入细胞呼吸的有机分子（葡萄糖、脂肪酸）。《细胞分裂》（第 4 单元）需要来自呼吸作用的 ATP 驱动染色体分离机制。《分子遗传学》（第 6 单元）以酶作为催化剂的概念为基础，应用于 DNA 聚合酶和 RNA 聚合酶。《生态学与生态系统》（第 9 单元）利用光合作用/细胞呼吸平衡来解释全球碳循环和生态系统生产力。《人体解剖与生理》（第 10 单元）将线粒体密度（在肌肉细胞、心肌细胞中）作为本单元能量学概念的直接应用。

Feeds into AP Biology and IB Biology.衔接 AP Biology 与 IB Biology。

This guide is the direct foundation for AP Biology Unit 3 (Cellular Energetics) and IB Biology HL Topic B3 (Cellular Respiration) and B4 (Photosynthesis). AP Biology extends chemiosmosis to include the exact proton-pumping complexes (I, III, IV) and adds the Calvin cycle carbon-counting in detail. IB Biology HL adds the inhibition of ATP synthase by oligomycin as an experimental context and requires electron micrograph identification of mitochondrial cristae and thylakoid stacks. Neither course exists yet in this repo; treat this guide as the prerequisite.本指南是 AP Biology 第 3 单元（细胞能量学）及 IB Biology HL Topic B3（细胞呼吸）和 B4（光合作用）的直接基础。AP Biology 将化学渗透扩展到包括精确的质子泵送复合体（I、III、IV），并详细增加卡尔文循环碳计数。IB Biology HL 增加了少霉素抑制 ATP 合酶作为实验背景，并要求通过电子显微照片识别线粒体嵴和类囊体堆叠。目前这两门课程均未收录于本仓库中；将本指南视为先修材料。

DINGRUI SCHOLARS

High School Biology · Cellular Energetics · v1高中生物 · 细胞能量学 · v1