📚 Biology Exam Prep: Mastering Biochemical Metabolic Pathways | 生物备考：生化代谢途径难点梳理
Metabolic pathways are the chemical engines of life. For A-Level and IB Biology students, understanding how molecules like glucose, pyruvate, and acetyl-CoA are transformed through glycolysis, the Krebs cycle, and oxidative phosphorylation is not merely a memorisation exercise — it is the foundation for answering demanding exam questions on respiration, photosynthesis, and homeostasis.
代谢途径是生命的化学引擎。对于 A-Level 和 IB 生物学生而言，理解葡萄糖、丙酮酸和乙酰辅酶A 等分子如何通过糖酵解、克雷布斯循环和氧化磷酸化被转化，不仅仅是一项记忆任务——它是回答呼吸作用、光合作用和稳态等难题的基础。
1. Glycolysis: The Universal Starting Point | 糖酵解：共同的起点
Glycolysis occurs in the cytoplasm of all living cells and does not require oxygen. One glucose molecule (6 carbon) is split into two molecules of pyruvate (3 carbon each). The process yields a net gain of 2 ATP and 2 NADH per glucose molecule.
糖酵解发生在所有活细胞的细胞质中，不需要氧气。一分子葡萄糖（6碳）被裂解为两分子丙酮酸（各3碳）。每分子葡萄糖净产生 2 个 ATP 和 2 个 NADH。
Key stages to remember:
需要牢记的关键阶段：
Phosphorylation: Glucose is phosphorylated by two ATP molecules to form fructose-1,6-bisphosphate.
磷酸化：葡萄糖被两分子 ATP 磷酸化，形成果糖-1,6-二磷酸。
Lysis: The 6-carbon fructose bisphosphate is split into two 3-carbon molecules (glyceraldehyde-3-phosphate, or GP).
裂解：6碳的果糖二磷酸被裂解为两个3碳分子（甘油醛-3-磷酸，即 GP）。
Oxidation and ATP generation: Each GP is oxidised, with NAD⁺ reduced to NADH, and substrate-level phosphorylation generates 4 ATP in total.
氧化与 ATP 生成：每个 GP 被氧化，NAD⁺ 被还原为 NADH，底物水平磷酸化共生成 4 个 ATP。
Net equation:
净反应方程：
Glucose + 2 NAD⁺ + 2 ADP + 2 Pi → 2 Pyruvate + 2 NADH + 2 H⁺ + 2 ATP + 2 H₂O
葡萄糖 + 2 NAD⁺ + 2 ADP + 2 Pi → 2 丙酮酸 + 2 NADH + 2 H⁺ + 2 ATP + 2 H₂O
2. The Link Reaction: Pyruvate to Acetyl-CoA | 连接反应：丙酮酸转化为乙酰辅酶A
In aerobic organisms, pyruvate is transported into the mitochondrial matrix. Here, pyruvate dehydrogenase catalyses the oxidative decarboxylation of pyruvate: one carbon is removed as CO₂, NAD⁺ is reduced to NADH, and the remaining two-carbon acetyl group binds to coenzyme A, forming acetyl-CoA.
在有氧生物中，丙酮酸被运入线粒体基质。在这里，丙酮酸脱氢酶催化丙酮酸的氧化脱羧：一个碳以 CO₂ 形式脱去，NAD⁺ 被还原为 NADH，剩余的两碳乙酰基与辅酶A结合，形成乙酰辅酶A。
Important exam points:
重要考点：
This step is irreversible — pyruvate cannot be regenerated from acetyl-CoA in animals.
此步骤不可逆——在动物中，丙酮酸不能从乙酰辅酶A再生。
It is the point of entry for the 'fate of pyruvate' decision: aerobic (link reaction → Krebs cycle) or anaerobic (fermentation).
这是"丙酮酸的命运"决定的入口点：有氧（连接反应→克雷布斯循环）或厌氧（发酵）。
Per glucose, this step occurs twice, producing 2 NADH and 2 CO₂.
每分子葡萄糖，此步骤发生两次，产生 2 个 NADH 和 2 个 CO₂。
3. The Krebs Cycle: The Central Hub | 克雷布斯循环：核心枢纽
Also called the citric acid cycle or TCA cycle, this series of reactions takes place in the mitochondrial matrix. Each turn processes one acetyl-CoA (2 carbons) and joins it with a 4-carbon oxaloacetate to form citrate (6 carbons). Over several steps, two CO₂ molecules are released, and the cycle regenerates oxaloacetate.
克雷布斯循环又称柠檬酸循环或三羧酸（TCA）循环，这一系列反应发生在线粒体基质中。每转一圈处理一分子乙酰辅酶A（2碳），与4碳的草酰乙酸结合形成柠檬酸（6碳）。经过若干步骤，释放两分子 CO₂，循环再生草酰乙酸。
Per turn of the cycle (i.e., per acetyl-CoA), the yield is:
每转一圈（即每分子乙酰辅酶A）的产物为：
| Product | 产物 | Number per turn | 每圈数量 |
| ATP (via substrate-level phosphorylation) | ATP（通过底物水平磷酸化） | 1 |
| NADH | 3 |
| FADH₂ | 1 |
| CO₂ (waste) | CO₂（废物） | 2 |
Per glucose molecule (two turns), the cycle produces 2 ATP, 6 NADH, 2 FADH₂, and 4 CO₂.
每分子葡萄糖（两圈）中，循环产生 2 个 ATP、6 个 NADH、2 个 FADH₂ 和 4 个 CO₂。
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