High School Biology

Biochemistry (Molecules of Life)生物化学（生命分子）

Carbon-based macromolecules are the chemical foundation of every living thing. This guide covers the unique properties of water that make life possible, then builds through the four classes of biological macromolecule — carbohydrates, lipids, proteins, and nucleic acids — explaining their monomers, polymers, structures, and roles. Enzymes and the factors that govern their reaction rate are treated in detail. The guide closes with dehydration synthesis and hydrolysis, the two universal reactions that build and break every polymer in the cell. All curricula mapped; worked examples and quizzes throughout.以碳为骨架的大分子是一切生命的化学基础。本指南从使生命成为可能的水（水）的独特性质出发，依次讲解四类生物大分子——碳水化合物（碳水化合物）、脂质（脂质）、蛋白质（蛋白质）与核酸（核酸）——阐明其单体、聚合物、结构与功能。酶（酶）及影响其反应速率的因素将作详细介绍。本指南以脱水缩合（脱水缩合）与水解（水解）作结，这两种反应是细胞中构建与分解每一种聚合物的通用反应。各大纲对照完整；全程设有例题与测验。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB SBI4U functional groups and enzyme mechanisms marked HonorsSBI4U 官能团与酶机制标为荣誉级
Read me first先读这里

How to use this guide如何使用本指南

Biochemistry is the chemistry of life — the four curricula agree on the core macromolecules (carbohydrates, lipids, proteins, nucleic acids) and enzymes, but diverge sharply on depth. US NGSS HS-LS1-6 is explicitly conceptual: its Assessment Boundary excludes naming macromolecules or detailing specific chemical reactions. Ontario SBI4U Strand B (B3.2–B3.5) is the most detailed: named functional groups (B3.3), enzyme structure and mechanism (B3.4), and all four biochemical reaction types (B3.5). BC Anatomy and Physiology 12 matches SBI4U depth for the enzyme section. Alberta Biology 20 Unit D GO1 names carbohydrates, lipids, proteins, and their enzymes and lists the enzyme factors (temperature, pH, substrate concentration, competitive and feedback inhibition). The table below maps each curriculum to these sections.生物化学是生命的化学——四套大纲在核心大分子（碳水化合物、脂质、蛋白质、核酸）与酶上高度一致，但在深度上差异显著。US NGSS HS-LS1-6 明确保持在概念层面：其评估边界排除命名大分子或细述具体化学反应。安大略 SBI4U B 单元（B3.2–B3.5）最为详细：命名官能团（B3.3）、酶的结构与机制（B3.4）以及四类生化反应（B3.5）。BC Anatomy and Physiology 12 在酶部分与 SBI4U 深度相当。阿尔伯塔 Biology 20 Unit D GO1 命名碳水化合物、脂质、蛋白质及其酶，并列举酶的影响因素（温度、pH、底物浓度、竞争性抑制与反馈抑制）。下表将各大纲与本指南各节对应。

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 (water), conceptual overviews of §2–§4 (carbon-based molecules), §7 (building/breaking polymers) — all at the HS-LS1-6 conceptual level§1（水）、§2–§4 概念性概述（碳基分子）、§7（聚合物的构建与分解）——均在 HS-LS1-6 概念层面 §5 enzyme mechanism detail and §6 nucleotide structure: Assessment Boundary excludes details of the specific chemical reactions or identification of macromolecules§5 酶机制细节与 §6 核苷酸结构：评估边界排除具体化学反应细节或大分子鉴别 NGSS HS Life Science — HS-LS1-6 PE and Assessment Boundary— HS-LS1-6 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, including functional groups (B3.3), enzyme mechanism and inhibition (B3.4), and biochemical reaction types (B3.5)全部 7 节完整学习，含官能团（B3.3）、酶机制与抑制（B3.4）以及生化反应类型（B3.5） Nothing — SBI4U Strand B is the primary biochemistry treatment in Ontario无 — SBI4U B 单元是安大略生物化学的主要学习内容 Ontario SBI3U/4U Biology — SBI4U Strand B B3.2–B3.5— SBI4U B 单元 B3.2–B3.5
🇨🇦 BC Anatomy & Physiology 12 HonorsBC Anatomy & Physiology 12 荣誉 §1–§7 all core; enzyme section (§5) is especially important: lock-and-key model, activation energy, allosteric inhibition all appear in the BC Content bullet verbatim§1–§7 全为核心；酶节（§5）尤为重要：BC 内容条目明确提及锁钥模型、活化能、变构抑制 Nothing — biochemistry and enzyme content maps directly to the Biological Molecules and Metabolism and Enzymes Content bullets无 — 生物化学与酶内容直接对应 BC 的生物分子与代谢与酶内容条目 BC Life Sciences 11 / Anatomy 12 — Anatomy and Physiology 12 Content— Anatomy and Physiology 12 内容
🇨🇦 AB Biology 20阿尔伯塔 Biology 20 §2–§5 core: Biology 20 D1.2k names carbohydrates, lipids, proteins and their enzymes; D1.3k lists enzyme factors (temperature, pH, substrate concentration, feedback inhibition, competitive inhibition)§2–§5 为核心：Biology 20 D1.2k 命名碳水化合物、脂质、蛋白质及其酶；D1.3k 列举酶的影响因素（温度、pH、底物浓度、反馈抑制、竞争性抑制） §6 nucleic acid structural detail is in Alberta Biology 30 Unit C GO3 (molecular genetics), not Biology 20§6 核酸结构细节在阿尔伯塔 Biology 30 Unit C GO3（分子遗传学）而非 Biology 20 中出现 Alberta Biology 20/30 — Biology 20 Unit D GO1 (20–D1.2k, 20–D1.3k)— Biology 20 Unit D GO1（20–D1.2k、20–D1.3k）

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

Know the four macromolecule classes (carbohydrate, lipid, protein, nucleic acid) and their monomers (monosaccharide, fatty acid/glycerol, amino acid, nucleotide). Know that dehydration synthesis builds polymers and hydrolysis breaks them. For enzymes: lower activation energy, lock-and-key model, and that temperature/pH affect rate. Read every cram-cheat box; skip the functional-group going-deeper sections.掌握四类大分子（碳水化合物、脂质、蛋白质、核酸）及其单体（单糖、脂肪酸/甘油、氨基酸、核苷酸）。了解脱水缩合构建聚合物，水解分解聚合物。关于酶：降低活化能、锁钥模型、温度/pH 影响速率。读每个速记框；跳过官能团的深入内容。

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

Link structure to function precisely: why are phospholipids amphipathic? Why do saturated fats have a higher melting point? Why do the four levels of protein structure matter? For enzymes, distinguish competitive inhibition (occupies active site) from allosteric inhibition (changes shape of active site) and predict the effect of each factor on a reaction-rate graph. For SBI4U, name the common functional groups (hydroxyl, carbonyl, carboxyl, amino, phosphate) and explain how each contributes to a macromolecule properties.精准将结构与功能挂钩：为何磷脂是双亲性的？为何饱和脂肪熔点更高？为何蛋白质的四级结构重要？对于酶，区分竞争性抑制（占据活性位点）与变构抑制（改变活性位点形状），并能预测各因素对反应速率图的影响。SBI4U 轨道还需命名常见官能团（羟基、羰基、羧基、氨基、磷酸基），并解释各官能团如何影响大分子的性质。

Honors flag.荣誉级标记。 The functional-group going-deeper in §2–§4, the enzyme mechanism detail in §5, and the four biochemical reaction types in §7 carry the Honors SBI4U chip, because Ontario SBI4U B3.3–B3.5 expects this depth while NGSS HS-LS1-6 explicitly excludes it. If your curriculum is NGSS, read these sections for conceptual grounding but do not memorize the named functional groups or reaction types.§2–§4 中的官能团深入内容、§5 的酶机制细节，以及 §7 的四类生化反应类型均标注 荣誉 SBI4U，因为安大略 SBI4U B3.3–B3.5 要求这一深度，而 NGSS HS-LS1-6 明确排除。若你的大纲是 NGSS，阅读这些部分以获得概念基础，但无需记忆命名官能团或反应类型。
Section 1 · Water and its properties第 1 节 · 水及其性质

Water and Its Properties水及其性质

Why water makes life possible — four key properties.为何水使生命成为可能 — 四大关键性质。
  • Polarity and hydrogen bonding:极性与氢键： Oxygen is more electronegative than hydrogen, giving the water molecule a partial negative charge at the oxygen end and partial positive charges at the two hydrogen ends. Adjacent water molecules form hydrogen bonds between the δ− oxygen of one molecule and the δ+ hydrogen of another. These bonds are individually weak but collectively give water its distinctive properties.氧的电负性强于氢，使水分子在氧端带部分负电荷，在两个氢端带部分正电荷。相邻水分子在一个分子的 δ− 氧与另一个分子的 δ+ 氢之间形成氢键。单条氢键较弱，但集体上赋予水独特的性质。
  • High specific heat capacity:高比热容： Water absorbs a large amount of heat energy for each degree of temperature rise. This stabilizes the temperature of cells and aquatic environments.水每升高一度温度需吸收大量热能。这稳定了细胞和水生环境的温度。
  • Universal solvent (cohesion/adhesion):通用溶剂（内聚力/附着力）： Polar and ionic substances dissolve in water because water molecules surround their charged or polar groups (hydration shells). Cohesion (water-to-water H-bonds) creates surface tension. Adhesion (water-to-other H-bonds) enables capillary action in plants. Alberta Biology 20 A2.2k names water as the "universal solvent" with hydrogen bonding verbatim.极性和离子性物质溶于水，因为水分子以水化层包围其带电或极性基团。内聚力（水-水氢键）产生表面张力。附着力（水-其他氢键）使植物的毛细作用成为可能。阿尔伯塔 Biology 20 A2.2k 明确将水命名为具有"氢键"的"通用溶剂"。
  • High heat of vaporization:高汽化热： Many hydrogen bonds must be broken for water to evaporate, so evaporation carries away a large amount of heat. This underlies sweating and transpiration as cooling mechanisms.水蒸发需打断大量氢键，因此蒸发会带走大量热量。这是出汗和蒸腾作为降温机制的基础。
Hydrophilic vs hydrophobic.亲水性与疏水性。

Substances that interact favourably with water (polar, ionic) are hydrophilic ("water-loving"). Substances that do not (nonpolar, e.g. oils) are hydrophobic ("water-fearing"). Amphipathic molecules have both a hydrophilic region and a hydrophobic region — phospholipids are the key example, and their amphipathic nature is what drives the spontaneous formation of the lipid bilayer.与水相互作用良好的物质（极性、离子性）称为亲水性（"喜水"）。不与水相互作用的物质（非极性，如油脂）称为疏水性（"怕水"）。双亲性分子同时具有亲水区和疏水区——磷脂是关键例子，其双亲性正是驱动脂质双分子层自发形成的原因。

Water molecules form hydrogen bonds with each other because water is:水分子之间形成氢键，原因是水是：
§1 · Q1
A nonpolar covalent molecule非极性共价分子
A polar molecule with partial charges at each end具有两端分部电荷的极性分子
An ionic compound离子化合物
An amphipathic molecule双亲性分子
Water is polar: the oxygen atom carries a partial negative charge (δ−) and each hydrogen carries a partial positive charge (δ+). The δ+ hydrogen of one molecule is attracted to the δ− oxygen of an adjacent molecule, forming a hydrogen bond.水是极性分子：氧原子带部分负电荷（δ−），每个氢原子带部分正电荷（δ+）。一个分子的 δ+ 氢被相邻分子的 δ− 氧吸引，形成氢键。
Hydrogen bonds between water molecules arise from water polarity: δ+ H of one molecule attracted to δ− O of another. Nonpolar molecules cannot form hydrogen bonds with water.水分子间的氢键来自水的极性：一个分子的 δ+ H 被另一个分子的 δ− O 吸引。非极性分子无法与水形成氢键。
A patient sweats heavily during a fever. Which property of water makes sweating an effective cooling mechanism?发烧时患者大量出汗。水的哪种性质使出汗成为有效的降温机制？
§1 · Q2
High density at 4°C4°C 时密度最大
Universal solvent properties通用溶剂性质
High heat of vaporization高汽化热
Adhesion to skin surfaces对皮肤表面的附着力
Water has a high heat of vaporization: converting liquid water to vapor requires breaking many hydrogen bonds, which absorbs a large quantity of heat. When sweat evaporates from skin, it carries that heat away, cooling the body.水具有高汽化热：将液态水转化为气态需打断大量氢键，这会吸收大量热量。汗水从皮肤蒸发时带走热量，从而冷却身体。
The cooling effect of evaporation is due to high heat of vaporization: evaporation requires energy to break hydrogen bonds, so heat is removed from the skin.蒸发降温效果源于高汽化热：蒸发需要能量打断氢键，因此热量从皮肤被带走。
Section 2 · Carbohydrates第 2 节 · 碳水化合物

Carbohydrates碳水化合物

Monomer: monosaccharide. Polymer: polysaccharide. General formula: (CH2O)n.单体：单糖。聚合物：多糖。通式：(CH2O)n。
  • Monosaccharides (simple sugars):单糖（简单糖）： glucose ($ ext{C}_6 ext{H}_{12} ext{O}_6$), fructose, galactose. The primary fuel for cellular respiration. Glucose is a 6-carbon ring; the number of carbons determines the name (triose = 3C, pentose = 5C, hexose = 6C).葡萄糖（$ ext{C}_6 ext{H}_{12} ext{O}_6$）、果糖、半乳糖。细胞呼吸的主要燃料。葡萄糖是六碳环；碳数决定名称（三碳糖 = 3C，五碳糖 = 5C，六碳糖 = 6C）。
  • Disaccharides:二糖： two monosaccharides joined by a glycosidic bond via dehydration synthesis. Examples: sucrose (glucose + fructose), lactose (glucose + galactose), maltose (glucose + glucose).两个单糖通过脱水缩合形成糖苷键相连。例：蔗糖（葡萄糖 + 果糖）、乳糖（葡萄糖 + 半乳糖）、麦芽糖（葡萄糖 + 葡萄糖）。
  • Polysaccharides:多糖： many monosaccharides linked. Starch (plants, energy storage), glycogen (animals, energy storage in liver and muscle), cellulose (plant cell walls, structural), chitin (fungi and arthropod exoskeletons, structural).许多单糖相连。淀粉（植物，储能）、糖原（动物，储能，存于肝脏和肌肉）、纤维素（植物细胞壁，结构性）、几丁质（真菌和节肢动物外骨骼，结构性）。
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Carbohydrate functions at a glance碳水化合物功能一览
Molecule分子Type类型Role作用
Glucose葡萄糖Monosaccharide单糖Primary energy fuel; substrate for respiration主要能量燃料；呼吸底物
Starch淀粉Polysaccharide多糖Energy storage in plants植物储能
Glycogen糖原Polysaccharide多糖Energy storage in animals (liver, muscle)动物储能（肝脏、肌肉）
Cellulose纤维素Polysaccharide多糖Structural support — plant cell walls结构支撑——植物细胞壁
Chitin几丁质Polysaccharide多糖Structural — fungal walls, insect exoskeletons结构性——真菌细胞壁、昆虫外骨骼
Starch and cellulose are both polysaccharides made of glucose monomers, yet one is digestible by humans and one is not. What structural difference explains this?淀粉和纤维素都是由葡萄糖单体组成的多糖，但人类只能消化其中一种。什么结构差异解释了这一现象？
§2 · Q1
The glycosidic bond type: α-1,4 bonds in starch vs β-1,4 bonds in cellulose糖苷键类型不同：淀粉为 α-1,4 键，纤维素为 β-1,4 键
Starch is made of fructose; cellulose is made of glucose淀粉由果糖组成；纤维素由葡萄糖组成
Cellulose has fewer glucose monomers than starch纤维素含有的葡萄糖单体比淀粉少
Starch has branches; cellulose does not淀粉有分支；纤维素没有
Both are glucose polymers, but the glycosidic linkage differs: starch uses α-1,4 (and α-1,6 for branches) bonds that human amylase can break. Cellulose uses β-1,4 bonds that humans lack the enzyme (cellulase) to cleave.两者均为葡萄糖聚合物，但糖苷键不同：淀粉使用 α-1,4（分支处为 α-1,6）键，人体淀粉酶可断裂。纤维素使用 β-1,4 键，人体缺乏相应酶（纤维素酶）来切断。
Both starch and cellulose are made of glucose. The difference is in the type of glycosidic bond (α vs β), not the monomer or chain length.淀粉和纤维素都由葡萄糖组成。差异在于糖苷键的类型（α 与 β），而非单体或链长。
Going deeper — functional groups on carbohydrates (SBI4U B3.3)深入 — 碳水化合物上的官能团（SBI4U B3.3）

Monosaccharides always carry hydroxyl groups (—OH) on most carbons and either an aldehyde group (—CHO, as in glucose: an aldose) or a ketone group (C=O, as in fructose: a ketose) on one carbon. The numerous —OH groups make monosaccharides highly polar and water-soluble. When two monosaccharides form a glycosidic bond, one —OH group from each monomer participates in the dehydration synthesis reaction, releasing water. Ontario SBI4U B3.3 expects students to identify the hydroxyl, carbonyl, and carboxyl groups within biological molecules and explain their contribution to the molecule's function.单糖在大多数碳上携带羟基（—OH），在某一碳上携带醛基（—CHO，如葡萄糖：醛糖）或酮基（C=O，如果糖：酮糖）。大量 —OH 基使单糖高度极性且水溶性强。两个单糖形成糖苷键时，每个单体各一个 —OH 参与脱水缩合反应，释放水分子。安大略 SBI4U B3.3 要求学生识别生物分子中的羟基、羰基和羧基，并解释其对分子功能的贡献。

Section 3 · Lipids第 3 节 · 脂质

Lipids脂质

Lipids are hydrophobic; built from fatty acids and glycerol (or other nonpolar components). No single monomer/polymer pattern — grouped by shared hydrophobic character.脂质是疏水性的；由脂肪酸和甘油（或其他非极性成分）构成。没有单一的单体/聚合物模式——以共有的疏水性归为一类。
  • Triglycerides (fats and oils):甘油三酯（脂肪与油）： one glycerol + three fatty acid tails joined by ester bonds via dehydration synthesis. Saturated fatty acids have no double bonds (all C—C single bonds, straight chains, solid at room temperature — e.g. butter). Unsaturated fatty acids have one or more C=C double bonds (kinked chains, liquid at room temperature — e.g. olive oil). Functions: long-term energy storage (twice the energy per gram as carbohydrates), insulation, organ cushioning.一个甘油 + 三条脂肪酸链，通过脱水缩合形成酯键。饱和脂肪酸无双键（全为 C—C 单键，直链，室温下为固态，如黄油）。不饱和脂肪酸含一个或多个 C=C 双键（弯折链，室温下为液态，如橄榄油）。功能：长期储能（每克能量是碳水化合物的两倍）、隔热、缓冲器官。
  • Phospholipids:磷脂： glycerol + two fatty acid tails + a phosphate head group. The phosphate head is hydrophilic; the fatty acid tails are hydrophobic. This amphipathic structure drives the spontaneous formation of the lipid bilayer of cell membranes in aqueous environments.甘油 + 两条脂肪酸链 + 磷酸头部基团。磷酸头部亲水；脂肪酸链疏水。这种双亲性结构在水性环境中驱动细胞膜脂质双分子层的自发形成。
  • Steroids:固醇： four fused carbon rings. Cholesterol is a membrane component that modulates fluidity. Steroid hormones (testosterone, estrogen, cortisol) are signaling molecules derived from cholesterol.四个融合碳环。胆固醇是膜组分，调节流动性。固醇激素（睾酮、雌激素、皮质醇）是由胆固醇衍生的信号分子。
Worked Example · Saturated vs unsaturated fats例题 · 饱和脂肪与不饱和脂肪

Explain why coconut oil (mostly saturated fat) is solid at room temperature, but canola oil (mostly unsaturated fat) is liquid.解释为何椰子油（主要为饱和脂肪）在室温下为固态，而菜籽油（主要为不饱和脂肪）为液态。

Saturated fatty acids have no C=C double bonds, so their chains are straight and can pack tightly together. The many van der Waals interactions between closely packed chains require more energy (higher temperature) to overcome, resulting in a higher melting point — solid at room temperature.饱和脂肪酸无 C=C 双键，故其链为直链，可紧密排列。紧密排列的链之间有大量范德华力，需要更多能量（更高温度）才能克服，导致熔点更高——室温下为固态。

Unsaturated fatty acids have one or more C=C double bonds, creating kinks (bends) in the chain. These kinks prevent tight packing. With fewer and weaker interactions between chains, less energy is needed to overcome them, resulting in a lower melting point — liquid at room temperature.不饱和脂肪酸含一个或多个 C=C 双键，在链上形成弯折。这些弯折阻止紧密排列。链间相互作用减少且较弱，克服它们所需能量更少，导致熔点更低——室温下为液态。

Which of the following best describes the role of phospholipids in cell membranes?下列哪项最准确地描述了磷脂在细胞膜中的作用？
§3 · Q1
They provide long-term energy storage for the cell为细胞提供长期储能
They act as enzymes to catalyze membrane reactions作为酶催化膜上的反应
They transport oxygen across the membrane跨膜运输氧气
Their amphipathic structure forms a selectively permeable bilayer其双亲性结构形成选择通透性双分子层
Phospholipids are amphipathic: the hydrophilic phosphate heads face the aqueous environments (inside and outside the cell), and the hydrophobic fatty acid tails face each other. This arrangement spontaneously forms the bilayer that serves as the cell membrane's structural foundation and creates selective permeability.磷脂是双亲性的：亲水磷酸头朝向水性环境（细胞内外），疏水脂肪酸链彼此相对。这种排列自发形成双分子层，构成细胞膜的结构基础并产生选择通透性。
The key role of phospholipids is forming the bilayer: their amphipathic nature (hydrophilic head, hydrophobic tail) organizes them into a double layer that is the structural basis of all cell membranes.磷脂的关键作用是形成双分子层：其双亲性（亲水头、疏水尾）使其自组织为双层，构成所有细胞膜的结构基础。
Section 4 · Proteins and their structure第 4 节 · 蛋白质及其结构

Proteins and Their Structure蛋白质及其结构

Monomer: amino acid. Polymer: polypeptide/protein. Bond joining monomers: peptide bond (dehydration synthesis).单体：氨基酸。聚合物：多肽/蛋白质。连接单体的键：肽键（脱水缩合）。
  • Amino acid structure:氨基酸结构： a central α-carbon bonded to: an amino group (—NH2), a carboxyl group (—COOH), a hydrogen, and a variable R group (side chain). There are 20 different R groups; the R group determines the amino acid's chemical properties (polar, nonpolar, charged, etc.).中心 α-碳与以下基团相连：氨基（—NH2）、羧基（—COOH）、氢，以及可变的 R 基（侧链）。共有 20 种不同的 R 基；R 基决定氨基酸的化学性质（极性、非极性、带电等）。
  • Four levels of protein structure:蛋白质结构的四个层次：
    1. Primary (°1): the sequence of amino acids in the polypeptide chain. Encoded by the gene.一级结构：多肽链中氨基酸的序列。由基因编码。
    2. Secondary (°2): local folding of the backbone into α-helices or β-pleated sheets, stabilized by hydrogen bonds between backbone groups.二级结构：主链局部折叠成 α-螺旋或 β-折叠，由主链基团间的氢键稳定。
    3. Tertiary (°3): the overall 3-D shape of a single polypeptide, determined by interactions between R groups (hydrophobic interactions, disulfide bonds, ionic bonds, hydrogen bonds). The active site of an enzyme is determined at this level.三级结构：单条多肽链的整体三维形状，由 R 基团间的相互作用决定（疏水相互作用、二硫键、离子键、氢键）。酶的活性位点在此层次确定。
    4. Quaternary (°4): the arrangement of two or more polypeptide subunits. Not all proteins have quaternary structure. Example: haemoglobin (four subunits).四级结构：两条或多条多肽亚基的排列。不是所有蛋白质都有四级结构。例：血红蛋白（四个亚基）。
  • Protein functions:蛋白质功能： enzymes (catalysis), structural (collagen, keratin), transport (haemoglobin, channel proteins), signaling (hormones, receptors), defense (antibodies), motor (actin, myosin).酶（催化）、结构性（胶原蛋白、角蛋白）、运输（血红蛋白、通道蛋白）、信号传导（激素、受体）、防御（抗体）、运动（肌动蛋白、肌球蛋白）。
Denaturation.变性。

Denaturation is the disruption of a protein's secondary, tertiary, or quaternary structure by heat, extreme pH, or chemical agents, without breaking peptide bonds. The primary structure remains intact, but the 3-D shape is lost, destroying function. Denaturation is usually irreversible (e.g. cooking an egg). Because enzyme function depends entirely on the specific 3-D shape of the active site, denaturation abolishes catalytic activity.变性是高温、极端 pH 或化学试剂破坏蛋白质二级、三级或四级结构的过程，不断裂肽键。一级结构保持完整，但三维形状丧失，导致功能破坏。变性通常不可逆（如煮熟鸡蛋）。由于酶的功能完全依赖活性位点的特定三维形状，变性会消除催化活性。

The specific 3-D shape of an enzyme's active site is primarily determined by which level of protein structure?酶活性位点的特定三维形状主要由蛋白质结构的哪个层次决定？
§4 · Q1
Primary structure — the amino acid sequence一级结构 — 氨基酸序列
Secondary structure — α-helices and β-sheets二级结构 — α-螺旋和 β-折叠
Tertiary structure — overall 3-D folding of the polypeptide三级结构 — 多肽链的整体三维折叠
Quaternary structure — arrangement of multiple subunits四级结构 — 多亚基的排列
The tertiary structure is the overall 3-D shape of a single polypeptide, and it is at this level that the active site is formed. The specific shape of the active site depends on the R-group interactions that hold the protein in its folded conformation. Denaturation (which disrupts tertiary structure) destroys enzyme activity.三级结构是单条多肽链的整体三维形状，活性位点在此层次形成。活性位点的特定形状取决于维持蛋白质折叠构象的 R 基团相互作用。变性（破坏三级结构）会消除酶活性。
The active site is part of the tertiary structure: the specific 3-D folding of the polypeptide. Primary structure encodes the sequence; secondary structure is local backbone folding; quaternary involves multiple subunits.活性位点是三级结构的一部分：多肽链特定的三维折叠。一级结构编码序列；二级结构是局部主链折叠；四级结构涉及多条亚基。
A student heats an enzyme solution to 95°C for 10 minutes, then tests its activity. What result is expected, and why?学生将酶溶液加热至 95°C 持续 10 分钟，然后测试其活性。预期结果是什么，原因是什么？
§4 · Q2
Activity doubles; high temperature speeds up enzyme reactions活性加倍；高温加速酶反应
Activity is abolished; high temperature denatures the protein, destroying the active site shape活性完全丧失；高温使蛋白质变性，破坏活性位点形状
Activity is unchanged; proteins are heat-stable活性不变；蛋白质耐热
Activity decreases slightly but recovers when cooled活性略有下降，冷却后恢复
Extreme heat (95°C for most enzymes) causes denaturation: the high thermal energy disrupts the weak bonds (hydrogen bonds, hydrophobic interactions) that maintain tertiary structure. The active site loses its specific shape, so the substrate can no longer bind. This is usually irreversible.极端高温（大多数酶在 95°C）导致变性：高热能破坏维持三级结构的弱键（氢键、疏水相互作用）。活性位点失去其特定形状，底物无法再结合。这通常是不可逆的。
At 95°C, enzyme denaturation occurs. The tertiary structure unfolds, destroying the active site. While moderate warming increases reaction rate, extreme heat causes irreversible loss of function.在 95°C 时，酶发生变性。三级结构展开，活性位点被破坏。虽然适度加热会提高反应速率，但极端高温会导致不可逆的功能丧失。
Section 5 · Enzymes第 5 节 · 酶

Enzymes: Function and Factors Affecting Rate酶：功能与影响速率的因素

Enzymes are biological catalysts — they speed up reactions by lowering activation energy without being consumed.酶是生物催化剂——通过降低活化能加速反应，自身不被消耗。
  • Lock-and-key model:锁钥模型： the substrate fits precisely into the active site of the enzyme (like a key in a lock). The active site has a specific shape complementary to the substrate. The enzyme–substrate complex forms, the reaction proceeds, products are released, and the enzyme is unchanged and available to catalyze another reaction.底物精确契合酶的活性位点（如锁与钥）。活性位点具有与底物互补的特定形状。形成酶-底物复合物，反应进行，产物释放，酶不变，可再次催化。
  • Induced-fit model (going deeper):诱导契合模型（深入）： the active site flexibly adjusts its shape as the substrate binds, improving the fit. More accurate than the rigid lock-and-key model.底物结合时活性位点灵活调整形状以改善契合。比刚性锁钥模型更准确。
  • Activation energy:活化能： the minimum energy input needed to start a reaction. Enzymes lower the activation energy barrier, so reactions proceed faster at normal body temperatures.启动反应所需的最低能量输入。酶降低活化能壁垒，使反应在正常体温下更快进行。
!
Factors affecting enzyme reaction rate (AB Biology 20 D1.3k; BC A&P 12)影响酶反应速率的因素（AB Biology 20 D1.3k；BC A&P 12）
Factor因素Effect on rate对速率的影响Reason原因
Temperature温度Increases up to optimum (~37°C for human enzymes), then drops sharply升至最适温度（人体酶约 37°C）后急剧下降Above optimum: denaturation destroys active site shape超过最适：变性破坏活性位点形状
pHpHPeak at optimum pH (e.g. pepsin ~2, catalase ~7); drops at extremes在最适 pH 时达峰（如胃蛋白酶约 2，过氧化氢酶约 7）；极端 pH 下降H+ / OH− alter ionic bonds and hydrogen bonds that maintain active site shapeH+ / OH− 改变维持活性位点形状的离子键与氢键
Substrate concentration底物浓度Increases rate until all active sites are occupied (saturation point), then plateaus速率增加直至所有活性位点被占满（饱和点），后趋于平稳At saturation, rate is limited by the number of enzyme molecules, not substrate availability饱和时，速率受酶分子数量限制，而非底物可用性
Competitive inhibition竞争性抑制Decreases rate; effect can be overcome by adding more substrate降低速率；可通过增加底物克服An inhibitor molecule with a shape similar to the substrate binds to the active site, blocking the substrate形状与底物相似的抑制剂分子结合活性位点，阻止底物结合
Feedback (allosteric) inhibition反馈（变构）抑制Decreases rate; cannot be overcome by adding more substrate降低速率；无法通过增加底物克服The product of a pathway binds to an allosteric site (not the active site), changing the enzyme's shape so the active site no longer functions代谢途径的产物结合变构位点（而非活性位点），改变酶的形状，使活性位点不再起作用
A poison blocks an enzyme by binding to its active site with a shape similar to the natural substrate. Adding more substrate restores enzyme activity. What type of inhibition is this?一种毒素以类似天然底物的形状结合酶的活性位点，阻断酶的活性。增加底物可恢复酶活性。这是哪种抑制类型？
§5 · Q1
Competitive inhibition竞争性抑制
Allosteric inhibition变构抑制
Feedback inhibition反馈抑制
Denaturation变性
Competitive inhibition: the inhibitor mimics the substrate and competes for the active site. Because both the inhibitor and substrate compete for the same site, increasing substrate concentration increases the probability of substrate binding over the inhibitor, restoring activity. Allosteric inhibition acts at a different (allosteric) site and cannot be overcome this way.竞争性抑制：抑制剂模拟底物，竞争活性位点。由于抑制剂和底物竞争同一位点，增加底物浓度可提高底物相对于抑制剂的结合概率，从而恢复活性。变构抑制作用于不同（变构）位点，无法通过此方式克服。
Competitive inhibition is overcome by excess substrate (both compete for the active site). Allosteric/feedback inhibition acts elsewhere on the enzyme and is NOT overcome by more substrate.竞争性抑制可被过量底物克服（两者竞争活性位点）。变构/反馈抑制作用于酶的其他位点，不能被更多底物克服。
The graph of enzyme reaction rate vs. temperature shows a peak at 37°C for a human enzyme. Above 50°C, the rate drops to near zero. What explains the sharp drop?某人体酶的反应速率对温度曲线在 37°C 时达到峰值。在 50°C 以上，速率骤降至接近零。什么解释了这一急剧下降？
§5 · Q2
Substrate molecules move too fast to bind the active site底物分子运动过快，无法结合活性位点
All substrate molecules are consumed at high temperature高温下所有底物分子被消耗
Enzyme molecules are broken down by protease enzymes酶分子被蛋白酶降解
The enzyme denatures: heat disrupts the bonds maintaining tertiary structure, destroying the active site shape酶发生变性：热量破坏维持三级结构的键，活性位点形状被破坏
High temperature breaks the weak bonds (hydrogen bonds, hydrophobic interactions) that maintain the protein's tertiary structure. The active site loses its specific 3-D shape, so the substrate cannot bind. This is denaturation — it explains why every enzyme has an optimum temperature and why reaction rate falls sharply above it.高温破坏维持蛋白质三级结构的弱键（氢键、疏水相互作用）。活性位点失去特定三维形状，底物无法结合。这就是变性——解释了为何每种酶都有最适温度，以及为何超过该温度后反应速率急剧下降。
The drop above the optimum temperature is caused by denaturation: thermal energy disrupts the weak bonds stabilizing tertiary structure, so the active site changes shape and can no longer bind substrate.最适温度以上的速率下降由变性引起：热能破坏稳定三级结构的弱键，活性位点形状改变，不再能结合底物。
Section 6 · Nucleic acids第 6 节 · 核酸

Nucleic Acids: DNA and RNA核酸：DNA 与 RNA

Monomer: nucleotide. Polymer: nucleic acid. Bond joining monomers: phosphodiester bond.单体：核苷酸。聚合物：核酸。连接单体的键：磷酸二酯键。
  • Nucleotide structure:核苷酸结构： each nucleotide has three components: a 5-carbon sugar (deoxyribose in DNA; ribose in RNA), a phosphate group, and a nitrogenous base. DNA bases: adenine (A), thymine (T), cytosine (C), guanine (G). RNA bases: A, uracil (U), C, G (uracil replaces thymine).每个核苷酸有三个组分：五碳糖（DNA 中为脱氧核糖；RNA 中为核糖）、磷酸基团和含氮碱基。DNA 碱基：腺嘌呤（A）、胸腺嘧啶（T）、胞嘧啶（C）、鸟嘌呤（G）。RNA 碱基：A、尿嘧啶（U）、C、G（尿嘧啶取代胸腺嘧啶）。
  • DNA:DNA： double-stranded helix; the two strands are antiparallel and held together by complementary base pairing (A=T via 2 H-bonds; G≡C via 3 H-bonds). Stores and transmits genetic information. Located in the nucleus (also in mitochondria and chloroplasts).双链螺旋；两条链反向平行，通过互补碱基配对结合（A=T，2 个氢键；G≡C，3 个氢键）。储存和传递遗传信息。位于细胞核中（线粒体和叶绿体中也有）。
  • RNA:RNA： single-stranded; ribose sugar; uracil instead of thymine. Three main types: mRNA (messenger — carries genetic code from DNA to ribosome), tRNA (transfer — brings amino acids to the ribosome), rRNA (ribosomal — forms ribosome structure). RNA is made in transcription and used in translation.单链；核糖糖；尿嘧啶代替胸腺嘧啶。三种主要类型：mRNA（信使——将遗传密码从 DNA 传递到核糖体）、tRNA（转运——将氨基酸带到核糖体）、rRNA（核糖体——构成核糖体结构）。RNA 在转录中生成，在翻译中使用。
  • ATP:ATP： adenosine triphosphate is a nucleotide derivative that acts as the cell's energy currency. Hydrolysis of the terminal phosphate bond releases energy for cellular work.三磷酸腺苷是核苷酸衍生物，充当细胞的能量货币。末端磷酸键的水解释放能量供细胞工作。
Which of the following correctly distinguishes DNA from RNA?下列哪项正确区分了 DNA 与 RNA？
§6 · Q1
DNA is single-stranded; RNA is double-strandedDNA 为单链；RNA 为双链
DNA contains deoxyribose and thymine; RNA contains ribose and uracilDNA 含脱氧核糖和胸腺嘧啶；RNA 含核糖和尿嘧啶
DNA contains uracil; RNA contains thymineDNA 含尿嘧啶；RNA 含胸腺嘧啶
DNA and RNA have identical base compositionsDNA 和 RNA 的碱基组成相同
DNA uses deoxyribose (one fewer oxygen on the sugar) and the base thymine (T). RNA uses ribose and replaces thymine with uracil (U). DNA is normally double-stranded; RNA is normally single-stranded. These are the two most testable structural differences.DNA 使用脱氧核糖（糖上少一个氧）和碱基胸腺嘧啶（T）。RNA 使用核糖，将胸腺嘧啶替换为尿嘧啶（U）。DNA 通常为双链；RNA 通常为单链。这是最常考的两个结构差异。
DNA = double-stranded, deoxyribose, thymine. RNA = single-stranded, ribose, uracil. Thymine is in DNA; uracil is in RNA.DNA = 双链，脱氧核糖，胸腺嘧啶。RNA = 单链，核糖，尿嘧啶。胸腺嘧啶在 DNA 中；尿嘧啶在 RNA 中。
A template DNA strand reads 3'-ATCGGT-5'. What is the complementary mRNA sequence produced during transcription?模板 DNA 链读作 3'-ATCGGT-5'。转录过程中产生的互补 mRNA 序列是什么？
§6 · Q2
3'-ATCGGU-5'3'-ATCGGU-5'
5'-ATCGGT-3'5'-ATCGGT-3'
5'-UAGCCA-3'5'-UAGCCA-3'
5'-TAGCCA-3'5'-TAGCCA-3'
mRNA is synthesized 5' to 3', complementary and antiparallel to the template strand. A pairs with U (not T, since this is RNA), T pairs with A, C pairs with G, G pairs with C. Template 3'-ATCGGT-5' gives mRNA 5'-UAGCCA-3'.mRNA 从 5' 合成至 3'，与模板链互补且反向平行。A 配 U（不是 T，因为是 RNA），T 配 A，C 配 G，G 配 C。模板 3'-ATCGGT-5' 产生 mRNA 5'-UAGCCA-3'。
In transcription, RNA polymerase reads the template 3' to 5' and synthesizes mRNA 5' to 3'. RNA uses uracil (U) not thymine (T). Pair each base: A→U, T→A, C→G, G→C.转录中，RNA 聚合酶从 3' 读至 5'，合成 mRNA 从 5' 至 3'。RNA 用尿嘧啶（U）而非胸腺嘧啶（T）。碱基配对：A→U，T→A，C→G，G→C。
Section 7 · Dehydration synthesis and hydrolysis第 7 节 · 脱水缩合与水解

Dehydration Synthesis and Hydrolysis脱水缩合与水解

Two universal reactions that build and break every polymer in the cell.构建与分解细胞中每种聚合物的两种通用反应。
  • Dehydration synthesis (condensation reaction):脱水缩合（缩合反应）： two monomers are joined by removing one molecule of water (—H from one monomer; —OH from the other). The covalent bond formed is the same type for each macromolecule class: glycosidic bond (carbohydrates), ester bond (lipids), peptide bond (proteins), phosphodiester bond (nucleic acids). Requires energy input (anabolic reaction).通过移除一个水分子（一个单体提供 —H，另一个提供 —OH）将两个单体连接。形成的共价键因大分子类别而异：糖苷键（碳水化合物）、酯键（脂质）、肽键（蛋白质）、磷酸二酯键（核酸）。需要能量输入（合成代谢反应）。
  • Hydrolysis:水解： a polymer is broken apart by adding water; one —H and one —OH from water are added across the broken bond, regenerating the original functional groups on each monomer. Hydrolysis is the reverse of dehydration synthesis. It is catalyzed by hydrolytic enzymes (e.g. amylase breaks starch, protease breaks proteins, lipase breaks fats). Releases energy (catabolic reaction).通过加水分解聚合物；水的一个 —H 和一个 —OH 加到断裂键的两端，重新生成每个单体上原有的官能团。水解是脱水缩合的逆反应。由水解酶催化（如淀粉酶分解淀粉、蛋白酶分解蛋白质、脂肪酶分解脂肪）。释放能量（分解代谢反应）。
Why these two reactions matter for all of biochemistry.为何这两种反应在整个生物化学中至关重要。

Every macromolecule in every cell is built by dehydration synthesis and broken down by hydrolysis. Digestion is hydrolysis: amylase, protease, and lipase enzymes in the gut add water to break the bonds of ingested polymers into their monomers, which are then absorbed. Protein synthesis (translation) is dehydration synthesis: the ribosome joins amino acids by peptide bonds with the release of water at each step. Recognizing the pattern (monomer + monomer → polymer + H2O; polymer + H2O → monomer + monomer) across all four macromolecule classes is the central organizing principle of biochemistry.每个细胞中的每种大分子都通过脱水缩合构建，通过水解分解。消化即水解：肠道中的淀粉酶、蛋白酶和脂肪酶向摄入的聚合物中加水，断裂其键，生成单体后被吸收。蛋白质合成（翻译）即脱水缩合：核糖体通过肽键连接氨基酸，每步均释放水。在四类大分子中识别这一模式（单体 + 单体 → 聚合物 + H2O；聚合物 + H2O → 单体 + 单体）是生物化学的核心组织原则。

Worked Example · Dehydration synthesis of a dipeptide例题 · 二肽的脱水缩合

Two amino acids, glycine and alanine, are joined by a peptide bond. (a) Identify the type of reaction. (b) Name the bond formed. (c) State what small molecule is released.甘氨酸和丙氨酸两个氨基酸通过肽键连接。(a) 判断反应类型。(b) 说明形成的键的名称。(c) 说明释放的小分子。

(a) Dehydration synthesis (condensation reaction).(a) 脱水缩合（缩合反应）。

(b) Peptide bond(b) 肽键 — forms between the carboxyl group (—COOH) of glycine and the amino group (—NH2) of alanine.— 形成于甘氨酸的羧基（—COOH）与丙氨酸的氨基（—NH2）之间。

(c) Water (H2O)(c) 水（H2O） — one —OH from the carboxyl group and one —H from the amino group combine to release H2O.— 羧基提供的一个 —OH 与氨基提供的一个 —H 结合，释放 H2O。

Amylase is a digestive enzyme that breaks down starch. What type of reaction does amylase catalyze, and what is added to each broken bond?淀粉酶是分解淀粉的消化酶。淀粉酶催化哪种类型的反应？每个断裂的键处加入了什么？
§7 · Q1
Dehydration synthesis; water is released脱水缩合；释放水
Hydrolysis; water is added across the broken glycosidic bond水解；水加到断裂的糖苷键两端
Oxidation; oxygen is added to each glucose monomer氧化；氧加到每个葡萄糖单体上
Phosphorylation; ATP donates a phosphate to each bond磷酸化；ATP 向每个键提供磷酸基
Starch is a polysaccharide. Breaking it down into glucose monomers is hydrolysis: water (H2O) is added across each glycosidic bond, splitting the bond and regenerating —OH on one glucose and —H on the next. Amylase is the enzyme that catalyzes this hydrolysis.淀粉是多糖。将其分解为葡萄糖单体是水解：水（H2O）加到每个糖苷键两端，断裂键，在一个葡萄糖上重新生成 —OH，在下一个葡萄糖上生成 —H。淀粉酶是催化这一水解的酶。
Breaking a polymer into monomers is hydrolysis (adding water). Dehydration synthesis builds polymers (removes water). Amylase catalyzes hydrolysis of the glycosidic bonds in starch.将聚合物分解为单体是水解（加水）。脱水缩合构建聚合物（脱水）。淀粉酶催化淀粉中糖苷键的水解。
Going deeper — the four biochemical reaction types (SBI4U B3.5)深入 — 四类生化反应类型（SBI4U B3.5）

Ontario SBI4U B3.5 identifies four main types of biochemical reactions: (1) Oxidation-reduction (redox) — electron transfer; e.g. NAD+ gains electrons to become NADH in cellular respiration. (2) Hydrolysis — bond cleavage by adding water, as above. (3) Condensation — bond formation with loss of water (= dehydration synthesis). (4) Neutralization — acid + base reaction producing a salt and water; relevant to pH buffering in cells (e.g. bicarbonate buffer in blood). Recognizing which reaction type underlies a given cellular process is an SBI4U-level expectation.安大略 SBI4U B3.5 确认四类主要生化反应：(1) 氧化还原（氧化还原）——电子转移；如细胞呼吸中 NAD+ 得到电子变为 NADH。(2) 水解——通过加水断裂键，如上所述。(3) 缩合——失去水形成键（即脱水缩合）。(4) 中和——酸碱反应产生盐和水；与细胞内 pH 缓冲相关（如血液中的碳酸氢盐缓冲）。识别哪种反应类型是特定细胞过程的基础，是 SBI4U 水平的期望。

Exam Tactics考试策略

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

Macromolecule identification questions大分子辨认题
  • Always name the monomer AND the bond type.始终同时说明单体名称和键的类型。 "Protein" alone rarely earns full marks; say "protein — polymer of amino acids joined by peptide bonds via dehydration synthesis."仅写"蛋白质"很少能得满分；应说"蛋白质——通过脱水缩合形成肽键连接的氨基酸聚合物"。
  • Distinguish storage from structural polysaccharides.区分储能多糖与结构多糖。 Starch and glycogen store energy; cellulose is structural. The key is the α-1,4 vs β-1,4 glycosidic bond type.淀粉和糖原储能；纤维素是结构性的。关键是 α-1,4 键与 β-1,4 键之分。
Enzyme questions酶相关题
  • Identify the type of inhibition first.先判断抑制类型。 Competitive inhibition: inhibitor binds the active site, overcome by more substrate. Allosteric inhibition: inhibitor binds elsewhere, NOT overcome by more substrate.竞争性抑制：抑制剂结合活性位点，可通过增加底物克服。变构抑制：抑制剂结合其他位点，不能通过增加底物克服。
  • Denaturation is not inhibition.变性不等于抑制。 Denaturation permanently unfolds the tertiary structure; inhibition is typically reversible. Extreme heat or extreme pH causes denaturation.变性永久展开三级结构；抑制通常可逆。极端高温或极端 pH 导致变性。
Dehydration synthesis vs hydrolysis脱水缩合与水解
  • Build = dehydration synthesis; break = hydrolysis.构建 = 脱水缩合；分解 = 水解。 Digestion is hydrolysis; protein synthesis is dehydration synthesis.消化是水解；蛋白质合成是脱水缩合。
  • The water rule:水的规则： dehydration synthesis releases H2O; hydrolysis consumes H2O. The molecule released when a peptide bond forms is water.脱水缩合释放 H2O；水解消耗 H2O。肽键形成时释放的分子是水。
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Why does water form hydrogen bonds?为何水能形成氢键？
Water is polar: the O end carries δ− and each H carries δ+. Adjacent molecules attract (δ+ H to δ− O), forming hydrogen bonds.水是极性分子：O 端带 δ−，每个 H 带 δ+。相邻分子相互吸引（δ+ H 吸引 δ− O），形成氢键。
Monomer of carbohydrates?碳水化合物的单体？
Monosaccharide (e.g. glucose, fructose, galactose). General formula: (CH2O)n.单糖（如葡萄糖、果糖、半乳糖）。通式：(CH2O)n。
Starch vs cellulose — key structural difference?淀粉与纤维素——关键结构差异？
Both are glucose polymers. Starch: α-1,4 glycosidic bonds (digestible by amylase). Cellulose: β-1,4 bonds (not digestible by humans).两者均为葡萄糖聚合物。淀粉：α-1,4 糖苷键（淀粉酶可消化）。纤维素：β-1,4 键（人类不能消化）。
Why are phospholipids amphipathic?为何磷脂是双亲性的？
The phosphate head is polar/hydrophilic; the two fatty acid tails are nonpolar/hydrophobic. One molecule has both a water-loving and a water-fearing region.磷酸头部是极性/亲水的；两条脂肪酸链是非极性/疏水的。一个分子同时具有亲水区和疏水区。
Why does saturated fat have a higher melting point than unsaturated fat?为何饱和脂肪的熔点高于不饱和脂肪？
Saturated: straight chains pack tightly → many van der Waals interactions → higher melting point. Unsaturated: C=C double bonds create kinks → loose packing → lower melting point.饱和：直链紧密排列 → 大量范德华力 → 熔点高。不饱和：C=C 双键产生弯折 → 排列疏松 → 熔点低。
Four levels of protein structure?蛋白质结构的四个层次？
1° amino acid sequence. 2° α-helices / β-sheets (H-bonds in backbone). 3° overall 3-D folding of polypeptide (R-group interactions). 4° arrangement of two or more subunits.一级结构：氨基酸序列。二级结构：α-螺旋/β-折叠（主链氢键）。三级结构：多肽整体三维折叠（R 基团相互作用）。四级结构：两条或多条亚基的排列。
What is denaturation?什么是变性？
Disruption of protein secondary/tertiary/quaternary structure by heat, extreme pH, or chemicals. Peptide bonds (primary structure) remain intact. Usually irreversible; destroys active site and function.高温、极端 pH 或化学物质破坏蛋白质二/三/四级结构。肽键（一级结构）保持完整。通常不可逆；破坏活性位点和功能。
How does an enzyme lower activation energy?酶如何降低活化能？
The active site binds the substrate (lock-and-key or induced-fit), forming an enzyme-substrate complex that stabilizes the transition state and reduces the energy barrier needed to start the reaction.活性位点结合底物（锁钥或诱导契合），形成酶-底物复合物，稳定过渡态，降低启动反应所需的能量壁垒。
Competitive vs allosteric inhibition — how to tell apart?竞争性抑制与变构抑制——如何区分？
Competitive: binds active site; more substrate overcomes it. Allosteric: binds different (allosteric) site; changes active site shape; NOT overcome by more substrate.竞争性：结合活性位点；增加底物可克服。变构：结合不同（变构）位点；改变活性位点形状；不能通过增加底物克服。
Monomer of nucleic acids; bond joining them?核酸的单体；连接单体的键？
Nucleotide (5-carbon sugar + phosphate + nitrogenous base). Linked by phosphodiester bonds via dehydration synthesis.核苷酸（五碳糖 + 磷酸 + 含氮碱基）。通过脱水缩合形成磷酸二酯键相连。
DNA vs RNA — three structural differences?DNA 与 RNA——三个结构差异？
1. DNA: deoxyribose sugar; RNA: ribose. 2. DNA: thymine (T); RNA: uracil (U). 3. DNA: usually double-stranded; RNA: usually single-stranded.1. DNA：脱氧核糖；RNA：核糖。2. DNA：胸腺嘧啶（T）；RNA：尿嘧啶（U）。3. DNA：通常双链；RNA：通常单链。
Dehydration synthesis vs hydrolysis — what is added or removed?脱水缩合与水解——加入还是去除什么？
Dehydration synthesis: removes H2O, forms covalent bond (builds polymer). Hydrolysis: adds H2O, breaks covalent bond (breaks down polymer).脱水缩合：去除 H2O，形成共价键（构建聚合物）。水解：加入 H2O，断裂共价键（分解聚合物）。
Mixed Practice综合练习

Practice Quiz综合测验

Which property of water explains why aquatic environments heat up and cool down more slowly than the surrounding air?水的哪种性质解释了为何水生环境升温和降温都比周围空气慢？
Q1
High density at 4°C4°C 时密度最大
Universal solvent properties通用溶剂性质
High specific heat capacity高比热容
Cohesion and surface tension内聚力与表面张力
High specific heat capacity means water absorbs a large amount of heat energy per degree of temperature rise (because hydrogen bonds must be disrupted). This thermal buffering stabilizes the temperature of aquatic environments and the bodies of organisms that are mostly water.高比热容意味着水每升高一度需吸收大量热能（因为需要打断氢键）。这种热缓冲稳定了水生环境及以水为主要成分的生物体的温度。
The thermal stability of water is due to its high specific heat capacity, a consequence of the hydrogen bonds between water molecules that must be broken to raise temperature.水的热稳定性来自其高比热容，这是水分子间氢键的结果——升温时必须打断这些键。
A molecule is described as having a hydrophilic phosphate head and two hydrophobic fatty acid tails. It spontaneously forms a bilayer in water. Which molecule is this?某分子具有亲水磷酸头部和两条疏水脂肪酸链，在水中自发形成双分子层。该分子是什么？
Q2
A triglyceride甘油三酯
A phospholipid磷脂
A steroid固醇
A glycoprotein糖蛋白
A phospholipid: glycerol backbone, two fatty acid tails (hydrophobic), and a phosphate-containing head group (hydrophilic). This amphipathic structure drives bilayer formation in aqueous environments, which is the structural basis of all cell membranes. Triglycerides have three fatty acid tails and no phosphate group, so they are not amphipathic and do not form bilayers.磷脂：甘油骨架、两条脂肪酸链（疏水）和含磷酸的头部基团（亲水）。这种双亲性结构在水性环境中驱动双分子层形成，构成所有细胞膜的结构基础。甘油三酯有三条脂肪酸链且无磷酸基，故非双亲性，不形成双分子层。
The amphipathic molecule with one phosphate head and two fatty acid tails that forms a bilayer is a phospholipid. Triglycerides have three fatty acid tails and no phosphate; steroids have a four-ring structure.具有一个磷酸头部和两条脂肪酸链并形成双分子层的双亲性分子是磷脂。甘油三酯有三条脂肪酸链且无磷酸；固醇具有四环结构。
An enzyme catalyzes the same reaction at pH 7 but barely functions at pH 2. An enzyme found in the stomach works best at pH 2. What does this tell us about enzyme active sites?某酶在 pH 7 时催化同一反应，但在 pH 2 时几乎不起作用。胃中的酶在 pH 2 时效果最佳。这说明酶活性位点的什么特征？
Q3
All enzymes have the same optimum pH of 7所有酶的最适 pH 均为 7
Stomach enzymes are made of different monomers than other enzymes胃中酶由与其他酶不同的单体构成
Extreme pH denatures all enzymes equally极端 pH 对所有酶的变性程度相同
Each enzyme has a specific optimum pH determined by its tertiary structure; different pH conditions affect the R-group interactions that maintain active site shape每种酶都有由其三级结构决定的特定最适 pH；不同 pH 条件影响维持活性位点形状的 R 基团相互作用
Enzyme active sites have specific 3-D shapes determined by R-group interactions (ionic bonds, hydrogen bonds). These interactions are sensitive to H+ and OH− concentration. Each enzyme has evolved an optimum pH where its active site shape is best maintained. Outside this range, R-group interactions are disrupted, the active site changes shape, and catalytic activity is lost.酶活性位点具有由 R 基团相互作用（离子键、氢键）决定的特定三维形状。这些相互作用对 H+ 和 OH− 浓度敏感。每种酶都进化出一个最适 pH，此时活性位点形状维持最佳。超出此范围，R 基团相互作用受到破坏，活性位点形状改变，催化活性丧失。
Each enzyme has its own optimum pH determined by the specific R-group interactions that maintain its tertiary structure and active site. Pepsin works at pH 2; most human enzymes work best near pH 7. pH denaturation is specific to the enzyme, not universal.每种酶都有其自身的最适 pH，由维持三级结构和活性位点的特定 R 基团相互作用决定。胃蛋白酶在 pH 2 时工作；大多数人体酶最适 pH 接近 7。pH 变性是酶特异性的，并非普遍适用。
During digestion, the enzyme lipase breaks down fats (triglycerides) into glycerol and fatty acids. What type of reaction is this, and what molecule is consumed?消化过程中，脂肪酶将脂肪（甘油三酯）分解为甘油和脂肪酸。这是哪种类型的反应？消耗了什么分子？
Q4
Hydrolysis; water (H2O) is consumed to break each ester bond水解；消耗水（H2O）来断裂每个酯键
Dehydration synthesis; water is released as the bond is broken脱水缩合；断键时释放水
Oxidation; oxygen is added across each ester bond氧化；氧加到每个酯键上
Phosphorylation; ATP is required to break each ester bond磷酸化；断裂每个酯键需要 ATP
Breaking a polymer (triglyceride) into its monomers (glycerol + fatty acids) is hydrolysis: a water molecule is added across each ester bond (—H to one fragment, —OH to the other), cleaving the bond. Lipase is the enzyme that catalyzes this hydrolysis reaction.将聚合物（甘油三酯）分解为单体（甘油 + 脂肪酸）是水解：一个水分子加到每个酯键两端（—H 加到一个片段，—OH 加到另一个），断裂键。脂肪酶是催化这一水解反应的酶。
Digestion is always hydrolysis (adding water to break bonds). Dehydration synthesis is the reverse (removing water to form bonds). Lipase catalyzes hydrolysis of ester bonds in triglycerides.消化始终是水解（加水断键）。脱水缩合是逆反应（脱水成键）。脂肪酶催化甘油三酯中酯键的水解。
A template DNA strand reads 3'-TACGGA-5'. What is the sequence of the mRNA produced during transcription?模板 DNA 链读作 3'-TACGGA-5'。转录产生的 mRNA 序列是什么？
Q5
3'-AUGCCU-5'3'-AUGCCU-5'
5'-TACGGA-3'5'-TACGGA-3'
5'-AUGCCU-3'5'-AUGCCU-3'
5'-ATGCCT-3'5'-ATGCCT-3'
mRNA is synthesized 5' to 3', complementary and antiparallel to the template. T in DNA pairs with A in RNA; A pairs with U; C pairs with G; G pairs with C. Template 3'-TACGGA-5' gives mRNA 5'-AUGCCU-3'.mRNA 从 5' 合成至 3'，与模板互补且反向平行。DNA 的 T 配 RNA 的 A；A 配 U；C 配 G；G 配 C。模板 3'-TACGGA-5' 产生 mRNA 5'-AUGCCU-3'。
Template is read 3' to 5'; mRNA grows 5' to 3'. Replace T with A, A with U, C with G, G with C: 3'-TACGGA-5' → 5'-AUGCCU-3'.模板从 3' 读至 5'；mRNA 从 5' 生长至 3'。将 T 替换为 A，A 替换为 U，C 替换为 G，G 替换为 C：3'-TACGGA-5' → 5'-AUGCCU-3'。
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

Biochemistry is the chemical vocabulary that every later biology unit speaks. The four macromolecule classes, dehydration synthesis, hydrolysis, and enzyme kinetics appear again directly in Cellular Energetics (Unit 3 — ATP as a nucleotide, glucose as a carbohydrate fuel, NADH as a nucleotide derivative), Molecular Genetics (Unit 6 — DNA and RNA structure, nucleotide monomers, base pairing), Cell Division (Unit 4 — the nucleotide building blocks of chromosomes), and Human Anatomy and Physiology (Unit 10 — digestive enzymes, hormone proteins, structural proteins).生物化学是后续所有生物学单元所使用的化学词汇。四类大分子、脱水缩合、水解和酶动力学将直接出现在《细胞能量学》（第 3 单元——ATP 作为核苷酸、葡萄糖作为碳水化合物燃料、NADH 作为核苷酸衍生物）、《分子遗传学》（第 6 单元——DNA 和 RNA 结构、核苷酸单体、碱基配对）、《细胞分裂》（第 4 单元——染色体的核苷酸构建块）与《人体解剖与生理》（第 10 单元——消化酶、激素蛋白、结构蛋白）中。

Within High School Biology.在 HS Biology 内部。

Cellular Energetics (Unit 3) uses the carbohydrate (glucose) and nucleotide (ATP, NADH) frameworks from this guide. Cell Division (Unit 4) opens with nucleic acid structure from Section 6. Molecular Genetics (Unit 6) deepens the DNA and RNA content introduced here into transcription, translation, and replication. Human Anatomy and Physiology (Unit 10) connects enzyme kinetics (Section 5) to digestive physiology and applies protein structure (Section 4) to hormones, antibodies, and structural proteins.《细胞能量学》（第 3 单元）使用本指南中碳水化合物（葡萄糖）和核苷酸（ATP、NADH）的框架。《细胞分裂》（第 4 单元）以第 6 节的核酸结构为开篇。《分子遗传学》（第 6 单元）将本指南介绍的 DNA 和 RNA 内容深化为转录、翻译和复制。《人体解剖与生理》（第 10 单元）将酶动力学（第 5 节）与消化生理相联系，并将蛋白质结构（第 4 节）应用于激素、抗体和结构蛋白。

No AP/IB Biology feeder product yet.目前尚无 AP/IB Biology 衔接产品。

Both AP Biology and IB Biology HL have dedicated biochemistry units that build directly on this content. AP Biology Unit 1 (Chemistry of Life) and IB Biology HL Topic B.1 (Molecules and Metabolism) assume the macromolecule structures and enzyme concepts covered here. Neither AP Biology nor IB Biology is available in this repo yet; this guide is the prerequisite foundation to revisit when those products ship.AP Biology 和 IB Biology HL 均有专门的生物化学单元，直接建立在本内容之上。AP Biology 第 1 单元（生命化学）和 IB Biology HL Topic B.1（分子与代谢）默认学生已掌握本指南涵盖的大分子结构和酶的概念。目前 AP Biology 和 IB Biology 在本仓库中均不可用；本指南是待相关产品上线时需回顾的先修基础。

DINGRUI SCHOLARS

High School Biology · Biochemistry (Molecules of Life) · v1高中生物 · 生物化学（生命分子）· v1