WJEC生物 生化 缩合水解 单体多聚 检测 | WJEC Biology Biochemistry Condensation Hydrolysis Monomers Polymers Tests

WJEC生物 生化 缩合水解 单体多聚 检测

生化是 WJEC 生物的基石 ：： 从分子层面理解生命，才能串联起细胞、生理和生态。Basic Biochemistry is the foundation of WJEC Biology: master the molecules, and the rest of the syllabus falls into place.

1. 单体 vs 聚合物 Monomers vs Polymers

🔹 单体 Monomer

单体是单个小分子，如葡萄糖(C₆H₁₂O₆)、氨基酸、核苷酸。它们可以重复连接形成长链。A monomer is a single small molecule, such as glucose (C₆H₁₂O₆), amino acids, or nucleotides. They can be repeatedly linked to form long chains.

🔸 聚合物 Polymer

聚合物是由多个单体通过缩合反应连接而成的大分子，如淀粉(葡萄糖聚合物)、蛋白质(氨基酸聚合物)。A polymer is a large molecule made of many monomers joined by condensation reactions, e.g. starch (glucose polymer) and proteins (amino acid polymers).

在 WJEC 考试中，describe the formation of a polymer 这类题常要求你写出缩合反应的具体过程 ：： 一个单体贡献 ：OH，另一个贡献 ：H，脱去一分子水，形成共价键。WJEC 评分标准特别强调关键词：condensation reaction / peptide bond / glycosidic bond / ester bond。In WJEC exams, you must name the specific bond formed (glycosidic, peptide, or ester) and mention that a water molecule is released. Vague answers like "they join together" will cost you marks.

2. 缩合反应 vs 水解反应 Condensation vs Hydrolysis

🔹 缩合 Condensation

两个单体通过共价键连接，释放出一分子水(H₂O)。需要酶的催化。Two monomers join via a covalent bond, releasing one water molecule (H₂O). Catalysed by enzymes.

🔸 水解 Hydrolysis

聚合物被分解为单体，需要消耗一分子水(H₂O)，在酶的催化下断裂共价键。A polymer is broken into monomers, consuming one water molecule (H₂O), with enzymes breaking the covalent bond.

缩合和水解是 WJEC Unit 1 最核心的生化概念。每次考到多糖或蛋白质的结构，几乎都会追问：what type of reaction forms this bond? 注意区分三个键型：glycosidic bond (碳水化合物), peptide bond (蛋白质), ester bond (脂质)。WJEC mark schemes frequently award one mark for naming the bond type and a second for stating "water is removed/added". 记忆口诀：缩 = 聚合 + 脱水；解 = 分解 + 加水。

3. α-葡萄糖 vs β-葡萄糖 α-Glucose vs β-Glucose

🔹 α-葡萄糖 α-Glucose

C1 上的 ：OH 指向下方(与 CH₂OH 同侧)。单体聚合形成淀粉和糖原。The ：OH on C1 points downwards (same side as CH₂OH). Polymerises to form starch and glycogen.

🔸 β-葡萄糖 β-Glucose

C1 上的 ：OH 指向上方(与 CH₂OH 异侧)。单体会形成纤维素。The ：OH on C1 points upwards (opposite side to CH₂OH). Polymerises to form cellulose.

仅 C1 羟基位置的微小差异，就决定了多糖的结构和功能完全不同。α-1,4 糖苷键产生螺旋结构(淀粉/糖原)，适合储存能量；β-1,4 糖苷键产生直链结构(纤维素)，每第二个葡萄糖翻转180°，允许大量氢键交联，赋予植物细胞壁极高的抗拉强度。WJEC 的 6-mark QWC (Quality of Written Communication) 题目常要求你 explain how the structure of starch, glycogen and cellulose relates to their functions。The tiny difference in ：OH orientation on C1 determines everything: α-1,4 bonds produce coiled structures ideal for energy storage, while β-1,4 bonds produce straight chains capable of extensive hydrogen bonding, giving cellulose its immense tensile strength. This is classic WJEC "structure explains function" territory.

4. 甘油三酯 vs 磷脂 Triglycerides vs Phospholipids

🔹 甘油三酯 Triglyceride

1 分子甘油 + 3 分子脂肪酸，通过酯键连接。完全疏水，用于长期能量储存、保温、保护器官。One glycerol + three fatty acids, joined by ester bonds. Fully hydrophobic, used for long-term energy storage, insulation, and organ protection.

🔸 磷脂 Phospholipid

1 分子甘油 + 2 分子脂肪酸 + 1 磷酸基团。具有亲水头部和疏水尾部(两亲性)，构成细胞膜磷脂双分子层。One glycerol + two fatty acids + one phosphate group. Amphipathic (hydrophilic head + hydrophobic tails), forming the phospholipid bilayer of cell membranes.

WJEC 最喜欢问的一个对比题是：explain how the structure of triglycerides and phospholipids relates to their different functions。关键得分点：triglycerides 有三条脂肪酸链，使它们完全疏水，适合储存而不干扰细胞渗透压；phospholipids 的磷酸头部带电，使其在水中自发形成双分子层(fluid mosaic model)。如果题目要求 evaluate, 还要提到胆固醇(cholesterol)在膜中的作用 ：： 调节膜的流动性，防止低温下膜硬化。A common WJEC comparison question: "explain how the structure of triglycerides and phospholipids relates to their different functions." Key marks: three fatty acid tails make triglycerides completely hydrophobic (safe storage without osmotic effects); the charged phosphate head of phospholipids drives spontaneous bilayer formation. For evaluate questions, add cholesterol's role in regulating membrane fluidity.

5. 淀粉 vs 糖原 vs 纤维素 Starch vs Glycogen vs Cellulose

🔹 淀粉 Starch

植物储能多糖。直链淀粉(amylose): α-1,4 键, 螺旋结构。支链淀粉(amylopectin): α-1,4 + α-1,6 支链。不溶于水，在细胞内形成淀粉粒，不改变渗透压。Plant energy storage. Amylose: α-1,4 bonds, helical. Amylopectin: α-1,4 + α-1,6 branches. Insoluble, forms starch grains in cells without affecting water potential.

🔸 糖原 Glycogen

动物储能多糖。α-1,4 主链 + 更多 α-1,6 支链(比支链淀粉更分支)。高度分支意味着更多末端可供酶同时水解，适合动物快速能量需求。Animal energy storage. α-1,4 backbone + more α-1,6 branches (more than amylopectin). Highly branched means more terminal ends for simultaneous enzyme hydrolysis, matching rapid energy demands.

🟢 纤维素 Cellulose

植物结构多糖。β-葡萄糖单体，β-1,4 键，每第二个葡萄糖翻转180°。直链，不分支，相邻链间大量氢键形成微纤维束，赋予细胞壁高抗拉强度。Plant structural polysaccharide. β-glucose, β-1,4 bonds, alternate 180° flip. Straight, unbranched, extensive inter-chain H-bonds form microfibrils, providing high tensile strength to cell walls.

WJEC 对此题的典型提问：compare and contrast the structure and function of starch and cellulose。答题框架：(1) 都含葡萄糖单体(相似点)；(2) 键型不同(α vs β)；(3) 链形态不同(螺旋 vs 直链)；(4) 功能不同(储能 vs 结构)。若涉及 glycogen, 需增加支链程度的对比。WJEC 要求用准确的生化术语描述，如 "1,4 glycosidic bonds" 和 "1,6 glycosidic bonds" 而不是泛泛的 bonds。Typical WJEC framing: "compare and contrast the structure and function of starch and cellulose." Answer framework: (1) both glucose polymers (similarity); (2) different bond types (α vs β); (3) different chain shapes (helical vs straight); (4) different functions (storage vs structural). For glycogen, add branching degree comparison. Precision matters: use "1,4 glycosidic bonds" not just "bonds."

6. 纤维状蛋白 vs 球状蛋白 Fibrous vs Globular Proteins

🔹 纤维状蛋白 Fibrous

长链，不溶于水，结构功能为主。胶原蛋白(collagen): 三条 α-螺旋链缠绕成三股螺旋(tropocollagen)，共价交联形成纤维，赋予肌腱和骨骼高强度。Long chains, insoluble, structural. Collagen: three α-helix chains coiled into a triple helix (tropocollagen), covalent cross-links form fibrils, giving high tensile strength to tendons and bones.

🔸 球状蛋白 Globular

紧凑球形，可溶于水，功能性为主。血红蛋白(haemoglobin): 四级结构, 4 条多肽链(2α+2β), 各含一个血红素辅基(Fe²⁺), 负责氧气运输。Compact and spherical, water-soluble, functional. Haemoglobin: quaternary structure, 4 polypeptide chains (2α+2β), each with a haem prosthetic group (Fe²⁺), responsible for oxygen transport.

WJEC 必考四级结构(quaternary structure)的区分：血红蛋白有四级结构(>1 条多肽链 + 辅基)，胶原蛋白也是四级结构(3 条多肽链缠绕)。注意：不是所有蛋白质都有四级结构 ：： 四级结构的定义是两条或更多多肽链的组合。Haemoglobin 的协同结合(cooperative binding)：O₂ 结合到第一个血红素后改变形状，使后续 O₂ 更易结合，形成 S 形(乙状)解离曲线。记住 protein structure 的四层：primary (氨基酸序列), secondary (α-helix / β-pleated sheet, 由 H 键维持), tertiary (3D 折叠, 由 H 键/离子键/二硫键/疏水作用维持), quaternary (>1 条多肽链)。WJEC expects you to distinguish quaternary structure precisely. Haemoglobin exhibits cooperative binding: binding of O₂ to the first haem group induces a conformational change, making subsequent O₂ binding easier, producing a sigmoidal (S-shaped) dissociation curve. Remember the four levels: primary (amino acid sequence), secondary (α-helix/β-pleated sheet, H-bonds), tertiary (3D folding, H-bonds/ionic bonds/disulfide bridges/hydrophobic interactions), quaternary (>1 polypeptide chain).

7. 生化检测对比 Biochemical Tests Comparison

Benedict's 还原糖 Reducing Sugars

样品 + Benedict's 试剂 → 沸水浴 5 分钟 → 蓝 → 绿 → 黄 → 橙 → 砖红沉淀(颜色越深，还原糖越多)。检测所有单糖和部分二糖(麦芽糖)。Sample + Benedict's reagent → boiling water bath 5 min → blue → green → yellow → orange → brick-red precipitate. Detects all monosaccharides and some disaccharides (maltose).

Benedict's 非还原糖 Non-Reducing

先用 HCl 酸水解(断裂糖苷键) → 加 NaHCO₃ 中和 → 再加 Benedict's 试剂沸水浴。蔗糖需此过程。Hydrolyse with HCl (break glycosidic bonds) → neutralise with NaHCO₃ → then Benedict's test as above. Required for sucrose.

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