OCR A-Level Biology: Biological Molecules Complete Guide — OCR A-Level 生物：生物分子完全指南

1. 生物分子的四大类别：糖类、脂质、蛋白质与核酸 | The Four Classes of Biological Molecules: Carbohydrates, Lipids, Proteins and Nucleic Acids

在 OCR A-Level 生物 A 课程中，2.2 模块”生物分子”是理解一切生命过程的基础。生物体由约 25 种元素构成，但其中四种元素碳、氢、氧、氮占据了细胞干重的绝大部分。由这些元素组成的有机分子可以被划分为四大类别：糖类、脂质、蛋白质和核酸。每一类分子都有独特的单体（monomer）和聚合物（polymer）结构，正是这些结构差异决定了它们在细胞中扮演的不同角色。

In OCR A-Level Biology A, Module 2.2 “Biological molecules” is the foundation for understanding all life processes. Living organisms are made from about 25 elements, but four of them – carbon, hydrogen, oxygen and nitrogen – account for the vast majority of the dry mass of a cell. The organic molecules built from these elements fall into four major classes: carbohydrates, lipids, proteins and nucleic acids. Each class has its own characteristic monomers and polymers, and it is precisely these structural differences that determine the distinct roles they play inside the cell.

糖类仅含碳、氢、氧三种元素，是细胞最主要的能量来源；脂质同样只含碳、氢、氧，但能量密度更高；蛋白质除碳、氢、氧外还含有氮，部分蛋白质还含硫；核酸则额外含有磷。从元素组成出发记忆四类分子，是考试中判断分子类别的第一步，例如题目给出”含氮元素”即可推断该分子为蛋白质或核酸。

Carbohydrates contain only carbon, hydrogen and oxygen and are the cell’s main energy source; lipids also contain only C, H and O but have a higher energy density; proteins contain nitrogen in addition to C, H and O, and some proteins also contain sulfur; nucleic acids additionally contain phosphorus. Starting from elemental composition is the first step in identifying molecular classes in exam questions – for example, if a question states that a molecule contains nitrogen, you can deduce that it is a protein or a nucleic acid.

2. 单糖与双糖：葡萄糖、果糖、蔗糖与乳糖的结构 | Monosaccharides and Disaccharides: Structure of Glucose, Fructose, Sucrose and Lactose

单糖是最简单的糖，不能再被水解为更小的糖分子。根据碳原子数目，单糖分为三碳糖、五碳糖和六碳糖，其中六碳糖（己糖）最为常见。葡萄糖、果糖和半乳糖都是己糖，分子式均为 C6H12O6，但它们的原子排列方式不同，因此互为同分异构体。葡萄糖以两种环状形式存在：α-葡萄糖和 β-葡萄糖，二者的区别在于第一位碳上的羟基方向，这一微小差异直接决定了后续多糖（淀粉与纤维素）的截然不同的结构。

Monosaccharides are the simplest sugars and cannot be hydrolysed into smaller sugar molecules. They are classified by their number of carbon atoms into trioses, pentoses and hexoses, of which the hexoses are the most common. Glucose, fructose and galactose are all hexoses with the molecular formula C6H12O6, but because their atoms are arranged differently they are isomers of one another. Glucose exists in two ring forms: alpha-glucose and beta-glucose, which differ in the orientation of the hydroxyl group on carbon 1. This tiny difference directly leads to the very different structures of the polysaccharides starch and cellulose.

两个单糖通过缩合反应（condensation reaction）连接，脱去一分子水，形成糖苷键（glycosidic bond），产物称为双糖。葡萄糖与葡萄糖缩合生成麦芽糖（maltose）；葡萄糖与果糖缩合生成蔗糖（sucrose）；葡萄糖与半乳糖缩合生成乳糖（lactose）。考试中常见的考点是：能够与 Benedict 试剂反应产生砖红色沉淀的糖称为还原糖，麦芽糖和乳糖都是还原糖，而蔗糖因为糖苷键连接了葡萄糖和果糖的两个还原端，属于非还原糖。

Two monosaccharides join through a condensation reaction, releasing one molecule of water and forming a glycosidic bond; the product is a disaccharide. Glucose + glucose gives maltose, glucose + fructose gives sucrose, and glucose + galactose gives lactose. A common exam point is: sugars that react with Benedict’s reagent to produce a brick-red precipitate are called reducing sugars. Maltose and lactose are reducing sugars, but sucrose is a non-reducing sugar because its glycosidic bond joins the reducing ends of both glucose and fructose.

3. 多糖结构比较：淀粉、糖原与纤维素 | Comparing Polysaccharides: Starch, Glycogen and Cellulose

多糖是由大量单糖通过糖苷键连接而成的聚合物。淀粉是植物储存能量的形式，由 α-葡萄糖构成，包含直链的直链淀粉（amylose）和带分支的支链淀粉（amylopectin）。直链淀粉呈螺旋状，结构紧凑且不溶于水，便于植物长期储存能量。糖原是动物和真菌储存能量的形式，也由 α-葡萄糖构成，但分支比支链淀粉更多、更短，使得糖原可以被迅速分解为葡萄糖，满足肌肉和肝脏快速释放能量的需求。

Polysaccharides are polymers formed from many monosaccharides joined by glycosidic bonds. Starch is the energy storage molecule of plants, made from alpha-glucose and consisting of unbranched amylose and branched amylopectin. Amylose coils into a helix, making it compact and insoluble in water, which suits long-term energy storage. Glycogen is the storage molecule of animals and fungi; it is also made of alpha-glucose but has many more, shorter branches than amylopectin, so it can be broken down quickly to release glucose for rapid energy supply in muscles and the liver.

纤维素则完全相反：它由 β-葡萄糖构成，每个 β-葡萄糖单元在连接时需要旋转 180 度，形成长的直链。相邻纤维素链之间通过大量氢键横向连接，聚合成微纤维（microfibrils），强度极高，因此纤维素是植物细胞壁的主要成分。三点对比是高频考题：淀粉和糖原由 α-葡萄糖构成、可被人体消化，而纤维素由 β-葡萄糖构成、人体缺乏相应酶而无法消化，但它提供了膳食纤维，促进肠道蠕动。

Cellulose is completely different: it is made of beta-glucose, and each beta-glucose unit must rotate 180 degrees when joining, producing long straight chains. Adjacent cellulose chains are cross-linked by numerous hydrogen bonds to form microfibrils of very high tensile strength, which is why cellulose is the main component of plant cell walls. A three-way comparison is a frequent exam question: starch and glycogen are made of alpha-glucose and can be digested by humans, while cellulose is made of beta-glucose and cannot be digested because humans lack the necessary enzyme; nevertheless it provides dietary fibre that promotes gut movement.

4. 食物检验实验：还原糖、非还原糖与淀粉的检测 | Food Tests: Detecting Reducing Sugars, Non-Reducing Sugars and Starch

生物分子实验是 A-Level 生物的必考内容。检验还原糖使用 Benedict 试剂：将待测液与 Benedict 试剂混合后水浴加热，若出现蓝色到绿色、黄色再到砖红色沉淀的颜色变化，说明存在还原糖，沉淀越多颜色越深，还能据此粗略比较还原糖含量。检验淀粉则使用碘液：滴加碘液后若变蓝黑色，说明存在淀粉，因为碘分子嵌入直链淀粉的螺旋结构中形成复合物。

Food tests are a compulsory part of A-Level Biology. Reducing sugars are detected with Benedict’s reagent: mix the sample with Benedict’s solution and heat in a water bath. A colour change from blue through green and yellow to a brick-red precipitate indicates a reducing sugar; the more precipitate, the deeper the colour, allowing rough comparison of sugar concentration. Starch is detected with iodine solution: a blue-black colour means starch is present, because iodine molecules slot into the helix of amylose to form a complex.

非还原糖（如蔗糖）的检验需要两步：先加入稀盐酸并加热，使蔗糖水解为葡萄糖和果糖，再用氢氧化钠中和酸，最后加入 Benedict 试剂并水浴加热。若此时出现砖红色沉淀，说明原来存在非还原糖。这一”水解-中和-检验”三步流程是实验题最爱考察的细节，尤其是”为什么必须先中和”这一步，答案是不能让酸与 Benedict 试剂反应或影响铜离子的还原。

Testing for a non-reducing sugar such as sucrose requires two extra steps: first add dilute hydrochloric acid and heat to hydrolyse sucrose into glucose and fructose, then neutralise the acid with sodium hydroxide, and finally add Benedict’s reagent and heat in a water bath. A brick-red precipitate at this stage shows that a non-reducing sugar was originally present. This three-step flow of hydrolyse – neutralise – test is a favourite detail in practical questions, especially “why must you neutralise first”: because the acid would otherwise react with Benedict’s reagent or interfere with the reduction of copper ions.

5. 甘油三酯与磷脂：脂质的结构和功能 | Triglycerides and Phospholipids: Structure and Functions of Lipids

脂质不溶于水，但溶于有机溶剂如乙醇。最重要的两类脂质是甘油三酯（triglycerides）和磷脂（phospholipids）。甘油三酯由一个甘油分子与三个脂肪酸分子通过酯键（ester bond）连接而成，形成过程同样是缩合反应，每个酯键形成时脱去一分子水。甘油三酯的主要功能是长期储能：相同质量下它释放的能量约为糖类的两倍，同时它不溶于水，不会像糖原那样改变细胞的渗透压，因此动物将多余能量以脂肪形式储存在脂肪细胞中。

Lipids are insoluble in water but soluble in organic solvents such as ethanol. The two most important classes are triglycerides and phospholipids. A triglyceride consists of one glycerol molecule joined to three fatty acid molecules by ester bonds, formed by condensation reactions in which one water molecule is released per ester bond. The main function of triglycerides is long-term energy storage: gram for gram they release about twice as much energy as carbohydrates, and because they are insoluble in water they do not affect the osmotic pressure of cells as glycogen would, which is why animals store surplus energy as fat in adipose cells.

磷脂的结构与甘油三酯相似，但第三个脂肪酸被一个含磷酸基团的头部取代。磷酸头部是亲水的（hydrophilic），两条脂肪酸尾部是疏水的（hydrophobic），这种”一头亲水、两头疏水”的两亲性（amphipathic）结构使磷脂在水环境中自动排列成双分子层：亲水头朝外接触水，疏水尾朝内相互靠拢。这一双分子层正是细胞膜的基本骨架，磷脂还参与形成肺表面活性物质，防止肺泡塌陷。

A phospholipid is similar to a triglyceride, except that the third fatty acid is replaced by a head group containing a phosphate group. The phosphate head is hydrophilic while the two fatty acid tails are hydrophobic, and this amphipathic structure – one hydrophilic head and two hydrophobic tails – makes phospholipids arrange themselves spontaneously into bilayers in water: heads face outward toward water and tails face inward away from it. This bilayer is the fundamental framework of the cell membrane, and phospholipids also form pulmonary surfactant, which prevents the alveoli from collapsing.

6. 饱和与不饱和脂肪酸：双键如何影响熔点和健康 | Saturated and Unsaturated Fatty Acids: How Double Bonds Affect Melting Point and Health

脂肪酸根据碳链中是否含有碳碳双键分为饱和与不饱和两类。饱和脂肪酸的碳链中所有碳原子都以单键相连，每个碳原子”饱和”地结合了最大数量的氢原子，碳链平直，分子之间可以紧密排列，分子间作用力强，因此熔点较高，在室温下通常呈固态，例如动物脂肪中的硬脂酸。

Fatty acids are classified as saturated or unsaturated according to whether their carbon chains contain carbon-carbon double bonds. In a saturated fatty acid every carbon atom is joined by single bonds and each carbon carries the maximum number of hydrogen atoms; the chains are straight and pack tightly together with strong intermolecular forces, so their melting points are higher and they are usually solid at room temperature, such as stearic acid in animal fats.

不饱和脂肪酸含有一个或多个碳碳双键，双键处碳链发生弯曲，形成”扭结”（kink），分子无法紧密排列，分子间作用力较弱，熔点因此降低，在室温下多为液态油，例如橄榄油和鱼油。含多个双键的称为多不饱和脂肪酸。健康方面，不饱和脂肪酸（尤其是顺式构型）有助于降低血液中的低密度脂蛋白，而人工氢化产生的反式脂肪酸会提高心血管疾病风险，这一联系是 OCR 考试中生物与健康结合题的常见素材。

An unsaturated fatty acid contains one or more double bonds, and at each double bond the chain bends to form a kink, so the molecules cannot pack closely, intermolecular forces are weaker, and the melting point is lower; these fatty acids are usually liquid oils at room temperature, such as olive oil and fish oil. Those with several double bonds are called polyunsaturated. For health, unsaturated fatty acids (especially in the cis configuration) help lower low-density lipoprotein in the blood, while trans fatty acids produced by artificial hydrogenation raise the risk of cardiovascular disease; this link is a common source of biology-and-health questions in OCR exams.

7. 氨基酸与肽键：蛋白质的单体如何连接 | Amino Acids and Peptide Bonds: How Protein Monomers Join

蛋白质由氨基酸构成，生物体内常见的氨基酸有 20 种。每个氨基酸分子都含有一个氨基（-NH2）、一个羧基（-COOH）、一个氢原子和一个可变的 R 基团，这四个部分都连接在同一个中心碳原子上。氨基酸之间的区别完全取决于 R 基团：R 基团可以是疏水性的、亲水性的、酸性的或碱性的，这些性质决定了氨基酸在蛋白质折叠时的行为。

Proteins are made of amino acids, and there are about 20 common types in living organisms. Every amino acid has an amino group (-NH2), a carboxyl group (-COOH), a hydrogen atom and a variable R group, all attached to the same central carbon atom. Amino acids differ only in their R groups: an R group can be hydrophobic, hydrophilic, acidic or basic, and these properties govern how the amino acid behaves during protein folding.

两个氨基酸通过缩合反应连接：一个氨基酸的羧基与另一个氨基酸的氨基反应，脱去一分子水，形成肽键（peptide bond）。两个氨基酸相连形成二肽，多个氨基酸相连形成多肽链。当多肽链较长或较复杂时便称为蛋白质。注意区分概念：蛋白质可以含有一条或多条多肽链，而多肽链只是氨基酸序列，尚未折叠成有功能的三维结构。

Two amino acids join by a condensation reaction: the carboxyl group of one reacts with the amino group of another, releasing a molecule of water and forming a peptide bond. Two amino acids linked together form a dipeptide, and many amino acids linked together form a polypeptide chain. Longer or more complex polypeptide chains are called proteins. Be careful with the distinction: a protein may contain one or more polypeptide chains, while a polypeptide is just the amino acid sequence and has not yet folded into a functional three-dimensional structure.

8. 蛋白质的四级结构：从氨基酸序列到三维构象 | Four Levels of Protein Structure: From Amino Acid Sequence to 3D Conformation

蛋白质的结构分为四个层次。一级结构（primary structure）是氨基酸在肽链中的排列顺序，由基因决定，任何一处氨基酸的改变都可能影响蛋白质功能，镰状细胞贫血正是血红蛋白中一个谷氨酸被缬氨酸替换所致。二级结构（secondary structure）是肽链通过氢键形成的局部折叠模式，主要是 α-螺旋和 β-折叠片，氢键存在于肽键的 N-H 与 C=O 之间。

Protein structure is described at four levels. The primary structure is the sequence of amino acids in the chain, determined by genes; changing even one amino acid can affect protein function, and sickle cell anaemia is caused by a single glutamic acid being replaced by valine in haemoglobin. The secondary structure is the local folding pattern formed by hydrogen bonds, mainly the alpha-helix and the beta-pleated sheet, with hydrogen bonds between the N-H and C=O groups of peptide bonds.

三级结构（tertiary structure）是整条多肽链在二级结构基础上进一步折叠形成的三维形状，由多种键共同维持：离子键（酸性与碱性 R 基之间）、氢键、二硫键（两个半胱氨酸的硫原子之间，是最强的键）以及疏水相互作用（疏水 R 基被包裹在分子内部）。四级结构（quaternary structure）则指两条或多条多肽链（亚基）组装成完整功能蛋白，例如血红蛋白由四条链组成，胶原蛋白由三条链拧成绳索状结构。

The tertiary structure is the overall three-dimensional shape formed when the whole chain folds on top of its secondary structure, held together by several types of bond: ionic bonds between acidic and basic R groups, hydrogen bonds, disulfide bridges (between the sulfur atoms of two cysteines, the strongest bonds), and hydrophobic interactions in which hydrophobic R groups are buried inside the molecule. The quaternary structure is the assembly of two or more polypeptide chains (subunits) into a complete functional protein: haemoglobin consists of four chains, and collagen is a rope-like structure of three chains twisted together.

9. 酶的作用机制：诱导契合模型与影响因素 | Enzyme Action: The Induced-Fit Model and Factors That Affect Rate

酶是生物催化剂，绝大多数酶是蛋白质。酶的活性位点（active site）形状与底物互补，底物与活性位点结合形成酶-底物复合物。现代”诱导契合”模型（induced fit model）认为，活性位点并非固定的锁孔，而是在底物结合时发生轻微形变，与底物更紧密地贴合，从而降低反应的活化能，使反应速率大幅提升。

Enzymes are biological catalysts, and the great majority are proteins. The active site of an enzyme is complementary in shape to its substrate, and the substrate binds to it to form an enzyme-substrate complex. The modern induced-fit model holds that the active site is not a rigid lock and key; instead it changes shape slightly when the substrate binds, moulding itself more closely around the substrate and lowering the activation energy of the reaction so that the rate increases dramatically.

温度和 pH 是影响酶活性的两大因素。温度升高时分子运动加快，反应速率上升，但超过最适温度后，高温破坏维持酶三级结构的氢键等化学键，酶的活性位点形状改变，发生不可逆的变性（denaturation），反应速率骤降。pH 同理：偏离最适 pH 会改变 R 基团的离子状态，破坏离子键和氢键，导致变性。考题常要求解释”为什么酶在高温下失活后冷却也无法恢复”，因为变性是永久性的结构破坏。

Temperature and pH are the two major factors affecting enzyme activity. As temperature rises, molecules move faster and the rate increases, but above the optimum temperature the heat breaks the hydrogen bonds and other bonds that maintain the enzyme’s tertiary structure; the active site changes shape and the enzyme undergoes irreversible denaturation, so the rate collapses. The same logic applies to pH: moving away from the optimum pH changes the ionisation state of R groups and disrupts ionic and hydrogen bonds, causing denaturation. A classic exam question asks why an enzyme denatured by high temperature cannot recover when cooled – because denaturation is a permanent destruction of structure.

10. 蛋白质的其他功能：抗体、转运与结构蛋白 | Other Protein Functions: Antibodies, Transport Proteins and Structural Proteins

除酶之外，蛋白质在生物体内承担着多种关键功能。抗体（antibodies）由 B 淋巴细胞产生，是与抗原特异性结合的免疫球蛋白，其 Y 形结构的两个臂部各有抗原结合位点，能够中和病原体或标记它们以供吞噬细胞清除。血红蛋白（haemoglobin）是转运蛋白的典型代表：四个亚基各含一个血红素基团，能够与氧可逆结合，在肺部高氧分压下结合氧，在组织低氧分压下释放氧。

Besides enzymes, proteins carry out many other vital functions. Antibodies are immunoglobulins produced by B lymphocytes that bind specifically to antigens; the two arms of their Y-shaped structure each carry an antigen-binding site, neutralising pathogens or marking them for destruction by phagocytes. Haemoglobin is a classic transport protein: each of its four subunits contains a haem group and binds oxygen reversibly, picking up oxygen where the partial pressure is high in the lungs and releasing it where the partial pressure is low in the tissues.

结构蛋白赋予组织强度和韧性：胶原蛋白（collagen）是结缔组织、骨骼和肌腱的主要成分，三条多肽链以甘氨酸为每第三个氨基酸缠绕成三股螺旋，再横向交联成纤维，抗拉强度极高；角蛋白（keratin）构成毛发、指甲和皮肤外层。此外，一些激素如胰岛素和胰高血糖素也是蛋白质，通过调节血糖浓度维持内环境稳定。功能多样性的根本原因在于蛋白质独特的氨基酸序列决定了独特的三维构象。

Structural proteins give tissues strength and elasticity: collagen is the main component of connective tissue, bone and tendons – three polypeptide chains with glycine as every third amino acid wind into a triple helix and cross-link into fibres of enormous tensile strength; keratin makes up hair, nails and the outer layer of skin. Some hormones such as insulin and glucagon are also proteins, maintaining homeostasis by regulating blood glucose concentration. The fundamental reason for this functional diversity is that each protein’s unique amino acid sequence determines its unique three-dimensional conformation.

11. 水的独特性质与生命意义 | The Unique Properties of Water and Their Biological Significance

水是含量最丰富的生物分子，约占细胞质量的 70% 以上。水分子是极性分子：氧原子电负性较强，吸引共用电子对，使氧端略带负电、氢端略带正电，相邻水分子之间形成氢键。单个氢键很弱，但大量氢键合在一起，赋予水一系列独特的性质。

Water is the most abundant biological molecule, making up over 70% of cell mass. The water molecule is polar: the oxygen atom is more electronegative and pulls the shared electrons toward itself, leaving the oxygen end slightly negative and the hydrogen ends slightly positive, so neighbouring molecules form hydrogen bonds. A single hydrogen bond is weak, but very large numbers of them together give water a set of unique properties.

这些性质包括：第一，水是极好的溶剂，离子化合物和极性分子（如葡萄糖、氨基酸）都能溶于水，使水成为代谢反应发生的介质；第二，水的比热容高，能吸收大量热量而自身温度变化小，帮助生物体维持稳定体温；第三，水的汽化热高，出汗散热是哺乳动物有效的降温机制；第四，水在 4 摄氏度时密度最大，冰浮在水面，隔绝下方水体与冷空气，使水生生物得以存活；第五，水几乎不可压缩，为植物细胞提供膨压支持。考试中经常要求”根据水的结构解释某性质”，答题时必须从氢键和极性入手。

These properties include: first, water is an excellent solvent – ionic compounds and polar molecules such as glucose and amino acids dissolve in it, making it the medium in which metabolic reactions take place; second, water has a high specific heat capacity, absorbing large amounts of heat with only a small temperature change and helping organisms maintain a stable body temperature; third, water has a high latent heat of vaporisation, so sweating is an effective cooling mechanism in mammals; fourth, water is densest at 4 degrees Celsius, so ice floats and insulates the water below, allowing aquatic life to survive; fifth, water is almost incompressible and provides turgor support to plant cells. Exams often ask you to “explain a property of water in terms of its structure”, and the answer must start from hydrogen bonding and polarity.

12. 蛋白质检验与食物能量：Biuret 试验与能量计算 | Testing for Proteins and Food Energy: The Biuret Test and Energy Calculation

检验蛋白质使用 Biuret 试验：先向样品中加入氢氧化钠溶液，再加入少量稀硫酸铜溶液，若溶液由蓝色变为紫色，说明存在蛋白质。原理是铜离子在碱性条件下与肽键形成紫色络合物，因此凡是含两个及以上肽键的分子（即二肽以上）都能给出阳性结果。注意顺序不能颠倒，且硫酸铜必须少量，过量会与碱反应生成蓝色沉淀干扰判断。

Proteins are detected with the Biuret test: add sodium hydroxide solution to the sample, then a little dilute copper(II) sulfate solution; a purple colour means protein is present. The principle is that copper ions form a purple complex with peptide bonds in alkaline conditions, so any molecule with two or more peptide bonds (a dipeptide or larger) gives a positive result. The order must not be reversed, and the copper sulfate must be added in small amounts – excess copper sulfate reacts with the alkali to form a blue precipitate that masks the result.

食物能量方面，可以用燃烧法测定：将食物样品干燥后完全燃烧，测量释放的热量使已知质量的水升高的温度，利用公式 能量(kJ) = 水的质量(g) x 4.2 x 温度变化(摄氏度) / 1000 计算。由于糖类和蛋白质每克约释放 17 kJ 能量，而脂质每克约释放 39 kJ，燃烧实验也常用来验证脂质能量密度更高。误差来源包括热量散失到周围环境、燃烧不充分等，这些误差分析同样是实验题的标准考点。

For food energy, a combustion method can be used: dry the food sample, burn it completely and measure how much the temperature of a known mass of water rises, then calculate using energy (kJ) = mass of water (g) x 4.2 x temperature rise (degrees Celsius) / 1000. Because carbohydrates and proteins release about 17 kJ per gram while lipids release about 39 kJ per gram, combustion experiments are also used to demonstrate that lipids have a higher energy density. Sources of error include heat lost to the surroundings and incomplete combustion, and these error analyses are standard points in practical questions.

Summary | 总结

本文系统梳理了 OCR A-Level 生物 A 模块 2.2 “生物分子”的核心内容：四大类生物分子的元素组成、单糖与双糖通过缩合反应形成糖苷键、多糖结构与功能的对应关系、Benedict 试验和碘液试验的检测原理、甘油三酯与磷脂的两亲性结构、饱和与不饱和脂肪酸对熔点和健康的影响、氨基酸通过肽键连接形成蛋白质的四个结构层次、酶的诱导契合模型与变性机制、蛋白质的多种功能、水的独特性质以及 Biuret 试验与能量计算。

This article has systematically reviewed the core content of Module 2.2 “Biological molecules” of OCR A-Level Biology A: the elemental composition of the four classes of biological molecules, glycosidic bond formation between monosaccharides and disaccharides by condensation, the structure-function relationships of polysaccharides, the principles of the Benedict’s and iodine tests, the amphipathic structures of triglycerides and phospholipids, the effects of saturated and unsaturated fatty acids on melting point and health, the four levels of protein structure built from amino acids joined by peptide bonds, the induced-fit model and denaturation of enzymes, the many functions of proteins, the unique properties of water, and the Biuret test with energy calculation.

掌握这些知识的关键是建立”结构决定功能”的思维框架：无论是糖类螺旋的紧凑性、纤维素氢键的强度、磷脂双分子层的形成，还是蛋白质四级结构的功能意义，都可以追溯到分子层面的结构差异。建议同学们在复习时亲手画出葡萄糖的两种环状结构、三种多糖的分支示意图以及氨基酸缩合反应的方程式，并用表格对比四类分子的元素组成、单体和检验方法，这样在考试中遇到实验设计题和结构分析题时就能快速定位考点。

The key to mastering this material is the “structure determines function” framework: whether it is the compactness of starch helices, the strength of cellulose hydrogen bonds, the formation of phospholipid bilayers, or the functional significance of quaternary protein structure, everything can be traced back to structural differences at the molecular level. When revising, draw the two ring forms of glucose, the branching diagrams of the three polysaccharides and the equation of amino acid condensation by hand, and use a table to compare the elemental composition, monomers and test methods of the four molecular classes; this way you can quickly locate the relevant points when you meet experimental design questions and structural analysis questions in the exam.

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