📚 Biochemical Polymers: Proteins, Polysaccharides and Nucleic Acids | 生化聚合物：蛋白质、多糖与核酸
Biochemical polymers are naturally occurring macromolecules formed by condensation polymerisation of small monomers. In A-Level Chemistry, three families dominate: proteins from amino acids, polysaccharides from monosaccharides, and nucleic acids from nucleotides. Their structure determines biological function, and their hydrolysis reactions return the original monomers.
生化聚合物是由小分子单体通过缩聚反应形成的天然大分子。在 A-Level 化学中，主要有三类：蛋白质（由氨基酸组成）、多糖（由单糖组成）和核酸（由核苷酸组成）。它们的结构决定生物功能，水解反应则可重新得到原始单体。
1. What Are Biochemical Polymers? | 什么是生化聚合物？
A polymer is a long molecule built from many repeating subunits called monomers. Biochemical polymers, or biopolymers, are synthesised in living organisms by condensation reactions. Each time two monomers join, a small molecule—usually water—is eliminated. The reverse reaction, hydrolysis, breaks the polymer apart by adding water.
聚合物是由大量重复单元（单体）组成的长分子。生化聚合物（生物高分子）是在生物体内通过缩合反应合成的。每两个单体连接时，都会脱去一个小分子——通常是水。相反的水解反应则通过加水将聚合物分解。
In A-Level assessments, you must be able to identify the repeating unit, name the linkage, and predict hydrolysis products for each biopolymer. You should also explain how structure relates to function in biological systems.
在 A-Level 考试中，必须能够识别重复单元、命名连接键并预测每种生物高分子的水解产物。还应能解释结构如何与生物系统中的功能相关。
2. Condensation Polymerisation in Biochemistry | 生物化学中的缩聚反应
Condensation polymerisation joins monomers that contain at least two functional groups. For example, amino acids contain both -NH₂ and -COOH groups, so they can link into long chains. Each new amide or peptide bond forms with the loss of one water molecule.
缩聚反应将至少含有两个官能团的单体连接起来。例如，氨基酸同时含有 -NH₂ 和 -COOH 基团，因此可以连接成长链。每形成一个酰胺键或肽键，就会脱去一个水分子。
This contrasts with addition polymerisation of alkenes, which involves only C=C bonds and no small-molecule loss. Biopolymers therefore belong to the condensation polymer family.
这与烯烃的加聚反应不同，加聚只涉及 C=C 键且不脱去小分子。因此，生物高分子属于缩聚类聚合物。
| Biopolymer | Monomer | Linkage | Small molecule lost |
|---|---|---|---|
| Protein | Amino acid | Peptide bond | H₂O |
| Starch / glycogen / cellulose | Monosaccharide | Glycosidic bond | H₂O |
| DNA / RNA | Nucleotide | Phosphodiester bond | H₂O |
The table summarises the three main classes of biochemical polymers that appear in Cambridge A-Level Chemistry.
上表总结了剑桥 A-Level 化学中出现的三大类生化聚合物。
3. Amino Acid Structure and Zwitterions | 氨基酸结构与两性离子
α-amino acids have the general formula H₂N-CHR-COOH, where R is a side chain. The central carbon is bonded to an amine group, a carboxylic acid group, a hydrogen atom, and a variable R group. In aqueous solution, an internal acid-base reaction creates a zwitterion: H₃N⁺-CHR-COO⁻.
α-氨基酸的一般式为 H₂N-CHR-COOH，其中 R 是侧链。中心碳原子连接一个氨基、一个羧基、一个氢原子和一个可变的 R 基。在水溶液中，内部酸碱反应生成两性离子：H₃N⁺-CHR-COO⁻。
The R group can be non-polar, polar, acidic, or basic. This explains why different amino acids have different isoelectric points and contribute to protein folding. The isoelectric point is the pH at which the zwitterion has no overall charge.
R 基可以是非极性、极性、酸性或碱性。这解释了为什么不同氨基酸具有不同的等电点，并对蛋白质折叠产生影响。等电点是指两性离子总电荷为零时的 pH。
Except glycine, where R = H, α-amino acids are chiral because the central carbon has four different groups. This gives two optical isomers. Proteins in living organisms contain mainly the L-form.
除甘氨酸 (R = H) 外，α-氨基酸具有手性，因为中心碳原子连接四个不同的基团。这产生两种光学异构体。生物体内的蛋白质主要含有 L 型氨基酸。
4. Peptide Bond Formation and Polypeptides | 肽键形成与多肽
When two amino acids react, the -COOH group of one amino acid and the -NH₂ group of the other form an amide linkage called a peptide bond. The reaction is a condensation reaction and releases one water molecule:
当两个氨基酸反应时，一个氨基酸的 -COOH 基团与另一个氨基酸的 -NH₂ 基团形成酰胺键，称为肽键。该反应为缩合反应，释放一分子水：
NH₂-CHR-COOH + NH₂-CHR’-COOH → NH₂-CHR-CO-NH-CHR’-COOH + H₂O
A dipeptide contains one peptide bond, a tripeptide contains two, and a polypeptide contains many. The peptide bond has partial double-bond character, which restricts rotation and influences protein shape.
二肽含有一个肽键，三肽含有两个，多肽含有许多个。肽键具有部分双键性质，这限制了旋转并影响蛋白质形状。
In a polypeptide chain, the repeating unit is usually written as -NH-CHR-CO- or -CO-NH-CHR-. Identifying this repeating unit from a given section of protein is a common examination skill.
在多肽链中，重复单元通常写作 -NH-CHR-CO- 或 -CO-NH-CHR-。从给定蛋白质片段中识别该重复单元是常见的考试技能。
5. Protein Structure: Primary to Quaternary | 蛋白质结构：一级到四级
The sequence of amino acids in a polypeptide is the primary structure. Secondary structure arises from hydrogen bonding between backbone N-H and C=O groups, giving α-helices and β-pleated sheets. Tertiary structure is the overall three-dimensional folding held by hydrogen bonds, ionic attractions, disulfide bridges, and hydrophobic interactions. Quaternary structure applies only to proteins with multiple polypeptide chains.
多肽中氨基酸的排列顺序是一级结构。二级结构由主链 N-H 和 C=O 基团之间的氢键形成，产生 α-螺旋和 β-折叠片。三级结构是由氢键、离子吸引、二硫桥和疏水相互作用维持的整体三维折叠。四级结构仅适用于由多条多肽链组成的蛋白质。
Disulfide bridges form by oxidation of two cysteine side chains:
二硫桥由两个半胱氨酸侧链氧化形成：
R-SH + HS-R’ → R-S-S-R’ + 2H⁺ + 2e⁻
In A-Level questions, you may be asked to classify these interactions or explain why denaturation destroys biological activity. Denaturation disrupts secondary and tertiary structure, breaking hydrogen bonds and hydrophobic interactions, but it does not break the primary sequence.
A-Level 题目可能要求对这些相互作用进行分类，或解释为什么变性会破坏生物活性。变性会破坏二级和三级结构，打断氢键和疏水相互作用，但不会破坏一级序列。
6. Polysaccharides from Monosaccharides | 由单糖形成的多糖
Monosaccharides such as glucose are monomers with several -OH groups. When two monosaccharides join by a condensation reaction, they form a disaccharide and a glycosidic bond. For example, two α-glucose molecules form maltose:
单糖（如葡萄糖）是含有多个 -OH 基团的单体。两个单糖通过缩合反应连接时，形成二糖和糖苷键。例如，两个 α-葡萄糖分子生成麦芽糖：
C₆H₁₂O₆ + C₆H₁₂O₆ → C₁₂H₂₂O₁₁ + H₂O
In α-glucose, the -OH on carbon 1 points below the plane of the ring; in β-glucose, it points above. This small difference leads to very different polymers: starch and glycogen are built from α-glucose, while cellulose is built from β-glucose.
在 α-葡萄糖中，1 号碳上的 -OH 指向环平面下方；在 β-葡萄糖中则指向上方。这一微小差异导致非常不同的聚合物：淀粉和糖原由 α-葡萄糖构成，而纤维素由 β-葡萄糖构成。
The glycosidic bond is named according to the carbon atoms it joins. For example, maltose contains an α-1,4-glycosidic bond because carbon 1 of one glucose joins carbon 4 of the next.
糖苷键根据它所连接的碳原子命名。例如，麦芽糖含有 α-1,4-糖苷键，因为一个葡萄糖的 1 号碳与下一个葡萄糖的 4 号碳相连。
7. Starch, Glycogen and Cellulose | 淀粉、糖原和纤维素
Starch is a storage polysaccharide in plants. It contains amylose, a mostly unbranched chain with α-1,4-glycosidic bonds, and amylopectin, which is branched through α-1,6-glycosidic bonds. Glycogen is the animal storage equivalent, with shorter but more frequent branches.
淀粉是植物中的储存多糖。它包含直链淀粉（主要由 α-1,4-糖苷键组成的近似无支链链）和支链淀粉（通过 α-1,6-糖苷键形成分支）。糖原是动物体内对应的储存多糖，分支更短但更频繁。
Cellulose is a structural polysaccharide in plant cell walls. It consists of β-glucose units linked by β-1,4-glycosidic bonds. Every other glucose unit is inverted, allowing straight chains to pack closely and form strong fibres through hydrogen bonding.
纤维素是植物细胞壁中的结构多糖。它由通过 β-1,4-糖苷键连接的 β-葡萄糖单元组成。每隔一个葡萄糖单元反转，使直链能够紧密排列，并通过氢键形成强韧的纤维。
| Polysaccharide | Monomer | Main linkage | Branching | Biological role |
|---|---|---|---|---|
| Starch | α-glucose | α-1,4 and α-1,6 | Some | Energy storage in plants |
| Glycogen | α-glucose | α-1,4 and α-1,6 | Extensive | Energy storage in animals |
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导