Carbohydrates: Structure and Function | 糖类：结构与功能

📚 Carbohydrates: Structure and Function | 糖类：结构与功能

Carbohydrates are one of the four major classes of biological macromolecules, alongside proteins, lipids, and nucleic acids. They serve as primary energy sources, structural components, and molecular recognition markers in living organisms. This article provides a systematic review of carbohydrate structure and function, tailored specifically to the CIE A-Level Biology syllabus.

糖类是四大类生物大分子之一，与蛋白质、脂质和核酸并列。它们是生物体的主要能量来源、结构组分和分子识别标志。本文系统梳理了糖类的结构与功能，专门针对CIE A-Level生物学考纲进行编写。

1. Monosaccharides: The Building Blocks | 单糖：基本构件

Monosaccharides are the simplest carbohydrates, typically containing 3 to 7 carbon atoms. Their general formula is (CH₂O)ₙ, where n ranges from 3 to 7. Glucose (C₆H₁₂O₆), the most important monosaccharide, has six carbon atoms and exists predominantly in a ring form in aqueous solution.

单糖是最简单的糖类，通常含有3至7个碳原子，通式为(CH₂O)ₙ，其中n为3至7。葡萄糖(C₆H₁₂O₆)是最重要的单糖，含有六个碳原子，在水溶液中主要以环状结构存在。

Monosaccharides are classified according to the number of carbon atoms: trioses (3C), tetroses (4C), pentoses (5C), and hexoses (6C). For A-Level purposes, the most significant are the hexoses (glucose, fructose, galactose) and pentoses (ribose and deoxyribose, which are components of nucleotides).

单糖按碳原子数目分类：丙糖(3C)、丁糖(4C)、戊糖(5C)和己糖(6C)。在A-Level考试中，最重要的是己糖（葡萄糖、果糖、半乳糖）和戊糖（核糖和脱氧核糖，它们是核苷酸的组成成分）。

Glucose ring structure (α-form): C₆H₁₂O₆ — a hexose monosaccharide
葡萄糖环状结构（α型）：C₆H₁₂O₆ —— 一种己糖单糖

Glucose is highly soluble in water due to the many hydroxyl (−OH) groups that form hydrogen bonds with water molecules. This solubility is essential for its transport in blood and its role in cellular respiration.

葡萄糖因含有多个羟基(−OH)能与水分子形成氢键而高度溶于水。这种溶解性对其在血液中的运输以及在细胞呼吸中的作用至关重要。

2. α and β Isomers of Glucose | 葡萄糖的α和β异构体

Glucose exists in two isomeric ring forms: α-glucose and β-glucose. These isomers differ only in the position of the hydroxyl group attached to carbon atom 1 (C1). In α-glucose, the −OH group on C1 is below the plane of the ring (in the standard Haworth projection); in β-glucose, it is above the plane.

葡萄糖以两种环状异构体形式存在：α-葡萄糖和β-葡萄糖。这两种异构体仅在第一碳原子(C1)上羟基的位置不同。在α-葡萄糖中，C1上的−OH位于环平面下方（标准Haworth投影中）；在β-葡萄糖中，它位于环平面上方。

This seemingly minor structural difference has profound functional consequences. α-glucose units form starch and glycogen, which serve as energy storage molecules, while β-glucose units form cellulose, a structural polysaccharide with high tensile strength. The difference in glycosidic bond orientation (α-1,4 vs β-1,4) dictates the three-dimensional folding and thus the physical properties of the resulting polymer.

这一看似微小的结构差异具有深远的功能影响。α-葡萄糖单位形成淀粉和糖原，作为能量储存分子；β-葡萄糖单位形成纤维素，这是一种具有高抗拉强度的结构多糖。糖苷键取向（α-1,4与β-1,4）的差异决定了聚合物的三维折叠方式，从而决定了其物理性质。

3. Disaccharides: Two Monosaccharides Joined | 二糖：两个单糖的连接

Disaccharides are formed when two monosaccharides undergo a condensation reaction, eliminating a water molecule and forming a glycosidic bond. This reaction is catalyzed by specific enzymes and requires energy input.

二糖由两个单糖通过缩合反应形成，反应中脱去一个水分子并形成糖苷键。该反应由特定酶催化，需要能量输入。

Three disaccharides are commonly tested in CIE A-Level Biology:

CIE A-Level生物学考试中常考三种二糖：

Maltose: α-glucose + α-glucose, linked by an α-1,4 glycosidic bond. Produced during starch digestion.
麦芽糖：α-葡萄糖 + α-葡萄糖，通过α-1,4糖苷键连接。在淀粉消化过程中产生。
Sucrose: α-glucose + fructose, linked by an α-1,2 glycosidic bond. Common table sugar, transported in plants.
蔗糖：α-葡萄糖 + 果糖，通过α-1,2糖苷键连接。日常食用糖，在植物中运输。
Lactose: β-galactose + α-glucose, linked by a β-1,4 glycosidic bond. Found in mammalian milk.
乳糖：β-半乳糖 + α-葡萄糖，通过β-1,4糖苷键连接。存在于哺乳动物乳汁中。

Condensation: glucose + glucose → maltose + H₂O
缩合反应：葡萄糖 + 葡萄糖 → 麦芽糖 + 水

Disaccharides can be hydrolysed back into their constituent monosaccharides by the addition of water, a reaction catalysed by enzymes. For example, maltase hydrolyses maltose into two glucose molecules in the small intestine.

二糖可以通过加水水解回其组成单糖，该反应由酶催化。例如，麦芽糖酶在小肠中将麦芽糖水解为两个葡萄糖分子。

4. Reducing and Non-Reducing Sugars | 还原糖与非还原糖

Reducing sugars are carbohydrates that can donate electrons and reduce other molecules. All monosaccharides and some disaccharides (such as maltose and lactose) are reducing sugars because they have a free aldehyde or ketone group that can be oxidised. Sucrose is a non-reducing sugar because both its anomeric carbon atoms are involved in the glycosidic bond, leaving no free aldehyde or ketone group.

还原糖是能够提供电子并还原其他分子的糖类。所有单糖和一些二糖（如麦芽糖和乳糖）都是还原糖，因为它们具有可被氧化的游离醛基或酮基。蔗糖是非还原糖，因为其两个异头碳都参与了糖苷键的形成，没有游离的醛基或酮基可用。

The Benedict’s test is used to detect reducing sugars: when heated with Benedict’s reagent, a reducing sugar causes a colour change from blue to green, yellow, orange, and finally brick-red precipitate. For non-reducing sugars, the sample must first be hydrolysed with dilute hydrochloric acid and then neutralised with sodium hydrogencarbonate before testing.

本尼迪特试验用于检测还原糖：与班氏试剂加热时，还原糖使颜色由蓝色变为绿色、黄色、橙色，最终形成砖红色沉淀。对于非还原糖，必须先使用稀盐酸水解样品，然后用碳酸氢钠中和，再进行测试。

5. Glycosidic Bonds in Detail | 糖苷键详解

A glycosidic bond is a covalent bond formed between the hydroxyl group of one monosaccharide and the hydroxyl group of another, with the elimination of water. The type of bond (α or β) depends on the configuration of the carbon involved in the bond.

糖苷键是一个单糖的羟基与另一个单糖的羟基之间形成的共价键，同时脱去一分子水。键的类型（α或β）取决于参与成键的碳的构型。

For α-glucose polymers, the glycosidic bond is formed between C1 of one glucose and C4 of the next, creating an α-1,4 bond. In β-glucose polymers, the bond is β-1,4. The key structural difference is that in β-1,4 linkages, every alternate glucose unit is rotated 180°, resulting in a straight, unbranched chain. This allows cellulose molecules to lie parallel and form hydrogen bonds between adjacent chains, creating microfibrils of exceptional strength.

在α-葡萄糖聚合物中，糖苷键形成于一个葡萄糖的C1与下一个葡萄糖的C4之间，形成α-1,4键。在β-葡萄糖聚合物中，键为β-1,4。关键的结构差异在于：β-1,4键中，每隔一个葡萄糖单元旋转180°，形成直链、无分支的链。这使得纤维素分子能够平行排列，并在相邻链之间形成氢键，从而产生具有极强张力的微纤维。

α-1,4 bond: C1–O–C4 (same orientation)
α-1,4键：C1–O–C4（同一方向）
β-1,4 bond: C1–O–C4 (alternate units rotated 180°)
β-1,4键：C1–O–C4（交替单元旋转180°）

6. Starch: Amylose and Amylopectin | 淀粉：直链淀粉和支链淀粉

Starch is the primary energy storage polysaccharide in plants, found in chloroplasts and amyloplasts. It consists of two components: amylose (10–30%) and amylopectin (70–90%).

淀粉是植物中主要的能量储存多糖，存在于叶绿体和淀粉体中。它由两种成分组成：直链淀粉（占10–30%）和支链淀粉（占70–90%）。

Amylose is a long, unbranched chain of α-glucose units joined by α-1,4 glycosidic bonds. The chain coils into a helical structure, making it relatively compact. Amylose is insoluble in cold water and gives a blue-black colour with iodine solution.

直链淀粉是由α-葡萄糖单位通过α-1,4糖苷键连接而成的长而无分支的链。该链卷曲成螺旋结构，因而相对紧密。直链淀粉不溶于冷水，遇碘液呈蓝黑色。

Amylopectin is a branched polymer: it has α-1,4 glycosidic bonds in the straight chains, with α-1,6 glycosidic bonds at branch points occurring every 24–30 glucose units. The branching creates many terminal glucose molecules, providing multiple sites for rapid enzyme action during digestion.

支链淀粉是一种支链聚合物：直链部分通过α-1,4糖苷键连接，分支点处为α-1,6糖苷键，每24–30个葡萄糖单元出现一个分支。这种分支产生许多末端葡萄糖分子，为消化过程中酶的快速作用提供了多个作用位点。

Starch is an ideal storage molecule because it is insoluble in water (so it does not affect cell osmotic pressure), compact, and easily hydrolysed to glucose when energy is needed.

淀粉是理想的储存分子，因为它不溶于水（因此不影响细胞渗透压）、结构紧凑、在需要能量时容易被水解为葡萄糖。

7. Glycogen: The Animal Storage Polysaccharide | 糖原：动物储存多糖

Glycogen is the main storage polysaccharide in animals, found primarily in the liver and skeletal muscle. Its structure is similar to amylopectin but more extensively branched: α-1,4 bonds in chains, with α-1,6 branch points occurring every 8–12 glucose units.

糖原是动物体内主要的储存多糖，主要存在于肝脏和骨骼肌中。其结构与支链淀粉相似，但分支更为密集：链内为α-1,4键，分支点α-1,6键每8–12个葡萄糖单元出现一次。

The high degree of branching in glycogen serves a functional purpose: it exposes more non-reducing ends (terminal glucose molecules) to the action of glycogen phosphorylase, enabling rapid release of glucose-1-phosphate during times of high energy demand. This is analogous to having many “exits” from which glucose can be quickly mobilised.

糖原的高度分支具有功能意义：它暴露更多的非还原端（末端葡萄糖分子）供糖原磷酸化酶作用，从而在能量需求高峰期能够快速释放1-磷酸葡萄糖。这类似于拥有许多”出口”，可以从多个位置快速动员葡萄糖。

Glycogen is more compact than starch, preventing it from solubilising in the cytoplasm and affecting osmotic balance. Its presence in liver cells maintains blood glucose concentration; in muscle cells, it provides a rapid source of ATP for contraction.

糖原比淀粉更紧凑，避免了在细胞质中溶解而影响渗透平衡。肝脏细胞中的糖原维持血糖浓度；肌细胞中的糖原则为肌肉收缩提供快速的ATP来源。

8. Cellulose: Structural Polysaccharide | 纤维素：结构多糖

Cellulose is the most abundant organic polymer on Earth, forming the main component of plant cell walls. Unlike starch and glycogen, cellulose is composed of β-glucose units linked by β-1,4 glycosidic bonds.

纤维素是地球上最丰富的有机聚合物，是植物细胞壁的主要组成成分。与淀粉和糖原不同，纤维素由β-葡萄糖单位通过β-1,4糖苷键连接而成。

Because the −OH group on C1 is above the ring in β-glucose, the glycosidic bond requires each successive glucose unit to rotate 180°. This produces a straight, unbranched chain rather than a coiled helix. Between 60 and 100 cellulose chains lie parallel to each other and are cross-linked by hydrogen bonds to form a microfibril, which provides tensile strength to withstand turgor pressure.

由于β-葡萄糖C1上的−OH位于环上方，糖苷键使每个连续的葡萄糖单元旋转180°，从而形成直链而非螺旋。60至100条纤维素链平行排列，通过氢键交叉连接形成微纤维，为抵御膨胀压力提供抗张强度。

Cellulose has several properties that make it suitable for structural support: it is chemically inert, insoluble, has very high tensile strength, and is resistant to digestion by most organisms. Ruminants and termites can digest cellulose only because symbiotic microorganisms in their guts produce cellulase.

纤维素具有多种适合提供结构支撑的理化性质：化学惰性、不溶性、极高的抗张强度，以及抵抗大多数生物的消化。反刍动物和白蚁能够消化纤维素，仅仅是因为其肠道内的共生微生物产生纤维素酶。

9. Comparison of Polysaccharides | 多糖结构-功能比较

The following table summarises the structural features and functions of the three major polysaccharides:

下表总结了三种主要多糖的结构特征与功能：

Property 性质	Starch 淀粉	Glycogen 糖原	Cellulose 纤维素
Monomer 单体	α-glucose	α-glucose	β-glucose
Glycosidic bond 糖苷键	α-1,4 and α-1,6	α-1,4 and α-1,6 (more branches)	β-1,4
Structure 结构	Helical (amylose) / branched (amylopectin)	Highly branched, compact	Straight, unbranched chains
Function 功能	Energy storage in plants	Energy storage in animals	Structural support in plant cell walls
Solubility 溶解性	Insoluble	Insoluble	Insoluble

10. Other Functions of Carbohydrates | 糖类的其他功能

Beyond energy storage and structural support, carbohydrates perform additional functions in biological systems:

除能量储存和结构支撑外，糖类在生物系统中还执行其他功能：

Cell recognition: Glycoproteins and glycolipids on cell surface membranes act as receptors and antigens. The specific carbohydrate sequences serve as recognition markers for cell-cell communication and immune responses. Blood group antigens (A, B, H) are determined by specific sugar sequences on red blood cell membranes.
细胞识别：细胞表面膜上的糖蛋白和糖脂作为受体和抗原。特异的糖序列充当细胞间通讯和免疫应答的识别标志。血型抗原（A、B、H）由红细胞膜上特定的糖序列决定。
Metabolic intermediates: Ribose (C₅H₁₀O₅) and deoxyribose (C₅H₁₀O₄) are components of RNA and DNA, respectively. They form the backbone of nucleic acids through phosphodiester bonds.
代谢中间产物：核糖(C₅H₁₀O₅)和脱氧核糖(C₅H₁₀O₄)分别是RNA和DNA的组分。它们通过磷酸二酯键形成核酸的骨架。
Lubrication and protection: Hyaluronic acid (a glycosaminoglycan) lubricates joints; mucins in mucus protect epithelial surfaces.
润滑与保护：透明质酸（一种糖胺聚糖）润滑关节；粘液中的粘蛋白保护上皮表面。

Carbohydrates also serve as precursors for the synthesis of other biomolecules, including amino acids and fatty acids, through metabolic pathways such as glycolysis and the pentose phosphate pathway.

糖类还作为其他生物分子合成的前体，包括通过糖酵解和磷酸戊糖途径等代谢途径合成氨基酸和脂肪酸。

11. Hydrolysis and Condensation: Key Reactions | 水解与缩合：关键反应

Two fundamental reactions govern carbohydrate chemistry in biological systems. Understanding these is essential for A-Level questions on digestion and synthesis:

在生物系统中，两个基本反应支配着糖类化学。理解这两者对A-Level考试中有关消化与合成的题目至关重要：

Condensation: Two monosaccharides join to form a disaccharide with the production of water. This is an anabolic, endergonic (energy-requiring) reaction. Example: glucose + fructose → sucrose + H₂O.

缩合反应：两个单糖结合形成二糖并生成水。这是一个合成代谢的、吸能（需要能量）的反应。例如：葡萄糖 + 果糖 → 蔗糖 + 水。

Hydrolysis: A disaccharide or polysaccharide is broken down into its constituent monosaccharides by the addition of water. This is a catabolic, exergonic (energy-releasing) reaction. Example: maltose + H₂O → glucose + glucose (catalysed by maltase).

水解反应：二糖或多糖通过加入水被分解为组成单糖。这是一个分解代谢的、释能（释放能量）的反应。例如：麦芽糖 + 水 → 葡萄糖 + 葡萄糖（由麦芽糖酶催化）。

(C₆H₁₂O₆)ₙ + nH₂O → nC₆H₁₂O₆ (hydrolysis of polysaccharide)
多糖水解通式：(C₆H₁₂O₆)ₙ + nH₂O → nC₆H₁₂O₆

In digestion, polysaccharides and disaccharides are hydrolysed by specific enzymes: salivary and pancreatic amylase hydrolyse starch to maltose; maltase, sucrase, and lactase hydrolyse their respective disaccharides on the brush border of the small intestine.

在消化过程中，多糖和二糖由特定酶水解：唾液淀粉酶和胰淀粉酶将淀粉水解为麦芽糖；小肠刷状缘上的麦芽糖酶、蔗糖酶和乳糖酶分别水解各自的二糖。

12. Summary and Examination Tips | 总结与考试要点

Carbohydrates are structurally diverse molecules whose specific functions are intimately linked to their chemical structure. For CIE A-Level Biology, the key points to remember are:

糖类是结构多样的分子，其特定功能与其化学结构密切相关。对于CIE A-Level生物学，需要记住的关键要点如下：

Monosaccharides are soluble, sweet-tasting reducing sugars; glucose is the primary respiratory substrate.
单糖是可溶、有甜味的还原糖；葡萄糖是呼吸作用的主要底物。
Disaccharides form by condensation reactions; maltose and lactose are reducing, sucrose is non-reducing.
二糖通过缩合反应形成；麦芽糖和乳糖为还原糖，蔗糖为非还原糖。
Starch (amylose + amylopectin) and glycogen are compact, insoluble energy stores with α-glycosidic bonds.
淀粉（直链淀粉+支链淀粉）和糖原是紧凑、不溶的能量储存物，含α-糖苷键。
Cellulose is a linear polymer of β-glucose with β-1,4 bonds; its hydrogen-bonded microfibrils provide remarkable tensile strength.
纤维素是β-葡萄糖的线性聚合物，含β-1,4键；其氢键连接的微纤维提供卓越的抗张强度。
Examiners often test the relationship between structure and function: for example, why branched glycogen allows rapid glucose release, or why cellulose’s linear structure suits its structural role.
考官经常考察结构与功能之间的关系：例如，为什么分支的糖原允许快速释放葡萄糖，或者为什么纤维素的线性结构适合其结构功能。
Be prepared to draw α-glucose and β-glucose and to identify the type of glycosidic bond (α-1,4, α-1,6, β-1,4) in given diagrams.
需能够画出α-葡萄糖和β-葡萄糖，并能识别给定图中的糖苷键类型（α-1,4、α-1,6、β-1,4）。

Command words often used in CIE papers include ‘explain how the structure of cellulose is related to its function’, ‘describe the difference between amylose and amylopectin’, and ‘compare glycogen and starch in terms of structure and function’. Practising these question types will consolidate your understanding and improve exam performance.

CIE试卷中常用的指令词包括”解释纤维素的结构如何与其功能相关”、”描述直链淀粉和支链淀粉的区别”以及”比较糖原和淀粉在结构和功能方面的异同”。练习这些题型将巩固理解并提高考试成绩。

Published by TutorHao | Biology Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Carbohydrates: Structure and Function | 糖类：结构与功能