Core Principles of Biochemistry | 生物化学原理核心知识

📚 Core Principles of Biochemistry | 生物化学原理核心知识

Biochemistry bridges chemistry and biology by examining the molecular machinery of living organisms. For A-Level Chemistry students, understanding the structure and function of biological molecules is not merely an extension of organic chemistry—it is a crucial application of bonding, equilibria, and thermodynamics in aqueous systems.

生物化学通过在分子层面审视生命有机体的运作机制，将化学与生物学紧密相连。对于A-Level化学学生而言，理解生物分子的结构与功能不仅仅是有机化学的延伸，更是化学键、化学平衡和热力学在水相体系中应用的关键体现。

1. Amino Acids and the Peptide Bond | 氨基酸与肽键

Amino acids are organic molecules containing both an amino group (−NH₂) and a carboxyl group (−COOH) attached to the same α-carbon. At physiological pH (~7.4), amino acids exist as zwitterions, where the amino group is protonated (−NH₃⁺) and the carboxyl group is deprotonated (−COO⁻). This dual ionic character gives amino acids high melting points and water solubility.

氨基酸是同时含有氨基（−NH₂）和羧基（−COOH）且两者连接在同一个α-碳上的有机分子。在生理pH（约7.4）条件下，氨基酸以两性离子形式存在，此时氨基被质子化（−NH₃⁺）而羧基去质子化（−COO⁻）。这种双重离子特性赋予氨基酸高熔点和良好的水溶性。

The isoelectric point (pI) is the pH at which an amino acid carries no net charge. At this point, its solubility is at a minimum, and its electrophoretic mobility is zero—a property exploited in protein purification techniques such as isoelectric focusing.

等电点（pI）是氨基酸净电荷为零时的pH值。在等电点处，氨基酸的溶解度最低，电泳迁移率为零——这一性质被广泛应用于等电聚焦等蛋白质纯化技术中。

When two amino acids condense, a peptide bond (an amide linkage, −CO−NH−) forms between the carboxyl group of one and the amino group of the next, releasing a water molecule. The chain of amino acids is called a polypeptide, and its sequence is the primary structure of the protein.

当两个氨基酸发生缩合反应时，一个氨基酸的羧基与另一个氨基酸的氨基之间形成肽键（酰胺键，−CO−NH−），同时释放一分子水。氨基酸链称为多肽，其氨基酸序列构成了蛋白质的一级结构。

2. Protein Structure: From Sequence to Function | 蛋白质结构：从序列到功能

Protein structure is organised into four hierarchical levels. The primary structure is the linear sequence of amino acids held together by covalent peptide bonds. The secondary structure arises from hydrogen bonding between backbone carbonyl and amide groups, producing regular patterns: the α-helix and the β-pleated sheet.

蛋白质结构分为四个层级。一级结构是由共价肽键连接的氨基酸线性序列。二级结构源于骨架羰基和酰胺基团之间的氢键作用，形成规律的构象：α-螺旋和β-折叠片层。

The tertiary structure is the overall three-dimensional folding of a single polypeptide chain, stabilised by four types of interaction:

三级结构是单条多肽链的整体三维折叠，由四种相互作用力稳定：

Hydrogen bonds between polar side chains / 极性侧链之间形成的氢键
Ionic bonds between charged side chains (e.g., −COO⁻ with −NH₃⁺) / 带电侧链之间的离子键（如−COO⁻与−NH₃⁺）
Hydrophobic interactions between nonpolar side chains in the core / 非极性侧链在分子内部形成的疏水相互作用
Disulfide bridges (−S−S−) formed by oxidation of two cysteine residues / 两个半胱氨酸残基氧化形成的二硫键（−S−S−）

The quaternary structure describes the assembly of multiple polypeptide subunits. A classic example is haemoglobin, a tetramer of two α-globin and two β-globin chains, each carrying a haem group with an Fe²⁺ ion that binds oxygen cooperatively.

四级结构描述多个多肽亚基的组装方式。经典例子是血红蛋白，它由两条α-珠蛋白和两条β-珠蛋白链组成四聚体，每条链含一个带有Fe²⁺离子的血红素基团，能够协同结合氧气。

3. Enzyme Kinetics and the Michaelis–Menten Model | 酶动力学与米氏方程模型

Enzymes are biological catalysts that accelerate reactions by lowering the activation energy (Eₐ). In A-Level Chemistry, the key model is the Michaelis–Menten mechanism:

酶是生物催化剂，通过降低活化能（Eₐ）来加速化学反应。在A-Level化学中，核心模型是米氏（Michaelis–Menten）机制：

E + S ⇌ ES → E + P

The enzyme (E) binds the substrate (S) reversibly to form the enzyme–substrate complex (ES), which then converts to product (P) and regenerates the enzyme. The Michaelis constant Kₘ is the substrate concentration at which the reaction rate reaches half of V_max. A low Kₘ indicates high enzyme affinity for its substrate.

酶（E）与底物（S）可逆结合形成酶-底物复合物（ES），随后复合物转化为产物（P）并再生出游离酶。米氏常数Kₘ是反应速率达到最大速率V_max一半时的底物浓度。Kₘ值越低，表示酶与底物的亲和力越高。

For the Lineweaver–Burk double reciprocal plot, the equation is linearised:

对于莱因韦弗-伯克（Lineweaver–Burk）双倒数图，米氏方程被线性化处理：

1/v = Kₘ/(V_max·[S]) + 1/V_max

Competitive inhibitors increase Kₘ but leave V_max unchanged—they compete with the substrate for the active site. Non-competitive inhibitors bind elsewhere, decreasing V_max while leaving Kₘ unaffected.

竞争性抑制剂使Kₘ增大但V_max不变——它们与底物竞争活性位点。非竞争性抑制剂结合在活性位点以外的位置，使V_max降低而Kₘ不受影响。

4. Carbohydrate Chemistry: Mono-, Di- and Polysaccharides | 碳水化合物化学：单糖、二糖与多糖

Carbohydrates are polyhydroxy aldehydes or ketones with the general formula Cₙ(H₂O)ₘ. Glucose (C₆H₁₂O₆) exists in equilibrium between its open-chain form and its cyclic hemiacetal form. In solution, the cyclic form predominates, creating α-glucose (OH on C1 trans to CH₂OH) and β-glucose (OH on C1 cis to CH₂OH).

碳水化合物是多羟基醛或多羟基酮，通式为Cₙ(H₂O)ₘ。葡萄糖（C₆H₁₂O₆）在开链形式和环状半缩醛形式之间达成平衡。在溶液中，环状形式占主导，产生α-葡萄糖（C1位OH与CH₂OH处于反式）和β-葡萄糖（C1位OH与CH₂OH处于顺式）两种异构体。

Monosaccharides are reducing sugars because the open-chain aldehyde can reduce Cu²⁺ to Cu⁺ in Benedict’s solution. Disaccharides form via glycosidic bonds: maltose (α-1,4), cellobiose (β-1,4), and sucrose (α-1,2), which is non-reducing because both anomeric carbons are involved in the linkage.

单糖是还原糖，因为其开链醛基能将本尼迪特（Benedict）试剂中的Cu²⁺还原为Cu⁺。二糖通过糖苷键连接形成：麦芽糖（α-1,4键）、纤维二糖（β-1,4键）和蔗糖（α-1,2键）。蔗糖是非还原糖，因为两个异头碳都参与了糖苷键的形成。

Polysaccharides are condensation polymers of monosaccharides. Starch (amylose and amylopectin) stores glucose in plants with α-1,4 and α-1,6 linkages; cellulose uses β-1,4 linkages that allow extensive hydrogen bonding between chains, giving plant cell walls their tensile strength.

多糖是单糖的缩合聚合物。淀粉（直链淀粉和支链淀粉）通过α-1,4和α-1,6糖苷键在植物中储存葡萄糖；纤维素使用β-1,4糖苷键，使得链间能够形成广泛的氢键网络，赋予植物细胞壁抗张强度。

5. Lipids and Cell Membrane Structure | 脂质与细胞膜结构

Lipids are a diverse class of hydrophobic biomolecules. Triglycerides consist of glycerol esterified with three fatty acids. Fatty acids are carboxylic acids with long hydrocarbon tails; they may be saturated (no C=C bonds) or unsaturated (containing one or more cis-C=C bonds that introduce kinks in the chain).

脂质是一类多样的疏水性生物分子。甘油三酯由甘油与三分子脂肪酸酯化而成。脂肪酸是具有长烃链尾部的羧酸，可以是饱和的（无C=C双键）或不饱和的（含一个或多个顺式C=C双键，使碳链产生弯曲）。

Phospholipids replace one fatty acid with a phosphate group, creating an amphiphilic molecule with a hydrophilic head and two hydrophobic tails. In aqueous environments, phospholipids self-assemble into a bilayer—the fundamental structure of all cell membranes.

磷脂将甘油三酯中的一个脂肪酸替换为磷酸基团，形成具有亲水头部和两条疏水尾部的两亲性分子。在水环境中，磷脂自发组装成双分子层——这是所有细胞膜的基本结构。

Hydrolysis of triglycerides with alkali (KOH or NaOH) produces soap—the process of saponification. The ionic carboxylate head of soap interacts with water, while the hydrocarbon tail dissolves grease, enabling emulsification.

甘油三酯与碱（KOH或NaOH）水解产生肥皂——这一过程称为皂化反应。肥皂的离子羧酸头部与水相互作用，而烃链尾部溶解油脂，从而实现乳化作用。

6. Nucleic Acids: DNA and RNA | 核酸：DNA与RNA

Nucleic acids are polymers of nucleotides, each composed of three components: a nitrogenous base (purine or pyrimidine), a pentose sugar (ribose in RNA, 2-deoxyribose in DNA), and a phosphate group. The nucleoside is formed by a β-N-glycosidic bond between the base and C1 of the sugar; phosphorylation at C5 produces the nucleotide.

核酸是核苷酸的聚合物，每个核苷酸由三个组分构成：含氮碱基（嘌呤或嘧啶）、戊糖（RNA中为核糖，DNA中为2-脱氧核糖）和磷酸基团。碱基与糖的C1之间通过β-N-糖苷键形成核苷；在C5位磷酸化即得到核苷酸。

DNA is a double helix held together by complementary base pairing: adenine pairs with thymine via two hydrogen bonds, and guanine pairs with cytosine via three hydrogen bonds. The antiparallel strands create major and minor grooves, while the hydrophobic bases stack in the interior—a stabilising force called base stacking.

DNA是由互补碱基配对维持的双螺旋结构：腺嘌呤与胸腺嘧啶通过两个氢键配对，鸟嘌呤与胞嘧啶通过三个氢键配对。两条反向平行的链形成大沟和小沟，疏水碱基在内部堆叠——这种稳定力称为碱基堆叠。

The negative charges of phosphate groups make DNA an acid, and at cellular pH the sugar-phosphate backbone is fully ionised. This is why DNA migrates toward the anode during gel electrophoresis, a separation technique routinely examined in practical assessments.

磷酸基团带负电使DNA呈酸性，在细胞pH条件下糖-磷酸骨架完全离子化。这就是DNA在凝胶电泳中向阳极迁移的原因——这是实验考核中常考的一种分离技术。

7. ATP: The Universal Energy Currency | ATP：通用能量货币

Adenosine triphosphate (ATP) is a nucleotide consisting of adenine, ribose, and a chain of three phosphate groups. The bonds between the phosphate groups are phosphoanhydride bonds, which store approximately 30.5 kJ/mol of free energy each under standard conditions.

三磷酸腺苷（ATP）是由腺嘌呤、核糖和三个磷酸基团连接而成的核苷酸。磷酸基团之间的化学键为磷酸酐键，在标准条件下每个键储存约30.5 kJ/mol的自由能。

The hydrolysis of ATP to ADP (adenosine diphosphate) and inorganic phosphate (Pᵢ) releases energy that drives endergonic reactions in the cell:

ATP水解为ADP（二磷酸腺苷）和无机磷酸（Pᵢ）时释放能量，驱动细胞内的吸能反应：

ATP + H₂O → ADP + Pᵢ ΔG°′ = −30.5 kJ·mol⁻¹

The hydrolysis is thermodynamically favourable due to relief of electrostatic repulsion between adjacent negative phosphate groups, increased resonance stabilisation of products, and the entropy gained upon release of phosphate. The regenerating pathway, oxidative phosphorylation, couples ATP synthesis to the exergonic flow of electrons down the electron transport chain.

ATP水解在热力学上是有利的，原因包括：相邻负电性磷酸基团间静电斥力的释放、产物共振稳定化增强、以及磷酸释放带来的熵增。ATP的再生途径——氧化磷酸化——将ATP合成与电子沿电子传递链流动的放能过程相偶联。

8. Glycolysis: The Central Catabolic Pathway | 糖酵解：核心分解代谢途径

Glycolysis is the metabolic pathway that converts one molecule of glucose (C₆H₁₂O₆) into two molecules of pyruvate (CH₃COCOO⁻), taking place in the cytoplasm. The overall reaction can be written as:

糖酵解是将一分子葡萄糖（C₆H₁₂O₆）转化为两分子丙酮酸（CH₃COCOO⁻）的代谢途径，发生在细胞质中。总反应可写为：

Glucose + 2NAD⁺ + 2ADP + 2Pᵢ → 2 Pyruvate + 2NADH + 2H⁺ + 2ATP + 2H₂O

The pathway consists of two phases: the investment phase consumes 2 ATP to phosphorylate glucose into fructose-1,6-bisphosphate; the pay-off phase cleaves this six-carbon molecule into two triose phosphates, which are oxidised and produce 4 ATP and 2 NADH. The net yield is therefore 2 ATP and 2 NADH per glucose.

此途径分为两个阶段：投入阶段消耗2个ATP，将葡萄糖磷酸化成果糖-1,6-二磷酸；产出阶段将该六碳分子裂解为两个三碳糖磷酸，后者被氧化并产生4个ATP和2个NADH。因此每分子葡萄糖的净产量为2个ATP和2个NADH。

The conversion of glyceraldehyde-3-phosphate involves the oxidation of an aldehyde to a carboxyl group and the uptake of inorganic phosphate—an example of substrate-level phosphorylation, in which ATP is generated directly from ADP and Pᵢ without an electron transport chain.

甘油醛-3-磷酸的转化涉及醛基氧化为羧基并摄取无机磷酸——这是底物水平磷酸化的例子，即不经过电子传递链而直接从ADP和Pᵢ生成ATP。

9. The Citric Acid Cycle and Redox Chemistry | 柠檬酸循环与氧化还原化学

The citric acid cycle (Krebs cycle) operates in the mitochondrial matrix, oxidising the acetyl group of acetyl-CoA (CH₃CO-S-CoA) to two molecules of CO₂. For each acetyl group entering the cycle, the following are produced:

柠檬酸循环（三羧酸循环）在线粒体基质中进行，将乙酰辅酶A（CH₃CO-S-CoA）的乙酰基氧化为两分子CO₂。每分子乙酰基进入循环，产生：

Product / 产物	Quantity / 数量	Oxidation state change / 氧化态变化
CO₂	2	Carbon fully oxidised (+4) / 碳被完全氧化（+4）
NADH	3	NAD⁺ reduced to NADH / NAD⁺被还原为NADH
FADH₂	1	FAD reduced to FADH₂ / FAD被还原为FADH₂
GTP (or ATP)	1	Substrate-level phosphorylation / 底物水平磷酸化

Three key dehydrogenases catalyse the oxidation: isocitrate dehydrogenase, α-ketoglutarate dehydrogenase, and malate dehydrogenase. These enzymes transfer hydride ions (H⁻) to NAD⁺ and protons to the medium, demonstrating classic redox chemistry in a biological context.

三个关键脱氢酶催化氧化反应：异柠檬酸脱氢酶、α-酮戊二酸脱氢酶和苹果酸脱氢酶。这些酶将氢负离子（H⁻）转移给NAD⁺，并将质子释放到介质中，展示了生物学情境中的经典氧化还原化学。

10. Oxidative Phosphorylation and Chemiosmosis | 氧化磷酸化与化学渗透

The electron transport chain (ETC) is embedded in the inner mitochondrial membrane. Electrons from NADH and FADH₂ pass through complexes I to IV, with each step involving a redox couple whose reduction potential becomes increasingly positive. The electrons ultimately reduce O₂ to water:

电子传递链（ETC）嵌于线粒体内膜中。来自NADH和FADH₂的电子依次通过复合物I至IV，每一步涉及还原电位逐渐变正的氧化还原电对。电子最终将O₂还原为水：

½O₂ + 2H⁺ + 2e⁻ → H₂O E°′ = +0.82 V

Complexes I, III, and IV pump protons from the matrix into the intermembrane space, generating a proton gradient (ΔpH) and an electrical potential (Δψ). Together these constitute the proton motive force (PMF). ATP synthase harnesses this PMF as protons flow back through its F₀ channel, driving rotational catalysis in the F₁ head to phosphorylate ADP.

复合物I、III和IV将质子从基质泵入膜间隙，产生质子梯度（ΔpH）和电位差（Δψ），两者共同构成质子动力势（PMF）。ATP合酶利用PMF——质子通过其F₀通道回流时驱动F₁头部发生旋转催化，使ADP磷酸化生成ATP。

The chemiosmotic theory, proposed by Peter Mitchell, explains how oxidation is coupled to phosphorylation: electron transport and ATP synthesis are spatially separated but energetically linked by the proton gradient. Uncoupling agents such as 2,4-dinitrophenol dissipate the proton gradient, releasing energy as heat rather than producing ATP.

化学渗透学说由彼得·米切尔提出，解释了氧化如何与磷酸化偶联：电子传递和ATP合成在空间上分离，但通过质子梯度实现能量上的紧密连接。解偶联剂如2,4-二硝基苯酚会使质子梯度消散，能量以热量而非ATP形式释放。

11. Metabolic Regulation at a Glance | 代谢调控速览

Cellular metabolism is regulated through allosteric control, covalent modification, and gene expression. A classic allosteric example is the feedback inhibition of phosphofructokinase-1 (PFK-1) by ATP—a high energy charge signals that glycolysis should slow down.

细胞代谢通过别构调控、共价修饰和基因表达三种方式进行调节。经典的别构调控例子是ATP对磷酸果糖激酶-1（PFK-1）的反馈抑制——高能量电荷信号指示糖酵解应放缓。

In A-Level exam questions, candidates should remember that metabolic pathways are compartmentalised: glycolysis occurs in the cytosol, the citric acid cycle in the mitochondrial matrix, and oxidative phosphorylation at the inner mitochondrial membrane. Redox couples such as NAD⁺/NADH and FAD/FADH₂ serve as electron carriers that link these compartments thermodynamically.

在A-Level考试中，考生应记住代谢途径具有区室化特征：糖酵解在细胞质中进行，柠檬酸循环在线粒体基质中进行，氧化磷酸化在线粒体内膜上进行。氧化还原电对如NAD⁺/NADH和FAD/FADH₂充当电子载体，在热力学上将各区室联系起来。

12. Exam Strategy: Data Interpretation and Common Traps | 备考策略：数据分析与常见陷阱

Biochemistry questions in A-Level chemistry exams often provide kinetic data or molecular structures and expect candidates to interpret them quantitatively. Key skills include calculating Kₘ and V_max from a Lineweaver–Burk plot, identifying hydrogen-bonding donors and acceptors in base pairs, and predicting whether a molecule is reducing based on the presence of a free anomeric carbon.

A-Level化学考试中的生物化学题目通常提供动力学数据或分子结构，期望考生进行定量解读。关键技能包括：从Lineweaver–Burk图中计算Kₘ和V_max，识别碱基对中的氢键供体和受体，以及根据是否存在游离异头碳来判断分子是否为还原糖。

Distinguish between Kₘ and V_max: Kₘ is a measure of affinity, not catalytic rate / 区分Kₘ与V_max：Kₘ衡量亲和力而非催化速率
Remember that enzymes do not alter ΔG or the equilibrium constant—they only lower Eₐ / 记住酶不改变ΔG或平衡常数——它们只降低Eₐ
Check the glycosidic bond position (α vs β, 1→4 vs 1→6) before classifying a disaccharide / 在分类二糖前，检查糖苷键位置（α或β，1→4或1→6）
For redox questions, track the oxidation number of carbon through each cycle step / 对氧化还原题目，追踪每一步中碳的氧化数变化

Finally, practice writing balanced equations for glycolysis and the citric acid cycle from memory. The ability to reproduce these pathways accurately with correct stoichiometry is a mark-scoring skill that distinguishes top candidates.

最后，建议练习凭记忆写出糖酵解和柠檬酸循环的平衡方程式。能够准确重现这些途径并具备正确的化学计量数，是区分顶尖考生的重要得分技能。

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Core Principles of Biochemistry | 生物化学原理核心知识