A-Level Biology: DNA Replication, Transcription & Translation — The Central Dogma | A-Level 生物：DNA 复制、转录与翻译——中心法则

The Central Dogma of Molecular Biology, first articulated by Francis Crick in 1958, describes the fundamental flow of genetic information within biological systems: DNA → RNA → Protein. This framework underpins virtually every process in living organisms — from how a single fertilised egg develops into a complex multicellular organism, to how bacteria respond to environmental stress by producing specific enzymes. For A-Level Biology students across all exam boards (AQA, Edexcel, OCR, CIE, WJEC), mastering the three core processes — DNA replication, transcription, and translation — is not merely a syllabus requirement but the conceptual key that unlocks topics ranging from gene expression and mutation to biotechnology and genetic engineering.

分子生物学的中心法则由 Francis Crick 于 1958 年首次提出，描述了生物系统中遗传信息的基本流向：DNA → RNA → 蛋白质。这一框架支撑着生物体中几乎每一个过程——从一个受精卵如何发育成复杂的多细胞生物，到细菌如何通过产生特定酶来应对环境压力。对于所有考试局（AQA、Edexcel、OCR、CIE、WJEC）的 A-Level 生物学生来说，掌握三个核心过程——DNA 复制、转录和翻译——不仅仅是课程大纲的要求，更是解锁从基因表达与突变到生物技术与基因工程等一系列主题的概念钥匙。

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1. DNA Structure: The Foundation | DNA 结构：基础

Before diving into the three processes, it is essential to be absolutely clear about the structure of DNA. DNA (deoxyribonucleic acid) is a double-stranded, antiparallel helix composed of nucleotides. Each nucleotide consists of three components:

• Deoxyribose sugar — a pentose (5-carbon) sugar lacking an –OH group on the 2′ carbon (hence deoxy).
• Phosphate group — links adjacent nucleotides via phosphodiester bonds between the 3′ carbon of one sugar and the 5′ carbon of the next, forming the sugar–phosphate backbone.
• Nitrogenous base — one of four: adenine (A), thymine (T), cytosine (C), or guanine (G).

在深入三个过程之前，必须对 DNA 的结构有绝对清晰的认识。DNA（脱氧核糖核酸）是由核苷酸组成的双链反平行螺旋。每个核苷酸由三个部分组成：

• 脱氧核糖——一种在 2′ 碳上缺少 –OH 基团的五碳糖（因此称为 脱氧）。
• 磷酸基团——通过一个糖的 3′ 碳与下一个糖的 5′ 碳之间的磷酸二酯键连接相邻核苷酸，形成糖–磷酸骨架。
• 含氮碱基——四种之一：腺嘌呤 (A)、胸腺嘧啶 (T)、胞嘧啶 (C) 或鸟嘌呤 (G)。

The two strands are held together by hydrogen bonds between complementary base pairs: A pairs with T via two hydrogen bonds, while C pairs with G via three hydrogen bonds. This complementary base pairing is the molecular basis for faithful DNA replication and accurate transcription. The strands run in opposite directions — one 5′ → 3′, the other 3′ → 5′ — a feature termed antiparallel that has profound consequences for both replication and transcription mechanisms.

两条链通过互补碱基对之间的氢键连接在一起：A 与 T 通过两个氢键配对，C 与 G 通过三个氢键配对。这种互补碱基配对是忠实 DNA 复制和准确转录的分子基础。两条链方向相反——一条是 5′ → 3′，另一条是 3′ → 5′——这一特征被称为反平行，对复制和转录机制都有着深远的影响。

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2. DNA Replication: Copying the Blueprint | DNA 复制：复制蓝图

2.1 The Semi-Conservative Model | 半保留复制模型

DNA replication follows the semi-conservative model, elegantly demonstrated by the Meselson–Stahl experiment (1958). In this landmark study, E. coli were grown in a medium containing the heavy nitrogen isotope ¹⁵N for many generations, then transferred to a ¹⁴N medium. After one round of replication, all DNA molecules were of intermediate density (one ¹⁵N strand + one ¹⁴N strand). After two rounds, half were intermediate and half were light — results that are consistent only with the semi-conservative model, ruling out both the conservative and dispersive alternatives.

DNA 复制遵循半保留模型，这一模型由Meselson–Stahl 实验（1958 年）精彩地证明。在这项里程碑式的研究中，大肠杆菌在含有重氮同位素 ¹⁵N 的培养基中培养了许多代，然后转移到 ¹⁴N 培养基中。经过一轮复制后，所有 DNA 分子都是中间密度（一条 ¹⁵N 链 + 一条 ¹⁴N 链）。经过两轮后，一半是中间密度，一半是轻密度——这些结果仅与半保留模型一致，排除了保留和分散两种替代模型。

2.2 Key Enzymes and Their Roles | 关键酶及其作用

DNA replication is a highly coordinated process involving a suite of enzymes. A-Level specifications require you to know the following:

Enzyme | 酶	Function | 功能
DNA Helicase | DNA 解旋酶	Unwinds the double helix by breaking hydrogen bonds between base pairs, creating a replication fork. | 通过断裂碱基对之间的氢键解开双螺旋，产生复制叉。
DNA Gyrase (topoisomerase) | DNA 旋转酶（拓扑异构酶）	Relieves the supercoiling tension ahead of the replication fork by introducing transient breaks in the DNA backbone. | 通过在 DNA 骨架中引入瞬时断裂来缓解复制叉前方的超螺旋张力。
Single-Strand Binding Proteins (SSBs) | 单链结合蛋白	Bind to and stabilise the separated single strands, preventing them from re-annealing. | 结合并稳定已分离的单链，防止它们重新退火。
DNA Primase | DNA 引物酶	Synthesises short RNA primers (≈10 nucleotides) that provide a free 3′–OH group for DNA polymerase to extend. | 合成短 RNA 引物（约 10 个核苷酸），为 DNA 聚合酶提供游离的 3′–OH 基团进行延伸。
DNA Polymerase III | DNA 聚合酶 III	The main replicative enzyme; adds DNA nucleotides to the 3′ end of the growing strand, always in the 5′ → 3′ direction. Also has 3′ → 5′ exonuclease proofreading activity. | 主要复制酶；将 DNA 核苷酸添加到生长链的 3′ 端，始终沿 5′ → 3′ 方向进行。还具有 3′ → 5′ 核酸外切酶校对活性。
DNA Polymerase I | DNA 聚合酶 I	Removes RNA primers and replaces them with DNA nucleotides (has 5′ → 3′ exonuclease activity). | 移除 RNA 引物并用 DNA 核苷酸替换（具有 5′ → 3′ 核酸外切酶活性）。
DNA Ligase | DNA 连接酶	Seals the nicks between Okazaki fragments on the lagging strand by catalysing the formation of phosphodiester bonds. | 通过催化磷酸二酯键的形成，封闭滞后链上冈崎片段之间的缺口。

2.3 The Replication Process: Leading and Lagging Strands | 复制过程：前导链与滞后链

Because DNA polymerase can only synthesise in the 5′ → 3′ direction, the two template strands are copied differently:

• Leading Strand (前导链): The template strand running 3′ → 5′ is copied continuously towards the replication fork. A single RNA primer is laid down by primase, and DNA polymerase III extends it uninterrupted.
• Lagging Strand (滞后链): The template strand running 5′ → 3′ must be copied discontinuously, in short segments called Okazaki fragments (≈100–200 nucleotides in eukaryotes, ≈1,000–2,000 in prokaryotes). Multiple RNA primers are synthesised, each extended by DNA polymerase III until it reaches the previous primer. DNA polymerase I then removes the primers and fills the gaps, and DNA ligase seals the remaining nicks.

由于 DNA 聚合酶只能沿5′ → 3′ 方向合成，两条模板链的复制方式不同：

• 前导链 (Leading Strand)： 沿 3′ → 5′ 方向运行的模板链被连续复制，朝向复制叉。引物酶放置一个 RNA 引物，DNA 聚合酶 III 不间断地延伸它。
• 滞后链 (Lagging Strand)： 沿 5′ → 3′ 方向运行的模板链必须不连续地复制，形成称为冈崎片段的短片段（真核生物中约 100–200 个核苷酸，原核生物中约 1,000–2,000 个）。合成多个 RNA 引物，每个由 DNA 聚合酶 III 延伸直到到达前一个引物。然后 DNA 聚合酶 I 移除引物并填补缺口，DNA 连接酶封闭剩余的切口。

Exam Tip: A common A-Level question asks you to explain why the lagging strand is synthesised discontinuously. The key phrase is “because DNA polymerase can only synthesise in the 5′ to 3′ direction, and the lagging strand template runs 5′ to 3′, so synthesis must occur away from the replication fork in short fragments.“

考试技巧：常见的 A-Level 考题要求你解释为什么滞后链是不连续合成的。关键表述是：”因为 DNA 聚合酶只能沿 5′ 到 3′ 方向合成，而滞后链模板沿 5′ 到 3′ 方向运行，所以合成必须以短片段的形式远离复制叉进行。“

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3. Transcription: DNA → mRNA | 转录：DNA → mRNA

Transcription is the process by which a specific segment of DNA — a gene — is copied into a complementary messenger RNA (mRNA) molecule. This is the first step in gene expression and occurs in the nucleus of eukaryotic cells.

转录是 DNA 的特定片段——一个基因——被复制成互补的信使 RNA (mRNA) 分子的过程。这是基因表达的第一步，发生在真核细胞的细胞核中。

3.1 Initiation | 起始

Transcription begins when RNA polymerase binds to a specific DNA sequence called the promoter region, located immediately upstream (5′) of the gene. In prokaryotes, the promoter contains conserved sequences at positions −10 (the Pribnow box, TATAAT) and −35. In eukaryotes, the TATA box (≈−25) is recognised by transcription factors (specifically TFIID, which contains the TATA-binding protein, TBP) that recruit RNA polymerase II — the polymerase responsible for mRNA synthesis.

转录开始时，RNA 聚合酶与位于基因紧邻上游（5′）的特定 DNA 序列——启动子区域结合。在原核生物中，启动子在 −10 位（Pribnow 盒，TATAAT）和 −35 位含有保守序列。在真核生物中，TATA 盒（约 −25 位）被转录因子（特别是含有 TATA 结合蛋白 TBP 的 TFIID）识别，这些转录因子招募 RNA 聚合酶 II——负责 mRNA 合成的聚合酶。

3.2 Elongation | 延伸

Once bound, RNA polymerase unwinds approximately 10–15 base pairs of DNA, forming a transcription bubble. It then reads the template strand (also called the antisense strand) in the 3′ → 5′ direction and synthesises a complementary RNA molecule in the 5′ → 3′ direction. The key difference from DNA replication is that RNA incorporates uracil (U) instead of thymine (T) — so wherever the template has an A, the RNA receives a U.

一旦结合，RNA 聚合酶解开约 10–15 个碱基对的 DNA，形成转录泡。然后它沿 3′ → 5′ 方向读取模板链（也称为反义链），并沿 5′ → 3′ 方向合成互补的 RNA 分子。与 DNA 复制的关键区别在于，RNA 掺入尿嘧啶 (U) 而不是胸腺嘧啶 (T)——因此模板上有 A 的地方，RNA 就得到 U。

3.3 Termination | 终止

Transcription terminates when RNA polymerase encounters a terminator sequence. In prokaryotes, this often involves a GC-rich hairpin loop followed by a run of uracils (rho-independent termination), which destabilises the RNA–DNA hybrid. In eukaryotes, termination is more complex and involves cleavage of the pre-mRNA at a specific site downstream of the polyadenylation signal (AAUAAA).

当 RNA 聚合酶遇到终止子序列时，转录终止。在原核生物中，这通常涉及一个富含 GC 的发夹环结构和随后的一串尿嘧啶（不依赖 rho 因子的终止），这会破坏 RNA–DNA 杂交体的稳定性。在真核生物中，终止更为复杂，涉及在聚腺苷酸化信号 (AAUAAA) 下游的特定位置切割前体 mRNA。

3.4 Post-Transcriptional Modification (Eukaryotes Only) | 转录后修饰（仅真核生物）

The primary transcript (pre-mRNA) in eukaryotes undergoes three major processing steps before it can be exported to the cytoplasm for translation:

1. 5′ Capping: A modified guanine nucleotide (7-methylguanosine) is added to the 5′ end. This 5′ cap protects the mRNA from degradation by exonucleases and is essential for ribosome binding during translation initiation.
2. 3′ Polyadenylation: Approximately 200 adenine nucleotides are added to the 3′ end, forming the poly-A tail. Like the cap, this protects against degradation and facilitates nuclear export.
3. Splicing: The pre-mRNA contains introns (non-coding sequences) interspersed between exons (coding sequences). A large ribonucleoprotein complex called the spliceosome removes introns and ligates exons together in a process known as RNA splicing. Alternative splicing allows a single gene to produce multiple different mRNA transcripts — and therefore multiple protein isoforms — by combining exons in different patterns.

真核生物中的初级转录本（前体 mRNA）在输出到细胞质进行翻译之前，需要经历三个主要的加工步骤：

1. 5′ 加帽： 在 5′ 端添加一个修饰的鸟嘌呤核苷酸（7-甲基鸟苷）。这个5′ 帽保护 mRNA 免受外切核酸酶的降解，并对翻译起始时的核糖体结合至关重要。
2. 3′ 聚腺苷酸化： 在 3′ 端添加约 200 个腺嘌呤核苷酸，形成poly-A 尾。与帽类似，它防止降解并促进核输出。
3. 剪接： 前体 mRNA 在外显子（编码序列）之间散布着内含子（非编码序列）。一个称为剪接体的大型核糖核蛋白复合体移除内含子并将外显子连接在一起，这一过程称为 RNA 剪接。可变剪接允许单个基因通过以不同模式组合外显子，产生多种不同的 mRNA 转录本——因此产生多种蛋白质亚型。

Exam Tip: When describing transcription in a prokaryote vs. eukaryote comparison question, remember that prokaryotic mRNA undergoes no post-transcriptional modification — transcription and translation are coupled in the cytoplasm, often with ribosomes beginning translation before transcription is even complete.

考试技巧：在描述原核生物与真核生物的转录比较题中，记住原核生物的 mRNA 不经历转录后修饰——转录和翻译在细胞质中偶联，核糖体通常在转录完成之前就开始翻译。

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4. Translation: mRNA → Protein | 翻译：mRNA → 蛋白质

Translation is the process by which the nucleotide sequence of mRNA is decoded into the amino acid sequence of a polypeptide chain. This occurs on ribosomes in the cytoplasm (or on the rough endoplasmic reticulum for secreted and membrane proteins).

翻译是将 mRNA 的核苷酸序列解码为多肽链的氨基酸序列的过程。这发生在细胞质中的核糖体上（对于分泌蛋白和膜蛋白，则发生在粗面内质网上）。

4.1 The Genetic Code | 遗传密码

The genetic code is the set of rules by which nucleotide triplets — codons — specify amino acids. Key features you must know for A-Level:

• Triplet code: Three nucleotides (one codon) specify one amino acid.
• Degenerate (redundant): Most amino acids are encoded by more than one codon (e.g., leucine has six codons: UUA, UUG, CUU, CUC, CUA, CUG). This degeneracy provides a buffer against point mutations — if the third base changes, the same amino acid may still be incorporated.
• Non-overlapping and commaless: The code is read in a continuous, non-overlapping manner from a fixed starting point.
• Universal: With very few exceptions (e.g., mitochondrial DNA, some protists), the same codons specify the same amino acids in all organisms — powerful evidence for common ancestry.
• Start codon: AUG, which codes for methionine, also signals the start of translation.
• Stop codons: UAA, UAG, and UGA do not code for any amino acid; they signal the termination of translation.

遗传密码是核苷酸三联体——密码子——指定氨基酸的规则集合。A-Level 必须掌握的关键特征：

• 三联体密码： 三个核苷酸（一个密码子）指定一个氨基酸。
• 简并性（冗余性）： 大多数氨基酸由多个密码子编码（例如，亮氨酸有六个密码子：UUA、UUG、CUU、CUC、CUA、CUG）。这种简并性为点突变提供了缓冲——如果第三个碱基发生变化，可能仍然掺入相同的氨基酸。
• 非重叠且无间隔： 密码从固定的起始点以连续、非重叠的方式读取。
• 通用性： 除极少数例外（如线粒体 DNA、某些原生生物），相同的密码子在所有生物中指定相同的氨基酸——这是共同祖先的有力证据。
• 起始密码子： AUG，编码甲硫氨酸，同时也标志着翻译的开始。
• 终止密码子： UAA、UAG 和 UGA 不编码任何氨基酸；它们标志着翻译的终止。

4.2 The Molecular Players | 分子参与者

Molecule | 分子	Role | 作用
mRNA (messenger RNA) | 信使 RNA	Carries the genetic code from the nucleus to the ribosome. Each codon specifies an amino acid. | 将遗传密码从细胞核携带到核糖体。每个密码子指定一个氨基酸。
tRNA (transfer RNA) | 转运 RNA	Adapter molecules with a cloverleaf secondary structure. Each tRNA has an anticodon (three nucleotides complementary to a specific mRNA codon) and carries the corresponding amino acid at its 3′ end. Aminoacyl-tRNA synthetases charge tRNAs by catalysing the ATP-dependent attachment of the correct amino acid.	具有三叶草二级结构的适配器分子。每个 tRNA 有一个反密码子（与特定 mRNA 密码子互补的三个核苷酸），并在其 3′ 端携带相应的氨基酸。氨酰-tRNA 合成酶通过催化 ATP 依赖的正确氨基酸附着来”充电”tRNA。
Ribosome | 核糖体	A large ribonucleoprotein complex composed of two subunits (60S + 40S = 80S in eukaryotes; 50S + 30S = 70S in prokaryotes). The ribosome has three tRNA binding sites: A site (aminoacyl), P site (peptidyl), and E site (exit). It catalyses peptide bond formation via its peptidyl transferase activity (ribozyme activity — the catalytic component is rRNA, not protein).	由两个亚基组成的大型核糖核蛋白复合体（真核生物：60S + 40S = 80S；原核生物：50S + 30S = 70S）。核糖体有三个 tRNA 结合位点：A 位点（氨酰基）、P 位点（肽基）和E 位点（出口）。它通过其肽基转移酶活性（核酶活性——催化成分是 rRNA，而不是蛋白质）催化肽键形成。

4.3 The Translation Mechanism | 翻译机制

Initiation (起始): The small ribosomal subunit binds to the 5′ cap of the mRNA and scans along until it encounters the start codon AUG. The initiator tRNA (carrying methionine) base-pairs with AUG via its anticodon UAC. The large ribosomal subunit then joins, positioning the initiator tRNA in the P site. The A site is now vacant, ready to accept the next aminoacyl-tRNA.

Elongation (延伸): This is a cyclic process with three steps:

1. Codon recognition: A charged tRNA whose anticodon is complementary to the codon in the A site enters and base-pairs. This requires elongation factor Tu (EF-Tu) and GTP hydrolysis.
2. Peptide bond formation: The peptidyl transferase centre of the large subunit catalyses the formation of a peptide bond between the amino acid in the P site and the amino acid in the A site. The growing polypeptide chain is transferred to the tRNA in the A site.
3. Translocation: The ribosome moves exactly three nucleotides (one codon) along the mRNA in the 5′ → 3′ direction. This shifts the tRNA from the A site to the P site, and the tRNA from the P site to the E site, from where it exits. This step requires elongation factor G (EF-G) and GTP hydrolysis. The A site is now vacant and the cycle repeats.

起始： 小核糖体亚基与 mRNA 的 5′ 帽结合，并沿 mRNA 扫描直到遇到起始密码子 AUG。起始 tRNA（携带甲硫氨酸）通过其反密码子 UAC 与 AUG 碱基配对。然后大核糖体亚基加入，将起始 tRNA 定位在 P 位点。A 位点此时空置，准备接受下一个氨酰-tRNA。

延伸： 这是一个包含三个步骤的循环过程：

1. 密码子识别： 一个反密码子与 A 位点密码子互补的带电荷 tRNA 进入并进行碱基配对。这需要延伸因子 Tu (EF-Tu)和 GTP 水解。
2. 肽键形成： 大亚基的肽基转移酶中心催化 P 位点氨基酸与 A 位点氨基酸之间肽键的形成。生长的多肽链被转移到 A 位点的 tRNA 上。
3. 转位： 核糖体沿 mRNA 的 5′ → 3′ 方向精确移动三个核苷酸（一个密码子）。这将 tRNA 从 A 位点移动到 P 位点，P 位点的 tRNA 移动到 E 位点并从那里退出。此步骤需要延伸因子 G (EF-G)和 GTP 水解。A 位点此时空置，循环重复。

Termination (终止): When a stop codon (UAA, UAG, or UGA) enters the A site, no tRNA can recognise it. Instead, a release factor (RF1 or RF2 in prokaryotes; eRF1 in eukaryotes) binds to the A site. This triggers the peptidyl transferase to hydrolyse the bond between the completed polypeptide and the tRNA in the P site, releasing the polypeptide. The ribosomal subunits, mRNA, and remaining tRNA then dissociate.

终止： 当一个终止密码子（UAA、UAG 或 UGA）进入 A 位点时，没有 tRNA 能够识别它。相反，一个释放因子（原核生物中为 RF1 或 RF2；真核生物中为 eRF1）结合到 A 位点。这触发肽基转移酶水解完整多肽与 P 位点 tRNA 之间的键，释放多肽。然后核糖体亚基、mRNA 和剩余的 tRNA 解离。

4.4 Polysomes and Efficiency | 多聚核糖体与效率

A single mRNA molecule can be translated simultaneously by multiple ribosomes, forming a structure called a polysome (or polyribosome). This dramatically increases the rate of protein synthesis — a single mRNA can produce many copies of a protein before it is degraded. Electron micrographs of polysomes are a classic A-Level image.

单个 mRNA 分子可以被多个核糖体同时翻译，形成称为多聚核糖体（polysome）的结构。这极大地提高了蛋白质合成的速率——单个 mRNA 在降解之前可以产生许多蛋白质拷贝。多聚核糖体的电子显微镜图像是经典的 A-Level 图像。

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5. Comparison Table: Replication vs. Transcription vs. Translation | 对比表：复制 vs. 转录 vs. 翻译

Feature | 特征	Replication | 复制	Transcription | 转录	Translation | 翻译
Template | 模板	Both DNA strands	Template strand only (antisense)	mRNA
Product | 产物	Two identical DNA double helices	Single-stranded mRNA	Polypeptide (protein)
Enzyme | 酶	DNA polymerase (I and III)	RNA polymerase (II for mRNA in eukaryotes)	Ribosome (peptidyl transferase — a ribozyme)
Monomer | 单体	dNTPs (dATP, dTTP, dCTP, dGTP)	NTPs (ATP, UTP, CTP, GTP)	Amino acids
Base pairing | 碱基配对	A–T, C–G	A–U, T–A, C–G, G–C	Codon–anticodon (mRNA–tRNA)
Location (eukaryotes) | 位置（真核）	Nucleus	Nucleus	Cytoplasm / rER
Direction of synthesis | 合成方向	5′ → 3′	5′ → 3′	N → C terminus
Primer required? | 需要引物？	Yes (RNA primer by primase)	No	No

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6. Common Exam Questions and Model Answers | 常见考题与标准答案

Q1: Describe how the structure of tRNA is related to its function. (4–6 marks)

Model Answer: tRNA has a cloverleaf secondary structure held together by hydrogen bonds between complementary bases (1). One loop contains an anticodon of three unpaired bases that is complementary to a specific mRNA codon, enabling codon–anticodon recognition (1). The 3′ end (CCA tail) serves as the amino acid attachment site, where a specific amino acid is covalently linked by an aminoacyl-tRNA synthetase (1). The small size of tRNA (≈75–90 nucleotides) allows it to fit into the ribosome’s A, P, and E sites during translation (1). Each tRNA is specific for one amino acid, ensuring the genetic code is accurately translated (1).

Q2: Explain why DNA replication is described as semi-conservative and describe the evidence that supports this model. (6 marks)

Model Answer: Semi-conservative replication means each new DNA molecule contains one original (parental) strand and one newly synthesised strand (1). The Meselson–Stahl experiment (1958) provided the evidence (1). E. coli were grown for many generations in a medium containing the heavy isotope ¹⁵N, so all DNA contained ¹⁵N (1). Bacteria were then transferred to a medium containing light ¹⁴N and allowed to divide once (1). DNA was extracted and centrifuged in a caesium chloride density gradient. After one generation, all DNA formed a single band at an intermediate density — consistent only with the semi-conservative model (1). After two generations, two bands appeared: one intermediate and one light, further confirming the model and ruling out the conservative and dispersive alternatives (1).

Q3: Compare and contrast DNA replication in prokaryotes and eukaryotes. (5 marks)

Feature | 特征	Prokaryotes | 原核生物	Eukaryotes | 真核生物
Genome structure	Single circular chromosome + plasmids	Multiple linear chromosomes
Origins of replication	Single origin (oriC)	Multiple origins per chromosome
DNA polymerases	DNA pol I, II, III, IV, V (pol III = main replicase)	DNA pol α, δ, ε (pol δ and ε = main replicases)
Okazaki fragment size	≈1,000–2,000 nucleotides	≈100–200 nucleotides
Replication rate	≈1,000 nucleotides/second	≈50 nucleotides/second
Telomeres	Not needed (circular chromosome)	Required; telomerase extends telomeres

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7. Key Terminology Summary | 关键术语总结

Term | 术语	Definition | 定义
Anticodon | 反密码子	Three-nucleotide sequence on tRNA complementary to an mRNA codon.
Codon | 密码子	Three-nucleotide sequence on mRNA that specifies an amino acid or a stop signal.
Exon | 外显子	Coding region of a gene that is retained in mature mRNA after splicing.
Intron | 内含子	Non-coding region of a gene removed during RNA splicing.
Okazaki fragment | 冈崎片段	Short DNA fragment synthesised discontinuously on the lagging strand during replication.
Phosphodiester bond | 磷酸二酯键	Covalent bond linking the 3′ carbon of one nucleotide to the 5′ carbon of the next via a phosphate group.
Polysome | 多聚核糖体	A complex of multiple ribosomes simultaneously translating a single mRNA molecule.
Promoter | 启动子	DNA sequence upstream of a gene where RNA polymerase binds to initiate transcription.
Proteome | 蛋白质组	The complete set of proteins expressed by a cell, tissue, or organism at a given time.
Spliceosome | 剪接体	Large RNP complex that catalyses the removal of introns from pre-mRNA.
Transcription factor | 转录因子	Protein that binds to specific DNA sequences to regulate the transcription of genes.

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8. Study Tips for A-Level Success | A-Level 成功的学习建议

1. Draw it out repeatedly. Being able to sketch and label the replication fork with all enzymes, the transcription bubble with RNA polymerase, and the ribosome with A/P/E sites is exam gold. Practice these diagrams until they become second nature.
2. Learn the enzyme names precisely. Examiners are strict about “DNA helicase” (not just “helicase”), “DNA polymerase” (not “polymerase”), and “RNA polymerase” (not “transcriptase”). Small distinctions carry marks.
3. Master the comparison tables. Questions asking you to “compare and contrast” replication, transcription, and translation are extremely common. A well-structured table can earn you full marks efficiently.
4. Understand the directionality. The concept of 5′ → 3′ synthesis explains almost everything — why the lagging strand exists, why Okazaki fragments are needed, why primers are required. This is the single most important concept to internalise.
5. Connect the processes. Don’t study replication, transcription, and translation as isolated topics. Understand how they form a continuous flow of information — the Central Dogma. When you see the big picture, the details make sense.

1. 反复画图。能够画出并标注带有所有酶的复制叉、带有 RNA 聚合酶的转录泡、以及带有 A/P/E 位点的核糖体，是考试得分的关键。反复练习这些图直到它们成为你的第二天性。
2. 精确学习酶的名称。考官对”DNA 解旋酶”（而不仅仅是”解旋酶”）、”DNA 聚合酶”（而不仅仅是”聚合酶”）和”RNA 聚合酶”（而不仅仅是”转录酶”）要求严格。细微的区别就能得分。
3. 掌握对比表。要求你”比较和对比”复制、转录和翻译的题目非常常见。一个结构良好的表格可以高效地为你赢得满分。
4. 理解方向性。5′ → 3′ 合成的概念几乎解释了一切——为什么存在滞后链，为什么需要冈崎片段，为什么需要引物。这是最需要内化的重要概念。
5. 串联各个过程。不要把复制、转录和翻译当作孤立的主题来学习。理解它们如何形成连续的信息流——中心法则。当你看到大局时，细节就变得合理了。

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This article covers the core A-Level Biology content on DNA replication, transcription, and translation. For further practice, consult your exam board’s specification, work through past paper questions, and practise drawing annotated diagrams of each process. Good luck with your studies!

本文涵盖了 A-Level 生物关于 DNA 复制、转录和翻译的核心内容。如需进一步练习，请查阅你的考试局大纲，完成历年真题，并练习绘制每个过程的注释图。祝学习顺利！