Transcription and Translation: The Central Dogma of Molecular Biology | 转录与翻译：分子生物学的中心法则

Every living organism — from a single-celled bacterium to a human being — relies on proteins to carry out virtually every biological function. But how does a cell know which proteins to make, and when? The answer lies in gene expression, a two-step process governed by the so-called Central Dogma of Molecular Biology: DNA makes RNA, and RNA makes protein. These two steps are known as transcription (copying DNA into messenger RNA) and translation (decoding mRNA into a polypeptide chain). Understanding this fundamental process is essential for A-Level Biology students — it appears on virtually every exam board’s specification, from CIE and AQA to Edexcel and OCR.

每一个生物体——从单细胞细菌到人类——都依赖蛋白质来执行几乎所有生物功能。但细胞是如何知道该制造哪些蛋白质、何时制造的呢？答案在于基因表达，这是一个由分子生物学中心法则控制的两步过程：DNA制造RNA，RNA制造蛋白质。这两个步骤分别被称为转录（将DNA复制成信使RNA）和翻译（将mRNA解码为多肽链）。理解这一基本过程对A-Level生物学学生至关重要——从CIE、AQA到Edexcel和OCR，它几乎出现在每个考试局的考纲中。

1. The Central Dogma — An Overview

The Central Dogma of Molecular Biology was first proposed by Francis Crick in 1958. It describes the directional flow of genetic information within a biological system:

DNA → RNA → Protein

In eukaryotic cells (such as animal and plant cells), DNA is safely stored inside the nucleus. However, protein synthesis occurs in the cytoplasm, on ribosomes. This spatial separation creates a logistical problem: the genetic code cannot leave the nucleus directly. The solution is transcription — creating a portable, single-stranded copy of a gene in the form of messenger RNA (mRNA). This mRNA molecule then exits the nucleus through nuclear pores and travels to the ribosomes, where translation takes place.

It is worth remembering the fundamental differences between DNA and RNA. DNA is double-stranded, contains the sugar deoxyribose, and uses the nitrogenous bases adenine (A), thymine (T), cytosine (C), and guanine (G). RNA is single-stranded, contains the sugar ribose, and replaces thymine with uracil (U). So whenever you see a U in RNA, think of it as the RNA equivalent of T in DNA.

值得记住的是DNA和RNA之间的基本区别。DNA是双链的，含有脱氧核糖，使用的含氮碱基是腺嘌呤(A)、胸腺嘧啶(T)、胞嘧啶(C)和鸟嘌呤(G)。RNA是单链的，含有核糖，并用尿嘧啶(U)替代了胸腺嘧啶。所以每当你在RNA中看到U，可以将其视为RNA中T的等价物。

2. Transcription — DNA to mRNA
转录——从DNA到mRNA

2.1 What is Transcription?

Transcription is the process by which the genetic information stored in a gene (a specific sequence of DNA) is copied into a complementary strand of messenger RNA. The enzyme responsible for this process is RNA polymerase.

转录是将基因（一段特定的DNA序列）中储存的遗传信息复制到一条互补的信使RNA链中的过程。负责这一过程的酶是RNA聚合酶。

2.2 The Stages of Transcription

Transcription occurs in three main stages: initiation, elongation, and termination.

Initiation (起始): Transcription begins when RNA polymerase binds to a specific DNA sequence called the promoter region, located just before (upstream of) the gene. In eukaryotes, transcription factors help RNA polymerase recognise and bind to the promoter. The section of DNA containing the gene then unwinds and the hydrogen bonds between the two strands break, exposing the template strand.

Elongation (延伸): RNA polymerase moves along the template strand (also called the antisense strand) in the 3′ to 5′ direction. As it moves, it reads the DNA bases and assembles a complementary mRNA strand in the 5′ to 3′ direction, using free RNA nucleotides present in the nucleoplasm. Remember the base-pairing rules: adenine (A) in DNA pairs with uracil (U) in mRNA (not thymine!), thymine (T) in DNA pairs with adenine (A) in mRNA, cytosine (C) pairs with guanine (G), and guanine (G) pairs with cytosine (C).

Termination (终止): RNA polymerase continues until it reaches a terminator sequence in the DNA. At this point, RNA polymerase detaches from the DNA, and the newly formed pre-mRNA molecule is released. The DNA double helix rewinds behind the enzyme. In prokaryotes, the mRNA is ready for translation immediately. In eukaryotes, however, the pre-mRNA must undergo post-transcriptional modification (see Section 3) before it can leave the nucleus.

起始: 转录开始时，RNA聚合酶与基因上游的特定DNA序列（称为启动子区域）结合。在真核生物中，转录因子帮助RNA聚合酶识别并与启动子结合。包含基因的DNA片段随后解旋，两条链之间的氢键断裂，暴露出模板链。

延伸: RNA聚合酶沿模板链（也称反义链）以3’到5’方向移动。它在移动过程中读取DNA碱基，利用核质中的游离RNA核苷酸，以5’到3’方向组装互补的mRNA链。记住碱基配对规则：DNA中的腺嘌呤(A)与mRNA中的尿嘧啶(U)配对（不是胸腺嘧啶！），DNA中的胸腺嘧啶(T)与mRNA中的腺嘌呤(A)配对，胞嘧啶(C)与鸟嘌呤(G)配对，鸟嘌呤(G)与胞嘧啶(C)配对。

终止: RNA聚合酶继续移动，直到遇到DNA中的终止序列。此时，RNA聚合酶从DNA上脱离，新形成的前体mRNA分子被释放。DNA双螺旋在酶后方重新缠绕。在原核生物中，mRNA立即就能进行翻译。但在真核生物中，前体mRNA必须经过转录后修饰（见第3节）才能离开细胞核。

2.3 The Template Strand vs. the Coding Strand

This is a common source of confusion in exam questions. Of the two DNA strands:

The template strand (antisense strand) is the strand that RNA polymerase actually reads. The mRNA produced is complementary to this strand.

The coding strand (sense strand) has the same sequence as the mRNA (with T replaced by U). It is NOT read by RNA polymerase.

Examiner’s tip: If you are given a DNA sequence and asked to write the mRNA sequence, always transcribe from the template strand, replacing T with U. Do NOT simply copy the coding strand — it may seem the same but the question expects you to demonstrate understanding of which strand is transcribed.

这是一个考试中常见的混淆点。在两条DNA链中：模板链（反义链）是RNA聚合酶实际读取的链。产生的mRNA与该链互补。编码链（有义链）与mRNA的序列相同（T替换为U）。它不被RNA聚合酶读取。考官提示：如果给你一条DNA序列并要求写出mRNA序列，总是从模板链转录，将T替换为U。不要简单地复制编码链——虽然看起来一样，但题目期望你展示对哪条链被转录的理解。

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

In eukaryotic cells, the primary transcript (pre-mRNA) must be processed before it can leave the nucleus. This processing involves three key modifications:

1. 5′ Capping (加帽): A modified guanine nucleotide (7-methylguanosine) is added to the 5′ end of the pre-mRNA. This 5′ cap protects the mRNA from enzymatic degradation in the cytoplasm and helps the ribosome recognise and bind to the mRNA during translation.

2. Polyadenylation (加尾): A long chain of adenine nucleotides — the poly-A tail — is added to the 3′ end of the pre-mRNA. This tail also protects against degradation and facilitates the export of mRNA from the nucleus to the cytoplasm. The poly-A tail gradually shortens over time, which determines the lifespan of the mRNA molecule.

3. Splicing (剪接): Eukaryotic genes contain introns (non-coding sequences) and exons (coding sequences). The introns must be removed and the exons joined together to form a continuous coding sequence. This process, called splicing, is carried out by a complex of RNA and proteins called the spliceosome. After splicing, the mature mRNA consists only of exons and is ready to leave the nucleus.

Alternative splicing is a remarkable phenomenon where the same pre-mRNA can be spliced in different ways, joining different combinations of exons to produce multiple different proteins from a single gene. This is one reason why the human genome (with approximately 20,000–25,000 genes) can produce over 100,000 different proteins.

在真核细胞中，初级转录本（前体mRNA）在离开细胞核之前必须经过加工。这一加工涉及三个关键修饰：5’端加帽（在5’端添加一个修饰的鸟嘌呤核苷酸，保护mRNA并帮助核糖体识别）、多聚腺苷酸化（在3’端添加poly-A尾，保护mRNA并促进其从核输出）和剪接（去除内含子、连接外显子）。可变剪接允许同一基因通过不同剪接方式产生多种不同的蛋白质——这就是为什么人类约20,000-25,000个基因能产生超过100,000种蛋白质的原因之一。

4. Translation — mRNA to Polypeptide
翻译——从mRNA到多肽

4.1 What is Translation?

Translation is the process by which the nucleotide sequence of mRNA is decoded into the amino acid sequence of a polypeptide chain. This process takes place on ribosomes in the cytoplasm (or on the rough endoplasmic reticulum).

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。这一过程发生在细胞质（或粗面内质网）的核糖体上。

4.2 The Genetic Code

The genetic code is a set of rules that maps each sequence of three nucleotides (called a codon) to a specific amino acid. There are 64 possible codons (4³ = 64), but only 20 standard amino acids. This means the code is degenerate — multiple codons can code for the same amino acid.

Key features of the genetic code:

Triplet code: Three nucleotides (one codon) code for one amino acid.

Non-overlapping: Each nucleotide belongs to only one codon. The ribosome reads the mRNA in groups of three, one after another.

Degenerate (简并性): Most amino acids are coded for by more than one codon. For example, the amino acid leucine is coded for by six different codons (UUA, UUG, CUU, CUC, CUA, CUG).

Universal (通用性): With very few exceptions, the genetic code is the same in all organisms — from bacteria to humans. This is powerful evidence for a common evolutionary origin.

Start and Stop Codons: The codon AUG codes for methionine and also serves as the start codon, signalling the beginning of translation. Three codons — UAA, UAG, and UGA — are stop codons. They do not code for any amino acid; instead, they signal the ribosome to terminate translation.

遗传密码的关键特征：三联体密码（三个核苷酸=一个密码子=一个氨基酸）、非重叠（每个核苷酸只属于一个密码子）、简并性（多个密码子编码同一氨基酸）、通用性（几乎所有生物共享同一套密码，这是共同进化起源的有力证据）。AUG既是甲硫氨酸的密码子，也是起始密码子；UAA、UAG和UGA是终止密码子，不编码任何氨基酸，而是发出翻译终止的信号。

4.3 The Machinery of Translation

Translation requires three key types of RNA, along with ribosomes:

mRNA (messenger RNA): Carries the genetic code from the nucleus to the ribosome. Each codon on the mRNA specifies one amino acid.

tRNA (transfer RNA): Small RNA molecules (approximately 80 nucleotides long) with a characteristic cloverleaf structure. Each tRNA molecule has two critical regions: the anticodon (a sequence of three nucleotides complementary to an mRNA codon) and an amino acid attachment site at the 3′ end. The enzyme aminoacyl-tRNA synthetase attaches the correct amino acid to its corresponding tRNA — a process called tRNA charging.

rRNA (ribosomal RNA): Along with proteins, rRNA forms the structure of ribosomes. Ribosomes have two subunits — a small subunit (which binds to the mRNA) and a large subunit (which catalyses peptide bond formation). The large subunit has three binding sites for tRNA: the A site (aminoacyl), the P site (peptidyl), and the E site (exit).

tRNA的特点：tRNA具有特征性的三叶草结构。每个tRNA分子有两个关键区域：反密码子（与mRNA密码子互补的三个核苷酸序列）和3’端的氨基酸附着位点。氨酰tRNA合成酶将正确的氨基酸连接到其对应的tRNA上——这一过程称为tRNA充电。核糖体有两个亚基——小亚基（结合mRNA）和大亚基（催化肽键形成）。大亚基有三个tRNA结合位点：A位点（氨酰位）、P位点（肽基位）和E位点（出口位）。

4.4 The Stages of Translation

Translation also proceeds through three stages: initiation, elongation, and termination.

Initiation (起始): The small ribosomal subunit binds to the mRNA near the 5′ cap and scans along until it reaches the start codon AUG. A tRNA molecule carrying methionine (with the anticodon UAC) binds to the start codon via complementary base pairing. This tRNA enters the P site directly. The large ribosomal subunit then attaches, forming a complete, functional ribosome.

Elongation (延伸): A second tRNA, carrying the amino acid specified by the next codon, enters the A site. A peptide bond forms between the amino acids in the P site and A site — catalysed by peptidyl transferase (an rRNA enzyme, a type of ribozyme). The ribosome then translocates (moves along the mRNA by one codon). The tRNA that was in the P site moves to the E site and exits; the tRNA that was in the A site moves to the P site. The A site is now empty and ready for the next tRNA. This process repeats, adding one amino acid at a time, with the polypeptide chain growing from the N-terminus to the C-terminus. Each elongation step requires energy from GTP hydrolysis.

Termination (终止): When a stop codon (UAA, UAG, or UGA) enters the A site, no tRNA can recognise it. Instead, a release factor protein binds to the stop codon. This triggers the ribosome to release the completed polypeptide chain and disassemble into its two subunits. The mRNA is also released and can be translated again.

起始：核糖体小亚基在5’帽附近与mRNA结合，沿mRNA扫描直到找到起始密码子AUG。携带甲硫氨酸的tRNA（反密码子UAC）通过互补碱基配对与起始密码子结合，直接进入P位点。随后大亚基附着，形成完整的核糖体。延伸：携带下一个密码子指定氨基酸的tRNA进入A位点。P位点和A位点的氨基酸之间形成肽键——由肽基转移酶（一种rRNA酶，属于核酶）催化。核糖体随后移位一个密码子。P位点的tRNA移至E位点并离开，A位点的tRNA移至P位点。A位点变空，为下一个tRNA做好准备。多肽链从N端向C端延伸。每次延伸需要GTP水解提供能量。终止：当终止密码子进入A位点，没有tRNA能识别它。取而代之的是释放因子蛋白与终止密码子结合，触发核糖体释放完整的多肽链并解体。

5. Comparison: Transcription vs. Translation

Feature 特征	Transcription 转录	Translation 翻译
Location 位置	Nucleus (eukaryotes) 细胞核	Cytoplasm / Ribosomes 细胞质/核糖体
Template 模板	DNA (template strand)	mRNA
Product 产物	mRNA (single-stranded)	Polypeptide chain
Enzyme 酶	RNA polymerase	Ribosome (peptidyl transferase)
Monomers 单体	RNA nucleotides (A, U, C, G)	Amino acids (20 types)
Base pairing 配对	DNA-RNA: A-U, T-A, C-G, G-C	mRNA-tRNA: A-U, U-A, C-G, G-C
Direction 方向	mRNA synthesised 5’→3′	Polypeptide synthesised N→C

6. Common Exam Questions and Pitfalls
常见考题与易错点

Q1: “Describe how the base sequence of a gene determines the primary structure of a protein.”

This is a classic 5–6 mark question. Structure your answer logically:

(1) During transcription, RNA polymerase binds to the promoter region of the gene. The DNA unwinds and RNA polymerase reads the template strand. (2) Free RNA nucleotides pair with complementary DNA bases — A with U, T with A, C with G, G with C. RNA polymerase joins adjacent nucleotides with phosphodiester bonds to form pre-mRNA. (3) The pre-mRNA undergoes splicing to remove introns, and a 5′ cap and poly-A tail are added. (4) The mature mRNA leaves the nucleus and binds to a ribosome in the cytoplasm. (5) During translation, the ribosome reads the mRNA three nucleotides at a time (codons). tRNA molecules bring specific amino acids, with their anticodons pairing with the mRNA codons. (6) Peptide bonds form between adjacent amino acids, creating a polypeptide chain. The order of amino acids in the polypeptide (the primary structure) is determined by the sequence of codons in the mRNA, which in turn is determined by the base sequence of the gene.

Q2: “Explain why the genetic code is described as degenerate.”

The genetic code is degenerate because most amino acids are coded for by more than one codon. For example, leucine is coded for by six different codons. This degeneracy provides a protective effect — if a point mutation changes the third base of a codon, there is a good chance it will still code for the same amino acid (this is called a silent mutation). The degeneracy also reduces the impact of errors during DNA replication and transcription.

Common Pitfalls 易错点

• Confusing template and coding strands: Always transcribe from the template strand. The coding strand has the same sequence as the mRNA (T→U).
• Forgetting that RNA uses uracil, not thymine: A common careless error — in RNA, A pairs with U, not T.
• Calling the product of translation a “protein”: Translation produces a polypeptide. It only becomes a functional protein after folding and, often, post-translational modification.
• Thinking the ribosome moves along the mRNA: It’s actually the ribosome that stays still relative to the growing polypeptide — the mRNA is threaded through the ribosome.
• Forgetting post-transcriptional modification in eukaryotes: Pre-mRNA must be capped, tailed, and spliced before translation. Prokaryotes skip all of this.
• 混淆模板链和编码链：总是从模板链转录。编码链与mRNA序列相同（T→U）。
• 忘记RNA用尿嘧啶而非胸腺嘧啶：常见粗心错误——RNA中A与U配对，不是T。
• 将翻译产物称为”蛋白质”：翻译产生的是多肽。只有经过折叠和翻译后修饰才能成为功能性蛋白质。
• 忘记真核生物中的转录后修饰：前体mRNA必须加帽、加尾和剪接才能进行翻译。原核生物跳过所有这些步骤。

7. Summary Table for Revision
复习总结表

Process 过程	Input 输入	Output 输出	Key Enzyme 关键酶	Location (Eukaryote) 位置
Transcription	DNA (gene)	pre-mRNA	RNA polymerase	Nucleus
RNA Processing	pre-mRNA	Mature mRNA	Spliceosome	Nucleus
Translation	mRNA + tRNA + amino acids	Polypeptide	Peptidyl transferase (rRNA)	Cytoplasm / RER

8. Key Vocabulary 核心词汇

English	中文	Definition
Transcription	转录	Synthesis of mRNA from a DNA template
Translation	翻译	Synthesis of a polypeptide from an mRNA template
RNA Polymerase	RNA聚合酶	Enzyme that catalyses transcription
Template Strand	模板链	The DNA strand read by RNA polymerase
Codon	密码子	Sequence of 3 mRNA nucleotides coding for 1 amino acid
Anticodon	反密码子	Sequence of 3 tRNA nucleotides complementary to a codon
Ribosome	核糖体	Organelle that carries out translation
Splicing	剪接	Removal of introns from pre-mRNA
Intron / Exon	内含子/外显子	Non-coding / Coding sequence in eukaryotic genes
Degenerate Code	简并密码	Multiple codons can code for the same amino acid
Peptide Bond	肽键	Covalent bond between adjacent amino acids
Polypeptide	多肽	Chain of amino acids — the product of translation
Gene Expression	基因表达	The process by which a gene’s information is used to synthesise a functional product

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This topic is a cornerstone of A-Level Biology. Master it, and you will have a solid foundation for understanding genetics, evolution, and the molecular basis of life. Good luck with your studies! 这一主题是A-Level生物学的基石。掌握它，你将拥有理解遗传学、进化和生命的分子基础的坚实基础。祝学习顺利！

Exam boards covered: CIE A-Level Biology (9700), AQA A-Level Biology (7402), Edexcel A-Level Biology B (9BI0), OCR A-Level Biology A (H420)