DNA and RNA Triplet Codes | DNA与RNA三联体密码

📚 DNA and RNA Triplet Codes | DNA与RNA三联体密码

The genetic information stored in DNA is expressed through a code that uses sequences of three nucleotides, known as triplet codes or codons. Each triplet of nucleotides specifies a particular amino acid, or signals the start or termination of protein synthesis. Understanding this code is fundamental to molecular biology and a core requirement of the Cambridge A-Level Biology syllabus.

DNA 中储存的遗传信息通过由三个核苷酸组成的序列——即三联体密码子——得以表达。每个核苷酸三联体指定一个特定的氨基酸，或发出蛋白质合成起始与终止的信号。理解这一密码是分子生物学的基础，也是剑桥 A-Level 生物考纲的核心要求。

1. The Triplet Nature of the Code | 三联体密码的本质

DNA contains only four different nitrogenous bases: adenine (A), thymine (T), cytosine (C) and guanine (G). Proteins, by contrast, are built from 20 different amino acids. If each base coded for one amino acid, only four amino acids could be specified; if pairs of bases coded for one amino acid, only 4² = 16 combinations would be possible, still insufficient to cover all 20 amino acids. A triplet code, however, provides 4³ = 64 possible combinations, more than enough to encode all 20 amino acids with redundancy to spare.

DNA 只含有四种不同的含氮碱基：腺嘌呤（A）、胸腺嘧啶（T）、胞嘧啶（C）和鸟嘌呤（G）。相比之下，蛋白质由 20 种不同的氨基酸构成。如果每个碱基编码一种氨基酸，只能指定 4 种氨基酸；如果每两个碱基编码一种氨基酸，只有 4² = 16 种组合，仍不足以覆盖全部 20 种氨基酸。而三联体密码则提供 4³ = 64 种可能的组合，足以编码所有 20 种氨基酸且有多余。

4³ = 64 possible triplet codons

4³ = 64 种可能的三联体密码子

Frameshift experiments by Francis Crick and Sydney Brenner in 1961 provided the first experimental evidence that the genetic code is read in triplets. By inserting or deleting one or two nucleotides in a gene, they observed that protein synthesis was completely disrupted; however, inserting or deleting three nucleotides restored partial function, confirming the triplet reading frame.

1961 年，弗朗西斯·克里克和悉尼·布伦纳通过移码突变实验首次获得遗传密码以三联体方式读取的实验证据。在基因中插入或删除一至两个核苷酸会使蛋白质合成完全中断；但插入或删除三个核苷酸则可部分恢复功能，从而证实了三联体阅读框的存在。

2. DNA vs RNA Codons | DNA与RNA密码子的区别

Although triplet codes are often discussed in the context of DNA, the codons actually used during translation are found on messenger RNA (mRNA). The DNA template strand is read during transcription to produce mRNA, but the mRNA codon sequence is complementary to the DNA template strand and identical to the DNA coding strand, except that uracil (U) replaces thymine (T).

虽然三联体密码常以 DNA 为背景讨论，但翻译过程中实际使用的密码子位于信使 RNA（mRNA）上。转录时以 DNA 模板链为模板合成 mRNA，mRNA 的密码子序列与 DNA 模板链互补，与 DNA 编码链一致，仅以尿嘧啶（U）替代胸腺嘧啶（T）。

Feature	DNA triplet	mRNA codon
Sugar	Deoxyribose	Ribose
Pyrimidines	T, C	U, C
Relationship	Template strand: complementary to mRNA	Coding strand equivalent (with U for T)

For example, if the DNA template strand reads 3′-TAC-5′, the corresponding mRNA codon is 5′-AUG-3′. On the other hand, if the DNA coding strand reads 5′-ATG-3′, the mRNA codon is also 5′-AUG-3′. This distinction is critical in exam questions that ask you to transcribe a given DNA sequence.

例如，若 DNA 模板链为 3′-TAC-5’，相应的 mRNA 密码子为 5′-AUG-3’。反之，若 DNA 编码链为 5′-ATG-3’，mRNA 密码子同样是 5′-AUG-3’。这一区别在考试中要求转录给定 DNA 序列的题目中至关重要。

3. Reading Frames and Codon Arrangement | 阅读框与密码子排列

The sequence of codons in mRNA is read in a fixed, non-overlapping manner from a start point. Each codon consists of three consecutive nucleotides, and codons do not share nucleotides. The reading frame is the specific grouping of nucleotides into codons, determined by the start codon.

mRNA 中密码子的读取方式是从起点开始固定、不重叠地进行。每个密码子由三个连续的核苷酸组成，密码子之间不共用核苷酸。阅读框是指核苷酸按特定方式分组为密码子的模式，由起始密码子决定。

Consider the mRNA sequence AUG GCA UCC UGA. It is read as four codons: AUG, GCA, UCC and UGA. If one nucleotide were inserted near the beginning, every subsequent codon would change — a frameshift mutation. This explains why insertions and deletions often have far more severe consequences than base substitutions.

考虑 mRNA 序列 AUG GCA UCC UGA，它被读取为四个密码子：AUG、GCA、UCC 和 UGA。如果在靠近起始处插入一个核苷酸，之后所有密码子都会改变——这就是移码突变。这解释了为什么插入和缺失通常比碱基替换造成的后果严重得多。

AUG GCA UCC UGA → Met-Ala-Ser-Stop

There is no punctuation or overlap in the code; translation simply proceeds codon by codon until a stop codon is encountered. This linear, non-overlapping arrangement ensures that a single mRNA molecule is translated accurately into a polypeptide chain of defined length.

遗传密码中没有标点，也不存在重叠；翻译只是逐密码子进行，直到遇到终止密码子。这种线性、不重叠的排列确保了单一 mRNA 分子能够准确翻译成确定长度的多肽链。

4. Start and Stop Codons | 起始密码子与终止密码子

Of the 64 codons, 61 encode amino acids and 3 are stop codons. The codon AUG serves a dual function: it encodes methionine and also acts as the start codon, setting the reading frame for translation. In prokaryotes, the start codon is sometimes GUG, encoding formylmethionine. The three stop codons are UAA, UAG and UGA; these do not correspond to amino acids but instead signal the release factor to terminate translation.

在 64 个密码子中，61 个编码氨基酸，3 个为终止密码子。密码子 AUG 具有双重功能：它编码甲硫氨酸，同时也作为起始密码子，为翻译设定阅读框。在原核生物中，起始密码子有时为 GUG，编码甲酰甲硫氨酸。三个终止密码子为 UAA、UAG 和 UGA；它们不对应任何氨基酸，而是向释放因子发出信号以终止翻译。

Codon	Amino acid / Function	氨基酸/功能
AUG	Methionine / Start	甲硫氨酸/起始
UAA	Stop (ochre)	终止（赭石型）
UAG	Stop (amber)	终止（琥珀型）
UGA	Stop (opal)	终止（蛋白石型）

In exam contexts, you must remember that the start codon AUG always codes for methionine at position one of the polypeptide, and that stop codons do not code for any amino acid. A polypeptide chain therefore begins with methionine, although this residue may be removed by post-translational modification.

在考试中，必须记住起始密码子 AUG 总是编码多肽链第一位的甲硫氨酸，而终止密码子不编码任何氨基酸。因此多肽链以甲硫氨酸开头，尽管该残基可能在翻译后修饰中被切除。

5. The Complete Genetic Code Table | 完整遗传密码表

The genetic code is conventionally presented as an RNA codon table, with the first, second and third bases of each codon arranged in rows, columns and sequence positions. The table below shows the standard code; you are expected to be able to use such a table to deduce amino acid sequences from given mRNA sequences.

遗传密码通常以 RNA 密码子表呈现，每个密码子的第一、第二和第三碱基分别排列为行、列和序列位置。下表展示了标准遗传密码；考试要求能够使用该表从给定的 mRNA 序列推导出氨基酸序列。

First base	Second base U	Second base C	Second base A	Second base G
U	UUU Phe, UUC Phe, UUA Leu, UUG Leu	UCU Ser, UCC Ser, UCA Ser, UCG Ser	UAU Tyr, UAC Tyr, UAA Stop, UAG Stop	UGU Cys, UGC Cys, UGA Stop, UGG Trp
C	CUU Leu, CUC Leu, CUA Leu, CUG Leu	CCU Pro, CCC Pro, CCA Pro, CCG Pro	CAU His, CAC His, CAA Gln, CAG Gln	CGU Arg, CGC Arg, CGA Arg, CGG Arg
A	AUU Ile, AUC Ile, AUA Ile, AUG Met/Start	ACU Thr, ACC Thr, ACA Thr, ACG Thr	AAU Asn, AAC Asn, AAA Lys, AAG Lys	AGU Ser, AGC Ser, AGA Arg, AGG Arg
G	GUU Val, GUC Val, GUA Val, GUG Val	GCU Ala, GCC Ala, GCA Ala, GCG Ala	GAU Asp, GAC Asp, GAA Glu, GAG Glu	GGU Gly, GGC Gly, GGA Gly, GGG Gly

Note that the table is read from the 5′ end. For example, the codon 5′-CAU-3′ has first base C, second base A and third base U, and therefore codes for histidine. Always confirm the direction of the mRNA sequence before consulting the table.

注意密码子表从 5′ 端开始读。例如，密码子 5′-CAU-3′ 的第一碱基为 C、第二碱基为 A、第三碱基为 U，因此编码组氨酸。查表前务必确认 mRNA 序列的方向。

6. Degeneracy of the Genetic Code | 遗传密码的简并性

The genetic code is described as degenerate because more than one codon can specify the same amino acid. For example, leucine is encoded by six codons: UUA, UUG, CUU, CUC, CUA and CUG. In most cases, the degeneracy arises from variation in the third base of the codon — the 3′ nucleotide — which is known as the wobble position.

遗传密码被称为简并的，因为多个密码子可编码同一种氨基酸。例如，亮氨酸由六个密码子编码：UUA、UUG、CUU、CUC、CUA 和 CUG。在大多数情况下，简并性源于密码子第三位碱基（3′ 端核苷酸）的变异，此位点称为摆动位点。

Degeneracy provides significant biological advantage: it minimises the harmful effects of point mutations. A single-base substitution at the third position of a codon often still codes for the same amino acid (a silent mutation), leaving the protein unchanged. This buffering effect increases the robustness of the genetic code against random mutation.

简并性提供了重要的生物学优势：它将点突变的有害效应降至最低。密码子第三位的单碱基替换通常仍编码相同的氨基酸（即沉默突变），蛋白质不受影响。这种缓冲效应增强了遗传密码对随机突变的鲁棒性。

7. Universality of the Genetic Code | 遗传密码的通用性

With very few exceptions, the same genetic code is used by all organisms — from bacteria to plants to humans. This universality strongly supports the theory of a common evolutionary origin for all life on Earth. It also explains why genes can be transferred between species, a principle exploited in genetic engineering.

除极少数例外，从细菌到植物再到人类，所有生物使用相同的遗传密码。这种通用性有力地支持了地球上所有生命具有共同进化起源的理论。它也解释了基因能够在物种间转移的原因——这一原理被广泛应用于基因工程。

Notable exceptions include the mitochondrial genetic codes of mammals, in which UGA codes for tryptophan rather than stop, and AUA codes for methionine rather than isoleucine. Some ciliates, such as Tetrahymena, use UAA and UAG to code for glutamine instead of as stop signals. These variations are minor but demonstrate that the code is not absolutely immutable.

值得注意的例外包括哺乳动物线粒体遗传密码：其中 UGA 编码色氨酸而非终止信号，AUA 编码甲硫氨酸而非异亮氨酸。某些纤毛虫（如四膜虫）使用 UAA 和 UAG 编码谷氨酰胺而非终止信号。这些变异虽小，却说明遗传密码并非绝对不可改变。

For the Cambridge A-Level syllabus, you should be able to state that the code is universal and non-overlapping, and discuss the evidence for universality and the evolutionary implications of exceptions.

对于剑桥 A-Level 考纲，你应当能够陈述遗传密码具有通用性和不重叠性，并讨论通用性的证据以及例外情况的进化意义。

8. Mutations and Triplet Codes | 突变与三联体密码

Mutations are changes to the nucleotide sequence of DNA, and their effects depend on how they alter the triplet code. Point mutations (base substitutions) can be classified as silent, missense or nonsense depending on the new codon formed. Insertions and deletions cause frameshifts, shifting the reading frame and altering all subsequent codons.

突变是 DNA 核苷酸序列的变化，其效应取决于它们如何改变三联体密码。点突变（碱基替换）根据新形成的密码子可分为沉默突变、错义突变或无义突变。插入和缺失引起移码，使阅读框发生移动并改变之后所有密码子。

Silent mutation: A base change that still codes for the same amino acid, e.g. CUU → CUC both code for leucine; no effect on the protein.
沉默突变：碱基改变后仍编码同一种氨基酸，如 CUU → CUC 均编码亮氨酸；对蛋白质无影响。
Missense mutation: A base change that codes for a different amino acid, e.g. GAA → GUA changes glutamic acid to valine, as in sickle-cell anaemia (GAG → GTG in the DNA coding strand).
错义突变：碱基改变后编码不同的氨基酸，如 GAA → GUA 将谷氨酸变为缬氨酸，镰状细胞贫血即由此引起（DNA 编码链中 GAG → GTG）。
Nonsense mutation: A base change that creates a premature stop codon, e.g. UAU → UAA, truncating the polypeptide.
无义突变：碱基改变产生提前的终止密码子，如 UAU → UAA，使多肽链截短。
Frameshift mutation: Insertion or deletion of 1 or 2 nucleotides shifts the reading frame; insertion or deletion of 3 nucleotides adds or removes a whole codon without shifting the frame.
移码突变：插入或缺失 1 至 2 个核苷酸使阅读框移动；插入或缺失 3 个核苷酸则增加或移除整个密码子而不改变阅读框。

Understanding the relationship between mutation type and codon change allows you to predict the likely impact on the polypeptide product — a common assessment objective in A-Level biology examinations.

理解突变类型与密码子变化之间的关系，使你能够预测多肽产物的可能影响——这是 A-Level 生物考试中常见的评估目标。

9. tRNA Anticodons and Base Pairing | tRNA反密码子与碱基配对

During translation, transfer RNA (tRNA) molecules carry amino acids to the ribosome. Each tRNA contains an anticodon, a triplet of nucleotides complementary to the mRNA codon. The anticodon binds to the codon via hydrogen bonds between complementary bases: A pairs with U, and G pairs with C.

在翻译过程中，转运 RNA（tRNA）分子将氨基酸携带至核糖体。每个 tRNA 含有一个反密码子，即与 mRNA 密码子互补的三核苷酸序列。反密码子通过互补碱基间的氢键与密码子结合：A 与 U 配对，G 与 C 配对。

For example, the mRNA codon 5′-AUG-3′ is recognised by the tRNA anticodon 3′-UAC-5′. Note the antiparallel orientation: the codon is written 5′ to 3′, while the anticodon is written 3′ to 5′. When writing anticodons, you must reverse and complement the codon sequence.

例如，mRNA 密码子 5′-AUG-3′ 由 tRNA 反密码子 3′-UAC-5′ 识别。注意反向平行方向：密码子以 5′ 到 3′ 书写，而反密码子以 3′ 到 5′ 书写。书写反密码子时，须将密码子序列反转并取互补碱基。

The wobble hypothesis explains why the third base of the codon shows relaxed pairing specificity. For instance, the tRNA anticodon containing inosine at the wobble position can pair with U, C or A at the third position of the codon, allowing a single tRNA to recognise multiple codons. This reduces the number of tRNA molecules required and contributes to the efficiency of translation.

摆动假说解释了为什么密码子第三位碱基的配对特异性较为宽松。例如，反密码子摆动位点含有肌苷时，可分别与密码子第三位的 U、C 或 A 配对，使单个 tRNA 能够识别多个密码子。这减少了所需 tRNA 分子的数量，并提高了翻译效率。

10. Examination Focus | 考试重点

In the Cambridge A-Level examination, questions on the triplet code commonly require you to: (1) transcribe a DNA base sequence into mRNA codons; (2) translate mRNA codons into an amino acid sequence using a provided codon table; (3) deduce the DNA coding and template strands from a given mRNA sequence; and (4) predict the consequences of specific mutations on the amino acid sequence.

在剑桥 A-Level 考试中，关于三联体密码的题目通常要求你：（1）将 DNA 碱基序列转录为 mRNA 密码子；（2）利用提供的密码子表将 mRNA 密码子翻译为氨基酸序列；（3）从给定的 mRNA 序列推断 DNA 编码链和模板链；（4）预测特定突变对氨基酸序列的影响。

Common examiner tips include: always mark the 5′ and 3′ ends of sequences, read the codon table from the correct end, remember that U replaces T in RNA, and be careful to distinguish between the template strand and the coding strand when transcribing. Practising past paper questions on the genetic code is the most reliable way to secure full marks on this topic.

常见的考官提示包括：始终标注序列的 5′ 和 3′ 端、从正确的一端读取密码子表、记住 RNA 中以 U 替代 T，以及在转录时仔细区分模板链与编码链。练习有关遗传密码的历年真题是在此考点获得满分的最可靠方法。

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DNA and RNA Triplet Codes | DNA与RNA三联体密码