BiologyCBSE Class 12 · Molecular Basis of Inheritance

Genetic Code and Translation

Updated 15 October 2026 · Fact-checked

The genetic code is the set of triplet codons in mRNA that specify amino acids: 61 code for amino acids and 3 (UAA, UAG, UGA) stop translation. It is unambiguous, degenerate, comma-less and nearly universal, with AUG as the start codon. In translation, charged tRNAs read codons on the ribosome, which joins amino acids by peptide bonds.

Understand Genetic Code and Translation

Why a triplet code? The physicist George Gamow argued that 4 bases must code for 20 amino acids. A doublet code gives only $4^{2} = 16$ combinations (too few); a triplet gives $4^{3} = 64$ (enough). Har Gobind Khorana developed chemical methods to make RNA with defined base combinations (homopolymers, copolymers); Marshall Nirenberg's cell-free system for protein synthesis and Severo Ochoa's enzyme (polynucleotide phosphorylase), which polymerises RNA without a template, helped decipher the code.

Properties of the genetic code.
1. The codon is a triplet. 61 codons code for amino acids; 3 (UAA, UAG, UGA) code for none and act as stop codons.
2. Unambiguous and specific: one codon codes for only one amino acid.
3. Degenerate: some amino acids are coded by more than one codon (e.g. phenylalanine by UUU and UUC).
4. Comma-less: codons are read contiguously with no punctuation.
5. Nearly universal: UUU codes for phenylalanine from bacteria to humans; exceptions occur in mitochondrial codons and some protozoans.
6. AUG has a dual function: it codes for methionine and acts as the initiator (start) codon.

Mutations and the reading frame.

Point mutation: a single base change, e.g. GAG → GUG in the β-globin gene gives sickle-cell anaemia (Glu → Val).
Frameshift: insertion or deletion of one or two bases changes the reading frame from that point onward, so every later codon (and amino acid) changes.
Insertion or deletion of three bases (or multiples) adds or removes one (or more) codons; the reading frame after that point is unchanged.

Analogy: RAM HAS RED CAP. Insert B after HAS: RAM HAS BRE DCA P (nonsense after the insertion). Insert BIG: RAM HAS BIG RED CAP (one extra word, rest intact).

tRNA, the adapter. tRNA has an anticodon loop with bases complementary to the codon, and an amino acid acceptor end where the amino acid binds. Each tRNA is specific for one amino acid; a special initiator tRNA starts translation; there are no tRNAs for stop codons. Its secondary structure looks like a clover leaf; its 3-D shape is an inverted L.

Translation steps.
1. Charging (aminoacylation) of tRNA: each amino acid is activated using ATP and attached to its own (cognate) tRNA. When two charged tRNAs are brought close, peptide bond formation is favoured.
2. The ribosome, the protein factory, is made of structural rRNAs and about 80 proteins, in two subunits (large and small). Translation begins when the small subunit meets the mRNA. The large subunit has two sites where successive aminoacyl-tRNAs bind close enough for a peptide bond. The ribosome also acts as a catalyst: in bacteria the 23S rRNA is the enzyme (a ribozyme).
3. Initiation: the ribosome binds the mRNA at the start codon AUG, which is recognised only by the initiator tRNA (carrying methionine).
4. Elongation: aminoacyl-tRNAs bind their codons by anticodon-codon pairing; the ribosome moves from codon to codon; amino acids are joined one by one by peptide bonds.
5. Termination: a release factor binds the stop codon, and the complete polypeptide is released.

The translational unit and UTRs. The translational unit is the stretch of mRNA from the start codon (AUG) to a stop codon; it codes for the polypeptide. The mRNA also has untranslated regions (UTRs) at the 5' end (before the start codon) and the 3' end (after the stop codon); they are needed for efficient translation.

Flow chart: amino acid + ATP + tRNA → charged tRNA → small subunit binds mRNA at AUG (5' UTR before it) → initiator tRNA-Met → large subunit joins → elongation (codon-anticodon pairing, peptide bonds, ribosome moves 5'→3') → stop codon → release factor → polypeptide released (3' UTR left untranslated).

Draw and label 1: tRNA (clover leaf). Checklist: a clover leaf with a stem at the top ending in a single-stranded 3' end; three loops; the bottom loop with three bases. Labels: amino acid acceptor end (3'), anticodon loop, anticodon, 5' end, (other loops can be labelled 'loops').

Draw and label 2: translation on a ribosome. Checklist: mRNA as a horizontal line (5' to 3'), small and large subunits over it, two tRNAs sitting on adjacent codons, a growing chain of beads on one tRNA, an arrow showing ribosome movement. Labels: mRNA, codon, anticodon, tRNA, amino acid, peptide bond, small subunit, large subunit, growing polypeptide, 5' UTR, 3' UTR, start codon.

Genetic Code and Translation: key facts

Number of codons: $4^{3} = 64 = 61 sense + 3 stop$; A doublet code ( $4^{2} = 16$ ) would be too few for 20 amino acids.
Amino acids in a translational unit: $amino acids = \frac{bases from AUG to stop (inclusive)}{3} - 1$; The stop codon codes for no amino acid; UTRs are not translated.
Start and stop codons: $start: A U G (M e t); stop: U AA, U A G, U G A$; No tRNA exists for stop codons.
Codon-anticodon pairing: $codon 5^{'} -AUG- 3^{'} \leftrightarrow anticodon 3^{'} -UAC- 5^{'}$; Antiparallel, complementary (A-U, G-C).
Frameshift rule: $\pm 1 or \pm 2 bases: frame shifts; \pm 3 k bases: frame kept$; Point substitution changes one codon only.

How to answer Genetic Code and Translation questions

Translation questions usually give a DNA or mRNA sequence. Convert to mRNA, find the frame, read codons.

1If DNA is given, write the mRNA (coding strand with U for T, or complement of the template), keeping 5'→3'.
2Find the first AUG reading 5'→3'; everything before it is the 5' UTR.
3Read triplets from AUG without gaps until the first in-frame stop codon (UAA, UAG, UGA); everything after it is the 3' UTR.
4Count amino acids = number of codons from AUG up to (not including) the stop codon.
5For mutations, rewrite the sequence and re-read codons from the change: substitution (one codon), ±1 or ±2 (frameshift), ±3 (codon added or lost).
6For anticodons, write the complement with opposite polarity (codon 5'-UUU-3' → anticodon 3'-AAA-5').

Genetic Code and Translation in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Translational unit, UTRs and amino acid count

Suppose a biotechnology intern receives this short mRNA from a sequencing report: 5'-GGAUCCAUGUUUGGCAAAUGGUAACCAUU-3', and must tell her supervisor which part will be translated.

Identify the 5' UTR, the translational unit and the 3' UTR, and give the amino acid sequence.

Try it first, then show the solutionSolution

Scan 5'→3' for the first AUG: GGAUCC is before it, so the 5' UTR is GGAUCC.
From AUG read triplets: AUG UUU GGC AAA UGG UAA. UAA is a stop codon, so the translational unit is AUGUUUGGCAAAUGGUAA.
After the stop codon: CCAUU is the 3' UTR.
Amino acids: AUG Met, UUU Phe, GGC Gly, AAA Lys, UGG Trp; UAA stop. So 5 amino acids: Met-Phe-Gly-Lys-Trp.

Answer: 5' UTR GGAUCC; translational unit AUG UUU GGC AAA UGG UAA; 3' UTR CCAUU; peptide Met-Phe-Gly-Lys-Trp.

Memory hook: Start at AUG, stop at the stop; the ends are UTR padding.

Tricks to remember Genetic Code and Translation

Short memory aids. Say each one aloud once, then use it on a question the same day.

U Are Annoying, U Are Gone, U Go Away
UAA, UAG, UGA: the three stop codons.
Use it when: Recalling stop codons.
AUG = A Useful Gateway
AUG opens translation and brings methionine.
Use it when: Dual role of AUG.
One-two shifts, three fits
Add or delete 1 or 2 bases: frame shifts; 3 bases: one codon fits in or out.
Use it when: Frameshift questions.

Genetic Code and Translation: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Degeneracy · 1 mark · Assertion-reasonBoard level

Assertion (A): The genetic code is degenerate. Reason (R): Some amino acids are coded by more than one codon. Choose the correct option.

Try it first, then show the solutionSolution

Both statements are true.
R is the definition of degeneracy, so it explains A.

Answer: Both A and R are true and R is the correct explanation of A.

Example 2: Codon arithmetic · 2 marks · Very short answerHarder board level

Suppose a planet's life uses 5 different bases in its nucleic acids and 24 amino acids. What is the minimum codon length? How many codons would a code of that length have?

Try it first, then show the solutionSolution

Singlet: 5 codons; doublet: $5^{2} = 25$ codons (1 mark).
25 ≥ 24, so a doublet code is enough; it would have 25 codons (1 mark).

Answer: Doublet; 25 codons.

Example 3: Aminoacylation and the ribosome · 3 marks · Short answerBoard level

Explain the aminoacylation of tRNA and its role in translation. How does the ribosome act as a factory for protein synthesis?

Try it first, then show the solutionSolution

Aminoacylation (charging): each amino acid is activated using ATP and attached to its own specific tRNA (1 mark).
Charged tRNAs bring amino acids to the codon by anticodon pairing; when two are close, peptide bond formation is favoured (1 mark).
The ribosome (rRNA + about 80 proteins, two subunits) binds mRNA, holds two aminoacyl-tRNAs in adjacent sites so peptide bonds form, moves codon by codon, and its rRNA (23S in bacteria) catalyses peptide bond formation (1 mark).

Answer: ATP-activated amino acids join tRNA; ribosome positions tRNAs and catalyses peptide bonds as a ribozyme.

Example 4: Two mutations compared · 5 marks · Long answerHOTS / competency

A DNA template strand reads 3'-TAC AAA AAG AAG AAA ATT-5'. (a) Write the mRNA and the peptide. (b) Mutation I changes the second codon of the template from AAA to AAT. Mutation II deletes the first A of the second codon. Give the new mRNA and peptide in each case. (c) Which mutation is more harmful and why?

Try it first, then show the solutionSolution

(a) mRNA 5'-AUG UUU UUC UUC UUU UAA-3'; peptide Met-Phe-Phe-Phe-Phe (stop) (1 mark).
(b) Mutation I (substitution): second codon becomes UUA: Met-Leu-Phe-Phe-Phe (stop); only one amino acid changes (1 mark).
Mutation II (deletion of one base): template 3'-TAC AAA AGA AGA AAA TT-5' read in triplets → mRNA 5'-AUG UUU UCU UCU UUU AA-3': Met-Phe-Ser-Ser-Phe, and the original stop codon is lost from the frame (1 mark).
Every codon after the deletion changes, and the stop is lost, so translation continues into downstream sequence (1 mark).
(c) Mutation II (frameshift) is more harmful: it changes all downstream amino acids, while the point mutation changes only one (1 mark).

Answer: Normal Met-Phe-Phe-Phe-Phe; I: Met-Leu-Phe-Phe-Phe; II: Met-Phe-Ser-Ser-Phe... (frameshift, stop lost); frameshift is more harmful.

Example 5: UTRs · 2 marks · Very short answerHarder board level

In the mRNA 5'-AGGAGGUAUGAUCUCGUAAAAUAAA-3', identify the translational unit and the UTRs. What is the significance of UTRs?

Try it first, then show the solutionSolution

First AUG is after AGGAGGU; reading AUG AUC UCG UAA: translational unit AUGAUCUCGUAA (codes Met-Ile-Ser) (1 mark).
5' UTR: AGGAGGU (before AUG); 3' UTR: AAUAAA (after UAA). UTRs are not translated but are needed for efficient translation (1 mark).

Answer: Translational unit AUG AUC UCG UAA; 5' UTR AGGAGGU, 3' UTR AAUAAA; needed for efficient translation.

Example 6: From DNA to oligopeptide · 3 marks · Short answerHarder board level

A DNA template segment 3'-TAC CCG AAA GGT ACT-5' is transcribed and translated. Write the mRNA, the anticodon of the tRNA for the second codon, and the peptide formed.

Try it first, then show the solutionSolution

mRNA (complement, antiparallel): 5'-AUG GGC UUU CCA UGA-3' (1 mark).
Second codon GGC; anticodon 3'-CCG-5' (1 mark).
Peptide: Met-Gly-Phe-Pro, then UGA stops translation (1 mark).

Answer: mRNA 5'-AUGGGCUUUCCAUGA-3'; anticodon 3'-CCG-5'; Met-Gly-Phe-Pro.

Where students lose marks in Genetic Code and Translation

The slips examiners see most often, and what to write instead.

Mistake: Counting the stop codon as an amino acid
Students divide the length by 3 and stop.
Wrong working
453 bases → 151 amino acids.
Right working
151 codons − 1 stop = 150 amino acids.
Do this: Always subtract the stop codon; exclude UTRs.
Usually costs: 1 mark.
Mistake: Reading from the wrong end
mRNA is given 3'→5' and students read left to right.
Wrong working
Reading 3'-AUCAGG...-5' from the left as codons.
Right working
Rewrite 5'→3' first, then read codons.
Do this: Translation always reads mRNA 5'→3'.
Usually costs: 1 mark.
Mistake: Calling degeneracy 'ambiguity'
Two different ideas are mixed.
Wrong working
The code is ambiguous because one amino acid has many codons.
Right working
Degenerate: one amino acid, many codons. Unambiguous: one codon, one amino acid.
Do this: Direction matters: amino acid → codons (degenerate); codon → amino acid (unambiguous).
Usually costs: 1 mark.
Mistake: Saying UTRs are translated at the ends
Students assume the whole mRNA is translated.
Wrong working
UTRs code for the first and last amino acids.
Right working
UTRs are before AUG (5') and after the stop codon (3') and are not translated; they aid efficient translation.
Do this: Translation is only AUG → stop.
Usually costs: 1 mark.

Exam tips

Learn a short codon list for practice items: AUG Met, UUU/UUC Phe, AAA/AAG Lys, GGC Gly, UGG Trp, GUG Val, GAG Glu, and the three stops.
Write mRNA 5'→3' before reading codons; mark codons with spaces.
Name the scientists: Gamow (triplet idea), Khorana (synthetic RNA), Nirenberg (cell-free system), Ochoa (polynucleotide phosphorylase).
In translation answers, cover charging, initiation, elongation and termination, with AUG, initiator tRNA and release factor.
Use the RAM HAS RED CAP analogy briefly to explain frameshifts.

Practice questions on Genetic Code and Translation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which codon acts as both the initiator and a code for an amino acid?
(a) UAA (b) AUG (c) UGA (d) GUG
Hint
Methionine.
Answer
(b) AUG.
Q2 · 1 mark · MCQBoard level
The smallest part of a DNA molecule that can be changed by a point mutation is:
(a) a codon (b) a gene (c) a nucleotide base (d) an amino acid
Hint
Single base change.
Answer
(c) a nucleotide base.
Q3 · 2 marks · Very short answerHarder board level
How many amino acids are coded by an mRNA with a 5' UTR of 30 bases, a translational unit of 453 bases (including the stop codon) and a 3' UTR of 60 bases?
Hint
Only the translational unit counts; the stop codon codes for none.
Answer
453 ÷ 3 = 151 codons; minus 1 stop codon = 150 amino acids. The UTRs (30 and 60 bases) are not translated.
Q4 · 3 marks · Short answerHOTS / competency
Suppose a researcher inserts three extra bases right after the start codon of a gene, and another researcher inserts one extra base at the same place. Predict the effect on the protein in each case.
Hint
Reading frame.
Answer
Three bases: one extra codon, so one extra amino acid is added after methionine; the reading frame and all later amino acids are unchanged (protein probably works, perhaps slightly altered). One base: the reading frame shifts after the start codon, so all following codons are read wrongly, giving a different amino acid sequence and often a premature stop: usually a non-functional protein (frameshift).
Q5 · 4 marks · Case-basedHarder board level
Read and answer.
Suppose a cell-free system (like Nirenberg's) is given a synthetic mRNA made only of U (poly-U) and, separately, one made of repeating AAG (AAGAAGAAG...).
(a) Which amino acid chain forms from poly-U? (1)
(b) What chain forms from (AAG)n read from the first base? (1)
(c) If reading started at the second base, what would be formed? What property of the code does this show? (2)
Hint
UUU = Phe; AAG = Lys; AGA = Arg.
Answer
(a) Polyphenylalanine (UUU = Phe). (b) Polylysine (AAG = Lys). (c) AGA AGA... = polyarginine. The same sequence gives different products depending on where reading starts: the code is read in contiguous, comma-less triplets in a fixed frame.
Q6 · 2 marks · Very short answerHarder board level
What is the role of tRNA as an adapter? Name its two key regions and state why there are no tRNAs for UAA, UAG and UGA.
Hint
Codon reading and amino acid carrying.
Answer
tRNA reads the codon on mRNA through its anticodon and carries the matching amino acid at its acceptor end, linking nucleic acid language to protein language. Key regions: anticodon loop and amino acid acceptor end. Stop codons code for no amino acid; they are recognised by release factors, so no tRNA exists for them.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.How many codons code for amino acids?Show answer
61.
2.Name the stop codons.Show answer
UAA, UAG, UGA.
3.What is aminoacylation?Show answer
Charging of tRNA with its amino acid using ATP.
4.Which rRNA acts as an enzyme in bacteria?Show answer
23S rRNA.
5.Where are UTRs located?Show answer
At the 5' end before the start codon and at the 3' end after the stop codon.
6.What does UUU code for?Show answer
Phenylalanine.

Genetic Code and Translation: frequently asked questions

What are the main properties of the genetic code?

It is triplet, unambiguous, degenerate, comma-less and nearly universal; AUG is the start codon and UAA, UAG, UGA are stop codons.

What is a frameshift mutation?

Insertion or deletion of one or two bases, which changes the reading frame so that every codon after the change is read differently.

Why is tRNA called an adapter molecule?

It reads a codon through its anticodon and carries the matching amino acid, connecting the mRNA sequence to the protein sequence.

What are the steps of translation?

Charging of tRNA, initiation at AUG with the initiator tRNA, elongation by codon-anticodon pairing and peptide bond formation, and termination at a stop codon by a release factor.

What is a translational unit?

The part of mRNA from the start codon (AUG) to the stop codon that codes for a polypeptide; UTRs flank it on both sides.