OCR A Level Biology A: The genetic code, transcription and translation (H420)

This guide is for OCR A Level Biology A (H420), written to the OCR specification Version 4.1 (April 2026), for first assessment in 2025. It covers module 2.1.3 learning outcomes (f) and (g): the nature of the genetic code, and the transcription and translation of genes to make polypeptides. DNA structure and replication are in other units, so they are only used here as background. For quick recall use the revision notes, and for exam-style questions use the practice set. Course pages: OCR A Level Biology hub and the printable checklist. Not sure where you stand? Try the free 10-minute diagnostics. The wider module is in Foundations in Biology.

What this unit covers

Specification	What you must be able to do
2.1.3(f)	Describe the genetic code as triplet, non-overlapping, degenerate and universal. Explain how a gene determines the sequence of amino acids in a polypeptide (the primary structure of a protein).
2.1.3(g)	Describe transcription and translation of genes to make polypeptides, including the roles of RNA polymerase, messenger RNA (mRNA), transfer RNA (tRNA) and ribosomal RNA (rRNA).

Be ready to move between a DNA base sequence, an mRNA sequence and an amino acid sequence.

From gene to polypeptide

A gene is a length of DNA whose base order fixes the amino acid order in a polypeptide. That order is the primary structure of the protein. The primary structure then determines how the chain folds, because the R groups of the amino acids interact with each other. So the order of bases in DNA indirectly sets the shape, and therefore the function, of the finished protein.

DNA stays in the nucleus of a eukaryotic cell, but proteins are made on ribosomes. RNA carries the message between them. The flow of information is:

DNA (gene) → mRNA → polypeptide

Making mRNA from the gene is transcription. Using the mRNA to build a polypeptide is translation.

The nature of the genetic code

The genetic code is the set of rules that links base sequences to amino acids. Four features are required.

Triplet

A group of three bases codes for one amino acid. In mRNA, a group of three bases is a codon. The reason for three is a counting argument:

• one base per amino acid gives only 4 possible codes
• two bases give 4 × 4 = 16 codes, fewer than the 20 amino acids used in proteins
• three bases give 4 × 4 × 4 = 64 codes, enough for all 20 amino acids and for signals

Of the 64 codons, one is the start signal (AUG, which also codes for methionine) and three are stop signals (UAA, UAG and UGA). Stop codons do not code for an amino acid.

Non-overlapping

Each base belongs to only one codon. The sequence is read in successive blocks of three, with no base used twice. A run of 12 bases is read as 4 codons. If the code were overlapping, the same 12 bases would give 10 codons, because a new codon would start at every base.

Degenerate

There are 64 codons but only 20 amino acids, so most amino acids have several different codons. For example, GGU, GGC, GGA and GGG all code for glycine. Often it is the third base that differs. Degenerate does not mean ambiguous: each codon codes for one amino acid only.

Universal

The same codon codes for the same amino acid in virtually all organisms. This is why a gene from one species can be read and used to make a protein in a different species.

Worked example 1: counting bases

A polypeptide has 212 amino acids. How many bases are in the mRNA coding region, including the stop codon?

1. Number of codons for amino acids = 212.
2. Add one stop codon: 212 + 1 = 213 codons.
3. Bases = 213 × 3 = 639 bases.

The template DNA strand over the same region also has 639 bases. Check by dividing: 639 ÷ 3 = 213 codons.

Worked example 2: why the reading frame matters

Take this mRNA: AUG AAA CCU GAU. Read in threes from the first base, it gives methionine, lysine, proline, aspartic acid. Now start one base along: UGA AAC CUG AU. The first codon is UGA, a stop codon, so synthesis would end at once. The code is read in one fixed frame set by the start codon, which is why the code must be non-overlapping and read from a fixed starting point.

Transcription

Transcription takes place in the nucleus of a eukaryotic cell. It makes a single-stranded mRNA copy of one gene.

1. RNA polymerase binds to the DNA at the start of the gene. It separates the two DNA strands over a short length by breaking the hydrogen bonds between base pairs, exposing the bases.
2. One strand is used as the template strand. Free RNA nucleotides pair with the exposed template bases by complementary base pairing: A in the template pairs with U (not T), T with A, C with G and G with C.
3. RNA polymerase joins the RNA nucleotides together by forming phosphodiester bonds between the sugar of one nucleotide and the phosphate of the next, so a strand of mRNA grows.
4. As the enzyme moves along the gene, the DNA behind it re-forms its double helix. At the end of the gene the mRNA is released.
5. The mRNA leaves the nucleus through a nuclear pore and goes to a ribosome in the cytoplasm.

The mRNA base sequence is complementary to the template strand. It is the same as the other DNA strand (the coding strand), except that U replaces T.

Worked example 3: transcription

The template strand reads (3’ to 5’): TAC TTT GGA CTA ATC

Complementary mRNA (RNA uses U, never T): AUG AAA CCU GAU UAG

Coding strand (5’ to 3’): ATG AAA CCT GAT TAG. This matches the mRNA with T in place of U.

Using the codons AUG = methionine, AAA = lysine, CCU = proline, GAU = aspartic acid and UAG = stop, the polypeptide is Met-Lys-Pro-Asp, four amino acids.

The three types of RNA

RNA	Structure	Role
mRNA	Single polynucleotide strand, a copy of one gene. Made in transcription. Length varies with the gene.	Carries the genetic code from the gene to the ribosome. Its codons are read in translation.
tRNA	A single strand folded by hydrogen bonding between some of its bases into a clover-leaf shape. One end carries a specific amino acid. A loop carries a three-base anticodon.	Brings amino acids to the ribosome. The anticodon pairs with the complementary codon on mRNA.
rRNA	Long RNA molecules with a defined folded shape, combined with proteins.	Makes up most of the ribosome, which has a large and a small subunit. Holds mRNA and tRNA in position and helps form peptide bonds.

Each tRNA type carries only one kind of amino acid. This match is what makes the code work: the anticodon fixes which amino acid is added at each codon.

Translation

Translation takes place on ribosomes in the cytoplasm. It builds a polypeptide in the order set by the mRNA codons.

1. The small ribosomal subunit binds to the mRNA. It moves to the start codon AUG.
2. A tRNA with the anticodon UAC, carrying methionine, pairs with AUG by complementary base pairing. The large subunit joins.
3. A second tRNA, whose anticodon is complementary to the next codon, enters the ribosome. The ribosome holds two tRNA molecules at a time.
4. The two amino acids are joined by a peptide bond in a condensation reaction. The first tRNA leaves, free to pick up another amino acid.
5. The ribosome moves along the mRNA by one codon, and the next tRNA enters. Steps 3 to 5 repeat, lengthening the chain.
6. At a stop codon no tRNA matches. The polypeptide is released and the ribosome subunits separate.

Several ribosomes can read one mRNA at the same time, so many copies of the polypeptide are made from one transcript.

Worked example 4: translation of a short message

The mRNA is AUG AAA CCU GAU UAG.

Codon	tRNA anticodon	Amino acid added
AUG	UAC	methionine
AAA	UUU	lysine
CCU	GGA	proline
GAU	CUA	aspartic acid
UAG	none	stop, chain released

Each anticodon is the complement of its codon. The polypeptide is Met-Lys-Pro-Asp.

Worked example 5: rate of translation

A ribosome adds 8 amino acids per second (data supplied). How long does it take to make a polypeptide of 240 amino acids?

Time = 240 ÷ 8 = 30 seconds.

Common errors

• Writing T in mRNA. RNA contains uracil, and no thymine.
• Pairing the codon with the wrong molecule. Codons are on mRNA, anticodons are on tRNA.
• Forgetting the stop codon when counting bases.
• Saying “degenerate” means a codon can code for several amino acids. It is the other way round: several codons can code for one amino acid.
• Saying translation happens in the nucleus.
• Describing rRNA as only a “store”. It is part of the ribosome structure and takes part in building the polypeptide.
• Stating that RNA polymerase copies both DNA strands. One strand is the template for each gene.

Next steps

Work through the practice questions and tick off the topic on the checklist.

Official syllabus

OCR, A Level Biology A (H420) specification, Version 4.1 (April 2026), first assessment 2025: module 2.1.3 Nucleotides and nucleic acids, learning outcomes (f) and (g).