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

These are condensed revision notes for OCR A Level Biology A (H420), aligned to the OCR specification Version 4.1 (April 2026), for first assessment in 2025. They cover module 2.1.3 learning outcomes (f) and (g). For full explanations read the study guide, then test yourself with the practice questions. Course pages: the OCR A Level Biology hub, the printable checklist and the free 10-minute diagnostics. Background is in Foundations in Biology revision notes.

The big picture

Step	Where (eukaryote)	Template	Product	Key enzyme or structure
Transcription	Nucleus	One strand of a gene	mRNA	RNA polymerase
Translation	Ribosome in cytoplasm	mRNA	Polypeptide	Ribosome (rRNA and protein), tRNA

Gene (DNA base sequence) determines mRNA codons, which determine the amino acid sequence. That sequence is the primary structure of the protein, and it decides how the chain folds.

2.1.3(f) The genetic code

Feature	Meaning	One-line evidence or example
Triplet	3 bases (one codon) code for 1 amino acid	4 × 4 = 16 is too few for 20 amino acids; 4 × 4 × 4 = 64 is enough
Non-overlapping	Each base is part of only one codon	12 bases make 4 codons, not 10
Degenerate	Most amino acids have more than one codon	GGU, GGC, GGA, GGG all code for glycine
Universal	The same codon codes for the same amino acid in virtually all organisms	A gene from one species can be read in another

Other facts to hold:

• 64 codons in total: 61 code for amino acids (including the start codon AUG, which codes for methionine) and 3 are stop codons (UAA, UAG, UGA).
• Stop codons code for no amino acid.
• Degenerate is not ambiguous: one codon never codes for two amino acids.
• The reading frame is fixed by the start codon, so shifting the frame changes every codon after it.

Base pairing rules

DNA template base	Pairs with (in mRNA)
A	U
T	A
C	G
G	C

mRNA matches the coding (non-template) strand, with U instead of T. A tRNA anticodon is complementary to its mRNA codon.

Method in steps: from template to polypeptide

1. Write the template strand in groups of three.
2. Write the complementary mRNA codons under it (A-U, T-A, C-G, G-C).
3. Find AUG, the start, and read codons in threes until a stop codon.
4. Use the codon list in the question to find each amino acid.
5. Give the anticodons if asked: complement each mRNA codon.

Worked reminder: template TAC GGA CTT TTT ACC gives mRNA AUG CCU GAA AAA UGG. With AUG = Met, CCU = Pro, GAA = Glu, AAA = Lys and UGG = Trp the polypeptide is Met-Pro-Glu-Lys-Trp.

2.1.3(g) Transcription in order

1. RNA polymerase binds at the start of the gene.
2. It breaks hydrogen bonds between the two DNA strands over a short length and exposes the template bases.
3. Free RNA nucleotides line up by complementary base pairing with the template.
4. RNA polymerase forms phosphodiester bonds between neighbouring RNA nucleotides, building the mRNA strand.
5. The DNA re-forms its double helix behind the enzyme.
6. mRNA is released and leaves the nucleus through a nuclear pore.

Translation in order

1. Small ribosomal subunit attaches to mRNA and finds the start codon AUG.
2. tRNA with anticodon UAC (carrying methionine) pairs with AUG. The large subunit joins.
3. A second tRNA pairs with the next codon. The ribosome holds two tRNA at a time.
4. The two amino acids are joined by a peptide bond (condensation). The first tRNA leaves.
5. The ribosome moves one codon along the mRNA. Repeat.
6. At a stop codon the polypeptide is released.

Many ribosomes can translate one mRNA at once, so many polypeptides are made from a single transcript.

Must-know distinctions

Compare	Point to state
Codon and anticodon	Codon: three bases on mRNA. Anticodon: three bases on tRNA, complementary to the codon.
Transcription and translation	Transcription makes mRNA from DNA (nucleus, RNA polymerase). Translation makes a polypeptide from mRNA (ribosome, tRNA).
mRNA, tRNA, rRNA	mRNA carries the code. tRNA brings specific amino acids. rRNA forms the ribosome with protein.
Template and coding strand	Template is copied by base pairing. Coding strand has the same sequence as the mRNA (T instead of U).
Degenerate and universal	Degenerate: many codons per amino acid. Universal: same code in nearly all organisms.

The three RNAs at a glance

• mRNA: single strand, copy of a gene, different length for each gene, carries codons.
• tRNA: single strand folded by hydrogen bonds into a clover-leaf, anticodon at one end, specific amino acid at the other.
• rRNA: forms most of the ribosome (large and small subunits, with proteins), holds mRNA and tRNA in place and takes part in peptide bond formation.

Key terms

Term	Definition to learn
Gene	A length of DNA whose base sequence codes for the amino acid sequence of a polypeptide
Codon	A triplet of bases on mRNA that codes for one amino acid or a stop signal
Anticodon	A triplet of bases on tRNA that is complementary to a codon
Template strand	The DNA strand whose bases are paired with RNA nucleotides during transcription
Start codon	AUG, which codes for methionine and sets the reading frame
Stop codon	UAA, UAG or UGA: no tRNA matches, so the polypeptide is released
Primary structure	The sequence of amino acids in a polypeptide
Polypeptide	A chain of amino acids joined by peptide bonds

Worked reminder: reading frame

mRNA: AUG GGU UUU GAA UGG UAA. In frame this gives Met-Gly-Phe-Glu-Trp, then stop. Start one base later (UGG GUU UUG AAU GGU) and none of the codons is the same, so the polypeptide would be completely different. The start codon fixes the frame, and the code is non-overlapping so there is only one correct frame.

Linking the gene to the protein

1. The base sequence of the gene fixes the order of codons in the mRNA.
2. The order of codons fixes the order of amino acids (primary structure).
3. The primary structure fixes the R group interactions, so it fixes the folding and the three-dimensional shape.
4. The shape fixes the function, for example the shape of an enzyme active site.

Quick calculations

Task	Method	Example
Bases in coding region	(amino acids + 1 stop) × 3	150 amino acids: 151 × 3 = 453 bases
Amino acids from mRNA length	bases ÷ 3, then subtract 1 for the stop	1260 bases: 420 codons, so 419 amino acids
Time to make a polypeptide	amino acids ÷ rate	300 amino acids at 15 per second = 20 s

Check that the base count is divisible by 3 before you divide.

Quick self-test

1. State what is meant by a triplet code.
2. Why can two bases per amino acid not code for all 20 amino acids?
3. How many codons does a 12-base sequence contain if the code is non-overlapping?
4. What does degenerate mean?
5. Which base is in mRNA but not in DNA?
6. Give the mRNA made from the template strand TAC CCT.
7. Name the enzyme that joins RNA nucleotides during transcription.
8. State the anticodon that pairs with the codon GGA.
9. A gene codes for 96 amino acids. How many bases code for the polypeptide and stop codon?
10. State two functions of rRNA.
11. Name the bond formed between amino acids during translation.
12. Why can a human gene be read by a bacterial cell?

Answers

1. Three bases (one codon) code for one amino acid.
2. Two bases give only 4 × 4 = 16 combinations, fewer than 20.
3. Four codons (12 ÷ 3).
4. More than one codon can code for the same amino acid.
5. Uracil.
6. AUG GGA.
7. RNA polymerase.
8. CCU.
9. (96 + 1) × 3 = 291 bases.
10. Any two from: forms part of the ribosome structure; holds mRNA and tRNA in position; takes part in forming peptide bonds.
11. Peptide bond (formed in a condensation reaction).
12. The code is universal: the same codons code for the same amino acids.

Where marks are usually lost

• Writing thymine in mRNA, or uracil in DNA.
• Confusing codon and anticodon, or saying the codon is on tRNA.
• Leaving out the stop codon when counting bases.
• Defining degenerate backwards, as one codon for several amino acids.
• Calling the code overlapping, or forgetting to say that each base is read as part of one codon only.
• Saying RNA polymerase “unzips” the DNA without naming the bonds: it breaks hydrogen bonds between bases, and forms phosphodiester bonds in the new strand.
• Describing the polypeptide as made in the nucleus.
• Leaving out complementary base pairing as the way tRNA matches mRNA.
• Describing tRNA as carrying “amino acids” in general: each type carries one specific amino acid.

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).