IB DP Biology: Protein synthesis: transcription, translation and the genetic code (DP Biology) – Revision Notes

These revision notes condense IB Diploma Programme Biology section D1.2 (Protein synthesis) from the IB Biology guide, first assessment 2025. D1.2.1 to D1.2.11 are SL and HL; D1.2.12 to D1.2.19 are additional higher level and are marked HL only. For full explanations and worked examples, read the study guide, then test yourself with the practice questions.

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Key terms

Term	Meaning
Transcription	Synthesis of RNA from a DNA template, by RNA polymerase
Translation	Synthesis of a polypeptide at a ribosome, using the base sequence of mRNA
Template strand	The DNA strand copied; the RNA is complementary to it
Codon	Three bases on mRNA that specify one amino acid or a stop
Anticodon	Three bases on tRNA that pair with a codon
Degeneracy	Most amino acids have more than one codon
Universality	Almost all organisms use the same genetic code
Point mutation	Change to a single base pair in a gene
Exon, intron (HL only)	Exons stay in mature mRNA; introns are removed
Proteasome (HL only)	Complex that breaks proteins down into amino acids

Transcription: D1.2.1 to D1.2.4

• RNA polymerase unwinds a short region of DNA and links RNA nucleotides in the order set by the template.
• Complementary pairing, held by hydrogen bonds: template A takes U, T takes A, G takes C, C takes G.
• The DNA base sequence is not changed by being copied, and the template can be used repeatedly.
• In non-dividing somatic cells the sequence must last the cell’s whole life.
• Not all genes are expressed at once. Transcription is the first stage of expression and a key point where a gene is switched on or off.

Method: coding strand to mRNA.

1. Write the template as the complement of the coding strand, or take the template as given.
2. Complement each base, using U for A on the template.
3. Check the mRNA is the same as the coding strand with T changed to U.
4. Label ends if you are HL: the mRNA is made 5’ to 3’.

The genetic code: D1.2.8 and D1.2.9

Idea	Key point
Triplet	4 bases give 4 single codes, 16 pairs, 64 triplets. Only triplets cover 20 amino acids
Count	61 codons code for amino acids; 3 are stop codons
Start	AUG codes for methionine and is the usual start
Degeneracy	Changing the last base of a codon often keeps the amino acid
Universality	Same code in bacteria and humans, so genes can be moved between species

Method: mRNA to polypeptide. Find AUG, split into threes, look up each codon, stop at a stop codon. The stop codon has no amino acid.

Translation: D1.2.5 to D1.2.7 and D1.2.10

Component	Fact to remember
mRNA	Binds to the small ribosomal subunit
Ribosome	Two tRNAs can bind at once to the large subunit
tRNA	One amino acid at one end, anticodon at the other

• Codon-anticodon pairing is complementary and uses hydrogen bonds.
• Elongation is a cycle: matching tRNA arrives, a peptide bond forms between its amino acid and the chain, the ribosome moves one codon along the mRNA, the empty tRNA is released.
• A peptide bond forms by condensation.

Mutations: D1.2.11

Type	Result in the polypeptide
Silent	Same amino acid, no change
Missense	One amino acid replaced; effect depends on the side chain and position
Nonsense	Stop codon appears early; truncated polypeptide
Insertion or deletion of one or two bases	Frameshift; every later codon changes

Example to remember: sickle cell disease. GAG (glutamic acid) becomes GUG (valine) in the haemoglobin β chain mRNA. The change in structure makes the haemoglobin aggregate in low oxygen.

HL only: D1.2.12 to D1.2.19

Statement	Remember
D1.2.12 Directionality	RNA is made 5’ to 3’; the ribosome reads mRNA 5’ to 3’
D1.2.13 Promoter	DNA next to a gene; transcription factors bind there and help RNA polymerase to attach
D1.2.14 Non-coding DNA	Regulators of expression, introns, telomeres, genes for rRNA and tRNA
D1.2.15 Processing	Introns removed, exons spliced; 5’ cap and poly-A tail stabilise the mRNA
D1.2.16 Alternative splicing	Different exon combinations, so one gene can give different polypeptides
D1.2.17 Initiation	Small subunit attaches at the 5’ end, moves to AUG, initiator tRNA binds, large subunit joins
D1.2.17 Ribosome sites	A site: tRNA arrives. P site: tRNA holding the chain. E site: tRNA exits
D1.2.18 Modification	Pre-proinsulin to proinsulin, then to insulin, in two stages
D1.2.19 Proteasomes	Constant breakdown and synthesis sustain a functional proteome

Worked reminders

Nonsense mutation. mRNA 5’-AUGGAUCCUUGGUAG-3’ reads AUG GAU CCU UGG UAG, which is Met-Asp-Pro-Trp then stop. If the fourth codon UGG changes to UGA, the stop arrives early and the polypeptide is Met-Asp-Pro, three amino acids instead of four.

Exon arithmetic (HL only). A gene has three exons of 120, 75 and 90 bases and two introns of 400 and 250 bases.

1. Pre-mRNA length: 120 + 75 + 90 + 400 + 250 = 935 bases.
2. Mature mRNA after splicing: 120 + 75 + 90 = 285 bases, which is 95 codons.
3. A variant that skips the 75 base exon: 120 + 90 = 210 bases, or 70 codons.

Elongation (HL only). Say what happens in order: aminoacyl tRNA enters the A site, a peptide bond forms between the chain on the P site tRNA and the new amino acid, the ribosome moves one codon, the empty tRNA moves to the E site and leaves.

Must-know distinctions

• Template strand and coding strand. The mRNA is complementary to the template and matches the coding strand.
• Codon and anticodon. Codon: mRNA. Anticodon: tRNA.
• Gene expression and transcription. Transcription is only the first stage of expression.
• Degeneracy and universality. Degeneracy: many codons for one amino acid. Universality: one code for nearly all life.
• Silent and missense. In both the DNA changes; only in missense does the amino acid change.
• Intron and exon (HL only). Introns are removed; exons are joined.
• Non-coding but transcribed (HL only). Genes for rRNA and tRNA make RNA, not polypeptides.

Quick self-test

1. The template strand is 3’-TACAAGGTCCTAATC-5’. Write the mRNA and the polypeptide.
2. Translate 5’-AUGAAAGGUUGGCCAUGA-3’.
3. In the mRNA in question 2, the fourth codon UGG changes to GGG. Name this type of mutation and state its effect.
4. In the same mRNA, the fourth base (the first A of the second codon) is deleted. State the first five amino acids produced and the type of mutation.
5. Give the anticodon for the codon GGU, written 3’ to 5’.
6. Why can two-base codons not code for all 20 amino acids?
7. How many bases are in the coding region, including the stop codon, of a polypeptide of 8 amino acids?
8. State two features of the genetic code that the guide expects you to use as terms.
9. HL only: name the three tRNA binding sites on a ribosome in the order tRNA passes through them.
10. HL only: state the purpose of adding a 5’ cap and a poly-A tail to pre-mRNA.
11. HL only: explain how alternative splicing lets one gene give more than one polypeptide.
12. Explain why a transcribed DNA template remains unchanged.

Answers

1. mRNA 5’-AUGUUCCAGGAUUAG-3’; codons AUG UUC CAG GAU UAG; Met-Phe-Gln-Asp then stop.
2. AUG AAA GGU UGG CCA UGA gives Met-Lys-Gly-Trp-Pro, then stop.
3. A missense (point) mutation: Trp is replaced by Gly, so the polypeptide has one different amino acid (Met-Lys-Gly-Gly-Pro).
4. Deleting that base gives AUG AAG GUU GGC CAU GA, so Met-Lys-Val-Gly-His; a frameshift deletion.
5. The anticodon is 3’-CCA-5’ (5’-ACC-3’).
6. Two bases give only 4² = 16 combinations, fewer than 20.
7. 8 × 3 + 3 = 27 bases.
8. Degeneracy and universality.
9. A site, P site, E site.
10. They stabilise the mRNA transcript, so it survives longer before it is broken down.
11. Different combinations of exons are spliced together from the same pre-mRNA, giving different mRNAs and so different polypeptides.
12. The polymerase reads the sequence by base pairing and builds a new RNA strand; it does not alter the DNA bases.

Where marks are usually lost

• Writing thymine into mRNA, or leaving uracil out of the template pairing.
• Giving the template-strand sequence when the question asks for mRNA.
• Adding an amino acid for the stop codon.
• Reading from the wrong AUG, or starting the codons in the wrong frame.
• Describing a mutation as harmful without stating how the amino acid sequence changes.
• Writing “the mutation changes the protein” with no link to structure or function.
• Mixing up codon and anticodon, or tRNA and mRNA.
• HL answers that name the sites A, P and E but give no role for each.
• Saying the ribosome moves “along the DNA”.
• Giving only one mark’s worth of detail for a question worth several points.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. This page is a paraphrase aligned to section D1.2.