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

This study guide covers IB Diploma Programme Biology section D1.2 (Protein synthesis), as set out in the IB Biology guide, first assessment 2025. Statements D1.2.1 to D1.2.11 are for standard level and higher level. Statements D1.2.12 to D1.2.19 are additional higher level, so every part marked HL only below is not needed at SL.

Use it with the revision notes and the practice questions. Course hub: IB DP Biology. Printable checklist. To find your weak spots first, try the free diagnostics.

What this unit covers

Statement	What you must be able to do	Level
D1.2.1 to D1.2.4	Describe transcription, RNA polymerase, base pairing, template stability and control of expression	SL and HL
D1.2.5 to D1.2.7	Describe translation, the roles of mRNA, ribosome and tRNA, codon-anticodon pairing	SL and HL
D1.2.8 and D1.2.9	Explain a triplet code, degeneracy and universality; use a codon table	SL and HL
D1.2.10	Describe elongation and peptide bond formation	SL and HL
D1.2.11	Explain how a point mutation can change protein structure	SL and HL
D1.2.12 to D1.2.14	Direction, promoters, non-coding DNA	HL only
D1.2.15 and D1.2.16	Cap, tail, intron removal, alternative splicing	HL only
D1.2.17 to D1.2.19	Initiation, A, P and E sites, modification, proteasomes	HL only

Transcription (D1.2.1 to D1.2.4)

Transcription is the synthesis of RNA using DNA as a template. The enzyme RNA polymerase does the work. It unwinds a short stretch of the double helix and builds a single strand of RNA by joining RNA nucleotides one at a time. Only one DNA strand, the template strand, is copied for a given gene. The other, the coding strand, has the same sequence as the RNA with T in place of U.

Base pairing. Each incoming RNA nucleotide lines up opposite a template base and is held there by hydrogen bonds between complementary bases. The pairs are:

Template DNA base	RNA nucleotide added
A	U
T	A
G	C
C	G

Adenine on the template pairs with uracil, because RNA contains no thymine. The hydrogen bonds are weak, so the new RNA peels away from the template and the DNA strands close up again behind the polymerase. RNA polymerase also forms the covalent bonds that link the RNA nucleotides.

Stability of the template (D1.2.3). Copying a strand does not alter it. The same template can be transcribed again and again. In somatic cells that no longer divide, such as mature neurons, the sequence has to be kept intact for the whole life of the cell, because there is no replication to replace it.

Transcription and gene expression (D1.2.4). Transcription is the first stage of gene expression. A cell does not express every gene at once. A pancreatic beta cell transcribes the insulin gene, for example, while most other cell types do not. Switching transcription on or off is a key point at which a cell controls which genes are expressed.

Worked example 1: from coding strand to mRNA

The coding strand of a short gene reads 5’-ATGGCTAACCGTTTATAA-3’. Find the template strand, the mRNA and the polypeptide.

1. The template strand is complementary and antiparallel: 3’-TACCGATTGGCAAATATT-5’.
2. Transcribe the template: A gives U, T gives A, C gives G, G gives C. The mRNA is 5’-AUGGCUAACCGUUUAUAA-3’. It matches the coding strand with U for T.
3. Split into codons: AUG GCU AAC CGU UUA UAA.
4. Read the codon table (below): Met, Ala, Asn, Arg, Leu, then a stop codon.

The polypeptide is Met-Ala-Asn-Arg-Leu, five amino acids. The stop codon codes for none.

The genetic code (D1.2.8 and D1.2.9)

The genetic code is the set of rules linking mRNA codons to amino acids. Each codon is three bases.

Why a triplet? Proteins use 20 different amino acids. With four bases:

• one base per amino acid gives only 4 codes,
• two bases give 4² = 16, still fewer than 20,
• three bases give 4³ = 64, enough for 20 amino acids plus signals.

Three is the smallest codon length that works.

Degeneracy means that most amino acids are coded by more than one codon. Of the 64 codons, 61 code for amino acids and 3 are stop codons, so several codons share an amino acid. This protects against mutation: altering the last base of a codon often still gives the same amino acid.

Universality means that nearly all organisms use the same code. Human mRNA translated by bacterial ribosomes gives the human protein, which is the basis of genetic engineering.

Reading the mRNA codon table

First base	Second U	Second C	Second A	Second G
U	Phe Phe Leu Leu	Ser Ser Ser Ser	Tyr Tyr Stop Stop	Cys Cys Stop Trp
C	Leu Leu Leu Leu	Pro Pro Pro Pro	His His Gln Gln	Arg Arg Arg Arg
A	Ile Ile Ile Met	Thr Thr Thr Thr	Asn Asn Lys Lys	Ser Ser Arg Arg
G	Val Val Val Val	Ala Ala Ala Ala	Asp Asp Glu Glu	Gly Gly Gly Gly

In each cell the four amino acids belong to third bases U, C, A, G in that order. AUG (row A, second base U, third base G) is both the start codon and methionine.

Method: find the first AUG, read non-overlapping groups of three, look up each, stop at a stop codon.

Translation (D1.2.5 to D1.2.7 and D1.2.10)

Translation is the synthesis of a polypeptide using the base sequence of mRNA. The mRNA base sequence is translated into the amino acid sequence.

Component	Role
mRNA	Carries the sequence copied from the gene; it binds to the small ribosomal subunit
Ribosome	Holds mRNA and tRNAs in position and catalyses peptide bond formation; two tRNAs can bind at once to the large subunit
tRNA	Carries one specific amino acid and has an anticodon that pairs with a codon

A codon is three bases on mRNA. An anticodon is the three unpaired bases on a tRNA that pair, by hydrogen bonding, with the codon. The anticodon 3’-CGA-5’ pairs with the codon 5’-GCU-3’ and the tRNA carries alanine.

Elongation (D1.2.10). The cycle repeats:

1. A tRNA whose anticodon matches the next codon binds, so two tRNAs sit side by side.
2. A peptide bond (a condensation reaction) forms between the amino acid on the new tRNA and the end of the growing chain, which moves onto the new tRNA.
3. The ribosome moves along the mRNA by exactly one codon. The first tRNA, now empty, is released, and the next codon is exposed.

The chain grows by one amino acid per cycle. The ribosome moves in a stepwise way, one codon at a time, until a stop codon is reached.

Worked example 2: how many bases?

A polypeptide has 150 amino acids. What is the least number of mRNA bases in the coding region?

1. 150 codons × 3 = 450 bases.
2. Add the stop codon: 450 + 3 = 453.

The coding region is at least 453 bases.

Mutations that change protein structure (D1.2.11)

A point mutation changes a single base pair in a gene. The effect depends on what the new codon codes for.

Use the mRNA from worked example 1:

Change in mRNA	New sequence	Effect
AAC to AAU (third base)	Met-Ala-Asn-Arg-Leu	Silent: Asn is still Asn, thanks to degeneracy
AAC to AAA	Met-Ala-Lys-Arg-Leu	Missense: Asn replaced by Lys
UUA to UAA	Met-Ala-Asn-Arg	Nonsense: stop codon appears early; shorter polypeptide

A famous example is sickle cell disease. In the gene for the β chain of haemoglobin, the codon GAG for glutamic acid becomes GUG for valine in the mRNA. Glutamic acid has a charged side chain; valine does not. In low oxygen the altered haemoglobin molecules stick together in long fibres that distort the red blood cell. One base change alters one amino acid, so the protein’s structure and function change.

If a base is inserted or deleted, every codon after that point is read in a new frame. Deleting the seventh base of the mRNA above gives AUG GCU ACC GUU UAU AA, which reads Met-Ala-Thr-Val-Tyr. This frameshift usually ruins the protein.

HL only: directionality, promoters and non-coding DNA (D1.2.12 to D1.2.14)

Directionality (D1.2.12). Nucleic acid strands have a 5’ end and a 3’ end. RNA polymerase adds each new nucleotide to the 3’ end of the growing RNA, so RNA is made in the 5’ to 3’ direction. The polymerase therefore moves along the template from its 3’ end towards its 5’ end. In translation, the ribosome moves along the mRNA from the 5’ end towards the 3’ end, reading codons in that order. The polypeptide is built from its amino end to its carboxyl end.

Promoters (D1.2.13). A promoter is a DNA sequence next to a gene where transcription begins. In eukaryotes, proteins called transcription factors bind to the promoter first. They help RNA polymerase attach. You do not need to name them.

Non-coding sequences (D1.2.14). Not all DNA codes for polypeptides. Examples:

• regulators of gene expression, such as promoters and other control sequences,
• introns, removed from the transcript,
• telomeres, the repeated sequences at the ends of eukaryotic chromosomes,
• genes for rRNA and tRNA, which are transcribed but whose products are RNA molecules, not polypeptides.

HL only: processing the transcript (D1.2.15 and D1.2.16)

In eukaryotes, the first transcript (pre-mRNA) is modified in the nucleus before it leaves:

• Introns are removed and exons are spliced together to give mature mRNA.
• A 5’ cap is added at the 5’ end and a poly-A tail at the 3’ end. These stabilise the mRNA.

Alternative splicing joins different combinations of exons from the same pre-mRNA, so one gene can code for different polypeptides. You only need this idea, not named examples.

Worked example 3: exons and introns

A gene has six exons of 84, 126, 63, 105, 141 and 99 bases, separated by five introns totalling 2150 bases. Ignore untranslated regions.

1. Total exon length = 618 bases. Transcript before splicing = 618 + 2150 = 2768 bases.
2. Introns make up 2150 / 2768 × 100 = 77.7 per cent of the pre-mRNA.
3. All six exons joined: 618 / 3 = 206 codons.
4. Variant that omits the 63 and 105 base exons: 618 − 63 − 105 = 450 bases, or 150 codons.

Every exon length divides by three, so omitting exons keeps the reading frame. One gene gives two polypeptides.

HL only: initiation, modification and recycling (D1.2.17 to D1.2.19)

Initiation of translation. The small ribosomal subunit attaches to the 5’ end of the mRNA and moves along to the start codon AUG. The initiator tRNA (carrying methionine) pairs with the start codon. A second tRNA then enters and the large subunit joins.

A, P and E sites. The ribosome has three tRNA binding sites. A new aminoacyl tRNA enters the A site. The tRNA holding the growing chain sits in the P site. Empty tRNAs leave from the E site (exit). After each peptide bond, the ribosome moves one codon and the tRNAs shift from A to P and from P to E.

Modification of polypeptides. Many polypeptides are not functional at first. Pre-proinsulin is modified in two stages: first a leading sequence is removed to give proinsulin, then a middle section is cut out, leaving two chains joined by bonds to form active insulin.

Proteasomes. These complexes break proteins down into amino acids. Sustaining a functional proteome needs constant synthesis and breakdown, and the amino acids are reused.

Common errors

• Putting T in RNA or U in DNA, or copying the template without complementing it.
• Adding an amino acid for the stop codon.
• Saying a silent mutation changes nothing: the DNA changed, the protein did not.
• Confusing codon (mRNA) with anticodon (tRNA).

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.