IB Biology · Theme D: Continuity and change · SL and HL
D1.2 Protein synthesis
A one-page summary of D1.2 Protein synthesis, the key terms to know, and sample exam questions with answers. For the full lesson, open the illustrated revision slides or practise in the app.
Guiding questions
How does a cell produce a sequence of amino acids from a sequence of DNA bases?
How is the reliability of protein synthesis ensured?
What D1.2 covers
Transcription — copying a gene into RNA
- D1.2.1Copying a gene into RNA
- D1.2.2Base pairing builds the RNA strand
- D1.2.3One template, many transcripts
- D1.2.4Switching genes on and off
Translation — building the polypeptide
- D1.2.5From RNA sequence to amino acids
- D1.2.6Three players, one ribosome
- D1.2.7Codon meets anticodon
- D1.2.9Reading the code from a table
- D1.2.8Reading three bases at a time
- D1.2.10The ribosome moves, the chain grows
- D1.2.11One base changed, one protein altered
HL — directionality, promoters and non-coding DNA
- D1.2.12 · HLBoth processes run 5' to 3'
- D1.2.13 · HLStarting transcription at the promoter
- D1.2.14 · HLDNA that doesn't code for protein
HL — processing mRNA and finishing the protein
- D1.2.15 · HLEditing the transcript before it's used
- D1.2.16 · HLOne gene, several proteins
- D1.2.17 · HLAssembling the ribosome to start
- D1.2.18 · HLA protein isn't finished at translation
- D1.2.19 · HLBreaking proteins down to build new ones
D1.2 Protein synthesis: summary
Transcription
- RNA polymerase synthesizes RNA from a DNA template, pairing A on DNA with U on RNA (no primer needed).
- The template is reusable and unchanged; transcription is the key step where a gene is switched on or off.
Translation basics
- mRNA binds the small subunit; two tRNAs bind the large subunit at once.
- Codon (mRNA triplet) pairs with the complementary anticodon (tRNA triplet) carrying one amino acid.
Genetic code & mutation
- Triplet code (4³=64); degenerate (multiple codons/amino acid) and universal across organisms.
- A point mutation can change one codon's amino acid and disrupt protein shape.
Ribosome mechanics
- The ribosome moves stepwise, one codon at a time, toward the mRNA's 3' end.
- Each step adds one amino acid via a new peptide bond to the growing chain.
HL · Directionality, promoters & non-coding DNA
- Both processes run 5'→3'; transcription factors + RNA polymerase bind the promoter to start transcription.
- Non-coding DNA includes regulators, introns, telomeres, and rRNA/tRNA genes.
HL · Processing & finishing the protein
- Introns spliced out, 5' cap + polyA tail added; alternative splicing makes multiple proteins from one gene.
- Initiation assembles the ribosome at the start codon (A/P/E sites); pre-proinsulin → insulin and proteasome recycling finish the story.
Key terms
- Codon
- A triplet of three consecutive bases on mRNA that specifies a particular amino acid or a start/stop signal.
- Anticodon
- The complementary triplet of bases on a tRNA molecule that pairs with a codon on mRNA.
- Degeneracy
- The genetic code's use of more than one codon to specify most amino acids.
- Universality
- The genetic code's property that the same codon specifies the same amino acid across essentially all organisms.
- Point mutation
- A change to a single base in a DNA sequence, which can alter the amino acid a codon specifies.
- Promoter HL
- A DNA sequence near the start of a gene where transcription factors and RNA polymerase bind to begin transcription.
- Intron HL
- A non-coding sequence within a gene that is removed from the transcript during post-transcriptional modification.
- Proteasome HL
- A cell structure that breaks down damaged or unneeded proteins, releasing amino acids for reuse.
- C-peptide HL
- The central section removed from proinsulin, leaving the linked A and B chains of mature insulin.
Sample exam questions
Three of the 57 multiple-choice questions for D1.2. Try each one before opening the answer.
Question 1. Which enzyme catalyses the synthesis of RNA using a DNA template during transcription?
- DNA polymerase
- RNA polymerase
- Helicase
- DNA ligase
Show the answer
Answer: B. RNA polymerase binds to the DNA template strand and catalyses the formation of an RNA molecule complementary to it, adding ribonucleotides one at a time.
Question 2. Which statement about transcription of a particular gene is correct?
- Both DNA strands are transcribed to produce two different mRNA molecules
- RNA polymerase requires an RNA primer to begin transcription
- Transcription occurs only in cells that are actively dividing
- Only one DNA strand — the template strand — is transcribed for a given gene
Show the answer
Answer: D. For any one gene, only the template strand is read by RNA polymerase; the resulting mRNA has the same sequence as the coding (non-template) strand, apart from U replacing T.
Question 3. A researcher finds that a single DNA template strand can be transcribed accurately many times over a cell's lifetime without any change in the base sequence read. What does this demonstrate?
- That a single strand of DNA is a stable, reusable template for transcription
- That transcription is a mutagenic process
- That RNA polymerase copies bases at random
- That DNA must be double-stranded to be transcribed at all
Show the answer
Answer: A. Because transcription does not alter the template strand, the same single strand can act as a template repeatedly and reliably, which is essential for genes that are transcribed many times during a cell's life.
Linking questions
Questions that connect D1.2 to other parts of the course, the kind that come up in Paper 2.
- Transcription uses DNA (A1.2, D1.1) as a template. Explain why only one DNA strand (the template strand) is transcribed for a given gene, and how DNA's double-stranded structure protects the template. (see A1.2, D1.1)
- Ribosomes are composed of rRNA and protein, and have A, P, and E sites (B1.2). How does the quaternary structure of the ribosome enable its function in translation? (see B1.2)
- Many proteins synthesised by ribosomes on the rough ER (B2.2) are destined for secretion or membrane insertion. Suggest how post-translational modification (D1.2.18) relates to a protein's eventual destination in the cell. (see B2.2)
- Mutations (D1.3) that change a single base can have very different consequences depending on the genetic code's degeneracy (D1.2.8). Explain why a substitution mutation is more likely to be silent than an insertion or deletion of one base. (see D1.3)
Practise D1.2
Study notes, every question and full markschemes for D1.2 are in the app with Pro. Two lessons are completely free to try: A1.1 Water and B1.1 Carbohydrates and lipids.