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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions practise IB Diploma Programme Biology section D1.2 (Protein synthesis) from the IB Biology guide, first assessment 2025. Questions 1 to 9 are SL and HL. Questions 10 to 12 are HL only. Use the codon table in the study guide where a question needs it.

Revise first with the revision notes. Course hub: IB DP Biology. Checklist: printable checklist. Diagnostics: all free 10-minute diagnostics.

Questions

1. Describe the role of hydrogen bonding in transcription. [3]

2. A segment of a gene has the template strand 3’-TACAAGCCTGGTACCTTTATC-5’.

(a) Deduce the base sequence of the mRNA transcribed from it. [2] (b) Deduce the amino acid sequence it codes for. [2]

3. A mature neuron does not divide, and it uses only some of its genes.

(a) Explain why the DNA template sequences must stay unchanged throughout the life of the neuron. [2] (b) Explain why transcription is a key stage for controlling which genes are expressed. [2]

4. The genetic code uses codons of three bases.

(a) Show by calculation that a two-base codon cannot specify every amino acid. [2] (b) State what is meant by degeneracy of the code and give one consequence of it. [2]

5. A human gene is inserted into yeast cells, and the yeast makes the human polypeptide. Explain what this shows about the genetic code. [3]

6. Translation occurs at ribosomes.

(a) State where mRNA binds on the ribosome and how many tRNAs can bind to the large subunit at one time. [2] (b) An mRNA codon reads 5’-ACC-3’. Write the anticodon of the tRNA that pairs with it, from 3’ to 5’. [1] (c) Explain how tRNA ensures that the correct amino acid is added. [2]

7. Describe the elongation stage of translation. [5]

8. The mRNA 5’-AUGUUCGGACCAUGGAAAUAG-3’ codes for a short polypeptide. For each change, deduce the effect on the polypeptide.

(a) The sixth base, C, is replaced by U. [2] (b) The eighth base, G, is replaced by A. [2] (c) The fifteenth base, G, is replaced by A. [2] (d) The seventh base is deleted. [2]

9. Sickle cell disease results from a point mutation in the gene for the β chain of haemoglobin. Explain how this mutation changes the structure of the protein. [4]

10. (HL only) Answer all parts.

(a) State the direction in which RNA polymerase adds nucleotides to the growing RNA, and which end of the template it moves towards. [2] (b) Describe the role of the promoter and of transcription factors. [3] (c) A gene for tRNA is transcribed but is described as non-coding. Explain. [2]

11. (HL only) A gene has five exons, of 96, 150, 72, 138 and 108 bases, separated by four introns of 520, 330, 880 and 410 bases.

(a) Calculate the length of the pre-mRNA, ignoring caps, tails and untranslated regions. [2] (b) Calculate the percentage of the pre-mRNA that is intron, to 1 decimal place. [1] (c) Calculate the number of codons in the mature mRNA when all five exons are retained, and when the 72 base exon is omitted. [2] (d) Explain how this gives two polypeptides from one gene. [1] (e) State one function of the 5’ cap. [1]

12. (HL only) Answer all parts.

(a) Describe the initiation of translation. [4] (b) Outline how pre-proinsulin is converted into insulin in two steps. [2] (c) In an experiment, the amount of protein Z in cultured cells was measured with a proteasome inhibitor. The amount was 100, 118, 141 and 166 arbitrary units at 0, 2, 4 and 6 hours. Calculate the percentage increase at 6 hours, and explain why protein Z accumulated. [2]

Answers

1. Hydrogen bonds form between template bases and complementary free RNA nucleotides [1]. Template A takes U, T takes A, G takes C and C takes G, so the RNA sequence follows the template [1]. They are weak, so the RNA can separate from the template and the DNA strands can pair up again [1]. [3] Examiner insight: Naming the pairs and the weakness of the bonds are separate points; “bonds hold them together” alone does not describe a role.

2. (a) Complement each template base, using U for A: 5’-AUGUUCGGACCAUGGAAAUAG-3’ [2]. [2] (b) Codons AUG UUC GGA CCA UGG AAA UAG [1]. Met-Phe-Gly-Pro-Trp-Lys, then stop [1]. [2] Examiner insight: Giving the template complement with T instead of U, or including the stop codon as an amino acid, loses the mark for that stage.

3. (a) The neuron does not divide, so its DNA is never replaced by copying [1]. Any lasting damage to a sequence would be kept, so the template must be conserved for the cell’s life [1]. [2] (b) Transcription is the first stage of gene expression [1]. Switching it on or off decides whether a gene is expressed, because only genes that are transcribed can have their mRNA translated [1]. [2] Examiner insight: Answers should link the cell’s non-dividing state to the template, rather than just saying DNA is stable.

4. (a) Two bases give 4 × 4 = 16 different codons [1]. This is fewer than the 20 amino acids [1]. [2] (b) Several different codons specify one amino acid [1]. A change to a codon’s last base can keep the same amino acid, so some mutations have no effect [1]. [2] Examiner insight: The calculation must be shown; stating that three is “enough” gains nothing without the figures.

5. Yeast ribosomes read the human mRNA codons and add the amino acids the codons specify [1]. A given codon has the same meaning in humans and yeast [1]. This shows that the genetic code is universal, shared by nearly all organisms [1]. [3] Examiner insight: Use the term “universal”, and explain it with the codon meaning being the same; a bare term may not be enough.

6. (a) mRNA binds to the small subunit [1]. Two tRNAs can bind at once to the large subunit [1]. [2] (b) 3’-UGG-5’ [1]. [1] (c) Each tRNA carries one specific amino acid [1]. Its anticodon pairs by complementary bases with the codon, so only the tRNA with the matching anticodon delivers its amino acid [1]. [2] Examiner insight: For part (b), check the direction label: writing the same letters in the wrong order is a common slip.

7. A tRNA with an anticodon complementary to the next codon binds to the ribosome [1]. Two tRNAs are bound side by side on the mRNA at this point [1]. A peptide bond forms between the amino acid on the new tRNA and the end of the growing chain [1]. The ribosome moves along the mRNA by one codon [1]. The empty tRNA is released and the next codon is exposed, so the cycle repeats [1]. [5] Examiner insight: The order of events matters; listing “peptide bond, tRNA, movement” without sequence is less likely to gain full credit.

8. (a) The codon changes from UUC to UUU, which still codes for Phe [1]. The polypeptide is unchanged, a silent mutation [1]. [2] (b) GGA becomes GAA, so Gly is replaced by Glu [1]. The polypeptide is Met-Phe-Glu-Pro-Trp-Lys, one amino acid different [1]. [2] (c) UGG becomes UGA, a stop codon [1]. Translation ends early, giving Met-Phe-Gly-Pro [1]. [2] (d) All codons after the deletion shift: AUG UUC GAC CAU GGA AAU [1]. The polypeptide begins Met-Phe-Asp-His-Gly-Asn and no longer matches the original after the second amino acid [1]. [2] Examiner insight: Name the type of change and quote the new amino acids; “the protein changes” without the sequence is too vague.

9. In the mRNA, a codon for glutamic acid, GAG, becomes GUG [1]. Valine replaces glutamic acid in the β chain [1]. Valine has a side chain without charge, unlike glutamic acid [1]. The altered haemoglobin molecules stick together in low oxygen and form fibres, changing the protein’s structure and function [1]. [4] Examiner insight: Link the single base change to the amino acid, and the amino acid to structure; skipping the middle step loses credit.

10. (a) RNA polymerase adds nucleotides to the 3’ end of the RNA, so the RNA is made 5’ to 3’ [1]. It moves along the template towards the template’s 5’ end [1]. [2] (b) The promoter is a DNA sequence near the start of a gene where transcription begins [1]. Transcription factors bind to the promoter [1]. They help RNA polymerase to attach and begin transcription [1]. [3] (c) The gene is transcribed into tRNA [1]. The product is RNA that is not translated, so the gene does not code for a polypeptide [1]. [2] Examiner insight: Part (c) rewards the distinction between coding for RNA and coding for a polypeptide.

11. (a) Exons: 96 + 150 + 72 + 138 + 108 = 564 [1]. Pre-mRNA: 564 + 2140 = 2704 bases [1]. [2] (b) 2140 / 2704 × 100 = 79.1 per cent [1]. [1] (c) 564 / 3 = 188 codons with all exons [1]; omitting 72 gives 492 / 3 = 164 codons [1]. [2] (d) Different combinations of exons are spliced together, giving different mRNAs and so different polypeptides [1]. [1] (e) The cap helps stabilise the mRNA [1]. [1] Examiner insight: Keep units (bases or codons) with each figure, and show the division by three.

12. (a) The small ribosomal subunit attaches to the 5’ end of the mRNA [1]. It moves along the mRNA to the start codon [1]. The initiator tRNA binds to the start codon, and a second tRNA binds [1]. The large subunit then attaches [1]. [4] (b) First stage: a leading sequence is removed from pre-proinsulin to give proinsulin [1]. Second stage: a middle section is cut out, leaving two chains that form active insulin [1]. [2] (c) (166 − 100) / 100 × 100 = 66 per cent increase [1]. The inhibitor blocked proteasomes, so protein Z was not broken down while synthesis continued [1]. [2] Examiner insight: A percentage change uses the starting value as the divisor; and explanations should refer to breakdown versus synthesis.

Where marks are usually lost

• Using T in mRNA or forgetting that translation halts at the stop codon.
• Giving template bases instead of mRNA bases.
• Giving an anticodon in the same direction as the codon without checking the 3’ and 5’ labels.
• Not stating the new amino acid in a mutation question.
• Using “mutation” without saying whether the DNA, mRNA or protein changes.
• Treating degeneracy as “one codon for many amino acids”, which reverses it.
• Describing elongation without the order of tRNA binding, bond formation and movement.
• HL: stating that introns “code for nothing” without saying they are removed from the transcript.
• HL: describing proteasomes without relating them to protein synthesis and breakdown.

Next steps

• Review the revision notes and the study guide.
• Course hub: IB DP Biology. Checklist: printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. These questions are aligned to section D1.2.