IB DP Biology: Gene expression and epigenetics (HL) (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 cover section D2.2, Gene expression, of the IB Biology guide, first assessment 2025. The whole section is HL only, so every question is HL only. Revise first with the study guide and the revision notes. The course hub, the printable checklist and the free 10-minute diagnostics are linked for planning. Data in the questions are supplied for the exercise.

Questions

1. (HL only) Explain what is meant by gene expression, and state the three stages that most commonly make up the process. [4]

2. (HL only) Distinguish between a promoter, an enhancer and a transcription factor. [3]

3. (HL only) The mRNA for a signalling protein has a half-life of 2.5 hours in a human cell. At time 0 there are 480 units of this mRNA and transcription of its gene stops.

(a) Calculate the amount of mRNA left after 10 hours. [2] (b) Explain one advantage to a cell of having mRNA with a short half-life. [2]

4. (HL only) Define epigenesis and explain why an epigenetic change alters the phenotype of a cell but not its genotype. [3]

5. (HL only) A genome contains 20 000 protein-coding genes. A pancreatic beta cell transcribes 8 800 different genes and contains insulin mRNA. A skeletal muscle fibre from the same person transcribes 9 600 different genes and contains no insulin mRNA.

(a) Calculate the percentage of genes transcribed in the beta cell. [1] (b) State what can be said about the genomes of the two cells. [1] (c) Explain why insulin mRNA is present in one cell type but not the other. [2] (d) Suggest one reason why the proteome of a cell is not an exact copy of its transcriptome. [1]

6. (HL only) (a) Explain the effect of methylating cytosine in the promoter of a gene. [2] (b) Describe the effect of methylating amino acids in histones. [2]

7. (HL only) Explain how a change in phenotype could be passed to daughter cells and to offspring without any mutation. [3]

8. (HL only) A study compared two districts. In 40 adults from each district, methylation of the promoter of gene F and the amount of its mRNA were measured. Mean methylation (± SD) was 64.2 ± 6.1% in the high-pollution district and 52.8 ± 5.8% in the low-pollution district. The mean mRNA was 0.58 in the high-pollution district and 1.00 (reference value) in the low-pollution district, in relative units.

(a) Calculate the difference in mean methylation between the districts. [1] (b) Calculate the percentage decrease in mRNA in the high-pollution district. [2] (c) Evaluate whether these data show that air pollution causes the lower expression of gene F. [3]

9. (HL only) Outline the epigenetic origins of the phenotypic differences between tigons and ligers. [4]

10. (HL only) Pairs of monozygotic twins were grouped by the number of years they had lived apart. The mean difference between the twins in methylation at one site was 2.8 percentage points for pairs who had lived apart 0 to 5 years and 10.4 percentage points for pairs who had lived apart more than 30 years.

(a) Calculate how many times greater the difference is in the second group (3 significant figures). [1] (b) Explain why monozygotic twins are suitable for studying environmental effects on gene expression. [2] (c) Suggest an explanation for the trend. [2] (d) State one limitation of this design. [1]

11. (HL only) In a bacterium, the enzymes that make tryptophan are encoded by genes controlled by a repressor. Enzyme activity was measured (arbitrary units).

Strain	No tryptophan in medium	Tryptophan in medium
Wild type	100	4
Mutant repressor	100	98

(a) Calculate the percentage decrease in activity in the wild type when tryptophan is added. [2] (b) Explain the result for the mutant strain. [2] (c) Describe how tryptophan regulates expression in the wild type. [3] (d) Outline how one named hormone can change gene expression in a target cell. [3]

Answers

1. Gene expression is the process by which information in a gene has effects on the phenotype [1]. The stages are transcription (mRNA made from the gene) [1], translation (polypeptide built from the mRNA) [1] and the function of the protein product, for example as an enzyme [1]. [4] Examiner insight: A definition that mentions phenotype is more complete than one that only says “genes are used”; list the three stages in order.

2. A promoter is the DNA sequence next to a gene where RNA polymerase binds [1]. An enhancer is a sequence where regulatory proteins bind to raise transcription of a gene [1]. A transcription factor is a protein that binds to a specific base sequence and helps or blocks transcription [1]. [3] Examiner insight: “Distinguish” asks for a difference for each term. Note that two are DNA sequences and one is a protein.

3. (a) 10 / 2.5 = 4 half-lives [1]. 480 × (1/2)⁴ = 30 units [1]. (b) A short half-life means the amount of mRNA falls quickly once transcription stops [1], so the cell can change protein output rapidly in response to a signal [1]. [4] Examiner insight: Show the number of half-lives; it makes the working easy to follow and easy to check.

4. Epigenesis is the process by which cells of a multicellular organism develop different patterns of differentiation [1]. Epigenetic tags change which genes are expressed, so the phenotype changes [1]. The tags do not alter the DNA base sequence, so the genotype is unchanged [1]. [3] Examiner insight: State clearly that the base sequence is not altered. This is the point that separates epigenetic change from mutation.

5. (a) 8 800 / 20 000 × 100 = 44% [1]. (b) They are the same (both cells hold the same genome) [1]. (c) Insulin is expressed in the beta cell but the gene is not transcribed in the muscle fibre [1]; the pattern of gene expression differs, because different regulation (such as different transcription factors) operates in each cell type [1]. (d) mRNA may be degraded quickly, or mRNA may not be translated, so some transcripts give little protein [1]. [5] Examiner insight: Do not write that the muscle cell “lacks” the insulin gene; its genome has it. The cell does not express it.

6. (a) A methyl group is added to cytosine in the promoter [1]. This represses transcription, so the gene downstream is not expressed [1]. (b) Methyl groups are added to amino acids in histones [1]. Depending on the site, transcription may be repressed or activated [1]. [4] Examiner insight: Part (b) needs “repressed or activated”. Saying histone methylation always represses loses the second point.

7. Epigenetic tags, such as methylated DNA or modified histones, can stay attached through mitosis [1], so daughter cells keep the same expression pattern with the base sequence unchanged [1]. If tags also stay attached through meiosis, the changed phenotype could be passed to offspring [1]. [3] Examiner insight: Mention both mitosis (daughter cells) and meiosis (offspring) when the question names both.

8. (a) 64.2 − 52.8 = 11.4 percentage points [1]. (b) (1.00 − 0.58) / 1.00 × 100 [1] = 42% [1]. (c) Higher methylation and lower mRNA in the high-pollution district fit a link between pollution, methylation and lower expression [1]. The mean ± SD ranges (58.1 to 70.3 and 47.0 to 58.6) only just overlap, which suggests a real difference, but the data are an association and cannot prove cause [1]. Other differences between districts, such as age or smoking, were not controlled, and only two districts were sampled [1]. [6] Examiner insight: An evaluation gains credit when it uses the data and also gives a limitation; a verdict with no evidence usually does not.

9. Most epigenetic tags are removed from the ovum and sperm [1], but not all are removed [1]. The retained tags on some genes differ between lion and tiger and depend on which parent supplied the gene [1]. So the same genes are expressed to a different extent in a tigon and a liger, giving different phenotypes, although the DNA sequences have not changed [1]. [4] Examiner insight: “Outline” needs the main points only. Write that this is a proposed explanation rather than a proven one.

10. (a) 10.4 / 2.8 = 3.71 [1]. (b) Monozygotic twins come from one zygote and share the same genome [1], so differences in expression are not due to sequence differences, and can be linked to the environment [1]. (c) Twins who live apart for longer meet more different environments [1], which can change methyl tags and so gene expression differently in each twin [1]. (d) Different pairs are compared, so other factors such as age, or the small number of pairs, may affect the results [1]. [6] Examiner insight: Link the trend to the environment and to tags. Avoid saying that “genes change”.

11. (a) (100 − 4) / 100 × 100 [1] = 96% [1]. (b) The mutant repressor is still active when tryptophan is present (98, close to 100) [1], so it probably cannot be switched by tryptophan and no longer blocks the genes [1]. (c) Tryptophan binds the repressor [1], which can then bind the operator [1] and block RNA polymerase, so the genes are transcribed far less [1]. (d) A steroid hormone (for example, testosterone) enters the cell and binds a receptor [1]; the complex enters the nucleus and binds a specific base sequence as a transcription factor [1]; transcription of target genes changes (here it rises) [1]. [10] Examiner insight: Name one hormone and describe the receptor and transcription factor link. A general statement that hormones “switch on genes” is too vague.

Where marks are usually lost

• Saying epigenetic changes alter the DNA sequence.
• Writing that all tags are erased from gametes.
• Saying a cell “lacks” a gene it simply does not express.
• Saying histone methylation always represses.
• Giving a percentage change without using the original value as the baseline.
• Not stating the number of half-lives in mRNA decay calculations.
• Claiming a correlation proves cause in pollution or twin data.
• Giving a hormone example with no receptor or transcription factor step.
• Confusing the effects of lactose and tryptophan on a repressor.

Next steps

• Re-read the revision notes and the study guide.
• Open the course hub and the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Biology guide, first assessment 2025, International Baccalaureate Organization, section D2.2 Gene expression (additional higher level).