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IB Biology 2025 · Theme D · Continuity and change

D2.2 Gene expression (HL). Practice questions with markscheme.

77 original IB-style questions on D2.2, written from the 2025 guide: 36 multiple-choice, 28 short-answer, 9 data-based, 2 extended-response parts, 1 drawing, 1 labelling. This subtopic is additional higher level in its entirety, so below is a 20-mark higher-level practice paper built from them. Hand one out as a class quiz or homework, or sit it yourself and mark against the scheme. Print it, project it, or build a fresh one on the same topic.

What the guide asks for

All 11 statements are additional higher level, so this subtopic is HL only.

  1. D2.2.1HL Gene expression as the mechanism by which information in genes has effects on the phenotype
  2. D2.2.2HL Regulation of transcription by proteins that bind to specific base sequences in DNA
  3. D2.2.3HL Control of the degradation of mRNA as a means of regulating translation
  4. D2.2.4HL Epigenesis as the development of patterns of differentiation in the cells of a multicellular organism
  5. D2.2.5HL Differences between the genome, transcriptome and proteome of individual cells
  6. D2.2.6HL Methylation of the promoter and histones in nucleosomes as examples of epigenetic tags
  7. D2.2.7HL Epigenetic inheritance through heritable changes to gene expression
  8. D2.2.8HL Examples of environmental effects on gene expression in cells and organisms
  9. D2.2.9HL Consequences of removal of most but not all epigenetic tags from the ovum and sperm
  10. D2.2.10HL Monozygotic twin studies
  11. D2.2.11HL External factors impacting the pattern of gene expression

In the bank for D2.2

  • 36 multiple-choice
  • 28 short-answer
  • 9 data-based
  • 2 extended-response parts
  • 1 drawing
  • 1 labelling
  • 77 higher level only

Every question is original and tagged to a guide statement. See the whole bank →

Make your own

The practice paper

Additional higher level only · 20 marks
Take it online → Build a fresh paper Paper code BbB-EAEAAAAAARQAO8gA
Biology · topic quiz
Higher level · topic practice, not an exam format
30 minutes20 marks

Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.

Covers D2.2 Gene expression · HL
Name:
1.

Most gene regulation in cells occurs at which stage?

[1]
  1. During DNA replication in S phase
  2. At transcription, when mRNA is made
  3. During mitosis and cytokinesis
  4. After the protein has been secreted
2.

Cells of three types were taken from one person: skin cells, developing red blood cells (which still have a nucleus) and β cells from the pancreas, which secrete insulin. For six genes, the table shows whether the mRNA transcribed from the gene and the protein coded for by the gene were detected in each type of cell (+ detected, – not detected). Two days before sampling, the β cells had been exposed to a signal that caused gene 6 to be transcribed for a few hours.

GeneProtein coded for by the geneSkin cell: mRNA / proteinDeveloping red blood cell: mRNA / proteinβ cell: mRNA / protein
1an enzyme of glycolysis+ / ++ / ++ / +
2keratin, a structural protein of skin+ / +– / –– / –
3globin, part of haemoglobin– / –+ / +– / –
4insulin– / –– / –+ / +
5rhodopsin, a light-absorbing protein of the eye– / –– / –– / –
6a protein that protects cells from stress– / –– / –– / +
(a)Deduce, with a reason, whether gene 5 is part of the genome of these three types of cell.[2]
(b)Using the data for genes 1 to 4, explain how the pattern of gene expression in each type of cell is related to the way it has differentiated.[3]
(c)Suggest why protein 6 was detected in the β cells although its mRNA was not.[2]
3.

A student studied three genes in leaf cells and root cells taken from one tomato plant (Solanum lycopersicum). Gene C codes for a protein that binds chlorophyll in chloroplasts, gene W codes for a membrane protein that takes up nitrate ions from the soil, and gene E codes for an enzyme of glycolysis. For each gene in each type of cell the table shows whether the gene was detected in the DNA, the percentage of the cytosines in the promoter of the gene that carried a methyl group, whether the histone proteins near the gene carried methyl groups, and the amount of mRNA transcribed from the gene. The student then wrote three claims.
Claim P: Root cells do not make the chlorophyll-binding protein because gene C was lost from root cells when they differentiated.
Claim Q: The promoter of gene W is methylated in leaf cells, so the base sequence of gene W in leaf cells must differ from its base sequence in root cells.
Claim R: Gene W is transcribed in root cells although the histones near it carry methyl groups. This result must be an error, because adding methyl groups always stops a gene from being transcribed.

Gene, type of cellGene detected in the DNAPromoter cytosines carrying a methyl group / %Methyl groups on nearby histonesmRNA / relative units
C, leafyes4yes100
C, rootyes81yes1
W, leafyes77yes0
W, rootyes6yes45
E, leafyes5yes30
E, rootyes3yes32
(a)Using the data, explain the error in claim P.[2]
(b)Explain the error in claim Q.[2]
(c)Explain the error in claim R.[2]
4.

Monozygotic (identical) twins develop from a single zygote. Researchers compared the methylation of the same set of cytosine sites in blood cells from each twin of a pair. They studied 40 twin pairs in each of three age groups and recorded the percentage of sites at which the two twins of a pair differed. The table shows the mean for each age group.

Age of twin pair / yearsSites with differing methylation between the twins / %
32.1
256.8
5011.4
(a)Describe the trend shown by the data.[1]
(b)Analyse what the trend suggests about the causes of these differences.[2]
(c)Evaluate whether these data show that the environment causes the diverging methylation.[3]

Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
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Markscheme BbB-EAEAAAAAARQAO8gA

One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.

1. D2.2.2 [1]
  • B: transcriptional control (transcription factors binding promoters/enhancers) is the main point of regulation; translation and protein degradation add further control;
2. D2.2.5
  • (a) [2 max]
    • yes, gene 5 is part of the genome of all three types of cell;
    • all the body cells of a person come from the zygote by mitosis, so they carry the same genes / the same genome;
    • it is not transcribed in any of the three, so it is missing only from their transcriptomes (and proteomes) / no cell expresses all of its genes;
    • it is probably expressed only in other cells, such as cells of the retina / eye, OWTTE;

    Do not award the first point without a reason.

  • (b) [3 max]
    • gene 1 is expressed in all three types of cell, because every cell needs the enzymes of glycolysis / to respire;
    • genes 2, 3 and 4 are each expressed in only one type of cell: keratin in skin cells, globin in developing red blood cells, insulin in β cells;
    • each of these proteins suits the cell that makes it to its role: keratin to the structure of skin, globin (haemoglobin) to the red blood cell, insulin to secretion by the β cell;
    • so a cell expresses the genes that its specialised role needs and not the others / its transcriptome and proteome reflect the type of cell it has become;
    • the pattern of gene expression determines how the cell differentiates, and the cells differ in which genes they express, not in which genes they carry, OWTTE;
  • (c) [2 max]
    • the mRNA was transcribed and translated in response to the signal two days earlier;
    • the mRNA has since been broken down by nucleases;
    • the protein lasts longer / is broken down more slowly than its mRNA, so it is still present;
    • so the proteome can include proteins whose mRNA is no longer part of the transcriptome, OWTTE;
3. D2.2.5, D2.2.6
  • (a) [2 max]
    • gene C was detected in the DNA of the root cells, so it has not been lost;
    • all the cells of the plant come from one zygote (by mitosis) and carry the same genes / genome, but no cell expresses all of its genes;
    • in root cells gene C is (almost) not transcribed (1 compared with 100 units of mRNA) because most of the cytosines in its promoter are methylated (81 % compared with 4 %);
    • so the protein is missing because the gene is silenced, not because the gene is absent, OWTTE;

    Award [1 max] for identifying the error without data or without the correct explanation.

  • (b) [2 max]
    • methylation adds methyl groups to cytosine bases, so the bases and their sequence are not changed;
    • it is an epigenetic tag, not a mutation, and it can be removed again / is reversible;
    • leaf and root cells of one plant have the same base sequence of gene W; the methylation only represses transcription of gene W in leaf cells (0 compared with 45 units of mRNA);
    • so the two types of cell differ in the expression of gene W / in phenotype, not in genotype, OWTTE;
  • (c) [2 max]
    • methylation of cytosines in a promoter represses transcription, but methylation of histones can either activate or repress transcription;
    • whether a methyl group on a histone activates or represses depends on which amino acid in the histone carries it;
    • so methylated histones near a gene that is being transcribed are not an error: the promoter of gene W in root cells is (almost) unmethylated (6 %) and the gene is transcribed (45 units);
    • every gene in the table has methylated histones, whether it is transcribed or not, so the histone data alone cannot show whether a gene is switched off, OWTTE;

    Do not accept "methylation never affects transcription".

4. D2.2.10
  • (a) [1]
    • older twin pairs differ in methylation at a greater percentage of sites (2.1 % at age 3 rising to 11.4 % at age 50);
  • (b) [2]
    • the twins have (essentially) identical genomes, so the growing differences cannot be genetic;
    • (so) epigenetic differences accumulate with age, consistent with environment/lifestyle progressively altering methylation (epigenetic drift), OWTTE;
  • (c) [3]
    • strength: using monozygotic twins controls for genotype, so differences are not due to DNA sequence; the sample of 40 pairs per group is reasonable;
    • limitation: the study is correlational/observational. No environmental factor was manipulated, and older twins may simply have lived apart longer (a confound);
    • limitation: stochastic (random) drift in methylation with cell divisions could also contribute, so the data show an association with age but do not by themselves prove environmental causation, OWTTE;

    Nature of Science: correlation vs causation, controls, sample.

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