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IB Biology D2.2: Gene expression worked solutions

Worked solutions to every practice question in D2.2: Gene expression of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

HL[1 mark]
A scientist lists every protein present in one muscle cell at a given moment. What is this list called? (A) The genome (B) The transcriptome (C) The proteome (D) The promoter
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C. The proteome is the complete set of proteins a cell holds at one time. The genome is all of its DNA, the transcriptome is all of its mRNA, and a promoter is a stretch of non-coding DNA.

Final answer

C

Question A2

HL[1 mark]
Which statement links nucleosomes to the control of transcription? (A) Nucleosomes carry amino acids to the ribosome. (B) Tight packing round nucleosomes makes a promoter harder to reach. (C) Nucleosomes join RNA nucleotides during transcription. (D) Nucleosomes digest mRNA after transcription.
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B. DNA wound tightly round histones is hard for RNA polymerase and transcription factors to reach, so transcription falls. Nucleosomes do not carry amino acids, join nucleotides or digest mRNA.

Final answer

B

Question A3

HL[1 mark]
A repressor protein is bound to a DNA sequence beside the promoter of a gene. What is the immediate effect? (A) The mRNA is broken down faster. (B) Ribosomes cannot read the mRNA. (C) DNA replication is blocked. (D) RNA polymerase cannot start transcription.
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D. The bound repressor blocks the promoter, so RNA polymerase cannot start transcription. mRNA breakdown, ribosome function and DNA replication are not what a repressor on the promoter controls.

Final answer

D

Question A4

HL[1 mark]
Cytosines in the promoter of a gene become methylated. What is expected? (A) Transcription falls. (B) The base sequence changes. (C) Transcription rises. (D) A shorter protein is made.
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A. Methylation of cytosines in a promoter represses transcription. It does not change the base sequence, raise transcription or shorten the protein.

Final answer

A

Question A5

HL[1 mark]
What happens to most epigenetic tags soon after an egg and a sperm fuse? (A) They are copied twice. (B) They are changed into mutations. (C) Most are removed. (D) All are kept unchanged.
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C. Most tags are removed so that the embryo can switch on the genes it needs. They are not duplicated, converted to mutations or all kept.

Final answer

C

Question A6

HL[1 mark]
Two identical twins show different activity of one gene. Their base sequences are the same. Which is the best explanation? (A) They have different gene orders. (B) Their environments changed the tags on the gene. (C) One twin has an extra chromosome. (D) Crossing over occurred between the twins.
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B. With the same base sequence, the difference must come from the environment changing epigenetic tags. Identical twins do not differ in chromosome number or gene order, and crossing over does not occur between two separate people.

Final answer

B

Question A7

HL[1 mark]
A drug blocks the nucleases that act on one mRNA. What is the most likely result? (A) Transcription of the gene stops. (B) The mRNA is no longer made. (C) Ribosomes cannot translate it. (D) The mRNA lasts longer and more protein is made.
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D. The mRNA is broken down more slowly, so it lasts longer and more of its protein is made. Transcription is not stopped and the mRNA is still made and translated.

Final answer

D

Question A8

HL[1 mark]
A gene is silenced by methylation of its promoter in a liver cell. What is expected in the cells formed from it by mitosis? (A) The gene stays silent. (B) The gene is switched on. (C) The gene is mutated. (D) The gene is deleted.
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A. The tags are copied when the cell divides, so the gene stays silent in the daughter cells. The sequence is not mutated or deleted, and silence is not reversed by division alone.

Final answer

A

Question A9

HL[1 mark]
Cell X contains mRNA for gene G but no protein G. Which statement is a possible explanation? (A) Gene G is missing from the genome of X. (B) Gene G was never transcribed. (C) The mRNA was broken down before it was translated. (D) Gene G has no base sequence.
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C. The mRNA can be broken down by nucleases or blocked from translation before protein is made. If the mRNA is present, gene G is in the genome and was transcribed.

Final answer

C

Question A10

HL[1 mark]
A gel of mRNA shows a band for gene H in nerve cells and no band in skin cells. What does this show? (A) Skin cells lack gene H. (B) Gene H is transcribed in nerve cells. (C) Nerve cells contain protein H. (D) Gene H has a different sequence in skin cells.
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B. A band means the gene was transcribed in that tissue. Skin cells still have the gene; the gel does not show protein and says nothing about a difference in sequence.

Final answer

B

Tutorial practice

Question B1

HL[4 marks]
In a bacterium the genes that make the amino acid tryptophan are controlled by a repressor protein. (a) State the role of the promoter. [1] (b) Explain how a high concentration of tryptophan stops the bacterium making more tryptophan. [3]
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(a) The promoter is the non-coding DNA sequence at which RNA polymerase binds to start transcription (1). (b) When tryptophan is plentiful it binds the repressor protein (1); the repressor then binds the DNA close to the promoter (1) and blocks RNA polymerase, so the genes for tryptophan synthesis are not transcribed and synthesis stops, which is negative feedback (1). The signal molecule acts on the protein, not on the DNA.

Final answer

(a) The promoter is the sequence where RNA polymerase binds to start transcription. (b) Tryptophan binds the repressor; the complex binds DNA beside the promoter and blocks RNA polymerase, so the genes are not transcribed and no more tryptophan is made: negative feedback.

Question B2

HL[5 marks]
Two cell lines have the same base sequence for gene R. The promoter methylation of gene R and its mRNA were measured. Cell line P: promoter methylation 8 percent, mRNA 100 units. Cell line Q: promoter methylation 71 percent, mRNA 12 units. (a) Calculate the percentage change in mRNA from P to Q. [2] (b) Explain the difference in mRNA between the two cell lines. [3]
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(a) \dfrac{12-100}{100}\times100 (1) =-88\%, that is a decrease of 88\% (1). (b) Methyl groups are added to cytosines in the promoter of gene R in Q (1); methylation of a promoter represses transcription, so RNA polymerase cannot easily start (1); less mRNA is made, 12 against 100 units (1). Since the base sequence is the same in both lines, the difference is epigenetic and not a mutation.

Final answer

(a) (12-100)/100\times100=-88\%, a fall of 88%. (b) Methyl groups on cytosines in the promoter CpG island repress transcription, so RNA polymerase makes less mRNA in Q. The base sequence is the same, so the tag, not a mutation, explains the difference.

Question B3

HL[4 marks]
The mRNA for protein K has a half-life of 4 h in a cell. Transcription of the gene then stops. (a) Calculate the percentage of the original mRNA that is left 12 h after transcription stops. [2] (b) Suggest why a short-lived mRNA lets a cell change its protein production quickly. [2]
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(a) 12 h is 12\div4=3 half-lives (1); 100\times\left(\tfrac12\right)^3=12.5\% (1). (b) Nucleases break down a short-lived mRNA quickly (1), so translation of it ends soon and the rate of making the protein changes quickly after transcription stops (1).

Final answer

(a) 12/4=3 half-lives, so 100\times(1/2)^3=12.5\%. (b) Nucleases break the mRNA down quickly, so little is left to translate soon after transcription stops, and the amount of protein K made falls quickly.

Question B4

HL[5 marks]
Methylation of DNA in a mammal before and after fertilisation (illustrative values). Methylation of DNA: sperm 85 percent, egg 70 percent; zygote at 8 h paternal 40 percent, maternal 62 percent; two-cell embryo paternal 25 percent, maternal 45 percent. (a) Compare the fall in methylation of paternal and maternal DNA. [2] (b) Explain why most tags are removed after fertilisation. [2] (c) State why some methylation remains. [1]
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(a) Paternal DNA: 85-25=60 points; maternal DNA: 70-45=25 points (1); the paternal fall is more than twice the maternal fall (1). (b) Most tags are removed from the gamete DNA so that the embryo adopts the unmethylated state (1) and genes needed in early development can be expressed rather than kept silent (1). (c) Imprinted genes keep their tags, so methylation does not fall to zero (1).

Final answer

(a) Paternal fell from 85% to 25%, a fall of 60 points; maternal fell from 70% to 45%, a fall of 25 points; the paternal fall is larger. (b) Removing the tags lets the embryo switch on the genes it needs for early development. (c) Some genes, imprinted genes, keep their tags.

Question B5

HL[5 marks]
Concordance is the percentage of pairs in which both twins have a condition when one does (200 pairs of each type). Concordance for condition X: identical twins 44 percent, non-identical twins 12 percent. (a) Compare the concordance of the two twin types. [2] (b) Deduce what the data show about the base sequence and the environment. [2] (c) State one limitation of twin studies. [1]
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(a) Identical 44% and non-identical 12% (1); 44\div12\approx3.7 times as high (1). (b) Identical twins share the same base sequence, so concordance below 100% shows that the environment, for example through epigenetic tags, changes expression (1). The much higher value for identical twins suggests that the sequence also contributes (1). (c) One limit: identical twins usually share their early environment, or only 200 pairs were studied, so the effect of the environment is not measured exactly (1).

Final answer

(a) Identical 44% against 12%: 3.7 times higher. (b) Identical twins share the base sequence, yet concordance is not 100%, so the environment, acting through tags, affects expression. The higher value suggests a genetic contribution. (c) Twins share much of their environment and the sample is small.

Question B6

HL[4 marks]
In a lineage of skin stem cells, gene G is silent because its promoter is methylated. (a) Predict the state of gene G in the daughter cells and explain your prediction. [2] (b) Explain why the silence of G is not a mutation. [1] (c) State one way the silence of G could end. [1]
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(a) Gene G stays silent in the daughter cells (1) because the methyl tags remain or are copied onto the new DNA at mitosis, so the promoter stays methylated and transcription stays repressed (1). (b) A mutation changes the base sequence; here the sequence is unchanged and only a tag has been added (1). (c) Removal of the methyl groups, for example by a drug, or reprogramming that removes tags (1).

Final answer

(a) Silent in the daughter cells: the methyl tags are copied at mitosis, so promoter methylation keeps transcription repressed. (b) The base sequence has not changed. (c) The tags could be removed, for example by a drug that removes methyl groups.

Question B7

HL[5 marks]
Blood samples from 60 children in each of two districts were tested for methylation of the promoter of gene R. Mean methylation of the promoter of gene R: low-pollution district 34 percent, high-pollution district 47 percent. (a) Calculate the percentage change in methylation from low to high pollution. [2] (b) Predict the effect on the expression of gene R. [2] (c) State why the data do not prove that pollution caused the change. [1]
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(a) \dfrac{47-34}{34}\times100 (1) =38\% (1). (b) Promoter methylation represses transcription (1), so less mRNA and less protein from gene R is expected in the high-pollution district (1). (c) Observational data show an association, not a cause; diet, income or other differences between districts could also change methylation (1).

Final answer

(a) (47-34)/34\times100=38\%. (b) Methylation of a promoter represses transcription, so less mRNA and less protein R is expected in district 2. (c) The data show an association only; other differences between the districts may explain it.

Question B8

HL[4 marks]
The presence of mRNA and protein for four genes was recorded in cell types X and Y of one animal (+ present, - absent). Gene G1: X mRNA and protein present, Y mRNA and protein present. G2: X present, Y absent. G3: X mRNA present but protein absent, Y absent. G4: absent in both. (a) State which gene is transcribed in X but not translated. [1] (b) Suggest two reasons for your answer to (a). [2] (c) State which of genome, transcriptome and proteome is the same in X and Y. [1]
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(a) Gene G3, because X has its mRNA but not its protein (1). (b) Two of: nucleases break the mRNA down before translation; translation of the mRNA is blocked or slowed; the mRNA is made too recently for protein to have been made (2). (c) The genome, because both cell types belong to one animal and hold all four genes (1).

Final answer

(a) Gene G3. (b) The mRNA is broken down by nucleases before it is translated; translation is blocked. (c) The genome: all four genes are in both cell types.

Advanced practice

Question C1

HL[7 marks]
Explain how gene expression is regulated in eukaryotic cells. Write seven separate points.
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Any seven of: gene expression is how the information in a gene has an effect on the phenotype, through transcription and translation to a working protein (1); transcription factors are proteins that bind specific base sequences (1); a repressor bound to a promoter blocks RNA polymerase (1); an activator bound to an enhancer promotes transcription (1); methylation of cytosine in the promoter represses transcription (1); methylation of histones, and so the packing of nucleosomes, changes access to the gene (1); nucleases break down mRNA, which limits translation (1); an external signal such as a hormone, or lactose or tryptophan in bacteria, acts through a regulatory protein (1); the tags are copied to daughter cells, so patterns of differentiation persist (1); the base sequence is not changed (1). A named example earns the point only if it is attached to one of these ideas.

Final answer

Gene expression links a gene to protein and phenotype. Transcription factors bind specific sequences at promoters or enhancers; repressors block RNA polymerase and activators help it. Methylation of promoter cytosines represses; histone methylation and nucleosome packing change access. Nucleases break down mRNA, limiting translation. A signal such as a hormone acts through a protein. Tags are copied at division; the base sequence is unchanged.

Question C2

HL[8 marks]
A signalling molecule was added to cultured cells at day 0. The methylation of the promoter of gene J, the amount of its mRNA and the amount of its protein were measured (arbitrary units for mRNA and protein). Day 0, 4, 8, 12: promoter methylation 62, 61, 30, 12 percent; mRNA of gene J 10, 11, 48, 90; protein J 8, 8, 15, 60. (a) Describe the relationship between methylation of the promoter and the amount of mRNA. [2] (b) Calculate the percentage change in mRNA from day 0 to day 12. [2] (c) Explain the relationship in (a). [2] (d) At day 8 the mRNA is 4.8 times, but the protein only about 1.9 times, the day 0 value. Suggest two reasons for the difference. [2]
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(a) Methylation falls from 62% at day 0 to 12% at day 12 while mRNA rises from 10 to 90 units, an inverse relationship (2: values for both variables, and the inverse pattern). (b) \dfrac{90-10}{10}\times100 (1) =800\% (1). (c) A less methylated promoter is more accessible, so RNA polymerase can bind and start transcription (1) and more mRNA is made (1). (d) At day 8 mRNA is 48\div10=4.8 times the day-0 value but protein is 15\div8\approx1.9 times (1: values); two of: translation takes time after transcription; protein accumulates slowly; nucleases break mRNA down before it is translated (1).

Final answer

(a) As methylation falls from 62% to 12%, mRNA rises from 10 to 90 units. (b) (90-10)/10\times100=800\%. (c) Less methylation of the promoter lets RNA polymerase bind and transcribe, so more mRNA is made. (d) Translation takes time after transcription and the protein has to accumulate; nucleases may break down mRNA, so protein rose by 1.9 times against 4.8 times for mRNA at day 8.

Question C3

HL[7 marks]
Methylation of the promoter of gene T was measured in a pair of identical twins at three ages. Twin 2 lived in a city with heavy air pollution; Twin 1 lived in the country. Methylation at ages 3, 30 and 60: Twin 1 52, 53, 55 percent; Twin 2 52, 61, 74 percent. (a) Calculate the difference in methylation between the twins at each age. [2] (b) Calculate the percentage change in methylation in Twin 2 from age 3 to age 60. [2] (c) Explain the change in the difference between the twins. [2] (d) A student concludes that pollution caused the change. State one reason this conclusion is not justified. [1]
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(a) At age 3: 52-52=0; at age 30: 61-53=8; at age 60: 74-55=19 (all three values, 2). (b) \dfrac{74-52}{52}\times100 (1) =42\% (1). (c) The twins have the same base sequence, so the divergence is not caused by a mutation (1); the environment changed the methylation, an epigenetic tag, so expression differs as the twins grow apart (1). (d) One pair only, and the lifestyles differ in many ways besides pollution (diet, exercise, disease), so the data show an association and not a cause (1).

Final answer

(a) Differences 0, 8 and 19 points. (b) (74-52)/52\times100=42\%. (c) The twins have the same base sequence, so a mutation cannot explain it; different environments changed the tags. (d) One pair only, and many other differences besides pollution.

Question C4

HL[6 marks]
In a bacterium, a repressor protein R binds DNA beside the promoter of the genes for enzyme E, which digests sugar S. Sugar S binds to R. In a mutant, R cannot bind DNA. Normal cell: sugar S absent, mRNA 2, enzyme 3; S present, mRNA 85, enzyme 120. Mutant: S absent, mRNA 82, enzyme 118; S present, mRNA 84, enzyme 121. (a) Explain the results for the normal cell with and without sugar S. [3] (b) Deduce what the results for the mutant show about the role of R. [2] (c) State which column of the table is part of the transcriptome. [1]
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(a) Normal cell without S: R is bound to the DNA and blocks RNA polymerase, so mRNA is only 2 units and enzyme 3 units (1); with S, S binds R, R leaves the DNA (1), and RNA polymerase transcribes, giving 85 units of mRNA and 120 units of enzyme (1). (b) In the mutant R cannot bind the DNA, so the promoter is never blocked (1); the gene is transcribed with or without S (about 82 and 84 units) (1). (c) The mRNA column is the transcriptome and the enzyme column is part of the proteome (1).

Final answer

(a) Without S, R binds the DNA and blocks RNA polymerase, so little mRNA and enzyme. With S, S binds R, R lets go, RNA polymerase transcribes, so mRNA and enzyme are high. (b) In the mutant R cannot block the promoter, so the gene is always on. (c) mRNA is the transcriptome; the enzyme is part of the proteome.

Question C5

HL[6 marks]
Female species 1 crossed with male species 2 gives hybrids that are 30% heavier at birth than the hybrids from female species 2 crossed with male species 1. Each hybrid has one set of chromosomes from each species. (a) Explain why the base sequence cannot explain the size difference. [2] (b) Explain the difference using genomic imprinting. [3] (c) State what happens to most tags after fertilisation. [1]
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(a) Both hybrids carry the same two sets of base sequences (1), so the difference cannot be caused by the sequence, for example by dominant alleles (1). (b) Imprinted genes keep their tags and only one copy is active, depending on the parent that supplied it (1); a growth gene that is methylated and silent when it comes from one species is active when it comes from the other (1); so the two crosses express different amounts of the growth gene, and the hybrids differ in size (1). (c) Most tags are removed soon after fertilisation, but imprinted genes keep theirs, which is why the parental origin still matters (1).

Final answer

(a) Both hybrids have the same combination of base sequences, so the sequence cannot explain it. (b) Imprinting: tags depend on the parent of origin, so a growth gene is silenced on the maternal or on the paternal copy; which copy is silent differs between the crosses, so expression of the growth gene differs. (c) Most tags are removed so the embryo can develop, but imprinted genes escape.

Connected questions

Question X1

Connected questions · HL[6 marks]
D2.2 + D1.2 + B2.3. A neurone and a muscle fibre from the same embryo hold the same genome. The muscle fibre makes the protein myosin; the neurone does not. (a) Explain how two cells with the same genome come to make different proteins. [3] (b) State the process in which the mRNA for myosin is used to make myosin. [1] (c) Suggest how the myosin gene is kept silent in the neurone and its daughter cells. [2]
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(a) Both cells have the same genome (1); different transcription factors are active in the two cells, so the myosin gene is transcribed in muscle but not in the neurone (1); an activator binds the enhancer or promoter of the myosin gene in muscle and mRNA is made (1). (b) Translation: a ribosome builds the myosin polypeptide from the mRNA (1). (c) In the neurone the myosin gene is silent because the activator is absent, or because its promoter is methylated and transcription is repressed (1); the pattern is kept when the cell divides, so the neurone lineage stays specialised (1).

Final answer

(a) Different transcription factors are active in each cell, so different genes are transcribed. In muscle a factor binds the myosin gene's promoter or enhancer and promotes transcription. (b) Translation at a ribosome. (c) The neurone may lack the activator, or the promoter may be methylated and silent.

Question X2

Connected questions · HL[6 marks]
D2.2 + C2.1 + D1.2. Oestradiol is a steroid hormone. It crosses the plasma membrane of a target cell, binds a receptor in the cytoplasm, and the pair enters the nucleus and binds DNA. (a) Explain why only target cells respond to oestradiol. [2] (b) Outline how the hormone changes the proteins made by a target cell. [3] (c) State why a steroid hormone can cross the plasma membrane directly. [1]
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(a) Only target cells contain the receptor for oestradiol, so only they respond (1); other cells have the same hormone passing by but nothing to bind (1). (b) The hormone–receptor pair enters the nucleus and binds to specific base sequences in the DNA (1), where it acts as a transcription factor and raises or lowers transcription of particular genes (1); the mRNA is then translated into more or less of a protein (1). (c) A steroid hormone is lipid-soluble (non-polar), so it passes through the phospholipid bilayer directly (1).

Final answer

(a) Only target cells have the receptor. (b) The hormone–receptor pair acts as a transcription factor: it binds DNA and changes transcription, so more or less mRNA is made and translated into protein. (c) Steroids are lipid-soluble and cross the phospholipid bilayer.

Question X3

Connected questions · HL[6 marks]
D2.2 + D2.1 + D1.3. In some tumours the promoter of a tumour-suppressor gene is heavily methylated, although its base sequence is normal. The gene's protein normally slows cell division. (a) Explain how methylation can have the same effect on the cell as deleting the gene. [2] (b) Explain why the daughter cells formed by mitosis also lack the protein. [2] (c) Distinguish this change from a mutation. [1] (d) Suggest how a drug could be used to treat the tumour. [1]
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(a) Methylation of the promoter represses transcription (1), so no tumour-suppressor protein is made, which has the same effect as losing the gene (1). (b) The tags remain or are copied at mitosis, so the daughter cells inherit the silent gene (1) and the cells keep dividing without the brake (1). (c) A mutation changes the base sequence, whereas the sequence here is normal and only a tag is present (1). (d) A drug that removes methyl groups from the promoter could restore transcription of the tumour-suppressor gene, so the protein is made again (1).

Final answer

(a) The methylated promoter represses transcription, so no mRNA or protein is made, which has the same result as a gene deletion. (b) The tags are copied at mitosis, so every daughter cell also lacks the protein and keeps dividing. (c) A mutation changes the sequence; here only a tag has been added. (d) A drug that removes methyl groups could let transcription restart.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Gene expression from the core ideas to full exam questions, with question types, worked examples and graded practice.

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