Using genome projects - AQA A-Level Biology (Paper 2) Exam Questions

Practise 17 exam-style questions on Using genome projects for AQA A-Level Biology (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Determining the genome of a pathogenic bacterium can be used to identify potential antigens for vaccines.

Give two other ways in which knowing the genome of a pathogenic bacterium could help to prevent or treat the disease it causes.

Model answer

1. Identify genes for antibiotic resistance, so doctors can choose an antibiotic that will work.
2. Identify enzymes found only in the bacterium, which can be targets for new drugs.

Explanation

A genome sequence has many uses besides finding antigens: it reveals resistance genes, unique proteins that drugs could target, sequences for diagnostic tests and differences between strains that show how an outbreak spreads. Each answer must be a different use, linked to preventing or treating the disease.

Mark scheme

1 mark
Identify genes for antibiotic resistance, so an effective antibiotic can be chosen
Max 2. List rule: if more than two ways are given, each wrong or contradictory extra way cancels one correct way. Points 1 to 4: 1 mark each.
1 mark
Identify proteins / enzymes found only in the bacterium as targets for new drugs / antibiotics
Accept 'design drugs that inhibit a bacterial enzyme not found in humans'.
1 mark
Develop tests to identify / diagnose the bacterium (e.g. with DNA probes or PCR)
Accept 'faster diagnosis'.
1 mark
Compare genomes from different patients / strains to trace the source or spread of an outbreak
Accept 'identify which strain is causing an outbreak'.

Question 2

3 marks

Sequencing projects have read the genomes of humans and of mice. On average, the protein-coding sequences of human genes and of the equivalent mouse genes are about 85% identical.

A newspaper article stated: 'Because mouse and human genes are so similar, a drug that cures a genetic disease in mice will also cure the same disease in humans.'

Evaluate this statement.

Model answer

Mice have equivalent genes that code for similar proteins, so a mouse with the faulty gene can be a good model of the human disease, and a drug that acts on the mouse protein may well act on the human one. However, about 15% of the coding sequence differs, so the human protein may have a different tertiary structure and the drug may not bind to it. Only coding sequences were compared: differences in non-coding DNA and regulatory genes mean the genes may be expressed differently in humans, and humans may break the drug down differently. So the statement is not justified: the drug must still be tested in clinical trials in humans.

Explanation

Genome projects allow genes to be compared between species, which is why mice are used as models of human genetic diseases. An evaluation needs a point for the statement and points against it. 85% identity in coding sequences still leaves differences in protein structure, and the presence of non-coding DNA and regulatory genes means that similar genes are not necessarily expressed in the same way.

Mark scheme

1 mark
(For) mice have equivalent genes coding for similar proteins (with similar functions), so mice with the faulty gene can be used as a model of the disease / a drug acting on the mouse protein may act on the human protein
Max 2 if the answer gives only points for or only points against the statement. Max 3. Points 1 to 5: 1 mark each.
1 mark
(Against) about 15% of the coding sequence differs, so the human protein may have a different amino acid sequence / tertiary structure, so the drug may not bind to it / may not work
Accept 'the drug's target protein may be a different shape in humans'.
1 mark
(Against) only protein-coding sequences were compared; non-coding DNA / regulatory genes may differ, so the genes may be expressed differently (in different tissues or at different times) in humans
1 mark
(Against) humans and mice differ in size / metabolism / physiology, so the drug may be broken down differently / need a different dose / cause different side effects
1 mark
(Against) 'will cure' is not justified: the drug must still be tested (in clinical trials) on humans for safety and effectiveness
Accept 'similar genes do not prove the drug will work in humans'.

Question 3

4 marks

The genome of a bacterium contains 1.8 × 10⁶ base pairs. Of these, 89% are in protein-coding genes. The mean length of a protein-coding gene in this bacterium is 960 base pairs.

Calculate the number of protein-coding genes in this genome. Give your answer to 3 significant figures.

Scientists used this genome sequence to determine the proteome of the bacterium. Explain what is meant by the genome and the proteome of an organism.

Model answer

Coding base pairs = 1.8 × 10⁶ × 0.89 = 1.602 × 10⁶
Number of genes = 1.602 × 10⁶ ÷ 960 = 1668.75 = 1670 (3 s.f.)

The genome is the complete set of genes (all the DNA) in a cell. The proteome is the full range of proteins that the cell is able to produce.

Explanation

Find the number of base pairs in protein-coding genes first (89% of the genome), then divide by the mean length of one gene. Because most of a bacterial genome codes for proteins and the genes have no introns, the number and sequences of its proteins can be predicted from the genome. Check the power of ten in standard form calculations. The genome is all the DNA (the complete set of genes) and the proteome is the full range of proteins a cell can make: definitions must convey 'complete' or 'full range'.

Mark scheme

1 mark
Correct method, e.g. 1.8 × 10⁶ × 0.89 ÷ 960; or a correct final answer of 1670 / 1.67 × 10³ with no working
A correct answer of 1670 (accept 1.67 × 10³) with no working scores 2 marks (points 1 and 2). Accept for 1 mark: 1668.75 or 1669 (not given to 3 significant figures) OR evidence of 1.602 × 10⁶ (coding base pairs) OR 1875 (89% not used) OR 167 or 16 700 (one power-of-ten error).
1 mark
1670 / 1.67 × 10³ (genes)
1 mark
Genome: the complete set of genes / all the DNA (base sequence) in a cell / of an organism
Accept 'all the genetic material of an organism'. Insufficient: 'the genes in a chromosome' or 'the DNA in the nucleus' (not all the DNA).
1 mark
Proteome: the full range of proteins that a cell / organism is able to produce / that are coded for by its genome
Insufficient: 'the proteins in a cell' with no idea of the full range / all the proteins.

Question 4

5 marks

A single-celled eukaryotic parasite causes a disease in cattle. It is carried by ticks, which transmit it to cattle when they feed. The parasite's genome contains about 4000 protein-coding genes.

Scientists identified the proteins present in the parasite at two stages of its life cycle. They found 1900 different proteins in the stage that lives in the tick and 2300 in the stage that lives in cattle blood. 1200 of these proteins were found in both stages.

Calculate the percentage of the parasite's protein-coding genes for which no protein was found in either stage.

Explain why the proteins found in the two stages are different, even though the parasite has the same genome in both stages.

Suggest which of the proteins found would be most suitable for a vaccine to protect cattle.

Model answer

Proteins found in at least one stage = 1900 + 2300 - 1200 = 3000
Genes with no protein found = 4000 - 3000 = 1000
Percentage = 1000 ÷ 4000 × 100 = 25%

Regulatory genes control which genes are transcribed. The conditions in the tick and in cattle blood are different, so different transcription factors are active and different sets of genes are expressed, producing different proteins.

Proteins on the surface of the stage that lives in cattle blood, because this stage is exposed to the cattle's antibodies.

Explanation

The 1200 proteins found in both stages must be counted only once, so the proteins found = 1900 + 2300 - 1200 = 3000. A eukaryotic genome contains regulatory genes that switch the transcription of other genes on and off, so the proteome changes from stage to stage - one reason why the genome of a complex organism cannot easily be translated into its proteome. A vaccine for cattle must target antigens present on the stage in the cattle's blood.

Mark scheme

1 mark
Correct method, e.g. (4000 - (1900 + 2300 - 1200)) ÷ 4000 × 100; or a correct final answer of 25 (%) with no working
Max 5. A correct answer of 25% with no working scores 2 marks (points 1 and 2). Accept for 1 mark: evidence of 3000 (proteins found in at least one stage) OR 1000 (genes with no protein found) OR 75 (percentage with a protein found). Do not credit -5 (1200 not subtracted).
1 mark
25 (%)
1 mark
Regulatory genes / transcription factors control which genes are transcribed / expressed
Points 3 and 4: 1 mark each. Accept epigenetic control for this point.
1 mark
Different genes are expressed in the different conditions of the tick and of the cattle blood / different proteins are needed at each stage
Ignore 'genes are switched on or off' unless transcription or expression is referred to.
1 mark
Proteins on the surface of the stage in cattle blood
Accept 'proteins of the blood stage that are exposed to the immune system / to antibodies'. Accept surface proteins found in both stages, or surface proteins of the form that passes from the tick into the cattle's blood. Insufficient: 'proteins of the blood stage' with no reference to the surface or to exposure to antibodies.

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