AQA A-level Biology: Genome Projects and Gene Technologies (7402) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow AQA A-level Biology (7402) sections 3.8.3 and 3.8.4, in the AQA AS and A-level Biology specification for first teaching 2015. All of this content is A-level only. Explanations are in the study guide and the revision notes. Course pages: AQA A-level Biology hub and the printable checklist. Weak spots? Try the free diagnostics. Use only the data given in each question. The marks shown are for practice and are not an official tariff.

Questions

1. Explain why the proteome of a bacterium can be predicted from its genome sequence more easily than the proteome of a human. [3]

2. The genome of a pathogenic bacterium has been sequenced. Explain how this information could be used to help produce a vaccine against it. [4]

3. A forensic laboratory starts a PCR with 6 DNA molecules. (a) Calculate the number of molecules after 7 cycles. [2] (b) Calculate the smallest whole number of cycles needed to produce at least 10 000 molecules. [2] (c) Explain why Taq polymerase is used. [2]

4. A human gene is to be expressed in bacteria. Explain why scientists make the gene from mRNA using reverse transcriptase instead of cutting it from human chromosomal DNA. [4]

5. A linear piece of DNA is 35 base pairs long. The top strand reads ACGGATCCTTAGCAGGATCCAATGCGGATCCTTCA. A restriction endonuclease recognises GGATCC and cuts between the first G and the second G on each strand. (a) Give the lengths of the fragments produced (top strand). [3] (b) Explain why the fragments from this DNA can join to a plasmid cut with the same enzyme. [2] (c) State the role of DNA ligase in this process. [2]

6. Bacteria were treated with a plasmid carrying a gene for a fluorescent protein. The plasmid also carried the human gene being cloned. Of 268 colonies on nutrient agar, 37 fluoresced under ultraviolet light. (a) Calculate the percentage of colonies that were transformed. [2] (b) Explain why the fluorescent gene is called a marker gene. [2] (c) Explain why a promoter and a terminator are placed on either side of the human gene. [2]

7. A normal allele has the region 5’ GCATTGACCTA 3’. A mutant allele has a single base change, giving 5’ GCATTGTCCTA 3’. (a) Write the 5’ to 3’ sequence of a probe that is complementary to the mutant region. [2] (b) Describe how this labelled probe is used to find whether a patient’s DNA contains the mutant allele. [3]

8. A couple are both unaffected, but their first child has a recessive genetic condition. (a) State the genotypes of the parents and the probability that their second child is affected. [2] (b) A man’s mother is a known carrier of a different recessive allele and his father is not. His partner is from a population where 1 in 50 people are carriers. Calculate the probability that their child is affected. [3]

9. A gene variant makes a patient break down a drug slowly. Evaluate the use of DNA screening for this variant before prescribing the drug. [4]

10. Fragments of DNA from a crime scene (C) and three suspects (S1 to S3) were separated by gel electrophoresis. Band sizes in kilobases are: C 7.2, 5.1, 3.8, 2.4; S1 7.2, 5.1, 4.4, 2.4; S2 7.2, 5.1, 3.8, 2.4; S3 6.0, 5.1, 3.8, 1.9. (a) Identify which suspect’s profile matches C and give the evidence. [2] (b) Explain why the 2.4 kb band is closer to the end of the gel than the 7.2 kb band. [3] (c) The chances of a coincidental match at four independent loci are 0.05, 0.1, 0.2 and 0.1. Calculate the chance of a coincidental match across all four. [2]

11. Explain how genetic fingerprinting can help a breeder of a rare animal species. [3]

12. A company proposes to sell crop seed carrying a gene transferred from another species, to farmers in a region with food shortages. Evaluate the ethical, financial and social issues. [6]

Answers

1. A bacterial genome has little non-coding DNA and few regulatory genes [1], so the base sequence can be read with the genetic code to give amino acid sequences [1]. A human genome has much non-coding DNA and many regulatory genes, so the genome cannot easily be translated into the proteome [1]. [3] Examiner insight: A comparison needs both organisms. Describing only the bacterium leaves the contrast unsaid.

2. Use the sequence to identify genes and predict the proteins [1]. Select proteins on the pathogen surface [1] that could act as antigens recognised by the immune system [1]. Produce this protein (or part of it) and use it in the vaccine to stimulate an immune response [1]. [4] Examiner insight: Naming the link from gene to protein to antigen is what separates a complete answer from a general statement about vaccines.

3. (a) 6 × 2⁷ [1] = 768 molecules [1]. (b) 6 × 2¹⁰ = 6144, which is below 10 000 [1]; 6 × 2¹¹ = 12 288, so 11 cycles [1]. (c) It is heat-stable [1], so it keeps working after the mixture is heated to about 95 °C in every cycle [1]. [6] Examiner insight: In part (b) show the value just below the target as well as the one above it; stating only the final number gives no evidence that 10 cycles falls short.

4. Human chromosomal DNA contains introns [1]. Bacteria cannot remove introns from the transcript [1]. mRNA has already had introns removed [1]. Reverse transcriptase makes cDNA with no introns, which bacteria can transcribe and translate [1]. [4] Examiner insight: “Bacteria cannot splice introns” is the explanation. Saying only that mRNA is “easier” does not answer the question.

5. (a) GGATCC occurs starting at bases 3, 15 and 26 [1]. Cuts fall after bases 3, 15 and 26 [1]. Fragments are 3, 12, 11 and 9 bases (sum 35) [1]. (b) The cut leaves single-stranded sticky ends [1] whose bases are complementary to the sticky ends on the plasmid, so they pair by hydrogen bonds [1]. (c) It joins the sugar-phosphate backbones [1] by forming phosphodiester bonds between the fragment and plasmid [1]. [7] Examiner insight: Show the cut positions and the check that the lengths add up to 35. Writing only a list of lengths gives no evidence of method.

6. (a) 37 ÷ 268 [1] × 100 = 13.8% [1]. (b) It lets you detect which cells took up the plasmid [1] because only transformed cells fluoresce [1]. (c) The promoter is where RNA polymerase binds to start transcription [1]; the terminator stops transcription at the end of the gene [1]. [6] Examiner insight: Give the percentage to a sensible number of significant figures and link the marker to its purpose of detection.

7. (a) The mutant strand is 5’ GCATTGTCCTA 3’. The complement is written in reverse: 5’ TAGGACAATGC 3’, with correct base pairing [1] and the correct 5’ to 3’ direction [1]. (b) Separate the patient’s DNA into single strands (denature) [1]. Add the labelled probe, which binds by complementary base pairing to the mutant allele only [1]. Wash off unbound probe and detect the label (radioactivity or fluorescence); a signal shows the mutant allele is present [1]. [5] Examiner insight: Writing the complement in the wrong direction is a common slip; label 5’ and 3’ ends so the direction is checkable.

8. (a) Both parents are heterozygous carriers [1]; probability that the second child is affected = 1/4 [1]. (b) The man is a carrier with probability 1/2 [1]; 1/2 × 1/50 × 1/4 [1] = 1/400 = 0.25% [1]. [5] Examiner insight: Lay out each probability being multiplied. A single final fraction without working earns less than a clear chain of steps.

9. For: identifies patients at risk of side effects [1]; allows a different drug or lower dose, which is personalised medicine [1]. Against: cost of screening [1]; privacy and possible discrimination with genetic data [1]. [4] Examiner insight: “Evaluate” needs points on both sides and not only a list of benefits.

10. (a) S2 [1]; all four bands match C (7.2, 5.1, 3.8, 2.4) [1]. (b) DNA carries a negative charge, so it moves towards the positive electrode [1]. The gel acts as a mesh, so smaller fragments pass through more easily [1]. A 2.4 kb fragment is shorter than a 7.2 kb fragment, so it moves further in the same time [1]. (c) 0.05 × 0.1 × 0.2 × 0.1 [1] = 0.0001, 1 in 10 000 [1]. [7] Examiner insight: Part (b) needs the link between size and distance, not only the statement that small fragments move further.

11. It can show which animals are closely related [1], so a breeder can avoid matings between close relatives [1] and measure genetic variability in the population, to keep diversity [1]. [3] Examiner insight: Give the biological consequence, such as avoiding inbreeding, rather than saying only that it “checks relatedness”.

12. Humanitarian and financial benefits: higher yield or nutrition could reduce hunger [1]; farmers might earn more [1]. Concerns: patents and seed prices could make farmers dependent on one company [1]; poor farmers may not afford the seed [1]. Environmental and ethical: gene flow to wild relatives or harm to other organisms, as environmentalists argue [1]. A judgement is given which weighs the two sides, for example that benefits depend on access and monitoring [1]. [6] Examiner insight: A final reasoned judgement is what turns a list of points into an evaluation.

Where marks are usually lost

• Not stating that mRNA has had its introns removed when explaining cDNA.
• Writing “bases” or “genes” where the question needs “amino acids” or “proteins”.
• Treating the heat-stable polymerase as the only reason PCR works and forgetting the primers.
• Showing a PCR answer with no working, or stopping at the first value above the target.
• Writing probe sequences in the wrong direction or identical to the allele.
• Confusing ligase (joins) with restriction endonuclease (cuts).
• Saying that larger DNA fragments move further through a gel.
• Treating a fingerprint match as proof instead of giving the chance of coincidence.
• Giving a one-sided evaluation with no judgement.

Next steps

• Review with the revision notes and the study guide.
• Course pages: AQA A-level Biology hub and the printable checklist.
• Take the free diagnostics.
• Book a free trial class.

Official syllabus

AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015, sections 3.8.3, 3.8.4.1, 3.8.4.2 and 3.8.4.3 (version 1.5), published by AQA.