AQA A-level Biology: Gene Mutations and Stem Cells (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 are based on sections 3.8.1 and 3.8.2.1 of the AQA specification for AS and A-level Biology (7401 and 7402), first teaching 2015. Neither section is part of the AS. Worked explanations are in the study guide and the revision notes. Codon information is supplied in each question.

Questions

1. A section of bases is removed from one gene and joined into a different chromosome. (a) State the name of this type of gene mutation. [1] (b) Explain how it could change the polypeptide coded for by the gene it was removed from. [2]

2. A gene is transcribed into this mRNA: AUG GCU UCC GAA GGG UGG UAA. Codon data: AUG Met; GCU and GCA Ala; UCC Ser; GAA and GAG Glu; GGG Gly; UGG Trp; UAA stop. (a) Give the amino acid sequence the mRNA codes for. [1] (b) The codon GAA changes to GAG. Explain the effect on the polypeptide. [2] (c) Instead, GAA changes to GCA. State the effect. [1] (d) Instead, GAA changes to UAA. Describe and explain the effect. [2]

3. Use the mRNA in question 2. Further codon data: AUC Ile; CGA and AGG Arg; GUG and GUA Val. (a) One A is inserted after the sixth base (after GCU). Write the new mRNA as triplets and give the amino acids it codes for. [2] (b) Instead, the three bases UCC are deleted. Give the amino acid sequence. [1] (c) Explain why the mutation in (a) has a greater effect than the one in (b). [2]

4. The template strand of a gene reads, from 3’ to 5’: TAC GAA TTT CCC GGT ATT. Codon data: AUG Met; CUU Leu; AAA and AAG Lys; GGG Gly; CCA Pro; UAA stop. (a) Write the mRNA transcribed from this template. [2] (b) The template triplet TTT changes to TTC. Explain why the polypeptide does not change. [2]

5. The coding strand of a gene reads ATG GAA TCC AAA GTC TGG TAA. Three mutant versions are shown. Codon data: AUG Met; GAA Glu; UCC Ser; AAA Lys; GUC Val; UGG Trp; GAC Asp; UUU Phe; AGU Ser; CUG Leu; GUA Val. Mutant 1: ATG GAA TCC AAA GTC GTC TGG TAA Mutant 2: ATG GAA TCC GAC TTT TGG TAA Mutant 3: ATG GAA TCC AAA AGT CTG GTA A For each mutant, name the type of mutation and describe its effect on the polypeptide. [6]

6. A polypeptide has 312 amino acids. A mutation produces a stop codon at what was codon 46. (a) Calculate the number of bases in the coding region of the original gene, including its stop codon. [1] (b) Calculate the percentage of the original amino acids missing from the mutant polypeptide. [2] (c) Explain why the mutant protein is unlikely to function. [1]

7. Totipotent cells occur in early mammalian embryos. (a) State what is meant by a totipotent cell. [1] (b) Name the type of stem cell described in each case: (i) found in embryos and able to divide in unlimited numbers; (ii) found in mature mammals and forms a limited number of cell types; (iii) found in mature mammals and forms only cardiomyocytes. [3] (c) All cells in an embryo have the same DNA. Explain how they can become different specialised cells. [2]

8. A student prepared 40 explants from cauliflower tissue for tissue culture. Five became contaminated. Of the rest, 31 formed callus. (a) Calculate the percentage of uncontaminated explants that formed callus. [2] (b) Describe two precautions to reduce contamination. [2] (c) Suggest what the formation of callus shows about the cells of an explant. [1]

9. (a) Describe how induced pluripotent stem (iPS) cells can be produced from adult skin cells. [3] (b) State one advantage of iPS cells compared with embryonic stem cells. [1]

10. In a trial, 45 patients with a heart disorder received a therapy using cells derived from pluripotent stem cells, and 45 received standard care. After 12 months, 27 treated patients and 15 control patients had improved. Tumours were found in 3 treated patients and in none of the control patients. (a) Calculate, for each group, the percentage of patients showing improvement. [2] (b) Evaluate the use of this therapy. [4]

11. A rare disorder is caused by a deletion of one base in the gene for an enzyme. (a) Explain how this deletion can stop the enzyme working. [4] (b) State one advantage of using iPS cells rather than embryonic stem cells to treat the patient. [1] (c) Skin cells from this patient are used to make iPS cells. Explain why the iPS cells would not by themselves cure the disorder. [2] (d) Give two concerns about using pluripotent stem cells in treatment. [2]

Answers

1. (a) Translocation. [1] (b) The removed bases are lost from the gene, as in a deletion. [1] This can change the amino acids coded for, and if the number of bases lost is not a multiple of three it causes a frame shift. [1] Examiner insight: Linking a translocation to what happens at the old site, in terms of deletions you already know, is a safer route than recalling a separate rule.

2. (a) Met-Ala-Ser-Glu-Gly-Trp, then stop. [1] (b) GAG and GAA both code for Glu. [1] The polypeptide is unchanged because the genetic code is degenerate. [1] (c) Glu is replaced by Ala, so one amino acid is different. [1] (d) UAA is a stop codon, so translation ends at this point. [1] The polypeptide is shorter: Met-Ala-Ser only. [1] Examiner insight: Naming the triplets and amino acids on each side of the change gains credit; “the protein changes” gives little to mark.

3. (a) The new mRNA reads AUG GCU AUC CGA AGG GUG GUA A. [1] The amino acids are Met-Ala-Ile-Arg-Arg-Val-Val. [1] (b) Met-Ala-Glu-Gly-Trp, then stop. [1] (c) In (a) one base is added, so the reading frame shifts. [1] Every triplet after the insertion changes, so many amino acids differ, whereas in (b) three bases are lost, the frame is kept and only Ser is missing. [1] Examiner insight: An explain-why question needs a comparison; mention both the frame and the number of amino acids affected.

4. (a) The mRNA is complementary to the template, with U in place of T. [1] AUG CUU AAA GGG CCA UAA. [1] (b) The template change TTT to TTC alters the mRNA codon from AAA to AAG. [1] Both code for Lys, because the genetic code is degenerate, so the polypeptide is unchanged. [1] Examiner insight: Check which strand you are given. The codon you compare must come from the mRNA, not from the template.

5. Mutant 1: a duplication of GTC. [1] Three bases are added, so the frame is kept, and the polypeptide has one extra Val: Met-Glu-Ser-Lys-Val-Val-Trp. [1] Mutant 2: an inversion of AAAGTC. [1] Lys-Val is replaced by Asp-Phe, so two amino acids change and the length stays the same. [1] Mutant 3: an addition of one base (A). [1] This is a frame shift: from the fifth triplet every amino acid changes (Met-Glu-Ser-Lys-Ser-Leu-Val) and the stop codon is no longer read. [1] Examiner insight: Name the mutation and state its effect separately. Noting whether the length has changed separates the types quickly.

6. (a) 312 amino acids plus one stop codon is 313 triplets, and 313 × 3 = 939 bases. [1] (b) The mutant has 45 amino acids, so 312 − 45 = 267 are missing. [1] 267 ÷ 312 × 100 = 85.6%. [1] (c) Most of the primary structure is missing, so the tertiary structure and any active site or binding site cannot form. [1] Examiner insight: Show the triplet count and the stop codon explicitly, so it is clear the stop codon was considered.

7. (a) A cell that can divide and produce any type of body cell. [1] (b) (i) Pluripotent. [1] (ii) Multipotent. [1] (iii) Unipotent. [1] (c) All the cells have the same DNA (genes) because they arise from the zygote by mitosis. [1] Each type of specialised cell translates only part of its DNA, so it makes different proteins. [1] Examiner insight: Say that cells translate only part of their DNA; saying specialised cells “lose” genes contradicts the idea being tested.

8. (a) 40 − 5 = 35 uncontaminated explants. [1] 31 ÷ 35 × 100 = 88.6%. [1] (b) Any two of: sterilise the explants; sterilise instruments and work near a flame; use sterile agar; seal the dish. [2] (c) The cells can divide and give rise to new cells (callus), as totipotent cells do. [1] Examiner insight: Decide the denominator before calculating; the question asks about the uncontaminated explants, not all 40.

9. (a) Take adult somatic cells, such as skin cells. [1] Treat them with appropriate protein transcription factors. [1] These switch on genes so that the cells become pluripotent. [1] (b) No embryo is destroyed, or the cells come from the patient so there is less risk of rejection. [1] Examiner insight: Mention adult somatic cells and transcription factors; “taken from embryos” does not describe iPS cells.

10. (a) Treated: 27 ÷ 45 × 100 = 60.0%. [1] Control: 15 ÷ 45 × 100 = 33.3%. [1] (b) More of the treated group improved, by 26.7 percentage points. [1] Tumours occurred in 3 treated patients and none in the control group, so there is a safety concern. [1] The groups are small and followed for only 12 months, so long-term benefit and risk are unknown. [1] Judgement: promising, but more and larger trials with longer follow-up are needed before the benefit outweighs the risk. [1] Examiner insight: An evaluation needs both sides, a limitation of the evidence and a conclusion; quoting figures from the data gains more than general statements.

11. (a) Deleting one base shifts the reading frame, because the number of bases lost is not a multiple of three. [1] Every triplet downstream of the deletion changes. [1] Many amino acids are different, or a stop codon appears early. [1] The primary structure and so the tertiary structure change, so the active site no longer has the shape to bind the substrate. [1] (b) No embryo is destroyed, or less risk of rejection because the cells come from the patient. [1] (c) The iPS cells come from the patient’s own cells, so they carry the same deletion. [1] Cells made from them would still make the faulty enzyme unless the gene is corrected. [1] (d) Any two of: pluripotent cells may form tumours; the process is inefficient; long-term safety is unknown; ethical objections if embryonic cells are used. [2] Examiner insight: Follow the chain: mutation, frame shift, amino acids, protein shape, function. Skipping a link leaves the argument incomplete.

Where marks are usually lost

• Forgetting that a degenerate code can leave the polypeptide unchanged.
• Calling a three-base deletion a frame shift.
• Describing a frame shift as affecting one amino acid only.
• Using the wrong strand when transcribing from a template.
• Dividing by the wrong total in a percentage calculation.
• Describing iPS cells as embryonic.
• Giving one side only in an evaluation, with no judgement.

Next steps

• Go back over weak areas in the study guide and the revision notes.
• Track progress with the printable checklist and browse the AQA A-level Biology hub.
• Find your next weak topic with the free 10-minute diagnostics.
• Want to talk these topics through? Book a free trial class.

Official syllabus

Source: AQA, AS and A-level Biology (7401 and 7402), specification for first teaching 2015, version 1.5, sections 3.8.1 and 3.8.2.1.