AQA A-level Biology: Mutation and Meiosis (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), section 3.4.3, Genetic diversity can arise as a result of mutation or during meiosis, in the AQA AS and A-level Biology specification for first teaching 2015. Explanations are in the study guide and the revision notes. Course pages: AQA A-level Biology hub and the printable checklist. Where a question gives codon data, use only that data. The marks shown are for practice and are not an official tariff.

Questions

1. A base substitution occurs in a gene, but the polypeptide it codes for is unchanged. Explain how this is possible. [3]

2. A short piece of mRNA reads AUG CCU GUA AAA CAC UGA. Codon data: AUG Met, CCU Pro, GUA Val, GUG Val, AAA Lys, AAC Asn, CAC His, ACU Thr, UGA stop. (a) Give the amino acid sequence it codes for. [1] (b) The third codon changes from GUA to GUG. Describe and explain the effect on the polypeptide. [2] (c) Instead, the first A of the fourth codon (AAA) is deleted. Give the new amino acid sequence for the codons that can be read and explain why the effect is greater than in part (b). [3]

3. Explain how non-disjunction during meiosis can lead to a child with 47 chromosomes in each body cell. [4]

4. A bacterial strain cannot grow without an added amino acid. Colonies that grow on plates lacking it have mutated. Plates without a test chemical gave 14, 16 and 12 colonies. Plates with the chemical gave 63, 70 and 77 colonies. (a) Calculate the mean number of colonies for each condition. [2] (b) Calculate the percentage increase caused by the chemical. [1] (c) Explain what the results suggest about the chemical. [1] (d) Suggest one change that would make the conclusion more reliable. [1]

5. A diploid cell has four chromosomes: one long pair and one short pair. DNA has been replicated. (a) State the number of chromosomes in each cell after meiosis I and how many chromatids each chromosome has. [2] (b) State the number of cells after meiosis II, and the number of chromosomes and chromatids per chromosome in each. [2]

6. Give four differences between mitosis and meiosis. [4]

7. A species has 2n = 18. Ignore crossing over. (a) Calculate the number of different gamete types one individual could make and the number of different zygotes from two such parents. [3] (b) Another species can produce 4096 different types of gamete. Calculate its diploid number. [2]

8. In a seaweed, haploid filaments release gametes. Two gametes fuse to form a zygote, which grows into a large diploid plant. Cells in this plant divide to form spores. Each spore grows into a haploid filament. (a) State where meiosis occurs. [1] (b) Explain how you know. [1] (c) Explain why the gametes are made by mitosis and not meiosis. [2]

9. Explain how independent segregation, crossing over and random fertilisation each increase genetic variation. [6]

10. A gene has 900 bases. Data supplied give a spontaneous mutation rate of 4 × 10⁻⁹ per base per replication. (a) Calculate the expected number of mutations in this gene over 3 × 10⁶ replications. [2] (b) A mutagenic agent increases the rate eight-fold. Calculate the new expected number. [1] (c) Explain why deleting one base near the start of the gene is usually more harmful than substituting one base at the same place. [3]

11. A species has 2n = 10. In one meiotic cell, the chromatids of one chromosome fail to separate during meiosis II in one of the two cells. (a) State the chromosome numbers of the four gametes. [2] (b) State the chromosome number of a zygote formed from the largest gamete and a normal gamete. [1] (c) Calculate the number of gamete types from independent segregation alone and the number of zygote types. [2] (d) Explain why one meiotic cell cannot make all of these gamete types. [3]

Answers

1. The genetic code is degenerate, so most amino acids have more than one codon [1]. The substitution changed a codon into another codon for the same amino acid [1]. The amino acid sequence, and so the protein, is therefore unchanged [1]. [3] Examiner insight: name degeneracy and say that the new triplet still codes for the same amino acid; “it doesn’t matter” is too vague to explain anything.

2. (a) Met-Pro-Val-Lys-His, then stop [1] (b) No effect [1]; GUA and GUG both code for Val, because the code is degenerate [1]. [2] (c) The mRNA becomes AUG CCU GUA AAC ACU GA, giving Met-Pro-Val-Asn-Thr [1]. A deletion shifts the reading frame [1], so every codon from the deletion onwards changes and the original stop codon is lost, whereas the substitution changed one codon only [1]. [3] Examiner insight: regroup the bases into triplets in your working so the shifted frame is visible; comparing both mutations completes the explanation.

3. The pair of chromosomes 21 fails to separate in meiosis I, or chromatids fail to separate in meiosis II [1]. This gives a gamete with 24 chromosomes (an extra chromosome 21) [1]. It fuses with a normal gamete of 23 [1]. The zygote has 24 + 23 = 47 chromosomes, which are copied by mitosis into all body cells [1]. [4] Examiner insight: give the arithmetic and say which gamete has the extra chromosome; “chromosomes don’t separate” alone is too vague.

4. (a) Without: (14 + 16 + 12) / 3 = 14; with: (63 + 70 + 77) / 3 = 70 [2] (b) (70 − 14) / 14 × 100 = 400% [1] (c) The chemical increases the rate of mutation, so it is a mutagenic agent [1] (d) Use more plates or repeat the experiment [1]. [5] Examiner insight: divide by the control value for percentage change; a conclusion should say the data suggest rather than prove.

5. (a) Two chromosomes (one long, one short) [1], each with two chromatids [1]. [2] (b) Four cells [1]; each has two chromosomes with one chromatid each [1]. [2] Examiner insight: the chromosome number halves at meiosis I, and the chromatid number changes at meiosis II; keep the two counts separate.

6. Any four from: mitosis has one division, meiosis two [1]; mitosis gives two cells, meiosis four [1]; mitosis keeps the chromosome number, meiosis halves it [1]; mitosis gives identical cells, meiosis genetically different cells [1]; homologous chromosomes pair and cross over in meiosis only (a further valid point). [4] Examiner insight: each difference needs both sides stated; “meiosis makes gametes” is a use, not a comparison.

7. (a) n = 9 [1]; 2⁹ = 512 gamete types [1]; 512 × 512 = 262 144 zygotes [1]. [3] (b) 2ⁿ = 4096, so n = 12 [1]; 2n = 24 [1]. [2] Examiner insight: write n = 9 before using the formula; using 18 in place of 9 is the usual error.

8. (a) In the diploid plant, where spores form [1] (b) The chromosome number halves from diploid to haploid there [1] (c) The filaments are already haploid [1], so mitosis keeps the number at n and fertilisation restores 2n [1]. [4] Examiner insight: argue from chromosome number at each stage; a life-cycle answer without ploidy labels is hard to justify.

9. Independent segregation: homologous pairs line up at random in meiosis I [1], so each gamete receives a different mix of maternal and paternal chromosomes [1]. Crossing over: non-sister chromatids of homologous chromosomes swap sections [1], giving new combinations of alleles on the chromatids [1]. Random fertilisation: which male gamete reaches and fuses with which female gamete is a matter of chance [1], giving many possible genotypes in the zygotes [1]. [6] Examiner insight: each process needs a place or timing and a result; “creates variation” repeats the question.

10. (a) 900 × 4 × 10⁻⁹ = 3.6 × 10⁻⁶ per replication [1]; × 3 × 10⁶ = 10.8 [1]. [2] (b) 10.8 × 8 = 86.4 [1] (c) A deletion moves every later base up one place [1], so all triplets after the deletion are changed [1]. A substitution alters at most one triplet, which may be the same amino acid because of degeneracy [1]. [3] Examiner insight: show the rate multiplied by bases first, then replications; quote the data in each step.

11. (a) The affected cell gives gametes of 6 and 4 chromosomes [1]; the other cell gives two gametes of 5 [1]. [2] (b) 6 + 5 = 11 [1] (c) n = 5, so 2⁵ = 32 gamete types [1]; 32 × 32 = 1024 zygote types [1]. [2] (d) After meiosis I, each of the two cells has one fixed combination of chromosomes [1]. Meiosis II only separates chromatids, so the two cells yield at most two types in total [1]. Many meiotic cells with different random orientations are needed [1]. [3] Examiner insight: in part (d) say that meiosis II does not reshuffle chromosomes; that reasoning is the core of the answer.

Where marks are usually lost

• Saying a silent mutation has “no mutation” rather than naming degeneracy.
• Forgetting that a deletion moves the reading frame, so every later codon changes.
• Writing triplets across the deletion without regrouping them.
• Giving the diploid number instead of the number of pairs when using 2ⁿ.
• Putting the halving of chromosome number at meiosis II.
• Writing “chromosomes” when chromatids are meant, and the reverse.
• Dividing by the new value when calculating a percentage increase.
• Treating all gametes as products of meiosis in plants and algae.

Next steps

• Revision notes
• Study guide
• AQA A-level Biology hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015, section 3.4.3 (version 1.5), published by AQA.