Mutations: Practice Questions (Biology 6.7)

Answer: (C). UAA is a stop codon. A single base was substituted, so this is a point mutation that creates a premature stop: a nonsense mutation. Only 39 amino acids are made instead of about 300.

• (A) Silent would need a codon for the same amino acid, such as UAU. Sharing the first two bases is not enough.
• (B) No amino acid replaces Tyr; the chain ends.
• (D) One base was replaced, not added or removed, so the frame is unchanged. Point mutations do not cause frameshifts.

Answer: (C). Removing three nucleotides removes exactly one codon’s worth of bases, so later codons are grouped exactly as before. The protein loses one amino acid (and, if the deletion spans two codons, the amino acid at the join may also change), but the rest of the sequence is normal.

• (A) and (B) change the number of bases by 1 or 2, so every later codon is regrouped.
• (D) Four is not a multiple of three: four bases shift the frame by one, just like a single insertion.

Answer: (B). The cell that divides normally gives two normal (n) gametes. In the other cell, both chromatids go to one gamete (n + 1) and none to the other (n − 1). So the gametes are n, n, n + 1, n − 1: one in four is n + 1.

• (A) ignores the failed separation.
• (C) is the fraction of gametes that are abnormal (n + 1 or n − 1) here, or the fraction that are n + 1 after an error in meiosis I.
• (D) would need every gamete to receive an extra copy, which is impossible: whatever one gamete gains, another loses.

(a) mRNA is complementary to the template: 5′-AUG CCU GAA AAC GGU CAU UGA-3′. Polypeptide: Met–Pro–Glu–Asn–Gly–His, then stop.

(b) New template: TAC GGA CTT TGC CAG TAA CT…, so the mRNA reads AUG CCU GAA ACG GUC AUU GA… Polypeptide: Met–Pro–Glu–Thr–Val–Ile… The UGA stop is no longer in frame, so translation continues past this stretch. This is a frameshift (deletion of one nucleotide).

(c) Template codon 4 becomes ATG, so the mRNA codon becomes UAC instead of AAC: Tyr replaces Asn. All other codons are unchanged. This is a missense point mutation.

(d) Mutation 1 is more likely to destroy function. Every amino acid after position 3 is different and the normal stop is lost, so the chain has a different sequence, folds differently and is unlikely to work. Mutation 2 changes one amino acid; the protein may still fold and work, unless position 4 is in the active site or is needed for correct folding.

Do not award the last point for “the frameshift denatures the protein”: the sequence changes; the protein is not denatured.

(a) Dark soil: dark 112 ÷ 200 = 56%, pale 58 ÷ 200 = 29%. Pale sand: dark 46 ÷ 200 = 23%, pale 104 ÷ 200 = 52%.

(b) The mutation is beneficial on dark soil and detrimental on pale sand. Whether it helps depends on the environment.

(c) On dark soil, nearly twice as many dark as pale beetles were recaptured (56% against 29%); on pale sand the pattern reverses (23% against 52%). A plausible explanation is that beetles matching the background are harder for predators to see, so more survive. The survivors reproduce, so the allele that matches the local background becomes more common in each population.

(d) The statement is not supported. Mutations arise at random, by replication errors or mutagens, not because of the colour of the soil. Both forms were present in the released groups at both sites. The soil selects among existing variants; it does not cause the mutation. (Recapture also assumes that both forms are equally easy to find and equally likely to stay near the release point; if not, recapture would not measure survival.)

(a) 1 transformation; 2 transduction; 3 conjugation; 4 transposition.

(b) A new mutation must arise by chance in each lineage, which is rare. In horizontal gene transfer, a gene that already exists passes directly between cells in the same generation, even between different strains. If an antibiotic is present, cells that receive the gene survive and reproduce, so resistance becomes common quickly.

(c) When both viruses copy their genetic material in the same cell, pieces can be exchanged or mixed when new particles are assembled. A new particle may carry genes from both parents: a recombinant strain with a new combination of traits.

(a) M1: 21 ÷ 20 = 1.05. M2: 3 ÷ 97 = 0.03. M3: 99 ÷ 102 = 0.97. (Wild type: 1.0.)

(b) M1 changes the amount: mRNA and protein both fall to about a fifth, but each molecule works normally (about 1.0 per unit). The mutation is likely in the promoter or another regulatory sequence, reducing transcription. M2 changes the type: normal amounts of mRNA and protein, but each molecule has almost no activity. Likely a missense change in the coding region, for example in the active site. M3 has no meaningful change in amount or activity: a neutral mutation.

(c) M3 could be a silent substitution (a new codon for the same amino acid), or a missense change to an amino acid with similar properties in a part of the protein that does not affect its function. The DNA changed, but the phenotype did not.

Accept “M1 could be a mutation that makes the mRNA less stable” if supported by the data.