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Modern Genetic Analysis Answers to Selected Problems
Chapter 7*

2. | Starting with a yeast strain that is pro-1, plate the cells on medium lacking proline. Only those cells that are able to synthesize proline will form colonies. Most of these cells will be revertants; however, some will have second-site suppressors. (Treating the cells with a mutagen before plating them will increase the yield.) See question

4. | Mutation rate is 4.25 × 105. You do not have to worry about revertants in this case, because the problem asks for only the net mutation frequency. See question

8. | Frameshift mutations arise from the addition or deletion of one or more bases that are not a multiple of three. When translated, this addition or deletion will alter the reading frame and therefore the amino acid sequence from the site of the mutation to the end of the protein product. In addition, frameshift mutations often result in premature stop codons in the new reading frame, leading to shortened protein products. A missense mutation changes only a single amino acid in the protein product. See question

11.

Depurination results in the loss of adenine or guanine from the DNA. Because the resulting apurinic site cannot specify a complementary base, replication is blocked. Under certain conditions, replication proceeds with a near-random insertion of a base opposite the apurinic site. In three-fourths of these insertions, a mutation will result.

Deamination of cytosine yields uracil. If left unrepaired, the uracil will be paired with adenine in replication, ultimately resulting in a transition mutation.

8-OxodG (8-oxo-7-hydrodeoxyguanosine) can pair with adenine, resulting in a transversion.

See question

15. | Leaky mutants are mutants with an altered protein product that retains a low level of function. Enzyme activity may, for instance, be reduced rather than abolished by a mutation. See question

17.

a. Because 5-UAA-3 does not contain G or C, a transition to GC pair in the DNA cannot result in 5-UAA-3.5-UGA-3 and 5-UAG-3 have the DNA antisense-strand sequence of 3-ACT-5 and 3-ATC-5, respectively. A transition to either of these stop codons occurs from the nonmutant 3-ATT-5. However, a DNA sequence of 3-ATT-5 results in an RNA sequence of 5-UAA-3, itself a stop codon.
b. Yes. An example is 5-UGG-3, which encodes Trp, to 5-UAG-3.
c. No. In the three stop codons, the only base that can be acted on is G (in UAG, for instance). Replacing the G with an A would result in 5-UAA-3, a stop codon.
See question

19.

a. A lack of revertants suggests either a deletion or an inversion within the gene.
b. Prototroph A: a reversion at the original mutant site may have occurred.

Prototroph B: an unlinked suppressor mutation occurred.

Prototroph C: a suppressor mutation occurred in a site very close [100% (4 × 2)/1000 = 0.8 m.u.] to the original mutation.

See question

23. | The allele for NF must have arisen spontaneously in one of the parents' germ lines. Their chance of having another affected child would range from 0 to 50% (the latter number if the entire germ line was mutant), depending on when this mutation happened (the size of the mutant clone of germ-line tissue). See question

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