Genetics in Auxotrophic Mutation Strains

1 Two new hair dyes A and B were tested in the Ames test. Both were tested with and without liver homogenate. Two auxotrophic strains of bacteria were used, one in which the auxotrophy was caused by a nonsense mutation (mutant 1) and the other by a frameshift mutation (mutant 2). After treatment the cells were plated on minimal medium. The + signs show experiments in which prototrophic colonies were observed. Explain the mutagenic action of these chemicals, and say which would probably be safe for human use.

2. In a tetraploid plant the A and B loci are centromere-linked and on separate chromosomes. The following cross is made:
A/A/A/A ; b/b/b/b X a/a/a/a ; B/B/B/B
And an F1 is obtained that is selfed to give an F2. If only one dominant allele is needed to give the dominant phenotypes at each locus, what phenotypes are expected in the F2 and in what proportions?

3.. In Drosophila, a cross is made involving two closely linked genes on the very small chromosome 4. a and b are recessive alleles.
Female a b+ / a+ b X male a b / a b
In the progeny, the expected common types were found, but there was also one rare wild type female.
(a) Explain what the common progeny phenotypes are expected to be, and give their proportions.
(b) Could the rare wild type female have arisen by crossing over? By nondisjunction? Explain your answers.
(c) The rare wild type female was test-crossed to a male tester of genotype a b / a b, and the progeny of the test cross were
1/6 wild type
1/6 expressed both recessives a and b

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...fied by the liver. You could argue that this is safe for human use, since it is made safe by the liver. However, I would say that this is also not suitable for huamn use, particularly an external one. Frameshift mutations are very dangerous (much more so than point mutations), and using it on your head would be dangerous.

There are many arguments to be made here, so just use your best judgment and justify what you say.

2. In a tetraploid plant the A and B loci are centromere-linked and on separate chromosomes. The following cross is made:
A/A/A/A ; b/b/b/b X a/a/a/a ; B/B/B/B
And an F1 is obtained that is selfed to give an F2. If only one dominant allele is needed to give the dominant phenotypes at each locus, what phenotypes are expected in the F2 and in what proportions?

I've never actually heard the term centromere-linked before. However, I believe it means that the gene in question is located very close to the centromere, which means that there will not be non-sister chromatid crossovers between these two genes, so they should sort Mendelainly.

F1: A/A/a/a ; B/B/b/b
F2: There are a lot of possibilities, but just think about the genes separately, assuming my hypothesis about non-sister chromatid crossing-over is correct.
For the A gene, we will have A/A, A/a, a/a gametes, in the proportions 1/4/1.
Same sort of thing for the B gene. Fortunately, only being interested in the phenotypes makes this slightly easier. There's a 1/6 chance of getting a/a. To get two of them for a/a/a/a is a 1/36 chance. Same thing for b/b/b/b. To get both at the same time is 1/36 * 1/36 = 1/1296.
So, we'd expect these phenotype proportions (using subtraction to get all the numbers right):
A / B - .94521 (about 1225/1296)
A / b - .02701 (about 35/1296)
a / B - .02701 (about 35/1296)
a / b - .00077 (about 1/1296)

3.. In Drosophila, a cross is made involving two closely linked genes on the very small chromosome 4. a and b are recessive alleles.
Female a b+ / a+ b X male a b / a b
In the ...