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Modern Genetic Analysis Chapter 8. Chromosome Mutations
Solved Problem

2.

We have lines of mice that breed true for two alternative behavioral phenotypes that we know are determined by two alleles at a single locus: v causes a mouse to move with a "waltzing" gait, whereas V determines a normal gait. After crossing the true-breeding waltzers and normals, we observe that most of the F1 is normal, but, unexpectedly, there is one waltzing female. We mate the F1 waltzer with two different waltzer males and note that she produces only waltzer progeny. When we mate her with normal males, she produces normal progeny and no waltzers. We mate three of her normal female progeny with two of their brothers and these mice produce 60 progeny, all normal. When, however, we mate one of these same three females with a third brother, we get six normals and two waltzers in a litter of eight. By thinking about the parents of the F1 waltzer, we can consider some possible explanations of these results:
a. A dominant allele may have mutated to a recessive allele in her normal parent.
b. There may have been a mutation to a dominant suppressor of V in one parent.
c. Meiotic nondisjunction of the chromosome carrying V in her normal parent may have given a viable aneuploid.
d. There may have been a viable deletion spanning V in the meiocyte from her normal parent.

Which of these explanations are possible, and which are eliminated by the genetic analysis? Explain in detail.

See answer

Solution

This question will force us to use concepts from Chapter 6 (suppression), Chapter 7 (mutation), and the present chapter (deletion and nondisjunction). The best way to answer it is to take the explanations one at a time and see if each fits the results given.
a. Mutation V to v

This hypothesis requires that the exceptional waltzer female be homozygous v /v. This is compatible with the results of mating her both to waltzer males, which wouldif she is v /vproduce all waltzer offspring (v /v), and to normal males, which would produce all normal offspring of genotype V /v. However, brothersister matings within this normal progeny should then produce a 3:1 normal-to-waltzer ratio. Because some of the brothersister matings actually produced no waltzers, this hypothesis does not explain the data.

b. Suppressor mutation s to S

Here the parents would be V /V.s /s and v /v.s /s, and a germinal mutation in one of them would give the F1 waltzer the genotype V /v.S /s. When we crossed her with a waltzer male, who would be of the genotype v /v.s /s, we would expect some V /v.s /s, which would be phenotypically normal. However, we saw no normal progeny from this cross, so the hypothesis is already overthrown. Linkage could save the hypothesis temporarily if we assumed that the mutation was in the normal parent, giving a gamete V S. Then the F1 waltzer would be V S /v s, and, if linkage were tight enough, few or no V s gametes would be produced, the type that are necessary to combine with the v s gamete from the male to give V s /v s normals. However, if this were true, the cross with the normal males would be V S /v s × V s /V s and this would give a high percentage of V S /V s progeny, which would be waltzers, none of which were seen.

c. Nondisjunction in the normal parent

This explanation would give a nullisomic gamete that would combine with v to give the F1 waltzer the hemizygous genotype v. The subsequent matings would be as follows:
v× v /v gives v /v and v progeny, all waltzers. This fits.
v× V /V gives V /v and V progeny, all normals. This also fits.
First intercrosses of normal progeny: V × V. This gives V and V /V, which are all normal. This fits.
Second intercrosses of normal progeny: V × V/v. This gives 25 percent each of V/V, V/v, V (all normals), and v (waltzers). This also fits.

This hypothesis is therefore consistent with the data.

d. Deletion of V in normal parent

Let's call the deletion D. The F1 waltzer would be D/v, and the subsequent matings would be as follows:
D/v × v/v. This gives v/v and D/v, all of which are waltzers. This fits.
D/v × V/V. This gives V/v and D/v, all of which are normal. This fits.
First intercrosses of normal progeny: D/V × D/v. This gives D/v and V/V, all normal. This fits.
Second intercrosses of normal progeny: D/v × V/v. This gives 25 percent of each of V/V, V/v, D/v (all normals), and D/v (waltzers). This also fits.

Once again, the hypothesis fits the data provided, so we are left with two hypotheses that are compatible with the results, and further experiments would be necessary to distinguish them. One obvious way would be to examine the chromosomes of the exceptional female under the microscope; aneuploidy should be relatively easy to distinguish from deletion.

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© 1999 by W. H. Freeman and Company. All rights reserved.