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Modern Genetic Analysis Answers to Selected Problems
Chapter 6*

3. | 50% See question

4. | The data indicate that there is a single gene with multiple alleles. The order of dominance is: black > sepia > cream > albino. b. bs × bc >> 3/4 black; 1/4 sepia See question

6.

a. The sex ratio is expected to be 1:1.
b. The female parent was heterozygous for an X-linked recessive lethal allele.
c. Half of the female progeny should be heterozygous for the lethal allele and half should be homozygous for the nonlethal allele. Individually mate the F1 females, and determine the sex ratio of their progeny.
See question

11. | a. Incomplete dominance. b. normal × woolly See question

16. | A / a ; R / r × A / a ; r / r See question

19. | a. Ab / Ab × aB / aB or AB / AB × ab / ab b. Ae / Ae × aE / aE or AE / AE × ae / ae c. Ad / Ad × aD / aD or AD / AD × ad / ad d. Be / Be × bE / bE or BE / BE × be / be See question

25. | 9/16 See question

26.

a. The data from the crosses indicate that the mutations are in different, unlinked genes and both are recessive. The data from the gel indicate that one mutation is in the gene that codes for the P protein. As a result, the mutant protein is truncated (smaller) and runs more rapidly in the gel. Possible mutations giving this result might include nonsense, frameshifts, deletions, or mutations leading to altered splicing. (Alternatively, you could conclude that the mutation caused a larger, but still nonfunctional, P protein to be made because of an insertion or splicing defect.) The other mutation is in a gene that encodes a regulatory protein required for P-gene expression. When mutant, P expression is greatly reduced.
b. Assume p+ = normal P protein; r+ = normal regulatory protein; both are required for normal function.
c. 1/8
d. Lane 1 and lane 3
e. Parents: p / p ; r+ / r+ × p+ / p+ ; r / r will look like lane 5 and lane 2, respectively.

F1: p+ / p ; r+ / r will look like lane 3.

See question

29. | A single gene showing incomplete dominance (1:2:1 ratio in F2): See question

33.

a. The results indicate that two unlinked genes are being studied. One gene determines whether the beetles have a diamond or a stripe. For these phenotypes, the diamond allele (D) is dominant to the stripe allele (d). The other gene determines whether the beetle has a spot. In this case, the spot allele (S) is dominant to the no-spot allele (s), and it is epistatic to D and d.
b. 3/4 spot; 9/64 stripe; 7/64 diamond.
See question

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