Week 5 – Homework – Answer Key
Due Feb. 23, 2013
A total of 20 points are possible for this homework
1. A black guinea pig is crossed with an albino guinea pig, producing 12 black offspring. When the very same albino is crossed with another black guinea pig, 7 black and 5 albinos are obtained. Explain this genetic outcome by writing out the genotypes for the parents, gametes, and offspring in both crosses.
First Cross: The fact that all F1 offspring are black suggests that the parents of the first cross were genotype BB x bb (where B=black and b=albino). The gametes produced by the black parent would have carried the B allele, while those produced by the albino parent carried the b allele. The F1 offspring of such a cross would be Bb, and since black is dominant over albino, all the F1 animals would have had black fur. +1 point
Second Cross: Since some of the F1 offspring are albino, that means that the black parent must have been heterozygous for fur color. So, the parents of the second cross were Bb x bb. The black parent would have produced gametes carrying either the B or the b allele, while the gametes of the white parent carried only the b allele. The white F1 offspring would have been genotype bb and the black F1 offspring were genotype Bb. +1 point
2. In Drosophila, three autosomal genes have alleles with the following dominance patterns: Gray body (G) is dominant over black (g)
Full wings (A) is dominant over vestigial wings (a)
Red eye (R) is dominant over sepia eye (r)
Two crosses were performed with the following results:
Parents: heterozygous red, full wings, crossed with sepia vestigial wings Offspring:
131 red, full
120 sepia, vestigial
122 red, vestigial
127 sepia, full
Parents: heterozygous gray, full wings, crossed with black, vestigial wings Offspring:
236 gray, full
253 black, vestigial
50 gray, vestigial
61 black, full
Are any of these three genes linked on the same chromosome? If so, what is the map distance (recombination rate) between them?
First cross: the genotypes of the two parents would have been RrAa x rraa. If the wing and eye color genes assort independently, we would expect a 1:1:1:1 ratio in the
offspring. That is exactly what the results of the cross show. Therefore, the eye color and wing shape genes are not linked.
Second Cross: Parental genotypes: GgAa x ggaa. Once again, we would expect a 1:1:1:1 ratio of F1 phenotypes from the cross. However, there are far more of the “parental” phenotypes (gray and full; black and vestigial). This indicates that the body color and wing shape genes are linked on the same chromosome. +1 point
Calculating the recombination rate: Total number of offspring = 600. Parental phenotypes = 489 (236+253). Recombinant phenotypes = 111 (50+61). 111/600 = 0.185 A recombination rate of 18.5% between the G and A genes. The R gene is unlinked.
3. Wild type Drosophila have gray bodies and straight wings, and mutants have been isolated with tan bodies (t) and curved wings (w). A cross between wild-type male and a tan female with curved wings gave the following results: 127 curved males; 131 curved, tan males; 136 tan females; 125 wild-type females. Explain these results giving the parental genotypes and the genotypes and ratios of the F1 offspring.
Answer: t is an autosomal gene since offspring of both sexes inherit the t+ (wild-type) allele from the male parent.
There are no curved female offspring but there are curved male offspring. That means that the male has no c allele, only a sex-linked c +(wild-type) allele. Since the female parent is tan, curved, and both markers are recessive, its genotype is tt cc .
Since some (one fourth) of the offspring are tan females, the male parent must have a t allele. Since it is phenotypically wild type, it must be heterozygous at the tan allele, i.e., t+t . Finally, since c is sex linked, the full genotype of the male parent is Yc + t+t . This analysis can be...
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