Week 5 Homework Answer Key Due Feb. 23, 2013

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1 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. 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. 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 Page 1

2 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. Calculating the recombination rate: Total number of offspring = 600. Parental phenotypes = 489 ( ). Recombinant phenotypes = 111 (50+61). 111/600 = 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 checked by a simple Punnett square: Yt + Yt c + t + c + t c t Yt + Yt c + t + c + t c t c t c t c t Another solution is possible, however. Tan could also be autosomal dominant. Under these circumstances the tan female parent would be heterozygous (Tt) and the wt male would be homozygous recessive (tt). A cross between these two would still produce 50:50 normal colored / tan offspring, of both sexes. Page 2

3 A Third solution: tan and curved are both sex linked. C = wt wings, c = curved wings (recessive) T = tan body, t = wt body color (recessive) Female Parent fly is XcT / Xct Male P fly is XCT / Y This works, too! +2 points for any workable solution 4. Clear wing, Black eye, and Hairless (c, b, and h) are linked, recessive traits carried on chromosome 3 of the Narragansett quahaug fly. A clear-winged, black-eyed fly is crossed with a hairless fly. A) Their F1 offspring were 97 wild type quahaug flies. What is the genotype of these F1 flies?? [4 points] Genotype: Cc Bb Hh or, [cbh / CBh], etc (heterozygous for all 3 traits) +2 points for genotype 10 of these flies were then mated with flies that exhibit all three traits (clear-winged, black-eyed, and hairless). They produced a total of 1,000 F2 offspring: clear-wings, black eyes 415 cbh hairless 411 CBh clear wings 61 cbh black eyes, hairless 58 Cbh wild type 22 CB clear wings, black eyes, hairless 26 cbh black eyes 4 CbH clear wings, hairless 3 cbh Use these data to construct a map of these three genes in two steps. Use these data to construct a map of these three genes, showing your work along the way. The map should show the order of the three genes, and indicate the recombination frequencies between them. Parental Allele Combinations: cbh and CBh Page 3

4 C-B B-H C-H = = = % 16.7% Grading: for each correct recombination frequency. 5.5% H C B 5.5% 12.6% Grading: +2 pts. for correct map. Award full credit for correct mapping of incorrect frequencies. 5) FlyLab Experiments (Note: 6 points total for this section) 1) Find one mutation from the list that is recessive to wild type. Show the cross(es) you performed to identify the recessive mutation and the results of each cross. (+2 points for the recessive mutation) 2) Find one mutation from the list that is dominant to wild type. Show the cross(es) you performed and the results of each cross. (+2 points for the dominant mutation) I went through the complete list of FlyLab mutations, classifying each of them as sexlinked, simple dominant, simple recessive, etc This is pretty easy to do. You design a mutation into the female fly first (that s the one of the left of the FlyLab window) and then cross her with a wt male fly. That way, if the mutation is sex-linked, you find out immediately (since it appears in the F1 males, but not the females). If the mutation is recessive, its phenotype doesn t appear in the F1; if it is dominant, all the F1 flies display the mutation. Note that several of the dominant alleles are actually lethal in the heterozygous condition. This can be confusing, because (without warning) FlyLab designs those mutants in the heterozygous condition when you select them. And the offspring you get in the F1 (as a result) are 1:1 dominant phenotype : wild type. Bristle Forked (sex-linked) Shaven (recessive) Singed (sex-linked) Spineless (recessive) Stubble (Dominant Lethal in homozygous condition) Page 4

5 Body Color Black (recessive) Ebony (recessive) Sable (Sex-linked) Tan (Sex-Linked) Yellow (Sex-Linked) Antennae Aristapedia (Dominant Lethal in homozygous condition) Eye Color Brown (recessive) Purple (recessive) Sepia (recessive) White (sex-linked) Eye Shape Bar (dominant) Eyeless (recessive) Lobe (dominant) Star (Dominant Lethal in homozygous condition) Wing Size Apterous (recessive) Miniature (sex-linked) Vestigial (recessive) Wing Shape Curly (Dominant Lethal in homozygous condition) Curved (recessive) Dumpy (recessive) Scallopped (sex-linked) Wing Vein Crossveinless (sex-linked) Incomplete (recessive) Wing Angle Dichaete (Dominant Lethal in homozygous condition) Page 5

6 Gene Linkage: 3) Find two mutations from the list that are linked to each other (occur on the same chromosome). Show the cross(es) you performed and the results of each cross. To find linked genes, you engineer a fly with two mutations. Then you cross that fly with a wt fly. If both mutations were recessive, all of the F1 flies should show the dominant wt phenotype. So, then cross the F1 flies with themselves, and look at the F2 progeny. If the genes you selected are not linked, you will get progeny in a 9:3:3:1 ratio (independent assortment). However, if the genes are linked, you will get something very different, with more of the parental phenotypes (wt and the double-recessive flies) that you would expect from independent assortment. Incidentally, you could also cross one of the F1 flies with a homozygous recessive fly. With independent assortment, you d get 1:1:1:1. Linked genes produce more of the parental types (wt and double mutant). (+2 points for a pair of linked genes) Here s an example of two gene pairs that do show 9:3:3:1 independent assortment: Incomplete veins Shaven Bristles Curved Wings Sepia Eyes I didn t check every conceivable combination, but I did find that the following gene pairs are linked, meaning that they do not show the 9:3:3:1 ratios: (other combinations are possible, of course) Vestigial Black Body Dumpy Black Body Incomplete veins Spineless (these genes are very closely linked!) Incomplete veins Ebony body Curved Wings Purple Eyes Curved Wings Brown Eyes Page 6

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