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Week 5 Homework Key

Decent Essays

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 …show more content…

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.

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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)

XcT / Xct
XCT / Y

Female Parent fly is
Male P fly is

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 hairless clear wings black eyes, hairless wild type clear wings, black eyes, hairless black eyes clear wings, hairless

cbH
CBh
cBH
Cbh
CB cbh CbH cBh 415
411
61
58
22
26
4
3

• 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

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