Question

1. At least 3 independently assorting loci are known to govern coat color in mice. The genotype C- will allow pigment to be produced by 2 other genes. The cc genotype does not allow pigment production resulting in albino mice. The striped “agouti” hair pattern depends on the genotype A- and the aa genotype is nonagouti. Finally, the color of the pigment may be black (B-) or brown (bb). Five different coat colors may be produced by the action of various alleles of these 3 loci. Wild type (agouti, black) A-B-C- Black (nonagouti) aaB-C- Brown (nonagouti) aabbc- Cinnamon (agouti, brown) A-bbc- Albino --------- CC a. What phenotypic frequencies are expected in the F2 generation from crosses of pure black with albinos of genotype AAbbcc? b. A cinnamon male is mated to a group of albino females of identical genotype and among their progeny were 43 wild type, 40 cinnamon, 39 black, 41 brown, and 168 albino. What are the most probable genotypes of the cinnamon male and the albino females to which he was mated? The answer should account for the observed ratios as well as the specific phenotypes.

Answer 1

Answer:

a).

AAbbcc (ablnino) x (black) aaBBCC –Parents

AaBbCc (wildtype)----------------------F1

AaBbCc x AaBbCc ----F1 xF1

Aa x Aa = A_ (3/4), aa (1/4)

Bb x Bb = B_ (3/4) & bb (1/4)

Cc x Cc = C_(3/4) & cc (1/4)

Wild type = A_B_C_ = ¾ * ¾ * ¾ = 27/64

Black = aaB_C_ = ¼ * ¾ * ¾ = 9/64

Cinnamon = aabbC_ = ¼ * ¼ * ¾ = 3/64

Brown = A_bbC_ = ¾ * ¼ * ¾ = 9/64

Total = 27+9+3+9 = 48/64

Albino 1 – (48/64) = 16/64

Phenotypic ratio in F2 generation =- 27:9:3:9:16

b).

aabbCc (cinnamon) x AaBbcc (albino)----Parents

	ABc	Abc	aBc	abc
abC	AaBbCc (wild type)	AabbCc (brown)	aaBbCc (black)	aabbCc (brown)
abc	AaBbcc (albino)	Aabbcc (albino)	aaBbcc (albino)	aabbcc (albino)

Observed and expected ratios = 1:1:1:1:4 = (wild: cinnamon : black : brown : albino)