3. In goats, a beard is produced by an autosomal allele that is domi- nant in ma
ID: 194484 • Letter: 3
Question
3. In goats, a beard is produced by an autosomal allele that is domi- nant in males and recessive in females. We'll use the symbol B for the beard allele and B+ for the beardless allele. Another indepen- dently assorting autosomal allele that produces a black coat (W) is dominant over the allele for white coat (w). Give the phenotypes and their proportions expected for the following crosses: (a) B+Bb Ww male x B+ Bb Ww female (b) B+Bb Ww male x Bt Bb ww female (c) B+B+ Ww male x Bb Bb Ww female (d) B+Bb Wu, male × Bb Bb uro femaleExplanation / Answer
a. B+BbWw x B+BbWw
Male gametes= B+W, B+w, Bb W and Bbw
Female gametes= B+W, B+w, BbW and Bbw
So making the Punnett square
Phenotype and expected proportions are 9 (Beard and black):3 (Beard and white): 3 (Beardless and black): 1 (Beardless and white).
b. B+BbWw male gametes=B+W, B+w, BbW and Bbw
B+Bbww female gametes=B+w, B+w,Bbw and Bbw
So Phenotype and expected proportions are 3 (beard and black): 3 (beard and white): 1 (beardless and black): 1 (beardless and white).
c. B+B+Ww male gametes= B+W,B+w, B+W, B+w
BbBbWw female gametes= BbW, Bbw, BbW, Bbw
So 3 beard and black:1 beard and white
d. B+BbWw male gametes=B+W, B+w, BbW and Bbw
BbBbww female gametes= Bbw, Bbw, Bbw,Bbw
So 1 beard and black: 1 beard and white: 1 beardless and black: beardless and white.
B+W B+w BbW Bbw B+W B+B+WW (Beard and black) B+B+Ww (Beard and black) B+BbW (Beard and black) B+BbWw (Beard and black) B+w B+B+Ww (Beard and black) B+B+ww (Beard and white) B+BbWw (Beard and black) B+Bbww (Beard and white) BbW B+BbWW (Beard and black) B+BbWw (Beard and black) BbBbWW (beardless and black) BbBbWw )beardless and black) Bbw B+BbWw (Beard and black) B+Bbww (Beard and white) BbWw (beardless and black) BbBbww (beardless and white)Related Questions
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