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A man who has male pattern baldness (X^bY) marries a woman who does not carry the allele (X^BX^B). What genotypic ratios would be expected of their children?

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Answer:

2
X^BX^b:2
X^BY

Step-by-step explanation:

The genotypic ratio of expected of their offspring would be 2
X^BX^b:2
X^BY.

From the illustration, the genotype of the man with male pattern baldness is
X^bY while the genotype of the woman without the baldness allele is
X^BX^B.

Crossing the two genotypes in marriage:


X^bY x
X^BX^B

Progeny:
X^BX^b
X^BX^b
X^BY
X^BY

Hence, the genotypic ratio of the offspring becomes:

2
X^BX^b:2
X^BY

User Ben Bolker
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