Question

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?

Answer 1

Answer:

2$X^BX^b$:2$X^BY$

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$X^BY$