contestada

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?

Respuesta :

Oseni

Answer:

2[tex]X^BX^b[/tex]:2[tex]X^BY[/tex]

Explanation:

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

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

Crossing the two genotypes in marriage:

                           [tex]X^bY[/tex]   x   [tex]X^BX^B[/tex]

Progeny: [tex]X^BX^b[/tex]    [tex]X^BX^b[/tex]    [tex]X^BY[/tex]       [tex]X^BY[/tex]

Hence, the genotypic ratio of the offspring becomes:

2[tex]X^BX^b[/tex]:2[tex]X^BY[/tex]