Question

With regard to pattern baldness in humans (a sex-influenced trait), a woman who is not bald and whose mother is bald has children with a bald man whose father is not bald. What is the probability for their first child being a daughter and bald?

Answer 1

let the alleles be B-baldness and b-normal

Baldness is a sex-influenced trait, males require only one B allele to be bald, but the woman has to be homozygous for the B allele to be bald.

The woman is not bald but her mother is bald, so she has inherited B allele from her mother so the genotype of the woman is Bb.

man is bald but his father is not bald, so his father is bb, so he has inherited b allele from his father, so the genotype of the man is Bb.

Bb * Bb

	B	b
B	BB	Bb
b	Bb	bb

probability of getting BB=number of BB/total number=1/4

XX * XY

	X	Y
X	XX ( female)	XY ( male)

probability of getting a female= 1/2

probability of getting female with genotype BB=probability of getting BB $\times$ probability of getting female

= 1/4$\times$1/2

=1/8

probability that their first child is a daughter and bald =1/8