Question

4. [15 pts total] Sam has hemophilia (a recessive, X-linked condition) but his daughter Margaret has normal phenotype. Margaret marries Mark who is not hemophiliac. A. [3 pts] If Margaret and Mark have a daughter, what is the probability that she will be a hemophiliac? B. [3 pts] If Margaret and Mark have a son, what is the probability that he will be a hemophiliac? C. 13 pts] If Margaret and Mark have four sons, what is the probability that all four will be born with hemophilia? D. [3 pts] If Margaret's and Mark's daughter marries a man with hemophilia, what is the probability that her daughter will be a hemophiliac? E. 13 pts] If Margaret's and Mark's daughter marries a man with hemophilia, what is the probability that her son will be a hemophiliac?

Answer 1

since Margaret's father is affected with the disease she is heterozygous for the trait, let the affected allele be Xh and normal allele be XH then  Margaret is XHXh and her husband Mark is not hemophilic so he is XHY.

A) XHXh*XHY

	XH	Xh
XH	XHXH (normal female)	XHXh (carrier female)
Y	XHY (normal male )	XhY (affected male)

probability of that girl is affected = number of affected girls/ number of girls=0/2=0

B) probability that boy is affected= number of affected boys/ number of boys=1/2

C) probability of that all 3 boys are affected= (probability of boy is affected)^4=(1/2)^4=1/16

D) Daughter can be either normal female or carrier female if the daughter is a normal female then all of her daughters will have at least one copy of normal allele so daughters are not affected but when the female is carrier XHXh*XhY

	XH	Xh
Xh	XHXh	XhXh (affected female)
Y	XHY	XhY

the probability that daughter is affected = number of affected girls/total number=1/2

E)Daughter can be either a normal female or carrier female if the daughter is a normal female then all of her sons will get a normal allele so no sons are affected.  but when the female is carrier XHXh*XhY

(from above punnet square )

the probability that boy is affected the = number of affected boy/number of boys=1/2