Question

Hemophilia B is an X-linked recessive trait that causes problems with blood-clotting.

1. (2 pts) If a normal man and a woman who is a carrier for hemophila decide to have children, what is the probability their first child will have hemophila? Please show your work using a Punnett Square.
2. (3 pts) How would X inactivation in the woman described above (a carrier for hemophilia) affect her phenotype for hemophilia symptoms?

Answer 1

Hemophilia is an X-linked recessive trait.
WT dominant allele = XH
Mutant recessive allele = Xh
Normal man = XHY
Carrier female = XHXh

Parental cross: XHXh X XHY
Progeny: XHXH XHY XHXh XhY
The probability for their first child to be affected = 1/4 = 25%

X-inactivation: If a genotype contains more than one X-chromosome, all the X-chromosomes except one will be inactivated by Xist noncoding RNA.
It is a random process
Barr body = Inactivated X-chromosome
XX = 1 Barr body
XY = 0 Barr body
XXX = 2 Barr bodies

Carrier female = XHXh
If the WT XH chromosome is inactivated, the carrier female would exhibit the mutant phenotype (Even though she is heterozygous).
If the mutant Xh chromosome is inactivated, the carrier female would not exhibit the mutant phenotype.