Question

PLESEE ANSWER!!! Hemophilia is a disease caused by a gene found on the X chromosome. Therefore, it is referred to as a sex-linked disease. The recessive allele causes the disease. A normal man marries a women that is heterozygous for the trait. What are the chances that their daughters will have hemophilia? What are the chances that their sons will have hemophilia?

Answer 1

Let H= No Hemophilia
h= Hemophilia
Now we will create the genotypes for the parents, it says that the mother is heterozygous which means she will carry the dominant and recessive allele, which would be Hh and it says the father is normal which would mean that he would be No Hemophilia dominant, so it would be HH

Cross the 2 created genotypes: Hh x HH
Now create your Punnett square using the 2 created genotypes above
H h
H HH Hh

H HH Hh

Looking at the Punnett square above, it appears that there’s a 2:2 ratio or a 50% chance of having Hemophilia and not having Hemophilia

So for a female child: 1/2 x 1/2= 1/4 of having a female with Hemophilia
The first one represents the female child having Hemophilia or not, that one can be obtained by simplifying the 2:2 ratio to a 1:1 ratio and the other one represents the chances of their child being a female. I assume that the steps above would be repeated again for the male child.


I hope that helps!