Question

Duchenne muscular dystrophy is a ressecive sex-linked disorder carried on the x-chromosome in humans. A woman who is heterozygous for this disorder marries a normal male. what is the percentage of offspring would be considered a carrier? (PLEASE EXPLAIN HOW YOU GOT YOUR ANSWER, BECAUSE I NEED TO KNOW HOW TO DO THE PUNNET SQUARE)

Answer 1

I did the Punnett square here. So, as we know, the trait is x linked recessive. This means that if you have two recessives, you have the trait. If you have only one recessive, you’re a carrier, and this is for women only (THIS is because women have two X chromosomes, so if one is defected the other one can cover up. Men only have one X chromosome, so if theirs has a defect, they have the trait).
We’ll start off our Punnett square with XHXh for mom and XHY for dad. For the top right box, bring down a X and a big H, then another X and big H. This girl is not a carrier, she has both big H’s so she is does not have the trait. For the box on the top right, well bring down X big H, then X little h. As you can see, she has a small recessive h... but she also has a big one. That means she’s a carrier. Out of female boxes (which for this Punnett square, is just the top two), only one box is a carrier. That means 50% of girls will have the trait.
For the boys, son 1 is clear for the trait, but son two has it completely.
So again your answer if 50% of the girls are carriers. Hope this helped :)