posted by Billy on .
Suppose that a mating between phenotypically normal parents produces two normal daughter and one son affected with hemophilia (x-linked disorder).
1)What is the probability that both of the daughters are heterozygous carriers?
so obviously the two parents had to be carriers on the disorder for the son to have hemophilia..or is that different with x-linked?..anyway i decided to do 2/3 x 2/3 = 4/9 but the actual answer is 1/2 x 1/2 x 1/2= 1/8...i don't get where all the 1/2s come from.
2)If one of the daughters and a normal male produce a son, what is the probability that the son will be affected.
for this one i thought the answer was o but i think that the situation changes for x-linked the answer is 1/2 x 1/2= 1/4 but im still not seeing where the one halves come in..
please explain i really appreciate it!!
1. Since dad is XY, and the characteristic is X-linked, he either has the disorder or not. He cannot be just a carrier. The problem indicates it is X-linked.
Therefore it is 1/2*1/2 = 1/4
I don't know where the book would get 1/8, since the question only concerns the two daughters.
2. Daughter had 50% chance of getting the recessive gene from mom, and she has a 50% chance of passing it down to any child.
I hope this helps you to understand.