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.

To answer your questions, let's start by understanding the inheritance pattern of X-linked disorders such as hemophilia. In humans, the X chromosome is one of the two sex chromosomes. Females have two X chromosomes, while males have one X and one Y chromosome.

1) Probability that both daughters are heterozygous carriers:
To determine the probability, we need to consider the genotype of the parents. Since hemophilia is an X-linked disorder, the affected son must have inherited the disease-causing allele from his mother. This means the mother must be a carrier of the disorder.

Since the father is phenotypically normal, he does not carry the disease-causing allele. Therefore, his genotype is XY (normal chromosome pair). The mother, being a carrier, must have one affected allele and one normal allele. Therefore, her genotype is XhX.

Now let's calculate the probability that both daughters are heterozygous carriers. Each daughter has a 1/2 chance of inheriting the affected X chromosome from her mother and a 1/2 chance of inheriting the normal X chromosome from her father.

The probability for the first daughter to be a heterozygous carrier is 1/2 (to inherit the affected X chromosome) multiplied by 1/2 (to inherit the normal X chromosome). This gives us 1/4.

Similarly, the probability for the second daughter to be a heterozygous carrier is also 1/4.

Since these two events are independent (meaning the outcome of one doesn't affect the other), we can multiply their probabilities together to get the overall probability:

Probability that both daughters are heterozygous carriers = (1/4) × (1/4) = 1/16.

Therefore, the correct probability is 1/16, not 4/9.

2) Probability that the son will be affected:
In this scenario, one daughter (who is a heterozygous carrier) and a normal male produce a son. To determine the probability of the son being affected, we need to consider the mother's genotype and the father's genotype.

The mother's genotype is XhX, where Xh represents the affected allele and X represents the normal allele. Therefore, she has a 1/2 chance of passing the affected X chromosome to her son.

The father's genotype is XY, which means he does not carry the affected allele. Therefore, he can only pass on a normal X chromosome to his son. This has a probability of 1/2.

Since these events are independent, we can multiply the probabilities together:

Probability that the son is affected = (1/2) × (1/2) = 1/4.

Thus, the correct probability is 1/4, not 0.

I hope this explanation clarifies the correct probabilities for both questions.