Genetics: Calculating probability

[TytoAlba95]

Homework Statement

What is the probability that the child will be affected and the parents will be carrier?

Relevant Equations

N.A.

My Attempt: For the child to be affected, both the parents must be carriers.
Hence P(child)= Probability of the father of passing down an affected allele x Probability of the mother of passing down an affected allele
= 1/2 x 1/2
=1/4
The answer is (a).
I'm completely lost. Please help.

 

[mjc123]

The probability is 1/4 if both parents carry the affected allele. But what is the probability of that? You know they don't have the disorder, but each has a sibling who does. What does that tell you about the probability of their being carriers?

 

[TytoAlba95]

Since the pedigree shows an autosomal recessive trait, the genotypes of the 1st generation parents are:

Thus, P(child being affected) = Probability of father being heterozygote x Probability of the father passing down the mutant allele x Probability of mother being heterozygote x Probability of the mother passing down the mutant allele
= 2/3 x 1/2 x 2/3 x 1/2

Reasons (chronologically):
a) 2/3 - From a cross between I-1 and I-2 heterozygote parents, there are only three possibilities of getting a phenotypically normal individual out of which 2 can be heterozygote.
b) 1/2 - The II-2 must be a heterozygote for III-1 to be affected, and out of his two allele, only one can be mutant, that can be passed on.
c) same as a
d) same as b

Now coming to the second question, here I have a problem accepting the key.
P(both the parents being heterozygotes)= 2/3 x 2/3 =4/9.

But the answer is 2/3 only?!

 

[mjc123]

It must mean the probability for each parent is 2/3. It is not very clearly expressed.