The probability that III-3 is a carrier (Rr)= 2/3

The probability that III-4 is a carrier (Rr)= 1/2

The probability that IV-1 will be affected (rr)= 1/12

Note: The original answer of 2/3 for III-3 was correct because the probability that III-3 is a carrier is calculated based on the available options. A punnet square is not entirely correct because the person is showing the dominant phenotype. The punnet square shows 3 options that will give the dominant phenotype, 2 of which are Rr (carrier).

Multiplying 2/3 x 1/2 x 1/4 (likelihood that the child will be rr from punnet square) = 1/12.

I just did this problem on masteringgenetics and got it correct.

You can calculate probabilities in pedigrees by considering the requirements for a specific outcome. In this case, the outcome that individual IV-1 will be affected has four requirements:

1.Individual III-3 is a carrier (probability = 2/3).

2.Individual III-4 is a carrier (probability = 1/2).

3.Individual III-3 passes the r allele to his child (probability = 1/2, assuming III-3 is a carrier, which is accounted for in requirement 1).

4.Individual III-4 passes the r allele to her child (probability = 1/2, assuming III-4 is a carrier, which is accounted for in requirement 2).

Because requirements 1 AND 2 AND 3 AND 4 must be met for the outcome in question (individual IV-1 to be affected), you calculate the answer by applying the multiplication rule (as implied by “AND”):

The probability that IV-1 will be affected (rr) = 2/3 x 1/2 x 1/2 x 1/2 = 1/12.