**If A and B have kids, what's the probability
of first child having PKU?**

Answer = P(A is het) X P (B is het) X P (if both hets, both pass on recessive trait)P(A is het) = P (A's father is het) X P (of passing recessive allele to A) = 1/2 X 1/2 = 1/4

P(B is het) = P (B's mother is het) X P (of passing recessive allele to B) = 2/3 X 1/2 = 2/6 = 1/3

Answer = P(A is het) X P (B is het) X P (if both hets both pass on recessive trait) = 1/4 X 1/3 X 1/4 = 1/48

**If first kid is unaffected, what's the probability
of second child having PKU?**

Answer = the same. Nothing has changed in terms of probability.

**If first kid is affected, what's the probability
of second child will not have PKU?**

The fact that the first kid had PKU tells you both parents were heterozygous, thus the chance of a normal child is the chance of the PP or Pp genotype = 3/4.