Under these circumstances assume the following Punnett square to be true.
Where NN or Nn = normal conditions and nn = PKU
a. all three of their children will be of normal phenotype
3/4 x 3/4 x 3/4 = 27/64
b. one or more of the three children will have the disease (x)
1  27/64 = 37/64











3 Combinations 







+ 3(3/4 x 1/4 x 1/4) 
+ 3(3/4 x 3/4 x 1/4) 

Note: the probability of the disease (x) = 1/4 & the probability of being normal (o) = 3/4
c. all three children will have the disease
1/4 x 1/4 x 1/4 = 1/64
d. at least one child out of three will be phenotypically normal
(Note: Remember that the probabilities of all possible outcomes always add up to 1)
1  1/64 = 63/64