Problem 8

Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers, what is the probability of each of the following?

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
All three have x
2 out of 3 has x
1 out of 3 has x

+
+
=

x x o
o o x

3 Combinations
x o x
o x o

o x x
x o o

+ 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