Question

1. Phenylketonuria (PKU) is a disease that results from a recessive gene. Two normal parents produce a child with PKU. After determining the parents genotype and performing a cross, answer to the following questions: [3 pts – 1 A, 1 B, and 1 Punnett square.

1. What is the probability of that event?

The probability is 1 out of 4 or 25% chance.

2. This couple wishes to have a large family and would like to have 5 children. What is the probability that the next 4 children won’t have PKU?

¼ * ¼ * ¼ * ¼ * ¼ = ? (I don't know if this is the direction to be going in)

Answer 1

Phenylketonuria (PKU) is inherited as autosomal recessive disorder

Let, A = Normal allele, a = Allele containing phenylketonuria trait

As both normal parents have a child with PKU, they must be heterozygous for this trait, i.e., Aa genotype. The cross between them is shown in below Punnett square.

Gametes	A	a
A	AA (Normal child)	Aa (Normal child)
a	Aa (Normal child)	aa (Affected child)

From the above Punnett square, we find that probability of occurring phenylketonuria is 1/4 or 25%.

So, probability of a child that will not develop phenylketonuria is 3/4. Now, born of a child is independent of born of another child.

So, probability that the next 4 children won’t have PKU = 3/4 x 3/4 x 3/4 x 3/4 = 81/256