Question

Phenylketonuria (PKU) is a disease that results from a recessive gene. Affected persons are unable to metabolize the amino acid phenylalanine. Two unaffected parents produce a child affected with PKU.

a. What is the probability that a sperm from the father will contain the PKU allele?
b. What is the probability that an egg from the mother will contain the PKU allele?
c. Their second child is unaffected. What is the probability that this child is a carrier?

Answer 1

A) 1/2

B) 1/2

C) 2/3

Step-by-step explanation:

Let's assume that the phenylketonuria is encoded by gene P, where the allele P is for normal individuals not affected by Phenylketonuria while the allele p is for individuals affected by it.

An unaffected individual will either be homozygous (PP) or heterozygous/Carrier (Pp). If two unaffected parents cross and produce an affected child, it means both parents are heterozygous or carriers of the recessive allele (p). Hence, they will possess a heterozygous (Pp) genotype.

Based on this explanation above,

a) the probability of the male parent producing a sperm that contains the PKU (recessive) allele will be 1/2 i.e. 1 recessive allele out of the pair (2). Since the Pp alleles of the male parent will separate into the sperms having either P or p.

B) The probability of the female parent producing an egg with PKU (recessive) allele will be 1/2 as well. The same explanation for male parent applies here.

C) According to the cross (see attached image), four possible offsprings will be produced with 3 unaffected phenotypically and 1 affected. The affected child will have a (pp) genotype. The unaffected children will have 1 homozygous (PP) and 2 heterozygous (Pp). These 2 heterozygous children are carriers of the recessive or PKU allele but will be phenotypically unaffected m Hence, out of 3 unaffected children, 2 will be carriers.

Therefore, the probability of the second born being a carrier (Pp) will be 2/3.