Solutions For An Introduction to Genetic Analysis Chapter 2 Problem 55P

b. If persons A and B marry, to have their first child to have PKU, both of them should be heterozygous to give the recessive allele to the child.

Probability of A having that allele should be considered form II-2. II-2 is P/-, because her father has the disease and is p/p. The probability of III-2 having the allele is 1/2 and probability of 'A' getting the allele from his father is 1/2. So, the probability of 'A', being a carrier of the allele is 1/2 X 1/2 = 1/4.

Probability of B having that allele should be considered from II-3 and II-4. Both II-3 and II-4 are heterozygous, as both their fathers are having the disease, as they are homozygous, for this recessive allele. So, the probability of III-3 being a carrier of the allele is 2/3 and the probability of 'B' receiving the allele from the mother is 1/2. So, the probability of B to be the carrier is 2/3 x 1/2 =1/3.

As both the parents are heterozygous, the probability of both the parents passing the recessive allele to the child is 1/4.

Thus, the probability of the child to have PKU is 1/4 X 1/3 X 1/4 =1/48.

c. If the first child is normal, the probability of the second child having PKU is 1/48. It is the same probability of the first child having PKU, as no additional information is gained.

d. The parents of the children are heterozygous, if the first child is to have PKU. The probability of having an affected child is 1/4.

So, the probability of having a unaffected child is 1-1/4 = 3/4