concerned about having a child with PKU as they are first cousins and the diseas

Question

concerned about having a child with PKU as they are first cousins and the disease runs in their family Phenylketonuria (PKU) is a rare autosomal recessive disorder, in which mutations on the gene for phenylalanine mental disability, if untreated. A couple, who have deci child with PKU provided, what is the probability that Xylase (PAH) result in elevated blood phenylalanine concentrations and severe as they are first cousins and the disease runs in their family. Based on the information ded to have children, is concerned about having a a) Their first child will be affected (3 pts) b) Now imagine that their first child has PKU. Determine the probability that the next three children will be affected as well. (3 pts) c) Determine the probability that among the next five children three will be normal and two affected in any order. (3 pts)

Explanation / Answer

As phenylketonuria is an autosomal recessive disorder and that couple had a pku affected son so the must be heterozygous carrier(Pp) for pku.

The chance of getting-

PP ( normal individual) is 1/4 = 25%

Pp ( normal but carrier , heterozygous individual) = 1/2 = 50%

pp ( affected individual) = 1/4 = 25%

a . So probability of first child will be affected is 1/4 .

b. First child is affected . So probability of next three children will be affected is 1/4 X 1/4 X 1/4x1/4 = 1/256

C. Here out of 5 children 3 will be normal and 2 Wil be affected. As we don't know the sequence it is unordered the affected coud came first or last so we use binomial expansion here.

Again probability of being normal ( heterozygous+ homozygous)= 3/4

Probability of being affected = 1/4

So probability of out of five two children will be affected is =

(5!/(3!X 2!)) X (3/4)3 X (1/4)2 = 135/512 = 0.264

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