PKU is a rare, fully penetrant, autosomal recessive condition in which an affected individual cannot break...

PKU is a rare, fully penetrant, autosomal recessive condition in which an affected individual cannot break down the amino acid phenylalanine. Neither Brian nor Mary has PKU, and they want to have children. They are concerned because Brian's father has PKU, and while neither of Mary's parents has PKU, her brother has PKU. Use symbols A and a.

1) What is the probability that Mary and Brian's first child will have PKU? Show work.
2) Mary and Brian have a child, and it does not have PKU. They decide to have a second child. Does knowing the phenotype of the first child cause you to change the probability you calculated in question 1 to predict the probability the next child will have PKU? If not, why?
3) Consider an alternate scenario where Mary and Brian have a child and it does have PKU. They decide to have another child. Does knowing the phenotype of the first child change the probability you calculated in 1 to predict the probability the next child will have PKU? If not, why?

Solutions

Expert Solution

Subpart 1:

As PKU is an autosomal recessive trait so the infected individuals must be homozygous for recessive alleles.

Let A is dominant allele and a is recessive allele for the trait'

As Brian's father has PKU it is must that his genotype is aa and Brian has no PKU means his genotype is Aa.

In Mary's family his brother has PKU and her perants dont have means both the parents are heterozygotes. So Mary's genotype could be either AA or Aa.

So the alleles generated from Brian would be A and a, where as Mary's possible alleles would be A and A if she is a homozygous and A & a considering her as a heterozygous for the trait.

the possibilities of genotypes of the children of Brian and Mary are as follows.

The total number of genotypic possibilities Brian and Mary's child may have is 8. Out of which only in 1 case the child will have PKU.

The probability of Brian and Mary's first child having PKU is (1/8)= 0.125

Subpart 2

If the first child does not have PKU then also Mary's genotype can either be homozygous or heterozygous and the probability for a second child to have PKU will remain same.

Subpart 3

If the first child does have PKU, it will be clear Mary's genotyp is heterozygous. So now the genotypic possibilities of Mary and Brian's child will change. Now the possible genotypes will be

So the total possible genotypes will be 4 in which only in one case the child will have PKU.

The probability of the next child having PKU will be 1/4= 0.25.

gladiator answered 1 year ago

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