Phenylketonuria is an autosomal recessive human disease Supp

Phenylketonuria is an autosomal recessive human disease. Suppose that both members of a couple are heterozygous for Pku and have children.

a. If they have two children, what is the probability that one or the other will have PKU but not both?

b. If they have two children what is the probability that they have a boy and a girl in either order , both without PKU?

c. If they have three children, what is the probability that at least one will have PKU?

d. If they have three children, what is the probability that they have both sexes and none is heterozygous.

I already know the answers for these. I need help getting to the answer.

Solution

Phenylketonuria is an autosomal recessive human disease. Suppose that both members of a couple are heterozygous for Pku and have children.

a). If they have two children, what is the probability that one or the other will have PKU but not both?

Assume that the gene coding for normal phenotype is “a” and the gene coding for normal phenotype is, “A;” A and dominant over a.

If two heterozygotes have children,

Aa* Aa --à A_ (3/4, unaffected), aa (1/4, affected)

The probability of one or the other will have PKU but not both is = The probability of A to have PKU + the probability of B to have PKU = ¼ + ¼ = 1/2

b. If they have two children what is the probability that they have a boy and a girl in either order, both without PKU?

The probability of that they have a boy and a girl in either order, both without PKU is, probability of being a boy and the probability of not having PKU multiplied by the probability of being a girl and the probability of not having PKU

= ½*3/4*1/2*3/4 = 9/ 64

c. If they have three children, what is the probability that at least one will have PKU?

Probability of atleast one will have PKU = 1- all having PKU = ¾ - (1/2)^3 = ¾ - 1/8 = 0.625

d. If they have three children, what is the probability that they have both sexes and none is heterozygous.

If they have three children, the probability that they have both sexes and none is heterozygous.

Is = probability of both sexes and the probability of homozygotes. (AA – ¼ and aa – 1/4)

= ½*1/2*1/4*1/4 = 1/1/64