Question

Galactosemia is a recessive human disease that is treatable by restricting lactose and glucose in the diet. Susan Smithers and her husband are both heterozygous for the galactosemia gene. If Susan and her husband have four children, what is the probability that:
a. none of the four will have galactosemia?b. at least one child will have galactosemia?c. only one child will have galactosemia?d. the first two will have galactosemia and the second two will not?e. two will have galactosemia and two will not, regardless of order?

Answer 1

Answer:

a. 81/256

b. 175/256

C. 1/4

D. 9/256

E. 54/256

Explanation:

Since both parents are heterozygous for galactosemia (Gg). The possibility of their children will be

GG Gg Gg gg. The probability for one child to be unaffected is 3/4 and affected is 1/4 since it is a recessive diseases:

A. Probability for none of the four children to be affected is 3/4*3/4*3/4*3/4=81/256

B.

Since the probability of unaffected children is 81/256

The probability that at least one child will be affected is 1-(3/4*3/4*3/4*3/4)=1-(81/256)

= 175/256

C. The probability that only one if the child will be affected is 1/4

D. The probability that the first two will have galactosemia and the second two will not is

1/4*1/4*3/4*3/4=9/256

E. The probability that two will have galactosemia and two will not, regardless of order.

Since the probability that two will have galactosemia and two will not is 9/256.

The total possible outcome that 2 among 4 children will be born regardless of order (first and second, first and third, first and fifth, second and third, second and fourth, third and fourth) is 6

Hence the probability two will have galactosemia and two will not, regardless of order is 9/256*6= 54/256