Question

Problem 2-20: 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: (1 point each)

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?

Please provide an explanation with the answer! Thank You!

Answer 1

0Gg x Gg

Offsprings =GG, Gg, Gg, gg

Probability of normal=3/4

Probability of disease =1/4

If four children

Then total outcomes=

All four have disease

None have disease

1 have disease and 3 are normal

2 have normal 2 have disease

3 have disease and 1 is normal

Total =5

Probability of none of four have galactosemia=1/5*3/4

=3/20

Probability of atleast one child have galactosemia=4/5*1/4

=1/5

Probability of only one child have galactosemia=1/5*1/4=1/20

Probability of 2 have disease and 2 are normal=1/5*(1/4+3/4)

=1/5