Galactosemia is a recessive human disease that is treatable by diet. susan smithers and her husband are both heterozygous for the galactosemia gene. if the couple has four children, what is probability that at least one of the children will have galactosemia?

The probability that a least one of the children will have Galactosemia is 0.6835

Explanation:

If Galactosemia is caused by recessive human g allele, then, the cross is Gg x Gg

GG = normal

Gg = normal

Gg = normal

gg = diseased

Here, the probability of suffering Galactosemia is 1/4 and be normal is 3/4

Probability of any one having Galactomia is 1/4

Probability of any one is a carrier and not suffers Galactosemia is 1/2

Probability of any onr is not a carrier and not suffers Galactosemia is 1/4

Probability that none have Galactomia is (3/4) * (3/4) * (3/4) * (3/4) = 81/256

The probability that a least one of the children will have Galactosemia is:

P = 1 - (81/256) = 175/256=0.6835