Question

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?

Answer 1

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

Step-by-step 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

Answer 2

The galactosemia is a condition, in which the affected person is not able to metabolize the galactose sugar. This is a genetic disorder, inherited as an autosomal recessive trait.

In this case, Susan and her husband both are heterozygote for the condition, which means they have one dominant and one recessive allele. So, the genotype of both the individuals are Gg and Gg (let’s say G codes for dominant allele and g is recessive allele).

The attached image represents the cross between two heterozygous individual.

The given cross shows 1 out of 4 children would be affected with galactosemia, so, there is 25% or 0.25 probability that one of the children would be affected.