Please explain the answers and show your work Thanks 3 Two h

Please explain the answers and show your work. Thanks

3. Two healthy parents know from blood tests that each carries a recessive allele responsible for Tay-Sachs disease. Tay-Sachs causes nerve cells to malfunction and results in death by age 4.
If their first three children have the disease, what is the probability that their fourth child will not?_________
What is the probability of parents like these having four diseased children?________
If their first three children are male, what is the probability that their fourth child will be male? ___________

Answers

¾; 1/256; ½

Solution

If R is dominant gene and r is recessive for the disease then Through punnet square two heterozygous parents will only be the carrier of the disease.

Rr x Rr -------->1RR:2Rr:1rr

Tay sachs disease require two recessive alleles to be present for disease to appear.It means only rr will have the disease. So, each time there is 1/4 chance that next child will have disease and there is 3/4 chance that next child will not have disease.

(a)Each time there is 3/4 chance that their next child, (let it be 4th) will not be affected.

(b)Probability of having 4 diseased children= 1/4 *1/4*1/4*1/4= 1/256

(c)Gender of earlier three children cannot determine the probability of having a male child fourth time. Each time there is 1/2 probability of having a male child.