Two linked genes, A and B, are separated by 18 cM. A man with genotype AB/ab marries a woman with genotype ab/ab. The man's father was AB/AB. a. What is the probability that their first child will have the genotype Ab/ab? b. What is the probability that their first two children will both have the genotype ab/ab?

a. The man's genotype is AB/ab, which indicates that his alleles for gene A and gene B could recombine due to crossing over during meiosis. Since the two genes are 18 centiMorgans (cM) apart, the probability that a recombination event occurs between them in a single meiotic event is 18%. The chance of getting Ab (from AB/ab) is the same as getting aB, which is half of 18%, because either one of the recombinations could happen. So the probability that their first child receives Ab from the father is 9%.

The woman's genotype is ab/ab, meaning she can only pass on the ab combination. To get Ab/ab in their child, the child must receive Ab from the father and ab from the mother. Since the mother can only give ab, we only need to consider the probability from the father which is 9%. Thus, the probability that their first child will have the genotype Ab/ab is 9%.

b. To find the probability that the first two children will both have the genotype ab/ab, we must consider that each child is an independent event. The man could pass on AB or ab with no recombination, or Ab or aB with recombination. However, for the children to have ab/ab, no recombination event should happen in the father's meiosis that will affect these offspring. The probability of the non-recombinant gamete (either AB or ab) is 1 - 0.18 = 0.82 (or 82%).

For each child to receive ab from the man, the probability is 0.82. To find the probability that both children receive ab from the man (and ab from the woman, which is certain), we multiply the probability of the first child getting ab with the probability of the second child getting ab.

Probability for the first child = 0.82 Probability for the second child = 0.82

The combined probability for both events is: 0.82 (for the first child) x 0.82 (for the second child) = 0.82^2

0.82^2 = 0.6724 or 67.24%

The probability that their first two children will both have the genotype ab/ab is 67.24%.