Question

1.) Two linked genes, A and B, are separated by 20cM. A man with genotype AB/ab marries a woman who is ab/ab. What is the probability that their first child will be Ab/ab?

2.) An individual has genotype Ab/aB. Gene loci A and B are 20cM apart. What is the proportion of expected progeny Ab/ab if a testcross is performed on this individual?

Answer 1

1) The recombination frequency between these gene loci is 20%. The man is expected to produce a total of 20% recombinant gametes, which includes two reciprocal products (10%A/b+ 10%a/B).The mother produces only one kind of gamete (ab). For the first child to be Ab/ab, he/she must inherit a specific recombinant gamete (A/b) from the father (probability = 0.1), and any gamete (a/b) from the mother (probability = 1). So, the final probability of the first child being Ab/ab = (0.1)(1) = 0.1 or this is 10%.

2) Ab= 40%

aB= 40%

AB= 10%

ab= 10%

So, the expected progeny of Ab/ab is

(0.4)(0.1) = 0.04 = 4%.