Genes A and B are 10 mu apart A man has an ABab genotype His

Genes A and B are 10 mu apart. A man has an AB/ab genotype. His father was AB/AB. He marries a woman who is ab/ab.
What is the likelihood they will have an Ab/ab child?

A. 25%

B. 10%

C. 50%

D. 5%

E. 45%

Solution

Answer:D. 5%

Reason:

Given the recombination frequency ---> 10% (or 10 mu apart)

Now, a man has an AB/ab genotype marries a woman who is ab/ab, then the likelyhood of the expected recombinant gametes are “10%” in which 5% A/b & 5% a/B. The father is going to produce 0.05% probability

The women (mother) is going to generate only “ab” type gametes (1% probability) so that the first child to be Ab/ab

Now the probability of the likelihood they will have an Ab/ab child --> Ab / ab = (0.05)(1) = 0.05 (or 5%).