Question

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%

Answer 1

The likelihood of a man with AB/ab genotype and a woman with ab/ab genotype having an Ab/ab child is 5%, as the father produces Ab gametes due to recombination at a rate of 5%.

Step-by-step explanation:

The question asks for the likelihood of a man with AB/ab genotype and a woman with ab/ab genotype having an Ab/ab child. First, we need to understand the concept of genetic recombination and Mendelian inheritance. We know that genes A and B are 10 map units (mu) apart, which means there is a 10% chance for a crossover event between them per meiosis.

Since the father has AB/ab genotype, and considering his gametes undergo recombination, 45% of his gametes would be AB or ab (non-recombinant), and 5% would be Ab or aB (recombinant), due to the 10% recombination rate between A and B. The mother can only provide ab gametes, as she is homozygous recessive (ab/ab). Therefore, the only way to produce an Ab/ab child is if the father provides an Ab gamete (recombinant) and the mother provides an ab gamete.

Since the father produces Ab gametes 5% of the time, and the mother always provides ab gametes, the probability of having an Ab/ab child is simply 5% times 100%, which equals 5%. So, the correct answer is D. 5%.