Question

If the a and b loci are 20 m.u. apart in humans and an A B/ a b woman mates with an a b/ a b man, what is the probability that their first child will be A b/ a b?

Answer 1

Answer:

The probability of having a Ab/ab child is 10%

Explanation:

The genes A/a and B/b are linked and 20 m.u. apart.

The parental cross is:

♀ AB/ab   X    ♂ ab/ab

Gametes:

The man only produces 1 type of gametes, so the probability of him producing an ab gamete is 1.

The woman produces 4: two parental (AB and ab) and two recombinant (Ab, aB).

Man: ab

Woman: AB, ab, Ab, aB

The formula to relate genetic distance with recombination frequency is:  

Genetic Distance (m.u.)= Recombination Frequency X 100.

Replacing the data in the formula, we have:

20 m.u. / 100 = Recombination Frequency

0.2 = Recombination Frequency

Because the Recombination Frequency is 0.2, the woman will generate recombinant gametes 20% of times, and parental gametes the other 80%. Each recombinant gamete will appear in 10% of the cases, and each parental gamete will appear in 40% of the cases.

The probailities for each possible genotype of the progeny resulting from that cross will be:

Parental: AB/ab 40%

Parental: ab/ab 40%

Recombinant: Ab/ab 10%

Recombinant: aB/ab 10%