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.
<u>The parental cross is:</u>
<h2>
♀ AB/ab X ♂ ab/ab</h2>
<u />
<u>Gametes:</u>
The man only produces 1 type of gametes, so the probability of him producing an <em>ab </em>gamete is 1.
The woman produces 4: two parental (<em>AB </em>and <em>ab</em>) and two recombinant (<em>Ab, aB</em>).
Man: ab
Woman: AB, ab, Ab, aB
<u>The formula to relate genetic distance with recombination frequency is: </u>
Genetic Distance (m.u.)= Recombination Frequency X 100.
<u>Replacing the data in the formula, we have:</u>
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%. <u>Each </u>recombinant gamete will appear in 10% of the cases, and <u>each </u>parental gamete will appear in 40% of the cases.
<u>The probailities for each possible genotype of the progeny resulting from that cross will be:</u>
Parental: AB/ab 40%
Parental: ab/ab 40%
<u>Recombinant: Ab/ab 10%</u>
Recombinant: aB/ab 10%