Answer:
d. 110
Explanation:
<u>Parental cross</u>: Gh/Gh x gH/gH
<u>F1</u>: Gh/gH
<u></u>
<u>The following test cross experiment is done:</u>
Gh/gH x gh/gh
<u>Gametes produced by these individuals:</u>
- gh/gh: gh (probability of 1)
- Gh/gH:
- Gh (parental)
- gH (parental)
- GH (recombinant)
- gh (recombinant)
The formula that relates genetic distance with recombination frequency is:
<h3>
Genetic Distance (m.u.)= Recombination Frequency X 100</h3>
In this problem:
22 m.u. / 100 = Recombination Frequency
0.22 = Recombination Frequency
The recombination frequency altogether is 0.22, but there are 2 possible types of recombinant gametes and when one is generated, the other one is generated as well. Therefore, each recombinant gamete has a frequency of half the total recombination frequency: 0.11
In an offspring of 1000 individuals, I would expect 110 to be GH/gh.