Two genes are 22 map units apart on the same chromosome. A true-breeding GGhh individual is crossed to a true-breeding ggHH indi

Question

3 years ago

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Two genes are 22 map units apart on the same chromosome. A true-breeding GGhh individual is crossed to a true-breeding ggHH indi

vidual. The F1 offspring are mated to gghh individuals. If this testcross produces 1000 offspring, how many of them would you expect to be GgHh?
Select one:
a. 390
b. 250
c. 220
d. 110

Biology

1 answer:

Oliga [24] · Answer 1 · 2022-04-29T04:32:58+03:00

Answer:

d. 110

Explanation:

Parental cross: Gh/Gh x gH/gH

F1: Gh/gH

The following test cross experiment is done:

Gh/gH x gh/gh

Gametes produced by these individuals:

- 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.