I was just wondering how they got this answer (they got D - 15%) 23) A plant of
ID: 150530 • Letter: I
Question
I was just wondering how they got this answer (they got D - 15%)
23) A plant of genotype A/A x b/b is crossed to a/a x B/B and an F1 is testcrossed to a/a x b/b. If the genes are linked, and 30 mu apart, the percentage of progeny with the recombinant A/a x B/a genotype will be?
A) 7.5%
B) 3.75%
C) 30%
D) 15%
E) 60%
Explanation / Answer
A/A x b/b crossed with a/a x B/B
a B
A b
A b / a B
F1 = A b / a B
F1 crossed with testcross (Linked & NON-recombination)
ab
Ab
A b / a b
aB
a B / a b
In case of non-recombination, probability of getting this genotype in progeny would be,
A b / a b = 1/2
a B / a b = 1/2
F1 crossed with testcross (Linked & Recombinant)
ab
Ab
A b / a b
aB
a B / a b
AB
A B / a b
ab
a b / a b
Highlighted gametes and their progeny are the recombinant, so the probability of getting the genotype A B / a b, among the recombinant would be 1/2. The same applies for the other recombinant, the probability of getting the genotype a b / a b, among the recombinant would be 1/2 . Here, while calculating the probability only recombinant is taken into the picture.
If the chromosomes are separated by 30 map units, then the resulting progeny are 30% recombinant and 70% parental.
Now multiply 30mu (i.e, 30% recombinant) with the probability of the recombinant A B / a b,
= 1/2 * 30%
= 0.5 * 30%
= 15%
a B
A b
A b / a B
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