Two gene in corn, c and d are normally 30 map units apart on chromosome 9. A str

Question

Two gene in corn, c and d are normally 30 map units apart on chromosome 9. A strain is found that has a paracentric inversion that occupies about a third of the region between c and d. The inversion does not include either locus. If and individual, which is heterozygous for both the the inversion and the two genes, is testcrossed, what would be the predicted recombination frequency for c and d. ( the answer is 20% how do you arive to that answer).

Two genes in corn, a and b are normally 10 map units apart on chromosome 10. A strain is found that has a paracentric inversion that occupies about half the region between a and b. The inversion does not include either locus. If an individual, which is heterozygous for both the inversion and the two genes, is testcrossed, what would be the predicted recombination frequency for a and b ( answer is 5% how doyou arive to that answer)

Explanation / Answer

Distance between c and d = 30 map units

Strain for paracentric inversion occurs in = 1/3 * 30 = 10 map units

So total map distance between c and d = 30 map units + 20 map units ( not used by strain) = 50 map units

Recombination frequency = (10 / 50 * 100 ) = 20%

Distance between a and b = 10 map units

Strain for paracentric inversion occurs in = 1/2 * 10 = 5 map units

So total map distance between cand d = [ 10 map units + 5 map units ( not used by strain) ] * 6 + 10 = 100 map units because it present on chromosome 10

Recombination frequency = (5 / 100 * 100 ) = 5%

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