Genes A , B , and C are linked on a chromosome and found in the order A-B-C. Gen

Question

Genes A, B, and C are linked on a chromosome and found in the order A-B-C. Genes A and Brecombine with a frequency of 8%, and genes Band C recombine at a frequency of 20%. For the cross a +b +c /abc+ × abc / abc, predict the frequency of progeny. Assume interference is zero.

Predict the frequency of a+b+c progeny.

Predict the frequency of abc+progeny.

Predict the frequency of a+bc+progeny.

Predict the frequency of ab+c progeny.

Predict the frequency of a+b+c+ progeny.

Predict the frequency of abc progeny.

Predict the frequency of a+bc progen

Explanation / Answer

One linkage map unit (LMU) is 1% recombination. Thus, the linkage map distance between two genes is the percentage recombination between those genes. To solve such problems you have to compare the parental genotypes with the recombinants. Wherever recombination takes place, it will be contributed by half of the chromatids. Hence, the total frequency will be divisible by 2.

frequency of a+b+c progeny = (1 - .08) (1 - .2) = 0.92 x 0.8 = 0.736

frequency of abc+ progeny = 0.736/2 = 0.355

frequency of a+ bc+ = (.08) (1 - .2) = 0.064 = 0.064/2 = 0.032, Here there is recombination between a+ and b. Hence the recombination frequency will be multiplied with the non recombined set of alleles. Since, this will happen only half the time the total value has been divided by half.

frequency of ab+c = 0.032, as this is the same case as the above one. Recomination between a single pair of alleles.

frequency of a+b+c+ = ( 1 - 0.08) (.2) = 0.184/2 = 0.092

frequency of abc progeny = 0.092, as the recombination event is the same as the upper genotype

frequency of a+bc progeny = (0.08) (.2) = 0.008, since double recombination takes place here. Compare with the parental genotype.

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