Genes A B and C are linked on a chromosome and found in the

Genes A, B, and C are linked on a chromosome and found in the order A-B-C. Genes A and B recombine with a frequency of 5%, and genes B and C recombine at a frequency of 22%. 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 progeny.

-Predict the frequency of ab+c+ progeny.

Solution

Recombination frequency between A and B = 5% = 0.05

Recombination frequency between B and C = 22% = 0.22

Parental genotype: a⁺b⁺c / abc⁺ and abc/abc.

Without recombination 50% of the progenies will have a⁺b⁺c / abc and the remaining 50% progenies will have abc⁺ / abc genotypes.

A) To produce a⁺b⁺c progeny, there is no need for recombination. Because it is already present in parent genotype. So,

Probability of having a⁺b⁺c progeny = (1-recombination between A and B)(1-recombination between B and C)/2

= (1-0.05)(1-0.22)/2 = (0.95*0.78)/2 = 0.741/2 = 0.3705

Probability of having a⁺b⁺c progeny = 0.3705

B) To produce abc⁺ progeny, there is no need for recombination. Because it is already present in parent genotype. So,

Probability of having abc⁺ progeny = (1-recombination between A and B)(1-recombination between B and C)/2

= (1-0.05)(1-0.22)/2 = (0.95*0.78)/2 = 0.741/2 = 0.3705

Probability of having abc⁺ progeny = 0.3705

C) To produce a⁺bc⁺ progeny, there is should be recombination between A and B alone. So,

Probability of having a⁺bc⁺ progeny = (recombination between A and B)(1-recombination between B and C)/2

= (0.05)(1-0.22)/2 = (0.05*0.78)/2 = 0.039/2 = 0.0195

Probability of having a⁺bc⁺ progeny = 0.0195

D) To produce ab⁺c progeny, there is should be recombination between A and B alone. So,

Probability of having ab⁺c progeny = (recombination between A and B)(1-recombination between B and C)/2

= (0.05)(1-0.22)/2 = (0.05*0.78)/2 = 0.039/2 = 0.0195

Probability of having ab⁺c progeny = 0.0195

E) To produce a⁺b⁺c⁺ progeny, there is should be recombinations between B and C alone. So,

Probability of having a⁺b⁺c⁺ progeny = (1-recombination between A and B)(recombination between B and C)/2

= (1-0.05)(0.22)/2 = (0.95*0.22)/2 = 0.209/2 = 0.1045

Probability of having a⁺b⁺c⁺ progeny = 0.1045

F) To produce abc progeny, there is should be no recombinations. Because it is already present in parent. So,

Probability of having abc progeny = (1-recombination between A and B)(1-recombination between B and C)/2

= (1-0.05)(1-0.22)/2 = (0.95*0.78)/2 = 0.741/2 = 0.3705

Probability of having abc progeny = 0.3705

G) To produce a⁺bc progeny, there is should be recombinations between A and B AND B and C. So,

Probability of having a⁺bc progeny = (recombination between A and B)(recombination between B and C)/2

= (0.05)(0.22)/2 = 0.011/2 = 0.0055

Probability of having a⁺bc progeny = 0.0055

H) To produce ab⁺c⁺ progeny, there is should be recombinations between A and B AND B and C. So,

Probability of having ab⁺c⁺ progeny = (recombination between A and B)(recombination between B and C)/2

= (0.05)(0.22)/2 = 0.011/2 = 0.0055

Probability of having ab⁺c⁺ progeny = 0.0055