Question

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 9%, and genes B and C recombine at a frequency of 26%. For the cross

a ⁺b ⁺c/abc⁺ × abc/abc,
predict the frequency of progeny. Assume interference is zero.

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

Enter your answer to four decimal places (example 0.2356 or 0.2300).

B) Predict the frequency of abc⁺ progeny.

Enter your answer to four decimal places (example 0.2356 or 0.2300)

C) Predict the frequency of a⁺ bc⁺ progeny.

Enter your answer to four decimal places (example 0.2356 or 0.2300)

D) Predict the frequency of ab⁺c progeny.

Enter your answer to four decimal places (example 0.2356 or 0.2300)

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

Enter your answer to four decimal places (example 0.2356 or 0.2300)

F) Predict the frequency of abc progeny.

Enter your answer to four decimal places (example 0.2356 or 0.2300)

G) Predict the frequency of a⁺bc progeny.

Enter your answer to four decimal places (example 0.2356 or 0.2300)

H) Predict the frequency of ab⁺c⁺ progeny.

Enter your answer to four decimal places (example 0.2356 or 0.2300)

Answer 1

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

= (1-0.09)(1-0.26)/2 = (0.91*0.74)/2 = 0.6734/2 = 0.3367

Probability of having a⁺b⁺c progeny = 0.3367

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

= (1-0.09)(1-0.26)/2 = (0.91*0.74)/2 = 0.6734/2 = 0.3367

Probability of having abc⁺ progeny = 0.3367

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

= (0.09)(1-0.26)/2 = (0.09*0.74)/2 = 0.0666/2 = 0.033

Probability of having a⁺bc⁺ progeny = 0.0333

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

= (0.09)(1-0.26)/2 = (0.09*0.74)/2 = 0.0666/2 = 0.033

Probability of having ab⁺c progeny = 0.0333

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

= (1-0.09)(0.26)/2 = (0.91*0.26)/2 = 0.2366/2 = 0.1183

Probability of having a⁺b⁺c⁺ progeny = 0.1183

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

= (1-0.09)(1-0.26)/2 = (0.91*0.74)/2 = 0.6734/2 = 0.3367

Probability of having abc progeny = 0.3367

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

= (0.09)(0.26)/2 = 0.0234/2 = 0.0117

Probability of having a⁺bc progeny = 0.0117

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

= (0.09)(0.26)/2 = 0.0234/2 = 0.0117

Probability of having ab⁺c⁺ progeny = 0.0117