Genes A , B , and C are linked on a chromosome and found in the order A-B-C. Gen
ID: 278191 • Letter: G
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)
Explanation / Answer
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
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