Two linked genes A and B are separated by 40 cM A man with g

Two linked genes, (A) and (B), are separated by 40 cM. A man with genotype AB/ab marries a woman who is ab/ab. What is the probability that their first two children are AB/ab?

30%

60%

9%

2%

e.20%

Assume that investigators crossed a strain of flies carrying the dominant eye mutation Lobe on the second chromosome with a strain homozygous for the second chromosome recessive mutations smooth abdomen and straw body. The F1 Lobe females were then backcrossed with homozygous smooth abdomen, straw body males and the following phenotypes were observed. What is the order of the genes?

            smooth abdomen, straw body   820

            smooth abdomen, Lobe                   42   

            smooth abdomen                          148  

        Lobe                                            780

            Lobe, straw body                          152

straw body                                     58

a.

smooth abdomen, straw body, lobe

straw body, lobe, smooth abdomen

straw body, smooth abdomen, lobe

From the previous problem, what is the closest value to the interference?

a.

1

0.2

0.5

0

0.8

Please show how you got the answer

Solution

1). c). 9%

Cross between AB/ab and ab/ab will have the progeny with the following genotypes:

AB/ab (1/4, parental), Ab/ ab (1/4, recombinant), aB/ab (1/4, recombinant), ab/ab (1/4, parental)

Given that the genes A and B are separated by 40 map units. Means, the percent of recombinants is 40% and the parental are 60% (each 30%)

The probability that their first two children are AB/ab is 9% (= 30*30/100)