Answer:
The progeny phenotypes and numbers are as follows:
+ + + 455-------- parental
a b c 470-------- parental
+ b c 35----------- recombinant
a + + 33---------- recombinant
+ + c 37------------ recombinant
a b + 35--------------- recombinant
+ b + 460---------- parental
a + c 475----------- parental
Total 2,000
+ + + 455-------- parental
a b c 470-------- parental
+ b + 460---------- parental
a + c 475----------- parental
It is two point testcross
So the parental ++ (455+460); ac (470+475)------------------++/ac
Gene order is ------------- ++/ac
+ b c 35----------- recombinant
a + + 33---------- recombinant
+ + c 37------------ recombinant
a b + 35--------------- recombinant
sum of recombinants--------------140
Frequency of recombintion= recombinants/total x100
Map distances between a-c = (140/2000) x 100 = 7 mu
a------------------7 mu -----------------c
Explanation:
