Question

A test cross of F1 flies +a/+b produced the following offspring :
++/ab = 9
ab/ab = 9
+b/ab = 41
+a/ab = 41
What will be the distance between the linked genes and how?
(1) 82 cM
(2) 18 cM (cis)
(3) 20 cM
(4) 18 cM (trans)

Answer 1

P: +a//+b X ab//ab
Organism +a//+b produces the following types of gametes:
Normal gametes: +a/, +b/
Gametes after crossover: ++/, ab/
Organism ab//ab produces the following types of gametes: ab/
G: +a/, +b/, ++/, ab/ X ab/
F1: +a//ab, +b//ab, ++//ab, ab//ab
Results:
++/ab = 9
ab/ab = 9
+b/ab = 41
+a/ab = 41
According to the results the crossover percentage – 18% (9+9).
The distance between the linked genes – 18 cM

Answer: (4) 18 cM (trans)