Homework Answers
Answer:
c. Y---2---X----38----Z
Explanation:
Hint: Non-recombinant (parental) combinations are produced more in number than recombinants. The parental (non-recombinant) genotypes is XY+/++Z
1).
If single crossover occurs between X&Y
Normal combination: XY/++
After crossover: X+/+Y
X+ progeny=10
+Y progeny = 0
Total this progeny = 10
Total progeny = 500
The recombination frequency between X&Y = (number of recombinants/Total progeny) 100
RF = (10/500)100 = 2%
2).
If single crossover occurs between Y&Z
Normal combination: Y+/+Z
After crossover: YZ/++
YZ progeny= 100
++ progeny = 90+10=100
Total this progeny = 200
The recombination frequency between Y&Z = (number of recombinants/Total progeny) 100
RF = (200/500)100 = 40%
3).
If single crossover occurs between X&Z
Normal combination: X+/+Z
After crossover: XZ/++
XZ progeny=100
++ progeny = 90
Total this progeny = 190
Total progeny = 500
The recombination frequency between X&Z= (number of recombinants/Total progeny) 100
RF = (190/500)100 = 38%
Recombination frequency (%) = Distance between the genes (mu)
Y--------2mu------X----38mu----Z
The order of genes = Y---X---Z
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