The genetic map between four loci is given: P—2cM—Q-1cM-R---6cM---S
The probability of recombination between P and S is
Select one:
a. 0.009
b. 0.012
c. 0.00012
d. 0.09
e. 0.9
Distance between P and S is 2+1+6= 9 cM.
Probability of producing recombinants= No. of recombinants/ No. of offsprings.
Also,
Genetic map distance= No. Of recombinants/Total no. of offsprings X 100
therefore from above 2 equation we can conclude that,
Genetic map distance= probability of producing recombinantsX 100
probability of producing recombinants= 9 / 100= 0.09
Answer- d. 0.09
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