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Question

A population of 200 individuals is in Hardy Weinberg equilibrium for a gene with only two alleles - B and b. If the gene frequency of an allele B is 0.8, the genotype frequency of Bb is 
  1.   0.64
  2. 0.4
  3. 0.04
  4. 0.32


Solution

The correct option is D 0.32
Frequnce of allele B = p = 0.8
Frequence of allele b = q =?
We know that, p + q = 1
q = 1 - 0.8 = 0.2 

According to Hardy Weinberg principle, 
p2+2pq+q2=1, where
p2 Frequency of BB 
q2 Frequency of bb
2pq Frequency of Bb
2pq=2×0.8×0.2
        = 0.32 

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