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In a population with two alleles, B and b, the allele frequency of B=0.59. What would be the frequency of the heterozygotes if the population is in Hardy-Weinberg equilibrium?

User Admirko
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2 Answers

5 votes

Answer:

0.483

Step-by-step explanation:

User Dbagnara
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3 votes

Answer:

0.483

Step-by-step explanation:

The given population is in Hardy-Weinberg equilibrium. If the gene has two alleles, the sum total of the frequencies of these two alleles will be one.

Therefore, the total of the frequency of allele B and frequency of allele b will be 1. f(B) + f(b)=1

If the frequency of allele "B" is 0.59, then the frequency of allele "b" will be=1-0.59= 0.41

The frequency of heterozygous genotype in the population= 2pq

p= frequency of the dominant allele

q= frequency of the recessive allele

So, 2pq= 2 x 0.59 x 0.41 = 0.483

User Brroshan
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