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There are two alleles, B and b, at a gene locus in a population at Hardy-Weinberg equilibrium. The relative frequency of the dominant (B) allele is 0.7. What is the expected number of heterozygous individuals if the total population size is 500 individuals?

Respuesta :

Answer:

210

Explanation:

Frequency of dominant allele is represented by "p"

and frequency of recessive allele is represented by "q"

Given

[tex]p = 0.7[/tex]

Thus,

[tex]q = 1-p\\q = 1 - 0.7 \\q = 0.3\\[/tex]

As per Hardy Weinberg's second equation of equilibrium -

[tex]p^2 + q^ 2 + 2pq = 1\\[/tex]

Substituting the given values in above equation, we get -

[tex]0.7^ 2+ 0.3^2 + 2pq = 1\\2pq = 1- 0.49-0.09\\2pq = 0.42[/tex]

So the total number of heterozygous individual is equal to

[tex]0.42 * 500\\= 210[/tex]

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