In a population that is in Hardy-Weinberg equilibrium, 16% of the individuals show the recessive trait. What is the frequency of the dominant allele in the population?

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In a population that is in Hardy-Weinberg equilibrium, 16% of the individuals show the recessive trait. What is the frequency of the dominant allele in the population?

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Answer 1 · 2018-02-04T18:56:12

0.6

Explanation:

Let's start with the basic Hardy-Weinberg equations first.
$p + q = 1$ $p+q=1$ and $p^{2} + 2 p q + q^{2} = 1$ $p^2+2pq+q^2=1$
With "p" being the dominant allele and "q" being the recessive allele

We know that 16% (or 0.16) show the recessive trait. This means that the fraction of the population with the recessive trait, $q^{2}$ $q^2$ , is 0.16

With the value for $q^{2}$ $q^2$ , $q$ $q$ can be calculated. What follows is that $q = 0.4$ $q=0.4$

With this knowledge "p" can be calculated. $1 - q = p$ $1-q=p$ , which results in "p" being 0.6

This 0.6 is the frequency at which the dominant allele is present in the population

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In a population that is in Hardy-Weinberg equilibrium, 16% of the individuals show the recessive trait. What is the frequency of the dominant allele in the population?

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Explanation:

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