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For a gene suspected of causing hypertension in humans, you observe the following genotype frequencies: A1A1 0.574; A1A2 0.339; A2A2 0.087. Is this gene in Hardy-Weinberg equilibrium? Why or why not? (Assume that a difference of three percent or more in any of the observed versus expected frequencies is statistically significant.) See Section 23.1 (Page 458) .

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Answer:

This gene is in Hardy-Weinberg equilibrium

Explanation:

As per the second equation of Hardy-Weinberg equilibrium, sum of genotypic frequencies of all types with in a population must be equal to one

Frequency for genotype A1A1 ([tex]p^{2}[/tex]) [tex]= 0.574[/tex]

Frequency for genotype A2A2 ([tex]q^{2}[/tex]) [tex]= 0.339[/tex]

Frequency for genotype A2A1 ([tex]2pq[/tex]) [tex]= 0.0.087[/tex]

Now,

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

Substituting the given values in above equation, we get -

[tex]0.574 + 0.339+0.087=1\\1=1[/tex]

Hence, this gene is in Hardy-Weinberg equilibrium

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