Question

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) .

Answer 1

This gene is in Hardy-Weinberg equilibrium

Step-by-step 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 (
`p^(2)`)
`= 0.574`

Frequency for genotype A2A2 (
`q^(2)`)
`= 0.339`

Frequency for genotype A2A1 (
`2pq`)
`= 0.0.087`

Now,

`p^2+q^2+2pq = 1\\`

Substituting the given values in above equation, we get -

`0.574 + 0.339+0.087=1\\1=1`

Hence, this gene is in Hardy-Weinberg equilibrium