Hardy-Weinburg Equlibrium?

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Hardy-Weinburg Equlibrium?

If a population that is in Hardy-Weinburg equilibrium has 500 individuals, with allele frequencies for p and q of 0.55 and 0.45, respectively, and 30% of the heterozygotes are killed by a natural disaster, how many heterozygotes are left? Will the population stay in genetic equilibrium?

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Answer 1 · 2018-02-28T14:28:44

Heterozygotes left: $0.346$ $0.346$ or $0.3465$ $0.3465$ (depending on if you need to round.)
Genetic Equilibrium: No

Explanation:

Population: $500$ $500$ individuals

$p = 0.55$ $p = 0.55$

$q = 0.45$ $q = 0.45$

To find the heterozygotes of the individual, you have to multiply.

$2 p q = 2 \times (0.55) \times (0.45) = 0.495$ $2pq = 2 xx (0.55) xx (0.45) = 0.495$

Now, $30 %$ $30%$ of $0.495$ $0.495$

$0.495 \times .30 = 0.1485$ $0.495 xx .30 = 0.1485$

Heterozygotes left:

$0.495 - 0.1485 = 0.3465$ $0.495 - 0.1485 = 0.3465$

or the rounded answer...

$0.495 - 0.149 = 0.346$ $0.495 - 0.149 = 0.346$

For genetic equilibrium to occur p and q must equal each other without any factors of mutation, random mating, migration, genetic drift, or for a species to have an infinitely large population.

For this problem, since some of the species died off; No, this species will not stay in genetic equilibrium.

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Hardy-Weinburg Equlibrium?

If a population that is in Hardy-Weinburg equilibrium has 500 individuals, with allele frequencies for p and q of 0.55 and 0.45, respectively, and 30% of the heterozygotes are killed by a natural disaster, how many heterozygotes are left? Will the population stay in genetic equilibrium?

1 Answer

Explanation:

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