Hardy-Weinberg

In 1908, Hardy and Weinberg present a law describing how the proportions of dominant and recessive genetic traits would be propagated in a large population. This establishes the mathematical basis for population genetics.

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HW Formula

Can someone break down how this formula works? I have some examples of it, but I have no idea how the math works. Thanks.

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Hardy-Weinberg equation

Hardy-Weinberg equation
p^2 + 2pq + q^2 = 1

Where p = the percent of the population with at least one of a certain allele (say, allele A),
and q = the percent of the population with at least one of the other allele (say, allele a).
p^2 = pp = the percent of the population homozygous for A (that is, genotype AA).
q^2 = qq = the percent of the population homozygous for a (that is, genotype aa).
2pq = pq + qp = the percent of the population that is heterozygous (that is, genotype Aa or aA. They are the same, hence 2pq instead of just pq).

1 on the right-hand side of the equal sign represents 100%: the sum of all the different percent-parts of the population (organisms that have AA or aa or Aa for their genotype) must equal 1, which is the same as 100%.

Example:
In a population of 100 mice under Hardy-Weinberg equilibrium and following Mendelian inheritance, there are 16 white mice. White is recessive, and black is dominant. How many heterozygotes and how many homozygous black mice are expected?

Solution:
Call the dominant black allele A and the recessive white allele a.
We know there are 16 white mice. That means that white mice constitute 16/100 = .16 = 16% of the population. This percent is q^2 from the Hardy-Weinberg equation, since white mice must be homozygous aa:

.16 = q^2
(now solve for q)
q = .4 = 40%

Thus 40 mice have at least one white allele.
This means that
1-.4 = .6
or
100% - 40% = 60%
or
100 - 40 = 60 mice have at least one black allele.
So p = .6 = 60%

Now that we know the percent values for p and q, we can find out how many heterozygotes (2pq) and how many homozygous black (p^2) mice are expected:

2pq = 2(.6)(.4) = .48 = 48% = 48 mice are heterozygotes
p^2 = (.6)(.6) = .36 = 36% = 36 mice are homozygous black

The total should add up to 100 mice (the size of the population):
16 white (which necessarily must be homozygous)
48 heterozygotes
+36 homozygous black
--------------------
100 mice

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