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%.

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?

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
100% - 40% = 60%
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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