Hardy-Weinberg Equilibrium

Allele frequencies (or percentages, if you prefer) in a population will remain in Hardy-Weinberg Equilibrium (HWE) from generation to generation if the following assumptions are met:

  1. Natural selection is not occurring
  2. Migration (Gene Flow) is not occurring
  3. Mutation is not occurring
  4. Genetic Drift is not occurring (i.e. the population remains very large)
  5. Mating occurs at random

Although these assumptions are rarely true in the natural world, they allow us to calculate an expected allele frequency. Significant differences between the observed and expected frequencies indicate that "something" (i.e. one or more of the above) is going on, and therefore tell us that "microevolution" is occurring.

In the following example, we have an organism with a gene locus that has two alleles p and q. Therefore the frequency of alleles p plus the frequency of alleles q in the population is 100%. In other words, p + q = 1. In any population of that organism, the possible genotypes are therefore pp, pq and qq.

Expected frequencies:

The expected genotype frequency of alleles p and q in a population is calculated as shown. This ought to look familiar: it's our old friend the Punnet's Square.

Expected genotype frequencies

p
q
p
p2
pq
q
pq
q2

In other words, p2 + pq + pq + q2 = 1, or 100%. The expected frequencies of the genotypes are therefore:

Genotype
Expected Frequency
pp
p2
pq
pq + pq (or 2pq)
qq
q2

Observed frequencies:

In a population of genotypes pp, pq and qq, the observed frequency of allele p equals the sum of all of the pp genotype plus half of pq genotype, The observed frequency of allele q is therefore half of pq (the q half) plus all of q2. If you know one value, you can of course just subtract it from 1 (100%) to get the value of the other.

Observed Genotype
pp
pq
qq

In other words, the observed frequency of p = 100%(pp) + 50%(pq) and q = 50%(pq) and 100%(qq)

 

Conclusion:

If observed and expected genotype frequencies are significantly different, the population is out of HWE.


Examples:

One common misconception is that dominant alleles will rise in frequency and recessive alleles will decline in frequency over time. In reality, allele frequencies will not change from one generation to the next if the assumptions listed above are not violated. A good example of this is human ABO blood type. Type O blood is recessive and is the most common.

In the hwe.xls Excel Spreadsheet, there are three examples to help make this more concrete.

Example 1: Allele A is dominant and allele a is recessive. Set the original frequencies of p (allele A) and q (allele a) at 0.6 and 0.4 in Generation 1. These are highlighted in blue. All other numbers are calculated from these two original data points. The frequency of genotype AA is determined by squaring the allele frequency A. The frequency of genotype Aa is determined by multiplying 2 times the frequency of A times the frequency of a. The frequency of aa is determined by squaring a. Try changing p and q to other values, ensuring only that p and q always equal 1. Does it make any difference in the results?

Example 2: Alleles A1 and A2 are co-dominant. In this case, A1 is at a frequency of 0.25 and A2 is at a frequency of 0.75.

Example 3: Alleles A and a are dominant and recessive. Note that allele A is at very low frequency despite being dominant. Does it increase in frequency?


Problem 1:

The second sometimes confusing thing about HWE is that after all of the examples above, you may wonder if it is possible for the observed and expected frequencies to differ. Here's an example where they do:

In a population of snails, shell color is coded for by a single gene. The alleles A1 and A2 are co-dominant. The genotype A1A1 makes an orange shell. The genotype A1A2 makes a yellow shell. The genotype A2 A2 makes a black shell. 1% of the snails are orange, 98% are yellow, and 1% of the snails are black.

Observed frequency of A1 allele = 0.01 + 0.5(.98) = 0.50 = 50%

p2 = Expected frequency of A1A1 = 0.25

2pq = Expected frequency of A1A2 = 0.5

q2 = Expected frequency of A2 A2 = 0.25

Phenotype
Genotype
Observed
Expected
Orange
A1A1
1%
25%
Yellow
A1 A2
98%
50%
Black
A2A2
1%
25%