Posted by Reed A. Cartwright on April 8, 2004 08:00 PM

In the first installment of EvoMath, I derived the Hardy-Weinberg Principle and discussed its significance to biology. In the second installment I will demonstrate how to test if a population deviates from Hardy-Weinberg equilibrium.

Recap:

A population is considered to be in Hardy-Weinberg equilibrium if the allele and genotype frequencies are as follows.

GenotypeFrequency
AAP = P^\prime = p^2
AaH = H^\prime =   2pq
aaQ = Q^\prime = q^2
AlleleFrequency
Ap = P + \frac{H}{2}
a q = 1-p = Q + \frac{H}{2}

evomath1_chart2.png

Test Procedure:

A goodness-of-fit test can be used to determine if a population is significantly different from the expections of Hardy-Weinberg equilibrium. If we have a series of genotype counts from a population, then we can compare these counts to the ones predicted by the Hardy-Weinberg model. We conclude that the population is not in Hardy-Weinberg equilibrium if the probability that the counts were drawn under the Hardy-Weinberg model is too small for the deviations to be considered due to random chance. The significance level that is typically used is \alpha = 0.05, i.e. the genotype counts have less than a one in twenty chance of being caused by a population in Hardy-Weinberg equilibrium.

In order to calculate this probability, we will use a test statistic, \chi^2, which was devised in 1900 by Karl Pearson and has a well characterized distribution. If O_i are the set of observed counts, and E_i are the set of expected counts, then

\chi^2 = \sum_i{\frac{ \left( O_i-E_i \right)^2 }{ E_i }}.

This test statistic has a "chi-square" distribution with \nu degrees of freedom. Since we are testing Hardy-Weinberg equilibrium with two alleles, \nu=1 (rationale not shown). Furthermore, it can be shown that if \chi^2 \ge 3.841 then \Pr{ \left\{ \chi^2 \right\} } \le 0.05. Therefore, if \chi^2 \ge 3.841 we will reject the null model and conclude that there is significant statistical support that the population is not in Hardy-Weinberg equilibrium.

Example 1:

Consider the following samples from a population.

GenotypeCount
AA30
Aa55
aa15
AlleleFrequency
A0.575
a0.425

Calculate the \chi^2 value.

GenotypeObservedExpected(O-E)2/E
AA30330.27
Aa55490.73
aa25180.50
Total1001001.50

Since \chi^2 = 1.50 < 3.841, we conclude that the genotype frequencies in this population are not significantly different than what would be expected if the population is in Hardy-Weinberg equilibrium.

Example 2:

Race and Sanger (1975) determined the blood groups of 1000 Britons as follows (from Hartl and Clarke 1997).

GenotypeObservedExpected
MM298294.3
MN489496.4
NN213209.3

This results in \chi^2 = 0.222 < 3.841. As in the previous example, the measured genotype frequencies are not significantly different from the expectations of Hardy-Weinberg equilibrium.

Example 3:

Matthijis et al. (1998) surveyed a group of 54 people suffering from Jaeken syndrome (from Freeman and Herron 2004).

GenotypeObservedExpected
OO1119.44
OR4325.92
RR08.64

This results in \chi^2 = 23.56 > 3.841. Unlike the previous two examples, the measured genotype frequencies are significantly different from the expectations of Hardy-Weinberg equilibrium. This indicates that one or more of the Hardy-Weinberg conditions are being violated; although, it does not tell us which ones.

Conclusion:

Although to derive the Hardy-Weinberg principle, we assumed that the size of the population was infinite, these statistical tests demonstrate that finite populations can approximately exist in Hardy-Weinberg equilibrium.

References:

  • Freeman S and Herron JC (2004) Evolutionary Analysis 3rd ed. Pearson Education, Inc (Upper Saddle River, NJ)
  • Hartl DL and Clarke AG (1997) Principles of Population Genetics 3rd ed. Sinauer Associates, Inc (Sutherland, MA)
  • Matthijis GE et al. (1998) Lack of homozygotes for the most frequent disease allele in carbohydrate-deficient-glycoprotein syndrome type 1A. American Journal of Human Genetics 62: 542-550
  • Race RR and Sanger R (1975) Blood Groups in Man 6th ed. JB Lippincott, Philadelphia

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