EvoMath 2: Testing for Hardy-Weinberg Equilibrium

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:Permalink

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

Test Procedure:Permalink

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

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:Permalink

Consider the following samples from a population.

Calculate the $\chi^2$ value.

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:Permalink

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

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:Permalink

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

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:Permalink

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.

• 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

Previous Installments:Permalink

• EvoMath 0: An Introduction
• EvoMath 1: The Hardy-Weinberg Principle