**Learning Objectives**

- Know and recognize the five assumptions of the Hardy-Weinberg principle
- Use the gene pool concept and the Hardy-Weinberg principle to determine whether a population is evolving at a locus of interest

**Measuring Evolutionary Change: the Hardy-Weinberg Equilibrium Principle**

How would a researcher know if selection or drift or even mutation were altering the allele frequencies for population? In other words, can we use the mechanisms of to detect evolution happening in real populations? To do that we’d need a null expectation or a baseline against which to measure change. We call that baseline the **Hardy-Weinberg equilibrium** (HWE). To calculate what the alleles frequencies (p and q in the example below) should be in the absence of any evolution, we need to assume that the population is undergoing no selection, no mutation, no drift, no gene flow, and that individuals are selecting mates at random.

Also recall that each individual is a diploid, carrying two copies (alleles) of each gene. Assume that the entire population only has two variants, or alleles, for a gene for pea color. Individuals that carry at least one *Y* allele have yellow coloration, while those who carry two copies of the *y* allele are green. If the frequency of the *y* allele is 0.1 = q, then the frequency of the normal allele is p = 1 – q = 0.9. Hardy-Weinberg equilibrium assumes those frequencies will not change from one generation to the next. To show that mathematically, we need to count the alleles in each generation. For instance, we can start by saying that if q = 0.1, then the green pea plants, who have two copies of *y*, have a genotype frequency of q^2 = 0.01. Likewise the yellow homozygotes have a frequency of p^2 = 0.81. But there’s a third type, the heterozygote, that has one copy of each. We can simply subtract the frequencies to see what the proportion of heterozygotes is: 1 – 0.81 – 0.01 = 0.18. Hardy-Weinberg gives us the simple equation to figure out where the alleles are, assuming no evolution: p^2 + 2pq + q^2 = 1. So, 0.18 = 2pq. Why is that? A heterozygote can inherit it’s green color *y* allele from either parent, mom or dad. Because there are two ways to arrive at the probability of a heterozygous state (pq), we need to add those two options together, and pq + pq = 2pq.

Below is a Crash Course Biology video on Population Genetics that explains Hardy-Weinberg equilibrium dynamically…using ear wax phenotype in humans.

Recommended Readings

Grant and Grant. 2002. Unpredictable Evolution in a 30-Year Study of Darwin’s Finches. Science 296: 707-711.

Hank Green’s videos are always entertaining, but the reason he gives for squaring the equation: p + q = 1 makes no sense to me. Instead, random mating of sexually reproducing individuals should produce homozygotes at frequencies of p^2 and q^2, and heterozygotes at frequencies of 2pq. The sum of these three genotype frequencies should be 1, as shown in the figure table on Hardy-Weinberg analysis.