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How can we apply the Hardy-Weinberg principle?

Sheep If a population is in Hardy-Weinberg equilibrium, allele frequencies will not change from generation to generation. Thus, we can use the allele frequencies of one generation to calculate the expected genotype frequencies of the next generation.

Let’s say that p is the frequency of the A allele in the population. If every organism in the population mates, the chance of getting two A alleles should be p x p. In other words, p2 should be the frequency of the AA genotype in the next generation. For example, if 80 out of 100 alleles for a gene are A, then p = 0.8. If every organism in the population mates, the frequency of the AA genotype in the next generation should be p2 = 0.82 = 0.64. If the frequency of the AA genotype does not equal 0.64 in the next generation, we can conclude that the population is not in Hardy-Weinberg equilibrium anymore, and it must be evolving!

We can do the same thing with the other allele, a. Let’s say that q is the frequency of the a allele. In the same way that we calculated the expected frequency of the AA genotype, q2 should be the frequency of the aa genotype. There are two ways to be heterozygous (pq or qp), so the expected frequency of the Aa genotype is 2(p x q). These are the “expected genotype frequencies.” They are what we would expect to see in the next generation based on allele frequencies if the population is not evolving.

If we measure the real genotype frequencies in the population and they do not match the expected genotype frequencies, the population is not in Hardy-Weinberg equilibrium. One or more of the assumptions of Hardy-Weinberg equilibrium is being violated, and now we can study the population further to find out how it is evolving. The Hardy-Weinberg principle can apply to either allele or genotype frequencies. Allele and genotype frequencies do not change unless evolutionary forces act on the population.

Because all the genotype frequencies in a population must add up to 1, we can write this as p2 + 2pq + q2 = 1, where p and q are the frequencies of the two alleles. This is called the Hardy-Weinberg equation. By convention, p is the frequency of the dominant allele (A), and q is the recessive allele (a). If we only know the frequency of one allele or genotype, we can use this equation to figure out the expected frequencies of the others.

Here are some practice problems for using allele frequencies to calculate expected genotype frequencies.

If a population is in Hardy-Weinberg equilibrium and the A allele has a frequency of 0.4, what will be the frequency of heterozygotes?

  1. 0.16
  2. 0.20
  3. 0.24
  4. 0.48

If the frequency of A is p = 0.4, the frequency of a must be q = 0.6. Thus, the expected frequency of heterozygotes (Aa) will be 2 x p x q = 2 x 0.4 x 0.6 = 0.48.

If the frequency of A is p = 0.4, the frequency of a must be q = 0.6. Thus, the expected frequency of heterozygotes (Aa) will be 2 x p x q = 2 x 0.4 x 0.6 = 0.48.

If the frequency of A is p = 0.4, the frequency of a must be q = 0.6. Thus, the expected frequency of heterozygotes (Aa) will be 2 x p x q = 2 x 0.4 x 0.6 = 0.48.

If the frequency of A is p = 0.4, the frequency of a must be q = 0.6. Thus, the expected frequency of heterozygotes (Aa) will be 2 x p x q = 2 x 0.4 x 0.6 = 0.48.

If a population is in Hardy-Weinberg equilibrium and the a allele has a frequency of 0.8, what will be the frequency of homozygous dominant individuals?

  1. 0.04
  2. 0.20
  3. 0.64
  4. 0.80

If the frequency of a is q = 0.8, the frequency of A must be p = 0.2. Thus, the expected frequency of homozygous dominant individuals (AA) will be p2 = 0.22 = 0.04.

If the frequency of a is q = 0.8, the frequency of A must be p = 0.2. Thus, the expected frequency of homozygous dominant individuals (AA) will be p2 = 0.22 = 0.04.

If the frequency of a is q = 0.8, the frequency of A must be p = 0.2. Thus, the expected frequency of homozygous dominant individuals (AA) will be p2 = 0.22 = 0.04.

If the frequency of a is q = 0.8, the frequency of A must be p = 0.2. Thus, the expected frequency of homozygous dominant individuals (AA) will be p2 = 0.22 = 0.04.

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