How are allele frequencies actually determined? The Hardy-Weinberg equation can be used to calculate allele frequencies. For example, let's apply the equation to PKU disease and designate the PKU allele "q," and the wild-type allele "p." What is the allele frequency for the PKU allele?
The frequency of individuals who are homozygous for the PKU allele in a population is q X q = q 2 . You should recognize this as the Rule of Multiplication. You should also know that q 2 represents the PKU-affected individuals in the population because they are homozygous for q. The frequency of individuals who are homozygous for the wild-type alleles (the p allele) in a population is p X p = p 2 . The frequency of individuals who are heterozygous in a population (PKU carriers) is p X q = pq. Note, there are two ways to get this last calculation (depending on which parent donates the q allele); therefore, the Rule of Addition is also used to compute the odds of being heterozygous (pq + pq or 2pq).
In a population, the frequency of PKU individuals who are homozygous for the allele + those who are heterozygous for the allele + those who are homozygous for the wild-type allele = 1. In other words, p 2 + 2pq + q 2 = 1. This is the Hardy-Weinberg equation for genotypes. (You may also recognize this as a binomial expansion.) Before going on, be sure you understand what these terms represent.Now, let's solve our original PKU problem. Restated, what is the allele frequency for the PKU allele (q)? We know that the phenotypic frequency of PKU disease is 1:10,000 or 0.0001. Remember that this is the phenotypic frequency, but we want to determine the allele frequency for the PKU allele (q). Recognize that q 2 is the phenotypic frequency (affected individuals) in the Hardy-Weinberg equation. Therefore, the square root of the phenotypic frequency will equal the allele frequency of the PKU allele. The square root of 0.0001 is 0.01.
Because the wild-type allele (p) + the PKU allele (q) must equal one (p + q = 1), we can determine the wild-type allele frequency. 1.0 - 0.01 = 0.99 = wild-type allele frequency.
Now let's try a problem using the squirrels from Kent State, which were described earlier in this tutorial. Take out a piece of paper and work along with the following description. The key to understanding the equation is writing it out yourself.
Suppose that you have decided to visit Kent State for the weekend to see these squirrels for yourself. You want to determine the frequency of black ( B ) and gray ( b ) alleles in the squirrel population. Since it is not possible to see all of the squirrels on campus, you decide to count 100 squirrels and use this as your sample to determine the allele frequencies. By the end of the day you have scored 100 squirrels; 84 were black and 16 were gray. Next you need to use the Hardy-Weinberg equation to determine the frequency of these two alleles in this population. At first you decide that the allele frequencies are 0.84 for the B allele and 0.16 for the b allele. Is this correct? Are you overlooking something? When you see a black squirrel, what is its genotype? Black squirrels are either BB or Bb . So some black squirrels are carrying the b allele. Black is a complete dominant, therefore, there is no way to tell which squirrels are heterozygous for this allele.
Since you do not know if a black squirrel has a BB or Bb genotype, you cannot use the number of black squirrels to determine the frequency of the black allele. But what about gray squirrels? Since gray is recessive to black, all of the gray squirrels are genotype bb , or they represent the q 2 value in the p 2 + 2pq + q 2 equation. Suddenly this is looking much easier. If q 2 = 16/100 = 0.16, then q = the square root of 0.16 = 0.4. Since p + q = 1, then 1 - q = p, so 1 - 0.4 = 0.6 = p. So the frequency of the black allele ( B ) is 0.6 and the frequency of the gray allele ( b ) is 0.4. Now that you have this information, you can estimate the frequencies of the different genotypes in the population: BB , Bb and bb .