The allele frequencies are the proportions of each allele among all alleles at a given locus. In this problem, the population is 300 individuals: 7 + 106 + 187 = 300. Since each individual is expected to have two alleles, the total number of alleles is 600. To find the...
The allele frequencies are the proportions of each allele among all alleles at a given locus. In this problem, the population is 300 individuals: 7 + 106 + 187 = 300. Since each individual is expected to have two alleles, the total number of alleles is 600. To find the allele frequencies, we need to find the number of each allele.
We are told 7 individuals have two copies of allele E (genotype EE) and 106 have a single copy (genotype ET). The total number of copies of E is thus 2*7 + 106 = 120. The allele frequency for E is this number divided by the total number of alleles—that is, 120/600, or 0.20 (20%). Similarly, 187 individuals have two copies of T, while 106 (the heterozygotes) have only one. The total number of T alleles is 2*187 + 106 = 480. The allele frequency for T is the fraction of the total number of alleles that are T, or 480/600 = 0.80 (80%).
So the allele frequencies are 0.20 and 0.80 for E and T, respectively. Let’s call these p and q. The expected numbers of each genotype are based on probabilities of each individual's receiving each combination of alleles. In a system of random mating, the probability of getting EE is the product of the independent probabilities of getting one copy of E from each parent. In random mating, the probability of each allele’s getting passed to an offspring is equal to its frequency in the population, which, for E, is 0.20. The probability of inheriting E from both parents is thus p^2 = 0.20*0.20 = 0.040. Similarly, the probability of inheriting T from both parents is q^2 = 0.80 * 0.80 = 0.64. This is also the proportion of individuals having the TT genotype. There are two different ways an individual can inherit ET: E from the male parent and T from the female, or the other way around. The probability of each is pq, so the overall frequency of heterozygotes in the population is 2pq = 2*0.20*0.80 = 0.32. These frequencies should add up to 1. We have 0.04 + 0.64 + 0.32 = 1.00.
If the population remains constant at 300 individuals and random mating is assumed, the population is expected to stabilize at 0.04*300 = 12 having genotype EE, 0.64*300 = 192 with genotype TT, and 0.32*300 = 96 with genotype ET. Notice that the initial sample did not contain this random distribution of alleles.