The Hardy-Weinberg Principle states that heredity alone cannot cause changes in the frequency of alleles in a population (Hardy, 1908). The frequency of alleles that make up a population will remain constant, generation after generation, so that the population remains in equilibrium, i.e. it is not evolving, allele frequencies do not change over time. Thus, evolution may be defined as a change in allele frequencies in a population over time. In order for the genetic makeup of a population to change over time (evolve), forces (sometimes referred to as evolutionary agents) must act to disrupt the Hardy-Weinberg equilibrium.
For example, we know that two alleles exist for tongue rolling in humans, which is inherited as a dominant. Using T for the dominant allele and t for the recessive allele, we know that TT and Tt individuals can roll the tongue, while tt individuals cannot roll the tongue. We can use p to represent the frequency of the T allele and q to represent the frequency of the other allele t. If the allele for tongue rolling is present 70% of the time, then the alternate allele for non rollers must be present 30% of the time, because p + q = 1.0. Note that we are discussing the frequency of alleles in the population, not genotype frequency. So if T is present 70% of the time, the allele would be found sometimes in homozygotes (TT) and sometimes in heterozygotes (Tt). Similar logic holds true for the t allele.
The Hardy-Weinberg formula for a two allele system is as follows:
p2 + 2pq + q2 = 1.0
In this equation, p2 is the frequency of the homozygote dominant condition, 2pq is the frequency of the heterozygote condition, and q2 is the frequency of the homozygote recessive condition in the population.
The Hardy-Weinberg formula can now be used to determine the genotype frequency of homozygote dominant individuals (TT), heterozygotes (Tt) and homozygote recessive (tt) individuals according to the equation. Since p = .7, then p2 = (.7)2 = .49. Thus 49% of the individuals in the population are homozygous for tongue rolling. The number of heterozygotes is calculated with 2pq = 2(.7)(.3) = .42; or 42% of the population can roll their tongue due to the heterozygous condition. Finally, the number of non rollers is calculated with q2 = (.3)2 = .09. So 9% of the population is homozygous recessive and cannot roll their tongue. All of these frequencies must add to 1.0 (100%), so always check your work.
p2 + 2pq + q2 = 1.0
.49 + .42 + .09 = 1.0
d. Using the Hardy-Weinberg formula, calculate the frequency of the homozygote dominant, heterozygote, and homozygote recessive individuals in the population.
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