Hardy - Weinberg

Recessive vs. Dominant

             The Hardy-Weinberg equilibrium was founded on the principle of populations and transfer of genes in the entire gene pool. However this does not account for the change in population, it only qualifies the populations that have large breeding populations, have random mating, have no mutations in alleles, have no selection, and have no differential migration. This is only a theoretical idea because every population creates its own selection when mating, population sizes change, populations move from place to place, breeding sizes change regularly, and mutations occur to advance the entire population. The Hardy-Weinberg equation dismisses the effect of evolution in a population.

            In Hardy-Weinberg, the variables p and q  represent the genotypes in the gene pool, p being  the dominant allele (A) and q being the recessive allele (a). The individuals of the population are represented by  p² and q²  p² represents the homozygous dominant individual and q² represents the homozygous recessive individual. We combine p and q (2pq) to get the heterozygous frequency of the entire population. With the conditions of Hardy-Weinberg existing in the problem, we can find all frequencies of each variable.

           For example: In a population of 1000 people, 190 individuals possess the homozygous recessive trait. If Hardy-Weinberg equilibrium exists in this population, what are all the variables in the entire population?

                         In equilibrium conditions, the frequency of homo recessive individuals is found first. 

                                            So, q² = .19 because 190/1000 total individuals is .19.

                   After  q² is found, one must take the square root of this number to get the recessive allele. 

                             So, q² = .19 | q= .44 because you square .19 to get .44 (the recessive allele).

                           Once q is found then it is subtracted from one to find p (the dominant allele). 

                   1 - .44= .56 this occurs because the allele frequencies together (p and q) should make 1.

                                 After p is found,  p² can be found from this number.

      So p = .56 |  p² = .31 this number occurs because when you square p and .56 which equals .31.

             When both p and q are founded, the frequency of the heterozygotes in the population.

     2pq = 2(.56)(.44) = .49 2 represents the two different alleles, p and q, and the are multiplied to get the frequency of heterozygotes of the 1000 people.