Population Genetics Allele Frequencies Genotype Frequencies The HardyWeinberg

Population Genetics Allele Frequencies Genotype Frequencies The Hardy-Weinberg Equation

Allele Frequency • Allele frequency = number of copies of an allele in a popuation ÷ total number of alleles in a population • Example: – A population of 100 pea plants have the following genotypes • 64 tall plants with the genotype TT • 32 tall plants with the genotype Tt • 4 dwarf plants with the genotype tt t = 32 + 2(4) ÷ 2 (64) + 2(32) + 2(4) = 40 ÷ 200 =. 2 or 20%

Genotype Frequency • Genotype frequency = number of individuals with a particular genotype in a population ÷ total number of individuals in a population – Example: tt = 4 ÷ 64 + 32 + 4 = 4 ÷ 100 =. 04, or 4%

The Hardy Weinberg Equation • Is a way to study whether allele and genotype frequencies will change over the course of many generations. • 1908 = Godfrey Harold Hardy and Wilhelm Weinberg independently discovered a mathematical expression that predicted the stability of allele and genotype frequencies from one generation to the next. • Hardy-Weinberg equilibrium = the allele and genotype frequencies do not change over the course of many generations (under a given set of conditions).

Factors that cause change in allele and genotype frequencies: • • • 1. Mutation 2. Random Genetic Drift 3. Migration 4. Natural Selection 5. Nonrandom Mating Assignment: describe each of the above factors on a sheet of paper (pg. 670).

Hardy-Weinberg Equation Example = If a gene exists as 2 different alleles (A and a), then the variable p represents the allele frequency of A and the variable q represents the allele frequency of a. The following equation shows that the allele frequency of A plus the allele frequency of a equals the 100% of the alleles for a particular gene: p+q=1

Hardy-Weinberg Equation p²+ 2 pq + q² = 1 • p² = the genotype frequency of AA • q² = the genotype frequency of aa • 2 pq = the genotype frequency of Aa

Example If: • p =. 8 (In other words, the allele frequency of A = 80% and the allele frequency of a is 20%. ) • q =. 2 Then: AA = p² = (. 8)² =. 64 The genotype frequency of AA is 64 % Aa = 2 pq = 2(. 8)(. 2) =. 32 The genotype frequency of Aa is 32 % The genotype frequency of aa is 4% aa = q² = (. 2)² =. 04

A. 8 a. 2 AA (. 8) =. 64 Aa (. 8)(. 2) =. 16 Aa (. 2) =. 4 A. 8 a. 2

The African Cheetah • This species has a relatively low level of genetic variation because the population was reduced to a small size approximately 10, 000 to 12, 000 years ago.