Modern Evolutionary Biology I. Population Genetics A. Overview Sources of Variation Agents of Change Mutation N.S. Recombination mutation (polyploidy) - crossing over - independent assortment VARIATION

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population G. Hardy and W. Weinberg 1. Definitions - Evolution: a change in the genetic structure of a population - Population: a group of interbreeding organisms that share a common gene pool; spatiotemporally and genetically defined - Gene Pool: sum total of alleles held by individuals in a population - Genetic structure: Gene array and Genotypic array - Gene/Allele Frequency: % of alleles at a locus of a particular type - Gene Array: % of all alleles at a locus: must sum to 1. - Genotypic Frequency: % of individuals with a particular genotype - Genotypic Array: % of all genotypes for loci considered; must = 1.

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population 1. Definitions 2. Basic Computations AA Aa aa Individuals 70 80 50 (200)

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population 1. Definitions 2. Basic Computations AA Aa aa Individuals 70 80 50 (200) Genotypic Array 70/200 = 0.35 80/200 = .40 50/200 = 0.25 = 1

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population 1. Definitions 2. Basic Computations AA Aa aa Individuals 70 80 50 (200) Genotypic Array 70/200 = 0.35 80/200 = .40 50/200 = 0.25 = 1 ''A' alleles 140 220/400 = 0.55

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population 1. Definitions 2. Basic Computations AA Aa aa Individuals 70 80 50 (200) Genotypic Array 70/200 = 0.35 80/200 = .40 50/200 = 0.25 = 1 ''A' alleles 140 220/400 = 0.55 'a' alleles 100 180/400 = 0.45

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population 1. Definitions 2. Basic Computations - Determining the Gene Array from the Genotypic Array a. f(A) = f(AA) + f(Aa)/2 = .35 + .4/2 = .35 + .2 = .55 b. f(a) = f(aa) + f(Aa)/2 = .25 + .4/2 = .25 + .2 = .45 KEY: The Gene Array CAN ALWAYS be computed from the genotypic array; the process just counts alleles instead of genotypes. No assumptions are made when you do this.

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population C. The Hardy-Weinberg Equilibrium Model 1. Goal: Describe what the genetic structure of the population would be if there were NO evolutionary change – if the population was in equilibrium.

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population C. The Hardy-Weinberg Equilibrium Model 1. Goal: Describe what the genetic structure of the population would be if there were NO evolutionary change – if the population was in equilibrium. For a population’s genetic structure to remain static, the following must be true: - random mating - no selection - no mutation - no migration - the population must be infinitely large

Initial genotypic freq. Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population C. The Hardy-Weinberg Equilibrium Model 2.Example: AA Aa aa Initial genotypic freq. 0.4 0.2 1.0 Gene freq. Genotypes, F1 Gene Freq's Genotypes, F2

Initial genotypic freq. Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population C. The Hardy-Weinberg Equilibrium Model 2.Example: AA Aa aa Initial genotypic freq. 0.4 0.2 1.0 Gene freq. f(A) = p = .4 + .4/2 = 0.6 f(a) = q = .2 + .4/2 = 0.4 Genotypes, F1 Gene Freq's Genotypes, F2

Initial genotypic freq. Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population C. The Hardy-Weinberg Equilibrium Model 2.Example: AA Aa aa Initial genotypic freq. 0.4 0.2 1.0 Gene freq. f(A) = p = .4 + .4/2 = 0.6 f(a) = q = .2 + .4/2 = 0.4 Genotypes, F1 p2 = .36 2pq = .48 q2 = .16 = 1.00 Gene Freq's Genotypes, F2

Initial genotypic freq. Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population C. The Hardy-Weinberg Equilibrium Model 2.Example: AA Aa aa Initial genotypic freq. 0.4 0.2 1.0 Gene freq. f(A) = p = .4 + .4/2 = 0.6 f(a) = q = .2 + .4/2 = 0.4 Genotypes, F1 p2 = .36 2pq = .48 q2 = .16 = 1.00 Gene Freq's f(A) = p = .36 + .48/2 = 0.6 f(a) = q = .16 + .48/2 = 0.4 Genotypes, F2 After one generation with these conditions, the population equilibrates

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population C. The Hardy-Weinberg Equilibrium Model 2.Example 3. Utility: If no populations meets these conditions explicitly, how can it be useful?

Initial genotypic freq. CONCLUSION:The real population is NOT in HWE. Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population C. The Hardy-Weinberg Equilibrium Model 2.Example 3. Utility: If no populations meets these conditions explicitly, how can it be useful? For comparison, like a “perfectly balanced coin” AA Aa aa Initial genotypic freq. 0.5 0.2 0.3 1.0 Gene freq. f(A) = p = .5 + .2/2 = 0.6 f(a) = q = .3 + .2/2 = 0.4 HWE expections p2 = .36 2pq = .48 q2 = .16 = 1.00 CONCLUSION:The real population is NOT in HWE.

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population C. The Hardy-Weinberg Equilibrium Model 3. Utility: - if a population is NOT in HWE, then one of the assumptions must be violated. Sources of Variation Agents of Change Mutation N.S. Recombination Drift - crossing over Migration - independent assortment Mutation Non-random Mating VARIATION So, if NO AGENTS are acting on a population, then it will be in equilibrium and WON'T change.

Modern Evolutionary Biology I. Population Genetics A. Overview B. The Genetic Structure of a Population C. The Hardy-Weinberg Equilibrium Model 3. Utility: - if a population is NOT in HWE, then one of the assumptions must be violated. -Also, If HWCE is assumed and the frequency of homozygous recessives can be measured, then the number of heterozygous carriers can be estimated. For example: If f(aa) = .01, then estimate f(a) = .1 and f(A) must be .9. f(Aa) = 2(.1)(.9) = 0.18.