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Gene Pool • The total collection of genes in a population at any given time. • Consists of all the alleles in all of the individuals of the population. • Each allele has a frequency in the population, notated by a p or a q • p represents the frequency of the dominant allele. • q represents the frequency of the recessive allele. Gene Pool Gene Pool • 2 alleles for color • p = dominant = black allele • q = recessive = white allele • Total alleles = 20 • p + q = 20 • p = 12 •q=8 Hardy-Weinberg Equilibrium • 1. 2. 3. 4. 5. Requirements: Large population. No gene flow (immigration or emigration). No mutations. Mating is completely random. No natural selection. The Hardy-Weinberg Equation • Uses • To calculate genotype and allele frequencies in a population. • To determine if a population is in Hardy-Weinberg equilibrium. • To estimate what % of a population is carrying the allele for a recessively inherited disease. The Hardy-Weinberg Equation p2 + 2pq + q2 = 1 where… p = frequency of the dominant allele q = frequency of the recessive allele p2 = frequency of the homozygous dominant genotype (AA) 2pq = frequency of the heterozygous genotype (Aa) q2 = frequency of the homozygous recessive genotype (aa) Example 1: As a class • The incomplete dominant trait for flower color is controlled by 2 alleles: • R = red • r = white • In a population of 500 flowers… • 320 are red (RR), 160 are pink (Rr), and 20 are white (rr) • Since each flower has two alleles for the trait, there are 1000 alleles in the population of flowers. • Find p and q. Example 2: Your Turn • What are the allele and genotypic frequencies for a population of 500 mice in which 245 are black (BB), 210 are brown (Bb), and 45 are white (bb)? Example 3: Slightly more challenging • What are the allele and genotypic frequencies for a population of 1000 pea plants in which 750 have blue flowers (B) and 250 have white flowers (b)? • • • • • f(bb) = 250/1000 = 0.25 = q2 f(b) = q2 = 0.25 = 0.5 Recall p + q = 1, f(B) = 1 – q = 1 – 0.5 = 0.5 f(BB) = p2 = (0.5)2 = 0.25 f(Bb) = 2pq = 2(0.5)(0.5) = 0.5