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Brachydactyly and evolutionary change We know the gene for brachydactyl fingers is dominant to normal fingers A man named Yule suggested that short-fingered people should become more common through time Godfrey Hardy showed this inference was wrong Wilhelm Weinberg derived the same solution to the problem independently The Hardy-Weinberg Law - the most important principle in population genetics The law is divided into three parts: a set of assumptions and two major results • In an infinitely large, randomly mating population, free from mutation, migration and natural selection (note there are five assumptions here) • the frequencies of the alleles do not change and • as long as the mating is random, the genotypic frequencies will remain in the proportions p2 (frequency of AA), 2pq (frequency of Aa) and q2 (frequency of aa) where p is the frequency of A and q is the frequency of a • The sum of the genotypic frequencies should be: p2 + 2pq + q2 = 1 Figure 22.3 DERIVING THE HARDY-WEINBERG PRINCIPLE-A NUMERICAL EXAMPLE 1. Suppose that the allele frequencies in the parental generation were 0.7 and 0.3. P1 = frequency of allele A1 = 0.7 P2 = frequency of allele A2 = 0.3 Gametes from parent generation A1 A1 2. 70% of the gametes in the gene pool carry allele A1 and 30% carry allele A2 . A1 A2 A2 A1 A2 .07 x 0.3=0.21 .03 x 0.7=0.21 A2 0.7 x 0.7 = 0.49 0.21 + 0.21 = 0.42 0.3 x 0.3 = 0.09 Homozygous Heterozygous Homozygous Gametes from offspring generation 49% of the gametes are from A1A1 parents. All of these carry A1 3. Pick two gametes at random from the gene pool to form offspring. Three genotypes are possible. 4. Calculate the frequencies of these three combinations of alleles. 5. When the offspring breed, imagine that their gametes go into a gene pool. 42% of the gametes are from A1A2 parents. Half of these carry A1 and half carry A2 9% of the gametes are from A2A2 parents. All of these carry A2 P1 = frequency of allele A1 = (0.49 + 1/2(0.42)) = (0.49 + 0.21) = 0.7 P2 = frequency of allele A2 = (1/2(0.42) + 0.09) = (0.21 + 0.09) = 0.3 Genotype frequencies will be given by: p12 : 2p1p2 : p22 as long as all Hardy-Weinberg assumptions are met 6. Calculate the frequencies of the two alleles in this gene pool. BEHOLD! The allele frequencies of A1 and A2 have not changed from parent generation to offspring generation. Evolution has not occurred. Genotypic frequencies under the Hardy-Weinberg Law The Hardy-Weinberg Law indicates: • At equilibrium, genotypic frequencies depend on the frequencies of the alleles • The maximum frequency for heterozygotes is 0.5 • If allelic frequencies are between 0.33 and 0.66, the heterozygote is the most common genotype Albinism in Hopi Native Americans The incidence of albinism is remarkably common (0.0043 or 13 in every 3000 Hopis) Assuming Hardy-Weinberg equilibrium, we can calculate q as the square root of 0.0043 = 0.066 p is therefore equal to 0.934 The frequency of heterozygotes in the population is 2pq = 0.123 In other words, 1 in 8 Hopis carries the gene for albinism! Take-home Lesson: For a rare allele, heterozygotes can be relatively common Extensions of the Hardy-Weinberg Law X-linked traits (two alleles) • Females have two copies of the gene but males are hemizygous • Female genotypes should be in the familiar p2: 2pq: q2 ratio • Male genotypes should be in a p:q ratio • This result shows why X-linked recessive traits are much more frequently observed in males Extensions of the Hardy-Weinberg Law Multiple alleles • Let p, q and r represent the frequencies of the three alleles • p + q + r = 1.0 • The Hardy-Weinberg model predicts the genotype frequencies will be (p + q + r)2 • The result is a ratio of p2: 2pq: q2: 2pr: r2: 2pq Testing for Hardy-Weinberg distributions If we know the frequency of genotypes in a population, we can test to see if those frequencies conform to the expectations from the Hardy-Weinberg model Let’s reconsider the data for scarlet tiger moths • Number of BB = 452 • Number of Bb = 43 • Number of bb = 2 Now we need to determine the allele frequency of B and b Let’s determine the frequency of allele B p = f(B) = [(2 x 452) + 43]/994 = 0.953 We can determine the value of q by difference: • q = 1 - p = 1 - 0.953 = 0.047 Next we determine the expected Hardy-Weinberg frequencies of genotypes (p2:2pq: q2) • That ratio is 0.908 BB: 0.090 Bb: 0.002 bb Now we can apportion the 497 individuals to their expected genotypes • Expected BB = 0.908 x 497 = 451 • Expected Bb = 0.090 x 497 =45 • Expected bb = 0.002 x 497 = 1 Compare the observed genotypes in the field: 453, 42 and 2 These moths clearly conform to the Hardy-Weinberg expectations