HardyWeinberg equilibrium HardyWeinberg equilibrium Is this a true

Hardy-Weinberg equilibrium

Hardy-Weinberg equilibrium Is this a ‘true’ population or a mixture? Is the population size dangerously low? Has migration occurred recently? Is severe selection occurring?

Quantifying genetic variation: Genotype frequencies red flowers: 20 = homozygotes = AA pink flowers: 20 = heterozygotes = Aa white flowers: 10 = homozygotes = aa

Quantifying genetic variation: Genotype frequencies red flowers: 20 = homozygotes = AA pink flowers: 20 = heterozygotes = Aa white flowers: 10 = homozygotes = aa N= # alleles = Genotype frequency =

Quantifying genetic variation: Genotype frequencies red flowers: 20 = homozygotes = AA pink flowers: 20 = heterozygotes = Aa white flowers: 10 = homozygotes = aa N = 50 # alleles = 100 Genotype frequency = 20: 10

Quantifying genetic variation: Genotype frequencies red flowers: 20 = homozygotes = AA pink flowers: 20 = heterozygotes = Aa white flowers: 10 = homozygotes = aa N = 50 # alleles = 100 Genotype frequency = 20: 10 Allelic frequencies: A= a=

Quantifying genetic variation: Genotype frequencies red flowers: 20 = homozygotes = AA pink flowers: 20 = heterozygotes = Aa white flowers: 10 = homozygotes = aa N = 50 # alleles = 100 Genotype frequency = 20: 10 Allelic frequencies: A = red (20 + 20) + pink (20) = 60 (or 0. 6) a = white (10 + 10) + pink (20 ) = 40 (or 0. 4)

Quantifying genetic variation: Genotype frequencies red flowers: 20 = homozygotes = AA pink flowers: 20 = heterozygotes = Aa white flowers: 10 = homozygotes = aa N = 50 # alleles = 100 Genotype frequency = 20: 10 Allelic frequencies: A = red (20 + 20) + pink (20) = 60 (or 0. 6) = “p” a = white (10 + 10) + pink (20 ) = 40 (or 0. 4) = “q”

Reduce these frequencies to proportions Genotype frequencies: AA = 20 or 0. 4 Aa = 20 0. 4 aa = 10 0. 2 Allelic frequencies: A = p = 0. 6 a = q = 0. 4

Check that proportions sum to 1 Genotype frequencies: AA = 20 or 0. 4 Aa = 20 0. 4 aa = 10 0. 2 Allelic frequencies: A = p = 0. 6 a = q = 0. 4 AA + Aa + aa = 1 p+q=1

Genotype frequencies: AA = 20 or 0. 4 Aa = 20 0. 4 aa = 10 0. 2 Allelic frequencies: A = p = 0. 6 a = q = 0. 4 AA + Aa + aa = 1 p+q=1 If we combined the alleles at random (per Mendel), then genotype frequencies would be predictable by multiplicative rule: AA = Aa = aa =

Genotype frequencies: AA = 20 or 0. 4 Aa = 20 0. 4 aa = 10 0. 2 Allelic frequencies: A = p = 0. 6 a = q = 0. 4 AA + Aa + aa = 1 p+q=1 If we combined the alleles at random (per Mendel), then genotype frequencies would be predictable by multiplicative rule: AA = p x p = p 2 Aa = p x q x 2 = 2 pq aa = q x q = q 2

Genotype frequencies: AA = 20 or 0. 4 Aa = 20 0. 4 aa = 10 0. 2 Allelic frequencies: A = p = 0. 6 a = q = 0. 4 AA + Aa + aa = 1 p+q=1 If we combined the alleles at random (per Mendel), then genotype frequencies would be predictable by multiplicative rule: AA = p x p = p 2 = 0. 36 Aa = p x q x 2 = 2 pq = 0. 48 aa = q x q = q 2 = 0. 16

Genotype frequencies: AA = 20 or 0. 4 Aa = 20 0. 4 aa = 10 0. 2 Allelic frequencies: A = p = 0. 6 a = q = 0. 4 AA + Aa + aa = 1 p+q=1 If we combined the alleles at random (per Mendel), then genotype frequencies would be predictable by multiplicative rule: AA = p x p = p 2 = 0. 36 Aa = p x q x 2 = 2 pq = 0. 48 aa = q x q = q 2 = 0. 16 check: sum of the three genotypes must equal 1

AA = p x p = p 2 = 0. 36 Aa = p x q x 2 = 2 pq = 0. 48 aa = q x q = q 2 = 0. 16 Thus, frequency of genotypes can be expressed as p 2 + 2 pq + q 2 = 1 this is also (p + q)2

AA Aa aa parental generation genotypes p, q gamete frequencies reproduction AA Aa aa offspring (F 1) genotypes

Hardy-Weinberg (single generation) observed genotypes AA Aa aa calculated parental generation genotypes allele frequencies p, q expected genotypes deduced gamete frequencies reproduction AA Aa aa offspring (F 1) genotypes

Under what conditions would genotype frequencies ever NOT be the same as predicted from the allele frequencies? observed AA Aa aa calculated expected p, q deduced AA Aa aa

Hardy-Weinberg equilibrium Observed genotype frequencies are the same as expected frequencies if alleles combined at random

Hardy-Weinberg equilibrium Observed genotype frequencies are the same as expected frequencies if alleles combined at random Usually true if - population is effectively infinite - no selection is occurring - no mutation is occurring - no immigration/emigration is occurring

Chi-squares, again! Observed data obtained experimentally Expected data calculated based on Hardy-Weinberg equation

Chi-squares, again! Observed data obtained experimentally Expected data calculated based on Hardy-Weinberg equation genotype AA Aa aa Total N observed 18 90 42 150 expected

Chi-squares, again! p = (2*18 + 90)/300 = 0. 42 q = (90 + 2*42)/300 = 0. 58 = # alleles in population of 150) genotype AA Aa aa Total N observed 18 90 42 150 expected

Chi-squares, again! p = (2*18 + 90)/300 = 0. 42 q = (90 + 2*42)/300 = 0. 58 genotype AA Aa aa Total N observed 18 90 42 150 expected p 2 = 0. 422 = 0. 176 q 2 = 0. 582 = 0. 336 2 pq = 0. 42*0. 58 = 0. 487

Chi-squares, again! p = (2*18 + 90)/300 = 0. 42 q = (90 + 2*42)/300 = 0. 58 p 2 = 0. 422 = 0. 176 q 2 = 0. 582 = 0. 336 2 pq = 0. 42*0. 58 = 0. 487 Multiply by N to obtain frequency genotype AA Aa aa Total N observed 18 90 42 150 expected 26. 5 73. 1 50. 5 150 (note: total observed frequencies must equal total expected frequencies)

Chi-squares, again! Chi square = Χ 2 = genotype AA Aa aa Total N observed 18 90 42 150 (observed – expected)2 expected 26. 5 73. 1 50. 5 150 dev. from exp. (obs-exp) -8. 5 16. 9 -8. 5 (obs-exp)2 exp 2. 70 3. 92 1. 42

Chi-squares, again! Chi square = Χ 2 = genotype AA Aa aa Total N observed 18 90 42 150 (observed – expected)2 expected 26. 5 73. 1 50. 5 150 degrees of freedom = 2 (= N – 1) dev. from exp. (obs-exp) -8. 5 16. 9 -8. 5 (obs-exp)2 exp 2. 70 3. 92 1. 42 sum = 8. 04 = X 2

* df = 2 X 2 = 8. 04 p <0. 05 (observed data significantly different) from expected data at 0. 05 level)