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 = 20:20:10

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 = 20:20:10 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 = 20:20:10 Allelic frequencies: A = red ( ) + pink (20) = 60 (or 0.6) a = white ( ) + 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 = 20:20:10 Allelic frequencies: A = red ( ) + pink (20) = 60 (or 0.6) = “p” a = white ( ) + pink (20 ) = 40 (or 0.4) = “q”

Reduce these frequencies to proportions Genotype frequencies: AA = 20 or 0.4 Aa = aa = Allelic frequencies: A = p = 0.6 a = q = 0.4

Check that proportions sum to 1 Genotype frequencies: AA = 20 or 0.4 Aa = AA + Aa + aa = 1 aa = Allelic frequencies: A = p = 0.6 p + q = 1 a = q = 0.4

Genotype frequencies: AA = 20 or 0.4 Aa = AA + Aa + aa = 1 aa = Allelic frequencies: A = p = 0.6 p + q = 1 a = q = 0.4 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 = AA + Aa + aa = 1 aa = Allelic frequencies: A = p = 0.6 p + q = 1 a = q = 0.4 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 = 2pq aa = q x q = q 2

Genotype frequencies: AA = 20 or 0.4 Aa = AA + Aa + aa = 1 aa = Allelic frequencies: A = p = 0.6 p + q = 1 a = q = 0.4 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 = 2pq = 0.48 aa = q x q = q 2 = 0.16

Genotype frequencies: AA = 20 or 0.4 Aa = AA + Aa + aa = 1 aa = Allelic frequencies: A = p = 0.6 p + q = 1 a = q = 0.4 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 = 2pq = 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 = 2pq = 0.48 aa = q x q = q 2 = 0.16 Thus, frequency of genotypes can be expressed as p 2 + 2pq + q 2 = 1this is also (p + q) 2

AA Aap, qAa aa parental generation gamete offspring (F1) genotypes frequenciesgenotypes reproduction

AA Aap, qAa aa calculated deduced observedallele expected genotypes frequencies genotypes parental generation gamete offspring (F1) genotypes frequenciesgenotypes reproduction Hardy-Weinberg (single generation)

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

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! genotype observedexpected AA 18 Aa 90 aa 42 Total N 150 Observed data obtained experimentally Expected data calculated based on Hardy-Weinberg equation

Chi-squares, again! genotype observedexpected AA 18 Aa 90 aa 42 Total N150 p = (2* )/300 = 0.42 q = (90 + 2*42)/300 = 0.58 = # alleles in population of 150)

genotype observedexpected AA 18 Aa 90 aa 42 Total N150 p = (2* )/300 = 0.42p 2 = = q = (90 + 2*42)/300 = 0.58q 2 = = pq = 0.42*0.58 = Chi-squares, again!

(note: total observed frequencies must equal total expected frequencies) Multiply by N to obtain frequency p = (2* )/300 = 0.42p 2 = = q = (90 + 2*42)/300 = 0.58q 2 = = pq = 0.42*0.58 = genotype observedexpected AA Aa aa Total N Chi-squares, again!

dev. from exp. (obs-exp) 2 genotype observedexpected (obs-exp) exp AA Aa aa Total N Chi-squares, again! Chi square = Χ 2 = (observed – expected) 2 expected 

dev. from exp. (obs-exp) 2 genotype observedexpected (obs-exp) exp AA Aa aa Total N sum = 8.04 = X 2 Chi-squares, again! Chi square = Χ 2 = (observed – expected) 2 expected  degrees of freedom = 2 (= N – 1)

p <0.05 (observed data significantly different) from expected data at 0.05 level) df = 2 X 2 = 8.04 *