hardy-weinberg equilibrium. is this a ‘true’ population or a mixture? is the population size...

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 frequenciesred flowers: 20 = homozygotes = AApink flowers: 20 = heterozygotes = Aawhite flowers: 10 = homozygotes = aa



N = # alleles = Genotype frequency =



N = 50 # alleles = 100 Genotype frequency = 20:20:10




Allelic frequencies: A = a =




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




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 + Aa + aa = 1 aa = 10 0.2

Allelic frequencies: A = p = 0.6 p + q = 1 a = q = 0.4



If we combined the alleles at random (per Mendel), then genotypefrequencies would be predictable by multiplicative rule:

AA =Aa = aa =




AA = p x p = p2

Aa = p x q x 2 = 2pq aa = q x q = q2




AA = p x p = p2 = 0.36Aa = p x q x 2 = 2pq = 0.48aa = q x q = q2 = 0.16





check: sum of the three genotypes must equal 1


Thus, frequency of genotypes can be expressed as

p2 + 2pq + q2 = 1 this is also (p + q)2

AA AAAa p, q Aa aa aa

parental generation gamete offspring (F1) genotypes frequencies genotypes

reproduction


calculated deduced

observed allele expectedgenotypes frequencies genotypes

parental generation gamete offspring (F1) genotypes frequencies genotypes

reproduction

Hardy-Weinberg (single generation)


calculated deduced

observed expected

Under what conditions would genotype frequencies ever

NOT be the same as predicted from the allele frequencies?


Observed genotype frequencies are the same as expected frequencies if alleles combined at random


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 experimentallyExpected data calculated based on Hardy-Weinberg equation

Chi-squares, again!

genotype observed expected AA 18 Aa 90aa 42Total N 150

Observed data obtained experimentallyExpected data calculated based on Hardy-Weinberg equation

Chi-squares, again!


p = (2*18 + 90)/300 = 0.42q = (90 + 2*42)/300 = 0.58

= # alleles in population of 150)


p = (2*18 + 90)/300 = 0.42 p2 = 0.422 = 0.176q = (90 + 2*42)/300 = 0.58 q2 = 0.582 = 0.336

2pq = 0.42*0.58 = 0.487

Chi-squares, again!

(note: total observed frequencies must equal total expected frequencies)

Multiply by N to obtain frequency

p = (2*18 + 90)/300 = 0.42 p2 = 0.422 = 0.176q = (90 + 2*42)/300 = 0.58 q2 = 0.582 = 0.336

2pq = 0.42*0.58 = 0.487

genotype observed expected AA 18 26.5Aa 90 73.1aa 42 50.5Total N 150 150

Chi-squares, again!

dev. from exp. (obs-exp)2

genotype observed expected (obs-exp) expAA 18 26.5 -8.5 2.70Aa 90 73.1 16.9 3.92aa 42 50.5 -8.5 1.42Total N 150 150

Chi-squares, again!

Chi square = Χ2 = (observed – expected)2

expected

dev. from exp. (obs-exp)2

genotype observed expected (obs-exp) expAA 18 26.5 -8.5 2.70Aa 90 73.1 16.9 3.92aa 42 50.5 -8.5 1.42Total N 150 150 sum = 8.04 = X2

Chi-squares, again!

Chi square = Χ2 = (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 = 2X2 = 8.04

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hardy-weinberg equilibrium. is this a ‘true’ population or a mixture? is the population size...

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