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Hardy-Weinberg equilibriumPowerPoint Presentation

Hardy-Weinberg equilibrium

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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 (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 = 20:20:10

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 + Aa + aa = 1

aa = 10 0.2

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

a = q = 0.4

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 genotype

frequencies would be predictable by multiplicative rule:

AA =

Aa =

aa =

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 genotype

frequencies would be predictable by multiplicative rule:

AA = p x p = p2

Aa = p x q x 2 = 2pq

aa = q x q = q2

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 genotype

frequencies would be predictable by multiplicative rule:

AA = p x p = p2 = 0.36

Aa = p x q x 2 = 2pq = 0.48

aa = q x q = q2 = 0.16

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 genotype

frequencies would be predictable by multiplicative rule:

AA = p x p = p2 = 0.36

Aa = p x q x 2 = 2pq = 0.48

aa = q x q = q2 = 0.16

check: sum of the three genotypes

must equal 1

AA = p x p = p2 = 0.36

Aa = p x q x 2 = 2pq = 0.48

aa = q x q = q2 = 0.16

Thus, frequency of genotypes can be expressed as

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

Aa p, q Aa

aa aa

parental generation gamete offspring (F1)

genotypes frequencies genotypes

reproduction

Hardy-Weinberg (single generation)

observed allele expected

genotypes frequencies genotypes

AA AA

Aa p, q Aa

aa aa

calculateddeduced

parental generation gamete offspring (F1)

genotypes frequencies genotypes

reproduction

Under what conditions would genotype frequencies ever NOT be the same as predicted from the allele frequencies?

observed expected

AA AA

Aa p, q Aa

aa aa

calculateddeduced

Hardy-Weinberg equilibrium the same as predicted from the allele frequencies?

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

Hardy-Weinberg equilibrium the same as predicted from the allele frequencies?

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! the same as predicted from the allele frequencies?

Observed data obtained experimentally

Expected data calculated based on Hardy-Weinberg equation

Chi-squares, again! the same as predicted from the allele frequencies?

Observed data obtained experimentally

Expected data calculated based on Hardy-Weinberg equation

genotype observed expected

AA 18

Aa 90

aa 42

Total N 150

Chi-squares, again! the same as predicted from the allele frequencies?

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

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

= # alleles in population of 150)

genotype observed expected

AA 18

Aa 90

aa 42

Total N 150

Chi-squares, again! the same as predicted from the allele frequencies?

p = (2*18 + 90)/300 = 0.42 p2= 0.422 = 0.176

q = (90 + 2*42)/300 = 0.58 q2= 0.582 = 0.336

2pq = 0.42*0.58 = 0.487

genotype observed expected

AA 18

Aa 90

aa 42

Total N 150

Chi-squares, again! the same as predicted from the allele frequencies?

p = (2*18 + 90)/300 = 0.42 p2= 0.422 = 0.176

q = (90 + 2*42)/300 = 0.58 q2= 0.582 = 0.336

2pq = 0.42*0.58 = 0.487

genotype observed expected

AA 18 26.5

Aa 90 73.1

aa 42 50.5

Total N 150 150

Multiply by N to obtain frequency

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

Chi-squares, again! the same as predicted from the allele frequencies?

Chi square = Χ2 = (observed – expected)2

expected

dev. from exp. (obs-exp)2

genotype observed expected (obs-exp) exp

AA 18 26.5 -8.5 2.70

Aa 90 73.1 16.9 3.92

aa 42 50.5 -8.5 1.42

Total N 150 150

Chi-squares, again! the same as predicted from the allele frequencies?

Chi square = Χ2 = (observed – expected)2

expected

dev. from exp. (obs-exp)2

genotype observed expected (obs-exp) exp

AA 18 26.5 -8.5 2.70

Aa 90 73.1 16.9 3.92

aa 42 50.5 -8.5 1.42

Total N 150 150 sum = 8.04 = X2

degrees of freedom = 2 (= N – 1)

* the same as predicted from the allele frequencies?

p <0.05 (observed data significantly different)

from expected data at 0.05 level)

df = 2

X2 = 8.04

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