- 202 Views
- Uploaded on
- Presentation posted in: General

Hardy-Weinberg equilibrium

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

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 = 20:20:10

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 0.4AA + 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 0.4AA + 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 0.4AA + 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 0.4AA + 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 0.4AA + 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 = 1this is also (p + q)2

AAAA

Aap, qAa

aa aa

parental generation gamete offspring (F1)

genotypes frequenciesgenotypes

reproduction

Hardy-Weinberg (single generation)

observedallele expected

genotypes frequencies genotypes

AAAA

Aap, qAa

aa aa

calculateddeduced

parental generation gamete offspring (F1)

genotypes frequenciesgenotypes

reproduction

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

observed expected

AAAA

Aap, qAa

aa aa

calculateddeduced

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 observedexpected

AA 18

Aa 90

aa 42

Total N 150

Chi-squares, again!

p = (2*18 + 90)/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

Chi-squares, again!

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

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

2pq = 0.42*0.58 = 0.487

genotype observedexpected

AA 18

Aa 90

aa 42

Total N150

Chi-squares, again!

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

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

2pq = 0.42*0.58 = 0.487

genotype observedexpected

AA 18 26.5

Aa 90 73.1

aa 42 50.5

Total N150 150

Multiply by N to obtain frequency

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

Chi-squares, again!

Chi square = Χ2 = (observed – expected)2

expected

dev. from exp. (obs-exp)2

genotype observedexpected (obs-exp) exp

AA 18 26.5-8.52.70

Aa 90 73.1 16.93.92

aa 42 50.5 -8.51.42

Total N150 150

Chi-squares, again!

Chi square = Χ2 = (observed – expected)2

expected

dev. from exp. (obs-exp)2

genotype observedexpected (obs-exp) exp

AA 18 26.5-8.52.70

Aa 90 73.1 16.93.92

aa 42 50.5 -8.51.42

Total N150 150 sum = 8.04 = X2

degrees of freedom = 2 (= N – 1)

*

p <0.05 (observed data significantly different)

from expected data at 0.05 level)

df = 2

X2 = 8.04