slide1
Download
Skip this Video
Download Presentation
Hardy-Weinberg equilibrium

Loading in 2 Seconds...

play fullscreen
1 / 29

Hardy-Weinberg equilibrium - PowerPoint PPT Presentation


  • 386 Views
  • Uploaded on

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

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Hardy-Weinberg equilibrium' - lazar


An Image/Link below is provided (as is) to download presentation

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 - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide2
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?

slide3
Quantifying genetic variation:

Genotype frequencies

red flowers: 20 = homozygotes = AA

pink flowers: 20 = heterozygotes = Aa

white flowers: 10 = homozygotes = aa

slide4
Quantifying genetic variation:

Genotype frequencies

red flowers: 20 = homozygotes = AA

pink flowers: 20 = heterozygotes = Aa

white flowers: 10 = homozygotes = aa

N =

# alleles =

Genotype frequency =

slide5
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

slide6
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 =

slide7
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)

slide8
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”

slide9
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

slide10
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

slide11
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 =

slide12
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

slide13
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

slide14
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

slide15
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

slide16
AA AA

Aa p, q Aa

aa aa

parental generation gamete offspring (F1)

genotypes frequencies genotypes

reproduction

slide17
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

slide18
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

slide19
Hardy-Weinberg equilibrium

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

slide20
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

slide22
Chi-squares, again!

Observed data obtained experimentally

Expected data calculated based on Hardy-Weinberg equation

slide23
Chi-squares, again!

Observed data obtained experimentally

Expected data calculated based on Hardy-Weinberg equation

genotype observed expected

AA 18

Aa 90

aa 42

Total N 150

slide24
Chi-squares, again!

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

slide25
Chi-squares, again!

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

slide26
Chi-squares, again!

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)

slide27
Chi-squares, again!

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

slide28
Chi-squares, again!

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)

slide29
*

p <0.05 (observed data significantly different)

from expected data at 0.05 level)

df = 2

X2 = 8.04

ad