1 / 29

Hardy-Weinberg equilibrium

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

lazar
Download Presentation

Hardy-Weinberg equilibrium

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Hardy-Weinberg equilibrium

  2. 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?

  3. Quantifying genetic variation: Genotype frequencies red flowers: 20 = homozygotes = AA pink flowers: 20 = heterozygotes = Aa white flowers: 10 = homozygotes = aa

  4. Quantifying genetic variation: Genotype frequencies red flowers: 20 = homozygotes = AA pink flowers: 20 = heterozygotes = Aa white flowers: 10 = homozygotes = aa N = # alleles = Genotype frequency =

  5. 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

  6. 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 =

  7. 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)

  8. 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”

  9. 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

  10. 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

  11. 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 =

  12. 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

  13. 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

  14. 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

  15. 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

  16. AA AA Aa p, q Aa aa aa parental generation gamete offspring (F1) genotypes frequencies genotypes reproduction

  17. 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

  18. 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

  19. Hardy-Weinberg equilibrium Observed genotype frequencies are the same as expected frequencies if alleles combined at random

  20. 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

  21. Chi-squares, again! Observed data obtained experimentally Expected data calculated based on Hardy-Weinberg equation

  22. 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

  23. 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

  24. 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

  25. 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)

  26. 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

  27. 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)

  28. * p <0.05 (observed data significantly different) from expected data at 0.05 level) df = 2 X2 = 8.04

More Related