1. Hardy-Weinberg equilibrium

2. 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 0.4AA + 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 0.4AA + 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 =

12. 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 = p 2 Aa = p x q x 2 = 2pq aa = q x q = q 2

13. 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 = p 2 = 0.36 Aa = p x q x 2 = 2pq = 0.48 aa = q x q = q 2 = 0.16

14. 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 = p 2 = 0.36 Aa = p x q x 2 = 2pq = 0.48 aa = q x q = q 2 = 0.16 check: sum of the three genotypes must equal 1

15. AA = p x p = p 2 = 0.36 Aa = p x q x 2 = 2pq = 0.48 aa = q x q = q 2 = 0.16 Thus, frequency of genotypes can be expressed as p 2 + 2pq + q 2 = 1this is also (p + q) 2

16. AA Aap, qAa aa parental generation gamete offspring (F1) genotypes frequenciesgenotypes reproduction

17. AA Aap, qAa aa calculated deduced observedallele expected genotypes frequencies genotypes parental generation gamete offspring (F1) genotypes frequenciesgenotypes reproduction Hardy-Weinberg (single generation)

18. AA Aap, qAa aa calculated deduced observed expected Under what conditions would genotype frequencies ever NOT be the same as predicted from the allele frequencies?

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

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

23. Chi-squares, again! genotype observedexpected AA 18 Aa 90 aa 42 Total N 150 Observed data obtained experimentally Expected data calculated based on Hardy-Weinberg equation

24. Chi-squares, again! genotype observedexpected AA 18 Aa 90 aa 42 Total N150 p = (2*18 + 90)/300 = 0.42 q = (90 + 2*42)/300 = 0.58 = # alleles in population of 150)

25. genotype observedexpected AA 18 Aa 90 aa 42 Total N150 p = (2*18 + 90)/300 = 0.42p 2 = 0.42 2 = 0.176 q = (90 + 2*42)/300 = 0.58q 2 = 0.58 2 = 0.336 2pq = 0.42*0.58 = 0.487 Chi-squares, again!

26. (note: total observed frequencies must equal total expected frequencies) Multiply by N to obtain frequency p = (2*18 + 90)/300 = 0.42p 2 = 0.42 2 = 0.176 q = (90 + 2*42)/300 = 0.58q 2 = 0.58 2 = 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 Chi-squares, again!

27. 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 = Χ 2 = (observed – expected) 2 expected 

28. 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 = X 2 Chi-squares, again! Chi square = Χ 2 = (observed – expected) 2 expected  degrees of freedom = 2 (= N – 1)

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