2. AP Biology 2007-2008 Measuring Evolution of Populations

3. AP Biology 5 Agents of evolutionary change MutationGene Flow Genetic DriftSelection Non-random mating

4. AP Biology Populations & gene pools  Concepts  a population is a localized group of interbreeding individuals  gene pool is collection of alleles in the population  remember difference between alleles & genes!  allele frequency is how common is that allele in the population  how many A vs. a in whole population

5. AP Biology Evolution of populations  Evolution = change in allele frequencies in a population  hypothetical: what conditions would cause allele frequencies to not change?  non-evolving population REMOVE all agents of evolutionary change 1. very large population size (no genetic drift) 2. no migration (no gene flow in or out) 3. no mutation (no genetic change) 4. random mating (no sexual selection) 5. no natural selection (everyone is equally fit)

6. AP Biology Hardy-Weinberg equilibrium  Hypothetical, non-evolving population  preserves allele frequencies  Serves as a model (null hypothesis)  natural populations rarely in H-W equilibrium  useful model to measure if forces are acting on a population  measuring evolutionary change W. Weinberg physician G.H. Hardy mathematician

7. AP Biology Hardy-Weinberg theorem  Counting Alleles  assume 2 alleles = B, b  frequency of dominant allele (B) = p  frequency of recessive allele (b) = q  frequencies must add to 1 (100%), so: p + q = 1 bbBbBB

8. AP Biology Hardy-Weinberg theorem  Counting Individuals  frequency of homozygous dominant: p x p = p 2  frequency of homozygous recessive: q x q = q 2  frequency of heterozygotes: (p x q) + (q x p) = 2pq  frequencies of all individuals must add to 1 (100%), so: p 2 + 2pq + q 2 = 1 bbBbBB

9. AP Biology H-W formulas  Alleles:p + q = 1  Individuals:p 2 + 2pq + q 2 = 1 bbBbBB BbBbbb

10. AP Biology What are the genotype frequencies? Using Hardy-Weinberg equation q 2 (bb): 16/100 =.16 0.4 q (b): √.16 = 0.4 0.6 p (B): 1 - 0.4 = 0.6 q 2 (bb): 16/100 =.16 0.4 q (b): √.16 = 0.4 0.6 p (B): 1 - 0.4 = 0.6 population: 100 cats 84 black, 16 white How many of each genotype? population: 100 cats 84 black, 16 white How many of each genotype? bbBbBB p 2 =.36 2pq=.48 q 2 =.16 Must assume population is in H-W equilibrium!

11. AP Biology Using Hardy-Weinberg equation bbBbBB p 2 =.36 2pq=.48 q 2 =.16 Assuming H-W equilibrium Sampled data bbBbBB p 2 =.74 2pq=.10 q 2 =.16 How do you explain the data? p 2 =.20 2pq=.64 q 2 =.16 How do you explain the data? Null hypothesis

12. AP Biology Application of H-W principle  Sickle cell anemia  inherit a mutation in gene coding for hemoglobin  oxygen-carrying blood protein  recessive allele = H s H s  normal allele = H b  low oxygen levels causes RBC to sickle  breakdown of RBC  clogging small blood vessels  damage to organs  often lethal

13. AP Biology Sickle cell frequency  High frequency of heterozygotes  1 in 5 in Central Africans = H b H s  unusual for allele with severe detrimental effects in homozygotes  1 in 100 = H s H s  usually die before reproductive age Why is the H s allele maintained at such high levels in African populations? Suggests some selective advantage of being heterozygous…

14. AP Biology Malaria Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells 1 2 3

15. AP Biology Heterozygote Advantage  In tropical Africa, where malaria is common:  homozygous dominant (normal)  die or reduced reproduction from malaria: H b H b  homozygous recessive  die or reduced reproduction from sickle cell anemia: H s H s  heterozygote carriers are relatively free of both: H b H s  survive & reproduce more, more common in population Hypothesis: In malaria-infected cells, the O 2 level is lowered enough to cause sickling which kills the cell & destroys the parasite. Hypothesis: In malaria-infected cells, the O 2 level is lowered enough to cause sickling which kills the cell & destroys the parasite. Frequency of sickle cell allele & distribution of malaria

16. AP Biology 2005-2006 Any Questions??

17. AP Biology 2006-2007 Hardy-Weinberg Lab Data Mutation Gene Flow Genetic DriftSelection Non-random mating

18. AP Biology Hardy Weinberg Lab: Equilibrium total alleles = 36.42 p (A): (4+4+7)/36 =.42.58 q (a): (7+7+7)/36 =.58 total alleles = 36.42 p (A): (4+4+7)/36 =.42.58 q (a): (7+7+7)/36 =.58 18 individuals 36 alleles 0.5 p (A):0.5 0.5 q (a):0.5 18 individuals 36 alleles 0.5 p (A):0.5 0.5 q (a):0.5 Original population AA4 Aa7 aa7 How do you explain these data? Case #1 F5 AA.25 Aa.50 aa.25 AA.22 Aa.39 aa.39

19. AP Biology Hardy Weinberg Lab: Selection total alleles = 30.80 p (A): (9+9+6)/30 =.80.20 q (a): (0+0+6)/30 =.20 total alleles = 30.80 p (A): (9+9+6)/30 =.80.20 q (a): (0+0+6)/30 =.20 15 individuals 30 alleles 0.5 p (A):0.5 0.5 q (a):0.5 15 individuals 30 alleles 0.5 p (A):0.5 0.5 q (a):0.5 Original population AA9 Aa6 aa0 How do you explain these data? Case #2 F5 AA.25 Aa.50 aa.25 AA.60 Aa.40 aa0

20. AP Biology Hardy Weinberg Lab: total alleles = 30.63 p (A): (4+4+11)/30 =.63.37 q (a): (0+0+11)/30 =.37 total alleles = 30.63 p (A): (4+4+11)/30 =.63.37 q (a): (0+0+11)/30 =.37 15 individuals 30 alleles 0.5 p (A):0.5 0.5 q (a):0.5 15 individuals 30 alleles 0.5 p (A):0.5 0.5 q (a):0.5 Original population AA4 Aa11 aa0 How do you explain these data? Case #3 F5 AA.25 Aa.50 aa.25 AA.27 Aa.73 aa0 Heterozygote Advantage

21. AP Biology Hardy Weinberg Lab: total alleles = 30.70 p (A): (6+6+9)/30 =.70.30 q ( a): (0+0+9)/30 =.30 total alleles = 30.70 p (A): (6+6+9)/30 =.70.30 q ( a): (0+0+9)/30 =.30 15 individuals 30 alleles 0.5 p (A):0.5 0.5 q (a):0.5 15 individuals 30 alleles 0.5 p (A):0.5 0.5 q (a):0.5 Original population AA6 Aa9 aa0 How do you explain these data? Case #3 F10 AA.25 Aa.50 aa.25 AA.4 Aa.6 aa0 Heterozygote Advantage

22. AP Biology Hardy Weinberg Lab: Genetic Drift total alleles = 12.83 p (A): (4+4+2)/12 =.83.17 q (a): (0+0+2)/12 =.17 total alleles = 12.83 p (A): (4+4+2)/12 =.83.17 q (a): (0+0+2)/12 =.17 6 individuals 12 alleles 0.5 p (A):0.5 0.5 q (a):0.5 6 individuals 12 alleles 0.5 p (A):0.5 0.5 q (a):0.5 Original population AA4 Aa2 aa0 How do you explain these data? Case #4 F5-1 AA.25 Aa.50 aa.25 AA.67 Aa.33 aa0

23. AP Biology Hardy Weinberg Lab: Genetic Drift total alleles = 10.4 p (A): (0+0+4)/10 =.4.6 q (a): (1+1+4)/10 =.6 total alleles = 10.4 p (A): (0+0+4)/10 =.4.6 q (a): (1+1+4)/10 =.6 5 individuals 10 alleles 0.5 p (A):0.5 0.5 q (a):0.5 5 individuals 10 alleles 0.5 p (A):0.5 0.5 q (a):0.5 Original population AA0 Aa4 aa1 How do you explain these data? Case #4 F5-2 AA.25 Aa.50 aa.25 AA0 Aa.8 aa.2

24. AP Biology Hardy Weinberg Lab: Genetic Drift total alleles = 10.6 p (A): (2+2+2)/10 =.6.4 q (a): (1+1+2)/10 =.4 total alleles = 10.6 p (A): (2+2+2)/10 =.6.4 q (a): (1+1+2)/10 =.4 5 individuals 10 alleles 0.5 p (A):0.5 0.5 q (a):0.5 5 individuals 10 alleles 0.5 p (A):0.5 0.5 q (a):0.5 Original population AA2 Aa2 aa1 How do you explain these data? Case #4 F5-3 AA.25 Aa.50 aa.25 AA.4 Aa.4 aa.2

25. AP Biology Hardy Weinberg Lab: Genetic Drift 5 individuals 10 alleles 0.5 p (A):0.5 0.5 q (a):0.5 5 individuals 10 alleles 0.5 p (A):0.5 0.5 q (a):0.5 Original population AAAaaapq.67.330.83.17 1.67.330.83.17 0.8.2.4.6 20.8.2.4.6 4.4.2.6.4 34.4.2.6.4 How do you explain these data? Case #4 F5 AA.25 Aa.50 aa.25

26. AP Biology 2007-2008 Any Questions??