Hardy-Weinberg Lab Data

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Presentation transcript:

Hardy-Weinberg Lab Data Mutation Gene Flow Genetic Drift Selection Non-random mating 2006-2007

Hardy Weinberg Lab: Equilibrium Original population Case #1 F5 18 individuals 36 alleles p (A): 0.5 q (a): 0.5 total alleles = 36 p (A): (4+4+7)/36 = .42 q (a): (7+7+7)/36 = .58 AA 4 Aa 7 aa 7 AA .25 Aa .50 aa .25 AA .22 Aa .39 aa .39 How do you explain these data?

Hardy Weinberg Lab: Selection Original population Case #2 F5 15 individuals 30 alleles p (A): 0.5 q (a): 0.5 total alleles = 30 p (A): (9+9+6)/30 = .80 q (a): (0+0+6)/30 = .20 AA 9 Aa 6 aa AA .25 Aa .50 aa .25 AA .60 Aa .40 aa How do you explain these data?

Hardy Weinberg Lab: Heterozygote Advantage Original population Case #3 F5 15 individuals 30 alleles p (A): 0.5 q (a): 0.5 total alleles = 30 p (A): (4+4+11)/30 = .63 q (a): (0+0+11)/30 = .37 AA 4 Aa 11 aa AA .25 Aa .50 aa .25 AA .27 Aa .73 aa How do you explain these data?

Hardy Weinberg Lab: Heterozygote Advantage Original population Case #3 F10 15 individuals 30 alleles p (A): 0.5 q (a): 0.5 total alleles = 30 p (A): (6+6+9)/30 = .70 q (a): (0+0+9)/30 = .30 AA 6 Aa 9 aa AA .25 Aa .50 aa .25 AA .4 Aa .6 aa How do you explain these data?

Hardy Weinberg Lab: Genetic Drift Original population Case #4 F5-1 6 individuals 12 alleles p (A): 0.5 q (a): 0.5 total alleles = 12 p (A): (4+4+2)/12 = .83 q (a): (0+0+2)/12 = .17 AA 4 Aa 2 aa AA .25 Aa .50 aa .25 AA .67 Aa .33 aa How do you explain these data?

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

Hardy Weinberg Lab: Genetic Drift Original population Case #4 F5-3 5 individuals 10 alleles p (A): 0.5 q (a): 0.5 total alleles = 10 p (A): (2+2+2)/10 = .6 q (a): (1+1+2)/10 = .4 AA 2 Aa 2 aa 1 AA .25 Aa .50 aa .25 AA .4 Aa .4 aa .2 How do you explain these data?

Hardy Weinberg Lab: Genetic Drift Original population Case #4 F5 5 individuals 10 alleles p (A): 0.5 q (a): 0.5 AA Aa aa p q 1 .67 .33 0 .83 .17 2 0 .8 .2 .4 .6 3 4 .4 .2 .6 .4 AA .25 Aa .50 aa .25 How do you explain these data?

Any Questions?? 2007-2008