Chapter 6

1. Chapter 6 Selection and Mutation

5. Adding Selection to the Hardy-Weinberg Analysis • If either of the following occurs then the population is responding to selection. 1. Some phenotypes allow greater survival to reproductive age. -or- 2. All individuals reach reproductive age but some individuals are able to produce more viable (reproductively successful) offspring. If these differences are heritable then evolution may occur over time.

6. Caution • It needs to be mentioned that most phenotypes are not strictly the result of their genotypes. • Environmental plasticity and • interaction with other genes may also be involved. • In other words it is not as simple as we are making it here but we have to start somewhere.

7. We will look at two possible effects of natural selection on the gene pool. • Selection may alter allele frequencies or violate conclusion #1 • Selection may upset the relationship between allele frequencies and genotype frequencies. Conclusion #1 is not violated but conclusion #2 is violated. In other words the allele frequencies remain stable but genotype frequencies change and can no longer be predicted accurately from allele frequencies.

8. FIRST: An example of what we might normally see happen to allele frequencies when natural selection is at work

9. Let B1 and B2 = the allele frequencies of the initial population with frequencies of B1 = .6 and B2 = .4 • After random mating which produces 1000 zygotes we get:

10. Selection Example

11. Selection Example

12. Selection Example

13. Selection Example

14. Selection Example

15. Selection Example

16. Selection Example

17. Selection Example

18. Selection Example

19. Selection Example

20. Selection Example

21. Selection Example

22. Selection Example

23. Selection Example

24. Selection Example

25. Selection Example

26. Selection Example

27. Selection Example

28. Selection Example

29. Selection Example

30. Selection Example

31. Selection Example

32. Selection Example

33. Sample analysis of a population for theeffects of selection • analyze the population on the basis of the fitness of the offspring produced. • The fittest individuals will survive the selection process and leave offspring of their own. • We are going to define fitnessas the survival rates of individuals which survive to reproduce.

34. Fitness formulas MEAN FITNESS • If : w11 = fitness of allele #1 homozygote (exp B1B1) w12 = fitness of the heterozygote (exp B1B2) w22 = fitness of allele #2 homozygote exp (B2B2) mean fitness of the population will be described by the formula: ŵ = p2w11 + 2pqw12 + q2w22 CAUTION! Use ONLY allele frequencies in these formulas NOT genotype frequencies!

35. For our previous example • B1= 0.6 and B2 = 0.4 and fitness of B1B1 = 1.0 (100% survived) fitness of B1B2 = .75 ( 75% survived) fitness of B2B2 = .50 (50% survived) • Figure the mean fitness now.

36. For our previous example • B1= 0.6 and B2 = 0.4 and fitness of B1B1 = 1.0 (100% survived) fitness of B1B2 = .75 ( 75% survived) fitness of B2B2 = .50 (50% survived) • Figure the mean fitness now. • ŵ= (.6)2(1)+

37. For our previous example • B1= 0.6 and B2 = 0.4 and fitness of B1B1 = 1.0 (100% survived) fitness of B1B2 = .75 ( 75% survived) fitness of B2B2 = .50 (50% survived) • Figure the mean fitness now. • ŵ= (.6)2(1)+(2(.6)(.4)(.75)) +

38. For our previous example • B1= 0.6 and B2 = 0.4 and fitness of B1B1 = 1.0 (100% survived) fitness of B1B2 = .75 ( 75% survived) fitness of B2B2 = .50 (50% survived) • Figure the mean fitness now. ŵ= (.6)2(1)+(2(.6)(.4)(.75)) + (.4)2 (.5) = .80

39. We can also use the fitness to calculate the expected frequency of each genotype in the next generation. We can do it the long way , if we know actual numbers OR……. B1B1 = P2w11ŵ B1B2 = 2pqw12ŵ B2B2 = q2w22ŵ We can use these formulas which can calculate the new expected genotype frequencies based on the fitness of each genotype and the allele frequencies in the current generation.

40. We can also use the fitness values and current allele frequencies to calculate the expected frequency of each allele in the next generation. B1 = p2w11+pqw12B2 = pqw12+q2w22ŵ ŵ

41. Finally, we can calculate the change () in the frequency of alleles B1 or B2 directly as follows: Δ B1 = Δp = p (pw11+qw12 – ŵ)ŵ Δ B2 = Δq = q (pw12+qw22 – ŵ)ŵ

42. Let’s do a sample problem. • Go back to the problem we did in class last time. Taking this current population that you have already analyzed, figure out what the new genotype and allele frequencies will be if the fitness of these individuals is actually as follows: • SS individuals 0.8 ; Ss individuals 1.0 and the ss individuals 0.6.

43. Last time we calculated S = .82 and s = .18 Now we set the fitnesses at w11(SS)=.8;w12(Ss)=1;w22(ss)=.6 Calculate the ŵ and B1B1; B1B2; and B2B2 values for the next generation now ŵ = p2w11 + 2pqw12 + q2w22 ŵ= (.82) 2 (.8) + 2(.82)(.18)(1.0) + (.18)2 (.6) ŵ = .537 + .295 + .019 = .85 B1B1 = P2w11ŵ B1B2 = 2pqw12ŵ B2B2 = q2w22ŵ ; SS = (.82)2(.8) / .85 = .633 ;Ss = 2(.82)(.18)(1.0) / .85 = .347 ;ss = (.18)2(.6) / .85 = 0.023 Hint: Do they add up to 1.0?

44. We an also calculate the new allele frequencies as well S = (.82)2(.8) + (.82)(.18)(1.0) = .806 .85 B1 = p2w11+pqw12 ŵ s = (.82)(.18)(1.0) + (.18)2(.6) = .196 .85 B2 = pqw12+q2w22ŵ

45. So…… B1B1 = .63 B1B2 = .35 B2B2 = .02 and B1 = .80 B2 = .20 Is this population in equilibrium? Have the allele frequencies changed? Can we predict the genotype frequencies from the allelic frequencies?

46. Experimental confirmation of loss of Hardy Weinberg equilibrium

47. Fruit fly experiments of Cavener and Clegg

48. Worked with fruit flies having two versions of the ADH (alcohol dehydrogenase) enzyme, F and S. (for fast and slow moving through an electrophoresis gel) • Grew two experimental populations on food spiked with ethanol and two control populations on normal, non-spiked food. Breeders for each generation were picked at random. • Took random samples of flies every few generations and calculated the allele frequencies for AdhF and AdhS

49. Figure 6.13 pg 185

50. Can we identify which assumption is being violated? • only difference is ethanol in food no migration assured random mating population size, drift? no mutation. • Must be selection for the fast form of gene. • Indeed studies show that AdhF form breaks down alcohol at twice the rate as the AdhS form. • Therefore offspring carrying this allele are more fit and leave more offspring and the make-up of gene pool changes.