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BIOE 109 Summer 2009 Lecture 5- Part II selection, mutation & migration.

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Presentation on theme: "BIOE 109 Summer 2009 Lecture 5- Part II selection, mutation & migration."— Presentation transcript:

1 BIOE 109 Summer 2009 Lecture 5- Part II selection, mutation & migration

2 Directional selection always maximizes mean population fitness!

3 Natural selection and mean population fitness

4 Question: How does w change during the process of directional selection?

5 Natural selection and mean population fitness Question: How does w change during the process of directional selection? Genotype: AA Aa aa Fitness: 1 1 0.50 Here, selection is favoring a dominant allele.

6 Directional selection always maximizes mean population fitness!

7 Natural selection and mean population fitness Question: How does w change during the process of directional selection? Genotype: AA Aa aa Fitness: 0.4 0.4 1 Here, selection is favoring a recessive allele.

8 Natural selection and mean population fitness Question: How does w change during the process of balancing selection?

9 Natural selection and mean population fitness Question: How does w change during the process of balancing selection? Genotype: AA Aa aa Fitness: 1-s 1 1-t

10 Natural selection and mean population fitness Question: How does w change during the process of balancing selection? Genotype: AA Aa aa Fitness: 1-s 1 1-t ^ Results in stable equilibrium point at p = t/(s + t)

11 Natural selection and mean population fitness Question: How does w change during the process of balancing selection? Example: suppose s = 0.40 and t = 0.60

12 Natural selection and mean population fitness Question: How does w change during the process of balancing selection? Example: suppose s = 0.40 and t = 0.60 Genotype: AA Aa aa Fitness: 0.6 1 0.4

13 Natural selection and mean population fitness Question: How does w change during the process of balancing selection? Example: suppose s = 0.40 and t = 0.60 Genotype: AA Aa aa Fitness: 0.6 1 0.4 ^ Stable equilibrium point at p = t/(s + t)

14 Natural selection and mean population fitness Question: How does w change during the process of balancing selection? Example: suppose s = 0.40 and t = 0.60 Genotype: AA Aa aa Fitness: 0.6 1 0.4 ^ Stable equilibrium point at p = t/(s + t) = 0.60

15 Balancing selection maximizes mean population fitness!

16 … but is only 0.75! Balancing selection maximizes mean population fitness!

17 Conclusion: Natural selection always acts to maximize mean population fitness

18 Mutation Let p = frequency of A 1 allele

19 Mutation Let p = frequency of A 1 allele Let q = frequency of A 2 allele

20 Mutation Let p = frequency of A 1 allele Let q = frequency of A 2 allele Let p = q = 0.50

21 Mutation Let p = frequency of A allele Let q = frequency of a allele Let p = q = 0.50 u A a Let u = 1 x 10 -5 (and ignore v)

22 Mutation Let p = frequency of A allele Let q = frequency of a allele Let p = q = 0.50 u A 1 a v Let u = 1 x 10 -5 (and ignore v)

23 Mutation denoting the change in a in one generation as  q,

24 Mutation denoting the change in a in one generation as  q,  q = u x p

25 Mutation denoting the change in a in one generation as  q,  q = u x p = (1 x 10 -5 )(0.5)

26 Mutation denoting the change in a in one generation as  q,  q = u x p = (1 x 10 -5 )(0.5) = 0.000005

27 Mutation denoting the change in a in one generation as  q,  q = u x p = (1 x 10 -5 )(0.5) = 0.000005 the a allele has increased in frequency to 0.500005.

28 Mutation denoting the change in a in one generation as  q,  q = u x p = (1 x 10 -5 )(0.5) = 0.000005 the a allele has increased in frequency to 0.500005. it would take another 140,000 generations to reach 0.875

29 Mutation denoting the change in a in one generation as  q,  q = u x p = (1 x 10 -5 )(0.5) = 0.000005 the a allele has increased in frequency to 0.500005. it would take another 140,000 generations to reach 0.875 Conclusion: The rate of change due to mutation pressure is extremely small!

30 Some comments on mutation 1. Mutation is the “raw material” that fuels all evolutionary change.

31 Some comments on mutation 1. Mutation is the “raw material” that fuels all evolutionary change. 2. Mutations occur randomly!

32 Some comments on mutation 1. Mutation is the “raw material” that fuels all evolutionary change. 2. Mutations occur randomly! 3. Mutations occur too infrequently to cause significant allele frequency change.

33 Some comments on mutation 1. Mutation is the “raw material” that fuels all evolutionary change. 2. Mutations occur randomly! 3. Mutations occur too infrequently to cause significant allele frequency change. 4. Most mutations are deleterious and experience “purifying selection”.

34 Some comments on mutation 1. Mutation is the “raw material” that fuels all evolutionary change. 2. Mutations occur randomly! 3. Mutations occur too infrequently to cause significant allele frequency change. 4. Most mutations are deleterious and experience “purifying selection”. 5. A small (but unknown) proportion of mutations are beneficial and lead to adaptation.

35 Migration (gene flow)

36 gene flow is simply the movement of genes among populations.

37 Migration (gene flow) gene flow is simply the movement of genes among populations. it can occur by the movement of gametes, or by the movement (and successful breeding) of individuals.

38 Migration (gene flow) gene flow is simply the movement of genes among populations. it can occur by the movement of gametes, or by the movement (and successful breeding) of individuals. its magnitude is determined by m

39 Migration (gene flow) gene flow is simply the movement of genes among populations. it can occur by the movement of gametes, or by the movement (and successful breeding) of individuals. its magnitude is determined by m m = the proportion of genes entering a population in individuals (genes) immigrating from a different population.

40 Sewall Wright’s Continent-Island model

41 A simple model of migration

42 let p I = frequency of A 1 allele on island

43 A simple model of migration let p I = frequency of A 1 allele on island let p C = frequency of A 1 allele on continent

44 A simple model of migration let p I = frequency of A 1 allele on island let p C = frequency of A 1 allele on continent let m = proportion of A 1 alleles moving to island each generation

45 A simple model of migration let p I = frequency of A 1 allele on island let p C = frequency of A 1 allele on continent let m = proportion of A 1 alleles moving to island each generation let p’ I = frequency of A 1 allele on island in the next generation

46 A simple model of migration let p I = frequency of A 1 allele on island let p C = frequency of A 1 allele on continent let m = proportion of A 1 alleles moving to island each generation let p’ I = frequency of A 1 allele on island in the next generation p’ I = (1-m)(p I ) + m(p c )

47 A simple model of migration let p I = frequency of A 1 allele on island let p C = frequency of A 1 allele on continent let m = proportion of A 1 alleles moving to island each generation let p’ I = frequency of A 1 allele on island in the next generation p’ I = (1-m)(p I ) + m(p c )  “resident” “immigrant”

48 A simple model of migration let  p I = change in frequency of A 1 allele on island from one generation to the next

49 A simple model of migration let  p I = change in frequency of A 1 allele on island from one generation to the next  p I = p’ I - p I

50 A simple model of migration let  p I = change in frequency of A 1 allele on island from one generation to the next  p I = p’ I - p I = (1-m)(p I ) + m(p c ) - p I

51 A simple model of migration let  p I = change in frequency of A 1 allele on island from one generation to the next  p I = p’ I - p I = (1-m)(p I ) + m(p c ) - p I = m(p c – p I )

52 An example:

53 let p c = 0.75

54 An example: let p I = 0.25let p c = 0.75

55 An example: let p I = 0.25let p c = 0.75 let m = 0.10

56 An example: let p I = 0.25let p c = 0.75 let m = 0.10 let’s ignore back-migration

57 An example: let p I = 0.25let p c = 0.75 let m = 0.10  p I = m(p c – p I )

58 An example: let p I = 0.25let p c = 0.75 let m = 0.10  p I = m(p c – p I ) = 0.10(0.75-0.25)

59 An example: let p I = 0.25let p c = 0.75 let m = 0.10  p I = m(p c – p I ) = 0.10(0.75-0.25) = 0.050

60 Now: p I = 0.30p c = 0.75 let m = 0.10

61 Now: p I = 0.30p c = 0.75 let m = 0.10  p I = m(p c – p I ) = 0.10(0.75-0.30) = 0.045

62 Conclusions

63 1. Gene flow can cause rapid evolutionary change.

64 Conclusions 1. Gene flow can cause rapid evolutionary change. 2. The long-term outcome will be the elimination of genetic differences between populations!

65 Conclusions 1. Gene flow can cause rapid evolutionary change. 2. The long-term outcome will be the elimination of genetic differences between populations! But… natural selection act to oppose gene flow

66 Example: Lake Erie water snakes (Nerodia sipedon)

67


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