1. BIOE 109 Summer 2009 Lecture 7- Part II Selection on quantitative characters

2. What is a quantitative (continuous) character?

3. Selection on quantitative characters What is a quantitative character? quantitative characters exhibit continuous variation among individuals.

4. Selection on quantitative characters What is a quantitative character? quantitative characters exhibit continuous variation among individuals. unlike discrete characters, it is not possible to assign phenotypes to discrete groups.

5. Examples of discrete characters

6. Example of a continuous character

7. Two characteristics of quantitative traits: 1. Controlled by many genetic loci

8. Two characteristics of quantitative traits: 1. Controlled by many genetic loci 2. Exhibit variation due to both genetic and environmental effects

9. Two characteristics of quantitative traits: 1. Controlled by many genetic loci 2. Exhibit variation due to both genetic and environmental effects the genes that influence quantitative traits are now called quantitative trait loci or QTLs.

10. What are QTLs?

11. QTLs possess multiple alleles, exhibit varying degrees of dominance, and experience selection and drift.

12. What are QTLs? QTLs possess multiple alleles, exhibit varying degrees of dominance, and experience selection and drift. some QTLs exhibit stronger effects than others – these are called major effect and minor effect genes, respectively.

13. What are QTLs? QTLs possess multiple alleles, exhibit varying degrees of dominance, and experience selection and drift. some QTLs exhibit stronger effects than others – these are called major effect and minor effect genes, respectively. the number and relative contributions of major effect and minor effect genes underlies the genetic architecture of the trait.

14. Mapping QTLs is expensive, labor intensive, and fraught with statistical problems!

15. Mapping QTLs is expensive, labor intensive, and fraught with statistical problems! QTL mapping can reveal: 1.Number of loci that influence a QT 2.Magnitude of their effects on phenotype 3.Their location on genome

16. Mapping QTLs is expensive, labor intensive, and fraught with statistical problems! QTL mapping can reveal: 1.Number of loci that influence a QT 2.Magnitude of their effects on phenotype 3.Their location on genome QTL mapping CANNOT reveal: 1.Identity of loci 2.Proteins they encode

17. What is heritability?

18. heritability is the proportion of the total phenotypic variation controlled by genetic rather than environmental factors.

19. What is heritability? heritability is the proportion of the total phenotypic variation controlled by genetic rather than environmental factors.

20. The total phenotypic variance may be decomposed: V P = total phenotypic variance

21. The total phenotypic variance may be decomposed: V P = total phenotypic variance V G = total genetic variance

22. The total phenotypic variance may be decomposed: V P = total phenotypic variance V G = total genetic variance V E = environmental variance

23. The total phenotypic variance may be decomposed: V P = total phenotypic variance V G = total genetic variance V E = environmental variance V P = V G + V E

24. The total phenotypic variance may be decomposed: V P = total phenotypic variance V G = total genetic variance V E = environmental variance heritability = V G /V P (broad-sense)

25. The total genetic variance (V G ) may be decomposed:

26. V A = additive genetic variance

27. The total genetic variance (V G ) may be decomposed: V A = additive genetic variance V D = dominance genetic variance

28. The total genetic variance (V G ) may be decomposed: V A = additive genetic variance V D = dominance genetic variance V I = epistatic (interactive) genetic variance

29. The total genetic variance (V G ) may be decomposed: V A = additive genetic variance V D = dominance genetic variance V I = epistatic (interactive) genetic variance V G = V A + V D + V I

30. The total genetic variance (V G ) may be decomposed: V A = additive genetic variance V D = dominance genetic variance V I = epistatic (interactive) genetic variance heritability = h 2 = V A /V P (narrow sense)

31. Estimating heritability

32. one common approach is to compare phenotypic scores of parents and their offspring:

33. Estimating heritability one common approach is to compare phenotypic scores of parents and their offspring: Junco tarsus length (cm) CrossMidparent valueOffspring value

34. Estimating heritability one common approach is to compare phenotypic scores of parents and their offspring: Junco tarsus length (cm) CrossMidparent valueOffspring value F 1 x M 1 4.34 4.73

35. Estimating heritability one common approach is to compare phenotypic scores of parents and their offspring: Junco tarsus length (cm) CrossMidparent valueOffspring value F 1 x M 1 4.34 4.73 F 2 x M 2 5.56 5.31

36. Estimating heritability one common approach is to compare phenotypic scores of parents and their offspring: Junco tarsus length (cm) CrossMidparent valueOffspring value F 1 x M 1 4.34 4.73 F 2 x M 2 5.56 5.31 F 3 x M 3 3.88 4.02

37.  Slope = h 2 Regress offspring value on midparent value

38. Heritability estimates from other regression analyses ComparisonSlope

39. Heritability estimates from other regression analyses ComparisonSlope Midparent-offspring h 2

40. Heritability estimates from other regression analyses ComparisonSlope Midparent-offspring h 2 Parent-offspring 1/2h 2

41. Heritability estimates from other regression analyses ComparisonSlope Midparent-offspring h 2 Parent-offspring 1/2h 2 Half-sibs 1/4h 2

42. Heritability estimates from other regression analyses ComparisonSlope Midparent-offspring h 2 Parent-offspring 1/2h 2 Half-sibs 1/4h 2 First cousins 1/8h 2

43. Heritability estimates from other regression analyses ComparisonSlope Midparent-offspring h 2 Parent-offspring 1/2h 2 Half-sibs 1/4h 2 First cousins 1/8h 2 as the groups become less related, the precision of the h 2 estimate is reduced.

44. Heritabilities vary between 0 and 1

45. Cross-fostering is a common approach Heritability of beak size in song sparrows

46. Q: Why is knowing heritability important?

47. A: Because it allows us to predict a trait’s response to selection

48. Q: Why is knowing heritability important? A: Because it allows us to predict a trait’s response to selection Let S = selection differential

49. Q: Why is knowing heritability important? A: Because it allows us to predict a trait’s response to selection Let S = selection differential Let h 2 = heritability

50. Q: Why is knowing heritability important? A: Because it allows us to predict a trait’s response to selection Let S = selection differential Let h 2 = heritability Let R = response to selection

51. Q: Why is knowing heritability important? A: Because it allows us to predict a trait’s response to selection Let S = selection differential Let h 2 = heritability Let R = response to selection R = h 2 S

52. Predicting the response to selection Example: the large ground finch, Geospiza magnirostris

53. Predicting the response to selection Example: the large ground finch, Geospiza magnirostris Mean beak depth of survivors = 10.11 mm

54. Predicting the response to selection Example: the large ground finch, Geospiza magnirostris Mean beak depth of survivors = 10.11 mm Mean beak depth of initial pop = 8.82 mm

55. Predicting the response to selection Example: the large ground finch, Geospiza magnirostris Mean beak depth of survivors = 10.11 mm Mean beak depth of initial pop = 8.82 mm S = 10.11 – 8.82 = 1.29

56. Predicting the response to selection Example: the large ground finch, Geospiza magnirostris Mean beak depth of survivors = 10.11 mm Mean beak depth of initial pop = 8.82 mm S = 10.11 – 8.82 = 1.29 h 2 = 0.72

57. Predicting the response to selection Example: the large ground finch, Geospiza magnirostris Mean beak depth of survivors = 10.11 mm Mean beak depth of initial pop = 8.82 mm S = 10.11 – 8.82 = 1.29 h 2 = 0.72 R = h 2 S = (1.29)(0.72) = 0.93

58. Predicting the response to selection Example: the large ground finch, Geospiza magnirostris Mean beak depth of survivors = 10.11 mm Mean beak depth of initial pop = 8.82 mm S = 10.11 – 8.82 = 1.29 h 2 = 0.72 R = h 2 S = (1.29)(0.72) = 0.93 Beak depth next generation = 10.11 + 0.93 = 11.04 mm

59. Modes of selection on quantitative traits

60. Directional selection on oil content in corn

61. Modes of selection on quantitative traits