1. Quantitative genetics: traits controlled my many loci Key questions: what controls the rate of adaptation? Example: will organisms adapt to increasing temperatures or longer droughts fast enough to avoid extinction? What is the genetic basis of complex traits with continuous variation?

2. Quantitative genetics vs. population genetics

3. Population genetics

4. Quantitative genetics

5. Breeder’s equation Breeder’s equation: R = h 2 S

6. Polygenic Inheritance Leads to a Quantitative Trait A B C D E F LOCUS TRAIT Z 1 # of individuals Z

7. S, the selection differential

8. R, the selection response,

9. Problems in predicting the evolution of quantitative traits -Dominance -Epistasis -Environmental effects

10. Environment alters gene expression - epigenetics yellow: no difference; red or green = difference age 3age 50

11. Overall: not all variation is heritable

12. Major questions in quantitative genetics How much phenotypic variation is due to genes, and how much to the environment? How much of the genetic variation is due to genes of large effect, and how much to genes of small effect?

13. Measuring heritability Variance: _ V p = Σ i (X i – X) 2 --------------- (N – 1)

14. Variance components V P = total phenotypic variance V P = V A + V D + V E + V GXE V A = Additive genetic variance V D = Dominance genetic variance (non- additive – can include epistasis) V E = Variance among individuals experiencing different environments V GXE = Variance due to environmental variation that influences gene expression (not covered in text)

15. Heritability = h 2 = V A / V P The proportion of phenotypic variance due to additive genetic variance among individuals h 2 = V A / (V A + V D + V E + V GXE ) Heritability can be low due to:

16. Additive vs. dominance variance Additive: heterozygote is intermediate Dominance: heterozygote is closer to one homozygote (Difference from line is due to dominance)

17. Dominance and heritability Dominance GenotypeToe len (cm) AA0.5 AA’1.0 A’A’1.0 f(A) = p = 0.5 f(A’) = q = 0.5 Start in HWE f(AA) = 0.25 f(AA’) = 0.50 f(A’A’) = 0.25 Codominant (additive) GenotypeToe len (cm) AA0.5 AA’0.75 A’A’1.0 f(A) = p = 0.5 f(A’) = q = 0.5 Start in HWE f(AA) = 0.25 f(AA’) = 0.50 f(A’A’) = 0.25

18. Starting Genotype frequencies Codominant (additive) Dominance and heritability II Dominance Starting Phenotype frequencies Genotype Phenotype (toe len – cm) Mean = 0.875 cm Mean = 0.75 cm

19. Starting Genotype frequencies Additive Select toe length = 1 cm Dominance Starting Phenotype frequencies Genotype Phenotype (toe len – cm) Mean = 1.0 cm S =

20. Genotype freq after selection, before mating Codominant (additive) Effects of dominance: genotypes after random mating Dominance Genotype Genotype freq after mating Genotype

21. Effects of dominance: phenotypes after random mating Phenotype freq after selection, before mating Codominant (additive)Dominance Genotype Phenotype freq after mating Phenotype mean = 0.954 mean = 1.0 R = 954 -.875 = 0.079 R=1 -.75 = 0.25

22. Effect of dominance on heritability Dominant S = R = R = h 2 S h 2 = R / S = Codominant (additive) S = R = R = h 2 S h 2 =

23. Second problem predicting outcome of selection: epistasis Example: hair color in mammals MC1- receptor MC1 agouti Agouti is antagonist for MC1R. If agouti binds, no dark pigment produced

24. Second problem predicting outcome of selection: epistasis Epistasis MC1 agouti MC1- receptor Normal receptorMutant: never dark pigment (yellow labs) Mutant: always dark pigment

25. Epistasis example GenotypePhenotype EE / AAdark tips, light band ee / --blond / gold / red E d - / --all dark EE / A d -blond / gold / red Want dark fur: population ee / AA Ee / A d A Ee / AA

26. Epistasis example ii Cross: Ee / AdA x Ee / AdA genotypephenotypefreq EE / A d A d 1/16 EE / A d A1/8 EE / AA1/16 Ee / A d A d 1/8 Ee / A d A1/4 Ee / AA1/8 ee / A d A d 1/16 ee / A d A1/8 ee / AA1/16

27. Measuring h 2 : Parent-offspring regression

28. Estimating h 2 Analysis of related individuals Measuring the response of a population, in the next generation, to selection

29. Heritability is estimated as the slope of the least-squares regression line

30. h 2 data: Darwin’s finches

31. h 2 example: Darwin’s finches mean before selection: 9.4 mean after selection: 10.1 S = 10.1 - 9.4 = 0.7

32. h 2 example: Darwin’s finches II mean before selection: 9.4 mean of offspring after selection: 9.7 Response to selection: 9.7 – 9.4 = 0.3 R = h 2 S; 0.3 = h 2 * 0.7 h 2 = R/S = 0.3 / 0.7 = 0.43

33. Controlling for environmental effects on beak size song sparrows: cross fostering Smith and Dhondt (1980) Offspring vs. biological parent (h 2 ); vs. foster parent (V E )

34. Cross fostering in song sparrows

35. Testing for environmental effects How can we determine the effect of the environment on the phenotype?

36. Two genotypes in two environments: possible effects on phenotype

37. Example of environmental effects: locusts

38. Environmental effects: carpenter ant castes queenmale major worker minor worker

39. Effect of GxE: predicting outcome of selection 7 yarrow (Achillea millefolium) genotypes 2 gardens Clausen, Keck, and Heisey (1948)

40. What happens to h 2 when selection occurs?

41. Modes of selection Gen. 2 trait, z freq. Gen. 0 Gen. 1

42. Mode of selection: directional trait, z freq. trait, z freq.

43. Mode of selection: stabilizing trait, z freq. trait, z freq.

44. Mode of selection: disruptive trait, z freq. trait, z freq.

45. Disruptive selection Distribution of mandible widths in juveniles that died (shaded) and survived (black)

46. Complications: correlations Darwin’s finches: beak width is correlated with beak depth Fitness Beak depth Beak width Beak depth Fitness Beak width

47. Detecting loci affecting quantitative traits (QTL)

48. QTLs and genes of major effects How important are genes of major effect in adaptation?

49. QTL analysis: Quantitative Trait Loci – where are the genes contributing to quantitative traits? Approach –two lineages consistently differing for trait of interest (preferably inbred for homozygosity) –Identify genetic markers specific to each lineage (eg microsatellite markers) –make crosses to form F1 –generate F2s and measure trait of interest –test for association between markers and trait –Estimate the effect on the phenotype of each marker

50. Example: Mimulus cardinalis and Mimulus lewisii Mimulus cardinalis Mimulus lewisii

51. Locate lineage-specific markers Legend M. lewisii specific M. cardinalis specific non-specific Mimulus cardinalis Mimulus lewisii Microsatellite length 250 (M. l.) or 254 (M. c.) Cross lines 

52. QTL Mapping: crosses F1 F2 recombinants intermediate phenotype scrambled phenotypes

53. QTL analysis: trait associations Homozygotes at marker 2 are closer to one parent Heterozygotes at marker 2 are intermediate in trait values

54. Trait analysis For each marker, ask whether changing genotype affects phenotype

55. QTL probabilities Estimate the probability of location of QTL based on association of markers and recombination probabilities (eg CD34B rarely associated with pH, CD34A nearly always, TG63 rarely) Markers

56. Example study: basis of floral traits in two Mimulus species (Schemske and Bradshaw, PNAS 1999) Pollination syndrome changes during the evolution of the Mimulus group: hummingbird or bee

57. F2 plants showed variation for most floral traits M. cardinalisM. lewisii F1 F2 plants Most traits had multiple QTL but one explaining > 25% of variation  Few genes of large effect important in this case

58. Case study: domestication traits in sunflowers Most traits showed many genes of small effect (< 10%) Problems: gene map resolution is low –35,000 genes per plant genome –100 markers on genetic map: 350 genes per marker A “weak” QTL can be due to –nearby gene of weak effect –more distant gene of strong effect (looks weak due to recombination between marker and locus)

59. Recap: quantitative genetics Are traits heritable? –Usually find heritability –However, environmental effects can be large Are genes controlling quantitative traits of large effect or small effect? –Some important genes for adaptation of large effect. –Overall pattern still unclear

60. Questions 1.Human height is highly heritable: among university students in the US, the heritability is 0.84. Yet, during the 1980s, when Guatemalan refugees fled the civil war to the United States, 12 year old Mayan children were four inches taller in the United States than in Guatemala. How is this possible? 2.Consider the following scenario: In October 2006 you banded and weighed all of the adult burrowing owls near Osoyoos. The average owl weighed 207.8 gm. The winter of 2006-2007 was especially cold, and many owls died before they bred in the spring. When you weighed those owls that survived to breed in April of 2007, their average weight was 211.8 gm. If the heritability of owl body weight is 0.25, what is the expected mean body weight in the NEXT generation of adults, assuming none die before they become adults? 3.Data demonstrating stabilizing selection for human head size at birth were collected in 1951 in the United States. If you were to collect the same data now, would you expect to find the same pattern? Why or why not? Would it matter where in the world you did your study? Also, see posted quantitative genetics practice questions for further practice.