Quantitative genetics: traits controlled my many loci

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 = h2S

6. LOCUS TRAIT A Z1 B # of individuals C D Z E F Polygenic Inheritance Leads to a Quantitative Trait

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 3 age 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: _ Vp= Σi (Xi – X)2 --------------- (N – 1)

14. Variance components VP = total phenotypic variance VP = VA + VD + VE + VGXE VA = Additive genetic variance VD = Dominance genetic variance (non- additive – can include epistasis) VE = Variance among individuals experiencing different environments VGXE = Variance due to environmental variation that influences gene expression (not covered in text)

15. Heritability = h2 = VA / VPThe proportion of phenotypic variance due to additive genetic variance among individuals h2 = VA / (VA + VD + VE + VGXE) 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 Genotype Toe len (cm) AA 0.5 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) Genotype Toe len (cm) AA 0.5 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 Dominance and heritability

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

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

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

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

22. Dominant S = R = R = h2S h2 = R / S = Codominant (additive) S = R = R = h2S h2 = Effect of dominance on heritability

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

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

25. Epistasis example Genotype Phenotype EE / AA dark tips, light band ee / -- blond / gold / red Ed- / -- all dark EE / Ad- blond / gold / red Want dark fur: population ee / AA Ee / AdA Ee / AA

26. Epistasis example ii Cross: Ee / AdA x Ee / AdA genotype phenotype freq EE / AdAd 1/16 EE / AdA 1/8 EE / AA 1/16 Ee / AdAd 1/8 Ee / AdA 1/4 Ee / AA 1/8 ee / AdAd 1/16 ee / AdA 1/8 ee / AA 1/16

27. Measuring h2: Parent-offspring regression

28. Estimating h2 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. h2 data: Darwin’s finches

31. h2 example: Darwin’s finches mean before selection: 9.4 mean after selection: 10.1 S = 10.1 - 9.4 = 0.7

32. h2 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 = h2S; 0.3 = h2 * 0.7 h2 = R/S = 0.3 / 0.7 = 0.43

33. Controlling for environmental effects on beak size • song sparrows: cross fostering Offspring vs. biological parent (h2); vs. foster parent (VE) Smith and Dhondt (1980)

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 major worker minor worker queen male

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

40. What happens to h2 when selection occurs?

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

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

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

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

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

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

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