1. Biometrical Genetics QIMR workshop 2013 Slides from Lindon Eaves

2. Biometrical Genetics from a BG point of view Focus is on individual differences and variation in a population Why are people different Why are family members similar

3. P G E Measured variable Latent variables Genotype Environment Phenotype he r

4. The Basic Model Phenotype=Genotype+Environment P=G+E {+f(G,E)} f(G,E) = Genotype-environment interaction and correlation

5. GENES (G) Contribution (“Heritability”) Type of Action (“Additive”, “Dominant”, Epistatic”) Number, location and function

6. Basic Model for Effects of a Single Gene on a Quantitative Trait Think about a variant with two alleles A and a… Mid-homozygote + Homozygous effect Dominance deviation Increasing Decreasing - Homozygous effect

7. Fisher (1918): Basic Ideas Continuous variation caused by lots of genes (“polygenic inheritance”) Each gene followed Mendel’s laws Environment smoothed out genetic differences Genes may show different degrees of “dominance” Genes may have many forms (“mutliple alleles”) Mating may not be random (“assortative mating”) Showed that correlations obtained by e.g. Pearson and Lee were explained well by polygenic inheritance

8. Mendelian Basis of Continuous Variation? Experimental Breeding Experiments

9. b. c. a.

10. Environment “E” Contribution (“1-heritability”) Type (Shared by family, unique to individual, remote, proximal,short-, long-term) Non-genetic inheritance Identification

11. Interactions and Correlations f(G,E) Mating system, population structure GxE interaction Multiple variables: Genetic and Environmental Correlation Direction of Causation and Causal networks G x E interaction G – E correlation Remembering, Forgetting, Development (GxAge, G x Time etc.)

12. Francis Galton (1822-1911) 1869: Hereditary Genius 1883: Inquiries into Human Faculty and its Development 1884-5: Anthropometic Laboratory at “National Health Exhibition”

13. Hereditary Genius (1869, p 317)

14. Galton’s Anthropometric Laboratory:

16. Karl Pearson (1857-1936) 1903: On the Laws of Inheritance in Man: I Physical Characteristics (with Alice Lee) 1904: II Mental and Moral Characteristics 1914: The Life, Letters and Labours of Francis Galton

17. Pearson and Lee’s diagram for measurement of “span” (finger-tip to finger-tip distance)

18. From Pearson and Lee (1903) p.378

20. From Pearson and Lee (1903) p.387

21. From Pearson and Lee (1903) p. 373

22. Modern Data The Virginia 30,000 (N=29691) The Australia 22,000 (N=20480)

23. ANZUS 50K: Extended Kinships of Twins Twins Parents of Twins Offspring of Twins Siblings of Twins Spouses of Twins © Lindon Eaves, 2009

24. Overall sample sizes Relationship # of pairs Parent-offspring Siblings 25018 18697 Spouses 8287 DZ Twins MZ Twins 5120 4623

25. Nuclear Family Correlations for Stature (Virginia 30,000 and OZ 22,000) © Lindon Eaves, 2009

26. Nuclear Family Correlations for Liberalism/Conservatism (Virginia 30,000 and Australia 22,000) © Lindon Eaves, 2009

27. The (Really!) BIG Problem Families are a mixture of genetic and social factors

28. Galton’s Solution: Twins (Though Augustine may have got there first – 5 th cent.)

29. One (ideal) solution Twins separated at birth

31. But separated MZs are rare

32. An easier alternative: Identical and non-identical twins reared together: Galton (Again!)

33. IDENTICAL TWINS MONOZYGOTIC: Have IDENTICAL genes (G) Come from the same family (C) Have unique experiences during life (E)

34. FRATERNAL TWINS DIZYGOTIC: Have DIFFERENT genes (G) Come from the same family (C) Have unique experiences during life (E)

35. Figure 9. The correlation between twins for stature (data from the Virginia Twin Study of Adolescent Behavioral Development). Data from the Virginia Twin Study of Adolescent Behavioral Development

37. Genotype Frequencies in Randomly Mating Population “Hardy-Weinberg Equilibrium” frequencies

38. What is the mean expected to be? Note: Effects measured from mid-homozygote (“m”)

39. With equal allele frequencies (easier!) put u=v= ½ And the mean is expected to be….

40. How does A/a affect the variance?

41. Equal allele frequencies u=v= ½ Additive component Dominance component

42. Q: What happens with lots of genes? A: The effects of the individual genes add up. IF… the genes are independent (“linkage equilibrium”) Requires random mating, complete admixture

43. So: Additive Genetic Variance Dominance Genetic Variance

44. Additive and Dominance Components: Unequal allele frequencies. Can show (see e.g. Mather, 1949) V A V D Q: What happens when u=v?

45. Bottom line: With unequal allele frequencies can still separate V A and V D but their definitions change

46. d<-1 # Homozygous effect ("additive") h<-1 # Heterozygous deviation ("dominance") u<-seq(0.01,0.99,by=.01) # Vector of frequencies of increasing allele v<-1-u # Frequencies of decreasing allele VA<-2*u*v*(d+(v-u)*h)^2 # Additive genetic variance VD<-4*u*u*v*v*h*h # Dominance genetic variance VP<-VA+VD # Total (genetic) variance # Plot results plot(u,VP,type="l", main="VA (red) and VD (green) as function of increasing allele frequency", xlab="Frequency of increasing allele",ylab="Variance component") # Add line for VA lines(u,VA,col="red") # Add line for VD lines(u,VD,col="green") Plotting Effect of Allele frequency on Genetic Variance Components (“R”)

47. VA+VD VA VD

48. What about the environment???

49. Two main sources of environment Individual experiences – not shared with siblings: V E “Family” environment – shared with siblings: V C

50. So: the TOTAL variance (Genes + Environment) is: V P = V A +V D +V E +V C

51. “Heritability” “Broad” heritability: h 2 b =(V A +V D )/V P Proportion of total variance explained by genes “Narrow” heritability: h 2 n =V A /V P Proportion of total variance explained by additive (homozygous) genetic effects (predicts response to selection – Fisher, 1930)

52. So far: have looked at effects on total variance… How do V A and V D affect the correlations between relatives?

53. Contribution of genes to correlation between relatives (r): r = C/V P Where C=Covariance between relative pairs “C” depends of kind of relationship (sibling, parent-offspring, MZ twin etc) But can also be expressed in terms of V A and V D

54. Approach 1.For a given relationship, work out expected frequencies of each type of pair (AA, aa etc.) 2.Write phenotypes of each type of relative 3.Compute cross-products of phenotypes of members of type of pair 4.Each cross-product by the corresponding frequency 5.Add the result of “4” across all pair types The answer is the covariance you want (if you have done the algebra right!)

55. For equal allele frequencies….

57. Contribution of one gene to covariance:

58. Notice that terms in d 2 and h 2 are separated – but their coefficients change as a function of relationship

59. Can add over all genes to get total contribution to covariance Cov(MZ) = V A + V D Cov(DZ) = ½V A + ¼V D Cov(U)= 0

60. Can use the same approach for other relationships

61. Relationship Contribution to Covariance V A VDVD Total variance 1 1 Sibling (DZ twin) ½ ¼ MZ twin 1 1 Half-sibling ¼ 0 First cousin 1 / 8 0 Parent-offspring ½ 0 Avuncular ¼ 0 Grand-parent 1 / 8 0 Unrelated 0 0 Contributions of V A and V D to covariances between relatives (ignoring environment)

62. Adding effects of Environment V P = V A + V D + V E + V C Cov(MZ) = V A + V D + V C Cov(DZ) = ½V A + ¼V D + V C Cov(UT) = V C Etc.

63. To get the expected correlations Just divided expectations by expected total variance Results are proportional contributions of V A, V D etc. to total variance

64. Practice (paper and pencil) Pick a “d” and “h” (e.g. d=1,h=1; d=1,h=0) Pick a frequency for the increasing (A) allele (e.g. u=0.2, u=0.7) Work out VA and VD Tabulate on board

65. Vienna