Lecture 3: Resemblance Between Relatives

1. Lecture 3: Resemblance Between Relatives

2. Heritability • Central concept in quantitative genetics • Proportion of variation due to additive genetic values (Breeding values) • h2 = VA/VP • Phenotypes (and hence VP) can be directly measured • Breeding values (and hence VA ) must be estimated • Estimates of VA require known collections of relatives

3. Ancestral relatives e.g., parent and offspring Collateral relatives, e.g. sibs

4. Half-sibs Full-sibs

5. Key observations • The amount of phenotypic resemblance among relatives for the trait provides an indication of the amount of genetic variation for the trait. • If trait variation has a significant genetic basis, the closer the relatives, the more similar their appearance

6. One allele IBD Both alleles IBD No alleles IBD Genetic Covariance between relatives Sharing alleles means having alleles that are identical by descent (IBD): both copies of can be traced back to a single copy in a recent common ancestor. Genetic covariances arise because two related individuals are more likely to share alleles than are two unrelated individuals.

7. Regressions and ANOVA • Parent-offspring regression • Single parent vs. midparent • Parent-offspring covariance is a intraclass (between class) covariance • Sibs • Covariances between sibs is an interclass (within class) covariance

8. ANOVA • Two key ANOVA identities • Total variance = between-group variance + within-group variance • Var(T) = Var(B) + Var(W) • Variance(between groups) = covariance (within groups) • Intraclass correlation, t = Var(B)/Var(T)

9. Situation 1 Situation 2 Var(B) = 0 Var(W) = 2.7 Var(T) = 2.7 Var(B) = 2.5 Var(W) = 0.2 Var(T) = 2.7 t = 0 t = 2.5/2.7 = 0.93

10. IBD allele Non-IBD alleles Parent-offspring genetic covariance Cov(Gp, Go) --- Parents and offspring share EXACTLY one allele IBD Denote this common allele by A1

11. • By construction, a and D are uncorrelated • By construction, a from non-IBD alleles are uncorrelated • By construction, D values are uncorrelated unless both alleles are IBD ) All red covariance terms are zero.

12. { Hence, relatives sharing one allele IBD have a genetic covariance of Var(A)/2 The resulting parent-offspring genetic covariance becomes Cov(Gp,Go) = Var(A)/2

13. The half-sibs share one allele IBD • occurs with probability 1/2 The half-sibs share no alleles IBD • occurs with probability 1/2 Half-sibs Each sib gets exactly one allele from common father, different alleles from the different mothers Hence, the genetic covariance of half-sibs is just (1/2)Var(A)/2 = Var(A)/4

14. Paternal allele not IBD [ Prob = 1/2 ] Maternal allele not IBD [ Prob = 1/2 ] -> Prob(zero alleles IBD) = 1/2*1/2 = 1/4 Paternal allele IBD [ Prob = 1/2 ] Maternal allele IBD [ Prob = 1/2 ] -> Prob(both alleles IBD) = 1/2*1/2 = 1/4 Full-sibs Each sib gets exact one allele from each parent Prob(exactly one allele IBD) = 1/2 = 1- Prob(0 IBD) - Prob(2 IBD)

15. Cov(Full-sibs) = Var(A)/2 + Var(D)/4 Resulting Genetic Covariance between full-sibs If both alleles IBD, cov(Gsib1, Gsib2 )= cov(ax+ay+dxy, ax+ay+dxy ) = cov(ax+ay, ax+ay) + cov(dxy, dxy) = Var(A) + Var(D)

16. Genetic Covariances for General Relatives Let r = (1/2)Prob(1 allele IBD) + Prob(2 alleles IBD) Let u = Prob(both alleles IBD) General genetic covariance between relatives Cov(G) = rVar(A) + uVar(D) When epistasis is present, additional terms appear r2Var(AA) + ruVar(AD) + u2Var(DD) + r3Var(AAA) +

17. Total environmental value Common environmental value experienced by all members of a family, e.g., shared maternal effects Specific environmental value, any unique environmental effects experienced by the individual The Environmental variance can thus be writtenin terms of variance components as VE = VEc + VEs Components of the Environmental Variance E = Ec + Es One can decompose the environment further, if desired. For example, plant breeders have terms for the location variance, the year variance, and the location x year variance.

18. Shared Environmental Effects contribute to the phenotypic covariances of relatives Cov(P1,P2) = Cov(G1+E1,G2+E2) = Cov(G1,G2) + Cov(E1,E2) Shared environmental values are expected when sibs share the same mom, so that Cov(Full sibs) and Cov(Maternal half-sibs) not only contain a genetic covariance, but an environmental covariance as well, VEc

19. Coefficients of Coancestry Suppose we pick a single allele each at random from two relatives. The probability that these are IBD is called Q, the coefficient of coancestry Qxy denotes the coefficient for relatives x and y Consider an offspring z from a (hypothetical) cross of x and y. Qxy = fz, the inbreeding coefficient of z

20. Qxx : The Coancestry of an individual with itself Self x, what is the inbreeding coefficient of its offspring? To compute Qxx, denote the two alleles in x by A1 and A2 Draw A1 Draw A2 fx IBD Draw A1 fx Draw A2 IBD Hence, for a non-inbred individual, Qxx = 2/4 = 1/2 If x is inbred, fx = prob A1 and A2 IBD, Qxx = (1+ fx)/2

21. fp Mother Offspring fo Parental allele qmf Qop = Parent & Offspring Offspring inbred Parent inbred 1/2 = Prob random offspring allele from father. Prob = qmf = fo that this allele is IBD to mother giving a contribution of fo/2

22. Full sibs (x and y) from parents m and f (1+ff)/2 1/2 1/2 (1+fm)/2 Q mf f f f m m m Q mf (1/2)(1/2) Q mf /4 Q mf Q = 1/8 + 1/8 = 1/4 Q =(2+fm+ff)/8 Q =(2+fm+ff +4 Q mf)/8 (1/2)(1/2)(1/2) (1/2)(1/2)(1/2) [(1 +ff )/2] (1/2)(1/2) [(1 +fm )/2] (1/2)(1/2)

23. Full sibs (x and y) from parents m and f Qxy = (2 + fm + ff + 4Qmf)/8 ff =Qsf,df fm =Qsm,dm Qxy = (2 + Qsm,dm + Qsf,df + 4Qmf)/8

24. ) ) ( ( Coefficient of coancestry of i Number of individuals (including x and y) in path Connecting x and y through i Number of individuals, including x and y on the path leading from two different (but related) ancestors j and k Coefficient of coancestry of j and k Computing qxy -- chain counting Paths through a single common ancestor (i) to both x and y Paths through two remove ancestors (j and k)

25. Qii = 1/2 Lord Raglan Lord Raglan Lord Raglan Qii = 1/2 Champion of England Champion of England Champion of England Qii = 1/2 Duchess of Gloster, 9th Mistletoe The Czar Qii = 1/2 Grand Duke of Gloster Mimulus Carmine Princess Royal Princess Royal Princess Royal Princess Royal Princess Royal Princess Royal Royal Duke of Gloster Royal Duke of Gloster Royal Duke of Gloster Royal Duke of Gloster Royal Duke of Gloster Royal Duke of Gloster Roan Gauntlet Path throughLord Raglan, n = 7 Contribution = (1/2)(1/2)6 Path throughChampion of England, n = 4 Contribution = (1/2)(1/2)3 Path through Lord Raglan, n = 7 Contribution = (1/2)(1/2)6 Path throughChampion of England, n = 4 Contribution = (1/2)(1/2)3 Compute q for Royal Duke of Gloster and Princess Royal q = (1/2) 7 + (1/2) 4 + (1/2) 4 + (1/2) 7 = 0.141 f for Roan Gauntlet = 0.141

26. Dxy, The Coefficient of Fraternity qfxmy qfxfy qmxmy qmxfy my fy fx mx x y Dxy = Prob(both alleles in x & y IBD) Dxy = qmxmyqfxfy +qmxfyqfxmy

27. Examples of Dxy Dxy = qmxmyqfxfy +qmxfyqfxmy (1) x and y are full sibs: mx = my = m, fx = fy = f Dxy = qmmqff +qmf2 If parents unrelated, qmf = 0 Dxy = 1/4 If parents not inbred, qmm =qff = 1/2 (2) x and y are paternal half-sibs: fx = fy = f Dxy = qmxmyqff +qmxfqmyf If parents unrelated, qmxf = qmyf = qmxmy = 0 Dxy = 0

28. Lord Raglan Champion of England Champion of England Duchess of Gloster, 9th fy Mistletoe The Czar q = 1/4 Grand Duke of Gloster Grand Duke of Gloster Mimulus Mimulus q = 1/4 Carmine Carmine mx fx q = (1/2)5 my q = (1/2)5 Princess Royal Princess Royal Royal Duke of Gloster Royal Duke of Gloster Roan Gauntlet What is D for Royal Duke of Gloster and Princess Royal y x Dxy = (1/2)5(1/4)+ (1/4)(1/2)5 = (1/2)6 Dxy = qmxmy(1/4)+qmxfyqfxmy Dxy = qmxmyqfxfy +qmxfyqfxmy Dxy = (1/2)5(1/4)+ (1/4)qfxmy Dxy = qmxmy(1/4)+ (1/4)qfxmy

29. D D D D D General Resemblance between relatives