Lecture 3: Resemblance Between Relatives

Heritability Estimates of VA require known collections of relatives 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

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

Half-sibs Full-sibs

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

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. Both alleles IBD One allele IBD No alleles IBD

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

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)

Situation 1 Situation 2 Var(B) = 0 Var(B) = 2.5 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

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

) All red covariance terms are zero. • 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.

{ 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

Half-sibs The half-sibs share no alleles IBD • occurs with probability 1/2 The half-sibs share one allele IBD • occurs with probability 1/2 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

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

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) Cov(Full-sibs) = Var(A)/2 + Var(D)/4

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) +

Components of the Environmental Variance Total environmental value Common environmental value experienced by all members of a family, e.g., shared maternal effects E = Ec + Es Specific environmental value, any unique environmental effects experienced by the individual VE = VEc + VEs The Environmental variance can thus be written in terms of variance components as 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.

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

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

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 Draw A1 IBD Draw A2 fx 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

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

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

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

Computing qxy -- chain counting 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 Coefficient of coancestry of i Paths through a single common ancestor (i) to both x and y Paths through two remove ancestors (j and k)

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

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

Examples of Dxy Dxy = qmxmyqfxfy + qmxfyqfxmy Dxy = qmmqff + qmf2 (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

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

General Resemblance between relatives D D D