Correlation for a pair of relatives 2 G 1 P h e i g gi = correlation in genetic values for the ith type of relative hi = correlation in environmental values for the ith type of relative

Two Generic Methods: (1) Genetic Epidemiology (2) Molecular approach Unmeasured genotypes Use correlations between informative relatives (e.g., twins adoptees) (2) Molecular approach Measured genotypes (almost always SNPs) Several approaches, two most common = GCTA analysis and LD score regression

Twin Method: 1 Three Structural Equations: 𝑅 𝑀𝑍 =1.0 ℎ 2 + 𝜂 𝑀𝑍 𝑒 2 𝑅 𝐷𝑍 = 𝛾 𝐷𝑍 ℎ 2 + 𝜂 𝐷𝑍 𝑒 2 1= ℎ 2 + 𝑒 2 (1) (2) (3) Problem: 3 Equations but 5 unknowns Solution: Make Assumptions Additive gene action, no assortative mating, therefore 𝛾 𝐷𝑍 =0.5 Equal environments assumption: 𝜂 𝑀𝑍 = 𝜂 𝐷𝑍 = 𝜂

Twin Method: 2 Three Structural Equations Rewritten: Solution: 𝑅 𝑀𝑍 =1.0 ℎ 2 +𝜂 𝑒 2 𝑅 𝐷𝑍 =0.5 ℎ 2 +𝜂 𝑒 2 1= ℎ 2 + 𝑒 2 (1) (2) (3) Solution: (1.A) Subtract Eq (2) from Eq (1) 𝑅 𝑀𝑍 − 𝑅 𝐷𝑍 = ℎ 2 +𝜂 𝑒 2 −0.5 ℎ 2 −𝜂 𝑒 2 =0.5 ℎ 2 (1.B) Multiply both sides by 2 2(𝑅 𝑀𝑍 − 𝑅 𝐷𝑍 )= ℎ 2

Twin Method: 3 Three Structural Equations Rewritten: Solution: 𝑅 𝑀𝑍 =1.0 ℎ 2 +𝜂 𝑒 2 𝑅 𝐷𝑍 =0.5 ℎ 2 +𝜂 𝑒 2 1= ℎ 2 + 𝑒 2 (1) (2) (3) Solution: (2) Substitute the estimate of h2 into Eq (3) 𝑒 2 =1− ℎ 2 (3) Substitite the estimates of h2 and e2 into either Eq (1) or (2) 𝜂=( 𝑅 𝑀𝑍 − ℎ 2 )/ 𝑒 2 𝜂=( 𝑅 𝐷𝑍 − 0.5ℎ 2 )/ 𝑒 2

Adoption Method (1) RBioSibs = .24 = .5h2 + he2 (2) RAdpSibs = .06 = he2 (3) h2 + e2 = 1 Solution: (1) Subtract Equation (2) from Equation (1): RBioSibs = .24 = .5h2 + he2 - RAdpSibs = -.06 = - he2 .18 = .5h2 (2) Multiply this result by 2: 2(.18) = 2(.5h2), so .36 = h2 (3) Substitute this quantity into Equation (3): .36 + e2 = 1, so e2 = .64 (4) Substitute the results from steps (2) and (3) into Equation (1) & solve for h: .06 = h(.64), so h = .06/.64 = .09

GCTA Analysis Select random individuals from the general population Genotype on a large number of loci Compute the genetic similarity for between each pair of individuals Those pairs with high genetic similarity should have more similar phenotypes than those with low genetic similarity 𝑿=𝑨 𝑉 𝐴 +𝑫 𝑋 𝑖𝑗 =( 𝑃 𝑖 −𝜇)( 𝑃 𝑗 −𝜇) 𝐴 𝑖𝑗 = correlation between additive genetic values for ijth pair D = diagonal matrix of residual effects VA = additive genetic variance

Linkage Disequilibrium (LD) Score Regression Logic = a causal variant in a haplotype block in strong disequilibrium is more more likely to have a high association with each loci than one in a block with weak disequilibrium. Block 3: High Probability Block 2: Medium Probability Block 1: Low Probability Bulaik-Sullivan et al. (2015), Nat. Genetics, 47, 291-297

Linkage Disequilibrium (LD) Score Regression So, compute a LD score for each locus. For the ith locus, the LD score equals the sum of the squared correlations with all the loci in the block, or ℓ 𝑖 = 𝑗 𝑟 𝑖𝑗 2 The larger the value of ℓ 𝑖 , the greater the chance of a causal variant, so regress the observed 𝜒 2 for each locus on its ℓ 𝑖 value. Bulaik-Sullivan et al. (2015), Nat. Genetics, 47, 291-297

Linkage Disequilibrium (LD) Score Regression 𝐸 𝜒 2 ℓ 𝑖 = 𝑁 ℎ 2 ℓ 𝑖 𝑀 +𝑁𝑎+1 N = sample size M = number of loci h2 = heritability a = confounding effects (e.g., pop stratification) ℓ 𝑖 = 𝑗 𝑟 𝑖𝑗 2 𝑟 𝑖𝑗 2 = squared correlation for all j loci in LD with the ith locus Bulaik-Sullivan et al. (2015), Nat. Genetics, 47, 291-297

Linkage Disequilibrium (LD) Score Regression 𝐸 𝜒 2 ℓ 𝑖 = 𝑁 ℎ 2 ℓ 𝑖 𝑀 +𝑁𝑎+1 Although it may not look like it, this equation is a linear regression equation. The dependent variable is the 𝜒 2 , the independent variable is ℓ 𝑖 , the intercept is Na + 1, and the slope is Nh2/M. Because we know N and M, we can calculate h2. Bulaik-Sullivan et al. (2015), Nat. Genetics, 47, 291-297