Partitioning the Genetic Variance

Partitioning the Genetic Variance 1 / 18

Partitioning the Genetic Variance In lecture 2, we showed how to partition genotypic values G into their expected values based on additivity (G A ) and deviations from the additivity as a result of dominance (δ), where G = G A + δ We will now focus on partitioning the genetic variance into additive and dominance variance components. 2 / 18

Partitioning the Genetic Variance In lecture 2, we showed how to partition genotypic values G into their expected values based on additivity (G A ) and deviations from the additivity as a result of dominance (δ) For this decomposition, individuals with alleles A i and A j at a locus have a mean genotypic value of G ij = G A ij + δ ij = µ G + α i + α j + δ ij We will now focus on partitioning the genetic variance into the additive and dominance components of variance: σg 2 = σ A 2 + σ D 2 σa 2 is the additive genetic variance. It is the genetic variance associated with the average additive effects of alleles σd 2 is the dominance genetic variance. It is the genetic variance associated with the dominance effects. 3 / 18

Heritability defined in terms of the variance components Note that for a quantitative phenotype Y, the quantity h 2 = σ A 2 is σy 2 referred to as the narrow sense heritability (or simply heritability) proportion of the total phenotypic variance that is due to additive effects. Can also be viewed as the extent to which phenotypes are determined by the alleles transmitted from the parents. In lecture 1, we defined that the quantity H 2 = σ G 2 σy 2 broad sense heritability = σ 2 A +σ 2 D σ 2 Y to be the proportion of the total phenotypic variance that is due to all genetic effects. 4 / 18

Decomposition of Genotypic Values We previously decomposed the genotypic value at a locus into additive effects and dominance deviations: G ij = G A ij + δ ij = µ G + α i + α j + δ ij For a locus with two allelic types, A 1 and A 2, we showed that the model can be given in terms of a linear regression of genotypic values on the number of copies of the A 1 allele such that: G ij = β 0 + β 1 X ij 1 + δ ij where X ij 1 is the number of copies of the type A 1 allele in genotype G ij, and with β 0 = µ G + 2α 2 and β 1 = α 1 α 2 = α, the average effect of allele substitution. 5 / 18

Components of Genetic Variance The original definition of the average effect of allele substitution, α, was actually given by Ronald Fisher (1918) in terms of linear regression! G ij = β 0 + β 1 X ij 1 + δ ij Note that Gij A = β 0 + β 1 X ij 1, the genotypic values predicted by the additive linear regression model for genotype A i A j, is the fitted value. We also have that δ ij, the deviation from the additive model due to dominance, is the residual. 6 / 18

Linear Regression Figure for Predicting Additive Genetic Values Falconer model for single biallelic QTL a d m -a bb Bb BB Var (X) = Regression Variance + Residual Variance = Additive Variance + Dominance Variance 15 7 / 18

Components of Genetic Variance Now Var(G) = Var(G A ) + Var(δ) + 2Cov(G A,δ) What is Cov(G A,δ)? 8 / 18

Components of Genetic Variance From the properties of least squares, the residuals are orthogonal to the fitted values, and thus Cov(G A,δ) = 0. So we have that or Let s first calculate σ 2 A = Var(G A ) Var(G) = Var(G A ) + Var(δ) σ 2 G = σ 2 A + σ 2 D 9 / 18

Representation of Genotypic Values We previously represented genotypic values for the three genotypes as follows a if genotype is A 2 A 2 Genotype Value = d if genotype is A 1 A 2 a if genotype is A 1 A 1 If p and q are the allele frequencies of the A 1 and A 2 alleles, respectively in the population, we previously showed that µ G = a(p q) + d(2pq) 10 / 18

Additive Variance Component For the additive genetic component, we have that µ G + 2α 2 if genotype is A 2 A 2 Gij A = µ G + α 1 + α 2 if genotype is A 1 A 2 µ G + 2α 1 if genotype is A 1 A 1 where and α 1 = pa + qd µ G = q[a + d(q p)] α 2 = qa + pd µ G = p[a + d(q p)] σ 2 A = E[( G A E[G A ]) 2] = E[(G A ) 2 ] ( E[G A ] ) 2, and since µg is a constant value that will be canceled in the variance calculation, we can essentially remove it by subtracting it out and calculating σ 2 A = E[(G A µ G ) 2 ] ( E[G A µ G ] ) 2. 11 / 18

Additive Variance Component Now, E[G A µ G ] is 2α 1 p 2 + (α 1 + α 2 )2pq + 2α 2 q 2 = 2α 1 p(p + q) + 2α 2 q(q + p) = 2pα 1 + 2qα 2 = 2(pα 1 + qα 2 ) = 2(0) = 0 12 / 18

Additive Variance Component For E[(G A µ G ) 2 ], we have E[(G A µ G ) 2 ] = p 2 (2α 1 ) 2 + 2pq(α 1 + α 2 ) 2 + q 2 (2α 2 ) 2 = p 2 (2α 1 ) 2 + 2pq(α 2 1 + 2α 1 α 2 + α 2 2 ) + q 2 (2α 2 ) 2 = α 2 1 (4p 2 + 2pq) + α 2 2 (4q 2 + 2pq) + 4α 1 α 2 pq = α 2 1 (2p(1 + p)) + α 2 2 (2q(1 + q)) + 4α 1 α 2 pq [since 4p 2 + 2pq = 4p 2 + 2p(1 p) = 4p 2 + 2p 2p 2 = 2p 2 + 2p = 2p(1 + p) ] 13 / 18

Additive Variance Component = 2α1 2 p(1 + p) + 2α2 2 q(1 + q) + 4α 1 α 2 pq = 2α1 2 p(q + 2p) + 2α2 2 q(p + 2q) + 4α 1 α 2 pq = 2α1 2 pq + 4α1 2 p 2 + 2α2 2 pq + 4α2 2 q 2 + 4α 1 α 2 pq = 2α1 2 pq + 2α2 2 pq + 4(α1 2 p 2 + α2 2 q 2 + α 1 α 2 pq) = 2α1 2 pq + 2α2 2 pq + 4(α1 2 p 2 + α2 2 q 2 + 2α 1 α 2 pq α 1 α 2 pq) = 2α1 2 pq + 2α2 2 pq 4α 1 α 2 pq + 4(α1 2 p 2 + α2 2 q 2 + 2α 1 α 2 pq) = 2α1 2 pq + 2α2 2 pq 4α 1 α 2 pq + 4(α 1 p + α 2 q) 2 = 2α1 2 pq + 2α2 2 pq 4α 1 α 2 pq + 0 = 2pq(α 1 α 2 ) 2 = 2pqα 2 14 / 18

Additive Variance Component So we have that σ 2 A = E[(G A µ G ) 2 ] (E[G A µ G ]) 2 = 2pqα 2 15 / 18

Dominance Variance Component We will now obtain σd 2 We have that δ ij = G ij Gij A where a if genotype is A 2 A 2 G ij = d if genotype is A 1 A 2 a if genotype is A 1 A 1 and Gij A = Now calculate σ 2 D = E(δ 2 ) (E(δ)) 2 µ G + 2α 2 if genotype is A 2 A 2 µ G + α 1 + α 2 if genotype is A 1 A 2 µ G + 2α 1 if genotype is A 1 A 1 16 / 18

Dominance Variance Component We have that δ 11 = a µ G 2α 1. Now a µ G = a a(p q) d(2pq) = a(1 p + q) + 2dpq = 2aq 2pdq = 2q(a dp) So, δ 11 = (a µ G ) 2α 1 = 2q(a dp) 2q[a + d(q p)] = 2q( dq) = 2q 2 d Now calculate δ 12 and δ 22. What is E[δ]? What is E[δ 2 ]? What is σd 2 = E(δ 2 ) (E(δ)) 2? 17 / 18

Dominance Variance Component δ 12 = 2pqd δ 22 = 2p 2 d E[δ] = 0 E[δ 2 ] = (2q 2 d) 2 p 2 + (2pqd) 2 2pq + (2p 2 d) 2 q 2 = (2pqd) 2 q 2 + (2pqd) 2 2pq + (2pqd) 2 p 2 = (2pqd) 2 So σ 2 D = (2pqd)2 18 / 18