Partitioning the Genetic Variance. Partitioning the Genetic Variance

Transcription

1 Partitioning the Genetic Variance

2 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.

3 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.

4 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.

5 Linear Regression Figure for Predicting Genetic Values (from Peter Visscher) Falconer model for single biallelic QTL a d m -a bb Bb BB

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

7 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 Var(G) = Var(G A ) + Var(δ) σ 2 G = σ2 A + σ2 D Let s first calculate σ 2 A = Var(G A )

8 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)

9 Additive Variance Component For the additive genetic component, we have that where and G A ij = 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 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 µ G 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.

10 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

11 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 α 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) ]

12 Additive Variance Component = 2α1p(1 2 + p) + 2α2q(1 2 + q) + 4α 1 α 2 pq = 2α1p(q 2 + 2p) + 2α2q(p 2 + 2q) + 4α 1 α 2 pq = 2α1pq 2 + 4α1p α2pq 2 + 4α2q α 1 α 2 pq = 2α1pq 2 + 2α2pq 2 + 4(α1p α2q α 1 α 2 pq) = 2α1pq 2 + 2α2pq 2 + 4(α1p α2q α 1 α 2 pq α 1 α 2 pq) = 2α1pq 2 + 2α2pq 2 4α 1 α 2 pq + 4(α1p α2q α 1 α 2 pq) = 2α1pq 2 + 2α2pq 2 4α 1 α 2 pq + 4(α 1 p + α 2 q) 2 = 2α1pq 2 + 2α2pq 2 4α 1 α 2 pq + 0 = 2pq(α 1 α 2 ) 2 = 2pqα 2

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

14 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 G A ij = 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 Now calculate σ 2 D = E(δ2 ) (E(δ)) 2

15 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(1 dp) So, δ 11 = (a µ G ) 2α 1 = 2q(1 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(δ2 ) (E(δ)) 2?

16 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