Lecture 2: Fisher’s Variance Decomposition

Contribution of a locus to a trait Basic model: P = G + E Phenotypic value -- we will occasionally also use z for this value Genotypic value Environmental value G = average phenotypic value for that genotype if we are able to replicate it over the universe of environmental values, G = E[P] Modifications: G - E covariance -- higher performing Animals may be disproportionately rewarded G x E interaction --- G values are different across environments. Basic model now becomes P = G + E + GE

Alternative parameterizations of Genotypic values Q1Q1 Q2Q1 Q2Q2 C C + a(1+k) C + 2a C C + a + d C + 2a C -a C + d C + a d measures dominance, with d = 0 if the heterozygote is exactly intermediate to the two homozygotes d = ak =G(Q1Q2 ) - [G(Q2Q2) + G(Q1Q1) ]/2 2a = G(Q2Q2) - G(Q1Q1) k = d/a is a scaled measure of the dominance

Example: Booroola (B) gene Genotype bb Bb BB Average Litter size 1.48 2.17 2.66 2a = G(BB) - G(bb) = 2.66 -1.46 --> a = 0.59 ak =d = G(Bb) - [ G(BB)+G(bb)]/2 = 0.10 k = d/a = 0.17

Fisher’s Decomposition of G One of Fisher’s key insights was that the genotypic value consists of a fraction that can be passed from parent to offspring and a fraction that cannot. G i j ° = ± b Dominance deviations --- the difference (for genotype AiAj) between the genotypic value predicted from the two single alleles and the actual genotypic value, G i j = π + Æ ± π G = X i j ¢ f r e q ( Q ) Mean value, with Average contribution to genotypic value for allele i Since parents pass along single alleles to their offspring, the ai (the average effect of allele i) represent these contributions b G i j = π + Æ The genotypic value predicted from the individual allelic effects is thus

G = π + Æ ± Fisher’s decomposition is a Regression G = π + 2 Æ ( ° ) N Residual error G i j = π + Æ ± Predicted value A notational change clearly shows this is a regression, Independent (predictor) variable N = # of Q2 alleles Regression slope Intercept G i j = π + 2 Æ 1 ( ° ) N ± Regression residual 2 Æ 1 + ( ° ) N = 8 > < : f o r ; e . g , Q

Allele Q1 common, a2 > a1 Slope = a2 - a1 Allele Q2 common, a1 > a2 Both Q1 and Q2 frequent, a1 = a2 = 0 1 2 N G G22 G11 G21

Æ = p a [ + k ( ° ) ] ± = G ° π Æ π = 2 p a ( 1 + k ) Genotype Q1Q1 Consider a diallelic locus, where p1 = freq(Q1) Genotype Q1Q1 Q2Q1 Q2Q2 Genotypic value a(1+k) 2a π G = 2 p a ( 1 + k ) Mean Allelic effects Æ 2 = p 1 a [ + k ( ° ) ] Dominance deviations ± i j = G ° π Æ

B V ( G ) = Æ + Average effects and Breeding Values i j B V = X ≥ Æ + The a values are the average effects of an allele Breeders focus on breeding value (BV) B V = n X k 1 ≥ Æ ( ) i + ¥ B V ( G i j ) = Æ + Why all the fuss over the BV? Consider the offspring of a QxQy sire mated to a random dam. What is the expected value of the offspring?

π ° = µ Æ + 2 ∂ B V ( S i r e ) G e n o t y p F r q u c V a l Q 1 / 4 x w 1 / 4 π + Æ ± z The expected value of an offspring is the expected value of µ For random w and z alleles, this has an expected value of zero ∂ µ ∂ For a random dam, these have expected value 0 π O = G + Æ x y 2 w z ± 4 π O ° G = µ Æ x + y 2 ∂ B V ( S i r e ) Hence,

B V ( S i r e ) = 2 π ° π ° = B V ( S i r e ) 2 + D a m G We can thus estimate the BV for a sire by twice the deviation of his offspring from the pop mean, B V ( S i r e ) = 2 π ° G More generally, the expected value of an offspring is the average breeding value of its parents, π ° G = B V ( S i r e ) 2 + D a m

æ = + Genetic Variances G = π + ( Æ ) ± 2 G A D æ ( G ) = X Æ + ± æ ( j = π g + ( Æ ) ± æ 2 ( G ) = π g + Æ i j ± As Cov(a,d) = 0 æ 2 ( G ) = n X k 1 Æ i + j ± Dominance Genetic Variance (or simply dominance variance) Additive Genetic Variance (or simply Additive Variance) æ 2 G = A + D

æ = ( p a k ) æ = p a [ + k ( ° ) ] æ = E [ Æ ] X p Q1Q1 Q1Q2 Q2Q2 0 a(1+k) 2a æ 2 A = E [ Æ ] m X i 1 p Since E[a] = 0, Var(a) = E[(a -ma)2] = E[a2] One locus, 2 alleles: Dominance effects additive variance æ 2 A = p 1 a [ + k ( ° ) ] When dominance present, asymmetric function of allele frequencies æ 2 D = E [ ± ] m X i 1 j p Equals zero if k = 0 One locus, 2 alleles: This is a symmetric function of allele frequencies æ 2 D = ( p 1 a k )

Additive variance, VA, with no dominance (k = 0) Allele frequency, p VA

Complete dominance (k = 1) VA VD Allele frequency, p

Overdominance (k = 2) VA VD Zero additive variance Allele frequency, p Allele frequency, p

Epistasis æ = + 2 G A D G = π + ( Æ ) ± A D DD Dominance value -- interaction between the two alleles at a locus Breeding value G i j k l = π + ( Æ ) ± A D DD Additive x Additive interactions -- interactions between a single allele at one locus with a single allele at another Additive x Dominant interactions -- interactions between an allele at one locus with the genotype at another, e.g. allele Ai and genotype Bkj Dominance x dominance interaction --- the interaction between the dominance deviation at one locus with the dominance deviation at another. These components are defined to be uncorrelated, (or orthogonal), so that æ 2 G = A + D