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Lecture 2: Fisher’s Variance Decomposition

Lecture 2: Fisher’s Variance Decomposition. Genotypic value. Phenotypic value -- we will occasionally also use z for this value. Environmental value. Contribution of a locus to a trait. Basic model: P = G + E. G = average phenotypic value for that genotype

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Lecture 2: Fisher’s Variance Decomposition

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  1. Lecture 2: Fisher’s Variance Decomposition

  2. Genotypic value Phenotypic value -- we will occasionally also use z for this value Environmental value Contribution of a locus to a trait Basic model: P = G + E 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

  3. C C C + a + d C + a(1+k) C + 2a C + 2a C -a C + d C + a 2a = G(Q2Q2) - G(Q1Q1) Alternative parameterizations of Genotypic values Q1Q1 Q2Q1 Q2Q2 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 k = d/a is a scaled measure of the dominance

  4. Example: Booroola (B) gene 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

  5. G = π + Æ + Æ + ± i j G i j i j The genotypic value predicted from the individual allelic effects is thus X Dominance deviations --- the difference (for genotype AiAj) between the genotypic value predicted from the two single alleles and the actual genotypic value, Average contribution to genotypic value for allele i Mean value, with b π = G ¢ f r e q ( Q Q ) G = π + Æ + Æ G i j i j i j G i j b G ° G = ± i j i j i j 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. Since parents pass along single alleles to their offspring, the ai (the average effect of allele i) represent these contributions

  6. G = π + Æ + Æ + ± i j G i j i j Residual error Predicted value G = π + 2 Æ + ( Æ ° Æ ) N + ± i j G i j 1 2 1 Intercept Independent (predictor) variable N = # of Q2 alleles Regression slope Regression residual 8 2 Æ f o r N = 0 ; e . g , Q Q > 1 1 1 < 2 Æ + ( Æ ° Æ ) N = Æ + Æ f o r N = 1 ; e . g , Q Q 1 2 1 1 1 1 2 > : 2 Æ f o r N = 2 ; e . g , Q Q 1 2 2 Fisher’s decomposition is a Regression A notational change clearly shows this is a regression,

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

  8. Mean π = 2 p a ( 1 + p k ) G 2 1 Allelic effects Æ = p a [ 1 + k ( p ° p ) ] 2 1 1 2 Æ = ° p a [ 1 + k ( p ° p ) ] 1 2 1 2 Dominance deviations ± = G ° π ° Æ ° Æ i j i j G i j Consider a diallelic locus, where p1 = freq(Q1)

  9. n ≥ ¥ ) ( X k k ( ) ( ) B V = Æ + Æ i k k = 1 B V ( G ) = Æ + Æ i j i j Average effects and Breeding Values The a values are the average effects of an allele Breeders focus on breeding value (BV) 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?

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

  11. B V ( S i r e ) = 2 ( π ° π ) G 0 B V ( S i r e ) B V ( D a m ) π ° π = + G 0 2 2 We can thus estimate the BV for a sire by twice the deviation of his offspring from the pop mean, More generally, the expected value of an offspring is the average breeding value of its parents,

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

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

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

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

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

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

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