PBG 650 Advanced Plant Breeding Module 5: Quantitative Genetics – Genetic variance: additive and dominance

The variance of a variable X is: V( X ) = E[( X i -  X ) 2 ] = E( X i 2 ) -  X 2 The covariance of variable X and variable Y is: Cov( X,Y ) = E[( X -  X )( Y -  Y )] = E( XY ) -  X  Y Variance and Covariance - definition

The variance of a constant is zero V( c ) = 0   V( c+X ) = V( X ) The variance of the product of a variable and a constant is the constant squared times the variance of the variable V( cX ) = c 2 V( X ) The variance of a sum of random variables is the sum of the variances plus twice the covariance between the variables V( X + Y ) = V( X ) + V( Y ) + 2Cov( X,Y ) Properties of variances

Application to a genetic model P = G + E G = A + D + I P = A + D + I + E Because there are no covariances among the components G ijkl =  + (  i +  j +  ij ) + (  k +  l +  kl ) + I ijkl

Additive genetic variance Variance of breeding values GenotypeFreq.Breeding Value (Breeding Value) 2 Freq. x Value 2 A1A1A1A1 p2p2 2q2q 4q224q22 4p2q224p2q22 A1A2A1A2 2pq (q - p)  (q - p) 2  2 2pq(q - p) 2  2 A2A2A2A2 q2q2 -2p  4p224p22 4p2q224p2q22 (No adjustment for the mean is necessary because the mean of breeding values is zero) When p=q =1/2 σ A 2 = (1/2) a 2 When d= 0 σ A 2 = 2 pqa 2

Dominance Variance Variance of dominance deviations GenotypeFreq. Dominance Deviations (Dominance Deviations) 2 Freq. x Dev. 2 A1A1A1A1 p2p2 -2q 2 d4q4d24q4d2 4p2q4d24p2q4d2 A1A2A1A2 2pq+2pqd4p2q2d24p2q2d2 8p3q3d28p3q3d2 A2A2A2A2 q2q2 -2p 2 d4p4d24p4d2 4p4q2d24p4q2d2 (No adjustment for the mean is necessary because the mean of dominance deviations is zero) When p=q =1/2 σ D 2 = (1/4) d 2 When d=0, σ D 2 = 0

Genetic variance For a single locus (It can be shown that the Cov(A,D) = 0)

Regression of genotypic values on allele number GenotypeFrequency Number of A 1 alleles ( X i ) Observed Genotypic Values ( Y i ) A2A2A2A2 q2q2 0 P-aP-a A1A2A1A2 2pq 1 P+dP+d A1A1A1A1 p2p2 2 P+aP+a Mean 2p2p  = P+ a(p-q)+2pqd Mean( X ) = (Σ f i X i ) = q 2 (0) + 2pq(1) + p 2 (2) = 2p(q+p) = 2p = p 2 (2 2 ) + 2pq(1 2 ) +q 2 (0 2 ) – (2p) 2 = 2pq P=MP=midparent value M

Covariance of genotypic values and allele number GenotypeFrequency Number of M 1 alleles ( X i ) Observed Genotypic Values ( Y i ) Adjusted Genotypic Values M2M2M2M2 q2q2 0 P-aP-a-a- M M1M2M1M2 2pq 1 P+dP+dd- M M1M1M1M1 p2p2 2 P+aP+aa- M Mean 2p2p P+ a(p-q)+2pqd 0 = p 2 (2)(P+ a ) + 2 pq (1)(P+ d ) + q 2 (0)(P- a ) – (2 p )(P+ a(p-q )+2 pqd ) = 2pq[a+d(q-p)] = 2pq  Same result with scaled values ( a, d, -a ) or the adjusted genotypic values: = p 2 (2)(a-M) + 2pq(1)(d-M) +q 2 (0)(-a-M)-(2p)(0)   2pq  M

Regression cont’d.

Genetic Variances - Example Options for estimating variances –Use formulas with known values of a and d –Calculate breeding values and dominance deviations, and estimate their variances –Regress observed values on number of Z 1 alleles Observed ValueFrequency Z2Z2Z2Z2 6 q2q2 Z1Z2Z1Z2 12 2pq Z1Z1Z1Z1 14 p2p2 Example from Falconer & Mackay p =0.6 q =0.4

Option 1 – use formula Observed ValueFrequency Genotypic Values Z2Z2Z2Z Z1Z2Z1Z Z1Z1Z1Z p =0.6 q =0.4 P = (6+14)/2 = 10 a =14-10=4 d =12-10=2

Option 2 – calculate variances directly GenotypeFrequency Genotypic value Breeding value Dominance deviations Z2Z2Z2Z Z1Z2Z1Z Z1Z1Z1Z Mean ( Σf i Y i ) 000 σ G 2 = 0.16(-5.76) (0.24) (2.24) = σ A 2 = 0.16(-4.32) (-0.72) (2.88) = σ D 2 = 0.16(-1.44) (0.96) (-0.64) =

Option 3 – Regress values on allele number GenotypeFrequency Number of Z 1 alleles ( X i ) Observed Genotypic Values ( Y i ) Z2Z2Z2Z2 q 2 = P- a = 6 Z1Z2Z1Z2 2pq = P +d = 12 Z1Z1Z1Z1 p 2 = P+ a =14 Mean 2p2p P+ a(p-q)+2pqd Mean( X ) = 0.16(0) (1) (2) = 1.20 = 0.16(0 2 ) (1 2 ) +0.36(2 2 ) – (1.20) 2 = = 2pq p =0.6 q =0.4

Option 3 – Regress values on allele number GenotypeFrequency Number of Z 1 alleles ( X i ) Observed Genotypic Values ( Y i ) Adjusted Genotypic Values Z2Z2Z2Z2 q 2 = P- a = 6 -a- M=-5.76 Z1Z2Z1Z2 2pq = P +d = 12 d- M=0.24 Z1Z1Z1Z1 p 2 = P+ a =14 a- M=2.24 Mean 2 p= 1.20 P+M= = 0.16(0)(6) (1)(12) (2)(14) – (1.20)(11.76) = = 2 pq  The result is the same if we use the adjusted genotypic values: = 0.16(0)(-5.76) (1)(0.24) (2)(2.24) – (1.20)(0) =

Regression cont’d.

Regression of genotypic values on allele number 0Z2Z20Z2Z2 1Z1Z21Z1Z2 2Z1Z12Z1Z1 genotypic value  = 3.6 breeding value Excel

Magnitude of genetic variances With no dominance, all genetic variance is additive and maximum genetic variance occurs when p=q=0.5 With complete dominance –maximum additive genetic variance occurs when the unfavorable allele has a frequency of q=0.75 –maximum dominance variance occurs when q=0.5 –maximum genetic variance occurs when q 2 =0.5 (q=0.71)

Effect of Inbreeding (selfing) on Variances Among Lines Within Lines Total Total genetic variance increases with selfing!! Hallauer, Carena and Miranda, 2010