Page 205 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

(Solberg et al., 2009). Liu et al. (2011) have demonstrated that the optimum level of
RP may differ for traits of different heritabilities but tends to vary between 10 and
20% of the genetic variance.
A mixed linear model with polygenic effects included is of this form:

y = Xb + Wu + Zg + e (11.13)
where u is the vector of random residual polygenic effects, W is the design matrix that
relates records to animals and other terms are defined as in Eqn 11.6. If a SNP-BLUP
model is fitted, the MME to be solved are:
æ XR X X ¢R W XR Z ö æ ö ˆ æ XR y ö
¢
¢
¢
-1
-1
-1
-1
b
ç ÷ ç ÷ ç ç ÷
¢
¢
ç WR X W ¢R W + A a 1 W ¢R Z÷ ç ÷ ˆ u = ç WR y÷ (11.14)
-1
-1
-1
-1
-1
ç ç -1 -1 -1 1 ÷ ç ÷ ç ç -1 ÷
¢
+
ø è ø
è ¢ ZR X ¢ Z R W ¢ ZR ZIa 2 ÷ ç ÷ ˆ g è ZR y ÷ ø
2
2
where a = s /s , with s equal to the chosen percentage of the additive genetic vari-
2
1 e u u
2
ance fitted as polygenic effect and a = s /s , with s calculated to account for the
2
2
2 e g g
percentage of additive genetic variance attributed to the polygenic effect. Thus a =
2
2
2
2
2
(s − s )/m with m = number of markers or 2Σp (1 – p )*[s /(s − s )].
2
a u j j e a u
However, if a GBLUP model is to be fitted, then the mixed linear model is:
y = Xb + Wu + Wa + e (11.15)
where a is the vector of DGVs and all other terms are as defined in Eqn 11.8. The
MME to be solved are:
⎛ XR X X ′R W XR Z* ⎞ ⎛ ⎞ ˆ b ⎛ XR y ⎞
−1
−1
′
−1
−1
′
′
⎜ −1 −1 −1 −1 ⎟ ⎜ ⎟ ⎜ −1 ⎟
′
ˆ u
⎜ W ′R X W ′RW + A a 1 W ′RW ⎟ ⎜ ⎟ = WR y ⎟ (11.16)
⎜ ⎜
⎜ −1 −1 − −1 −1 ⎟ ⎜ ⎟ ˆ a ⎜ −1 ⎟
′
′
′
⎝ WR X W ′R W WR W + G a 2 ⎠ ⎝ ⎠ ⎝ WR y ⎠
where:
2
2
2
2
2
a = s /s and a = s /(s − s )
1 e u 2 e a u
Example 11.5
The data in Example 11.1 is analysed assuming the same genetic parameters to
compute DGVs for the reference animals without using weights. It is also assumed
that 10% of the additive genetic variance is due to residual polygenic effect in the
model. The analysis has been carried out using both Eqns 11.14 and 11.16 without
any weights.
2
2
Given that s = 35.241, then s = 0.1*35.241 = 3.5241. Therefore, for both
a u
2
Eqns 11.14 and 11.16, a = s /s = 245/3.5241 = 69.521. However, for Eqn 11.14,
2
1 e u
a = s /s and now equals 2Σp(1 – p)*[s /(s − s )] = 3.5383*(245/(35.241 – 3.5241) =
2
2
2
2
2
2 e g j j e a u
2
2
2
27.332, while in Eqn 11.16, a = s /(s − s ) = 7.725.
2 e a u
The matrix Z in Eqn 11.14 is as defined in Example 11.2, while W in Eqns 11.14
and 11.16 have been set up in Example 11.3. The matrix A is for the eight reference
−1
animals. All matrices for Eqns 11.14 and 11.16 have therefore been defined. The mean
and SNP solutions from solving the MME in Eqn 11.14 are given in Table 11.3.
Computation of Genomic Breeding Values and Genomic Selection 189