Page 252 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

Thus the heritabilities for BW and CD are 0.14 and 0.18, respectively, with a
genetic correlation of 0.62 between the two traits.
The model in Eqn 13.13 was used for the analysis of BW, thus b is the vector of
1
solutions for origin of heifer, calving season and sex of calf and u is the vector of
1
solutions for sire effects. The same effects were fitted for CD, with t being the vector
of solutions for the fixed effects and n for the sire effects. Let q be as follows:
b′ = (d , d , s , s , f , f )
1 1 2 1 2 1 2
u′ = (uˆ , uˆ , uˆ , uˆ , uˆ , uˆ )
1 11 12 13 14 15 16
t′ = (d′, d′, s′, s′, f′, f′)
1 2 1 2 1 2
n′ = (n , n , n , n , n , n )
1 2 3 4 5 6
where d (d′), s (s′) and f (f′) are level i of the effects of heifer origin, calving season and
i i i i i i
ˆ
sex of calf, respectively; for BW (CD), u and n are the solutions for the sire j for BW
1j j
and CD, respectively.
The matrix X , which relates records for BW to the effects of heifer origin, calving
1
season and sex of calf, can be set by principles already outlined in previous chapters.
For the example data, all fixed effects affecting BW also affect CD; therefore, H is an
identity matrix and X = X . Similarly, the matrix Z = Z . The remaining matrix in
2 1 1 2
Eqn 13.17 can be obtained through matrix multiplication and addition.
Equation 13.17 needs starting values for t and n to commence the iterative process.
(0)
(0)
The starting values used were solutions (t and n ) from Eqn 13.17 with W [i−1] = I,
q [i−1] = a vector of (0,1) variables (1, difficulty; 0, otherwise) and n [i−1] = 0. The solu-
tions to Eqn 13.17 using these starting values are shown in Table 13.3, with equa-
tions for the second levels of calving season and sex of calf effects set to zero because
of dependency in the systems of equations. Using these solutions, the calculation of
(0)
(0)
q and W in the next round of iteration are illustrated for the first and last two
animals in the example data.
First, m in Eqn 13.18 is calculated for these animals using Eqn 13.24.
For animals 1 and 2:
X t + Z n = (d′ + sˆ′ + f′ + n ˆ ) = 0.1873 + −0.0874 + 0.2756
2 2 1 1 1 1
+ (−0.1180) = 0.2575
Therefore, from Eqn 13.24, using the mean of birth weight, m is:
1
m = 0.2575 + 0.1155(41 − 43.02) = 0.0242
1
and:
m = 0.2575 + 0.1155(37.5 − 43.02) = −0.3800
2
For animals 46 and 47:
X t + Z n = (d′ + sˆ′ + f′ + n ˆ ) = 0.1484 + −0.0874 + 0.0 + 0.0079 = 0.0690
2 2 2 1 2 6
Therefore, from Eqn 13.22:
m = 0.0690 + 0.1155(43.5 − 43.02) = 0.1244
46
and:
m = 0.0690 + 0.1155(36.5 − 43.02) = −0.6841
47

236 Chapter 13