Page 109 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

The matrices of weights (W ) using Eqn 5.19 are:
i
⎛ 0 55884 − . − 1 ⎛ 0 4620 − 0 62436 ⎞ ⎛ 0 4229 − 0 3614⎞
0 62436⎞
.
.
.
.
.
W = =
1 ⎜ ⎝ − 0 62436 0 99693⎠ ⎟ ⎜ ⎝ − 0.662436 0 99693 ⎠ ⎟ ⎜ ⎝ − 0 3614 1 0000⎠ ⎟
.
.
.
.
.
and:
0 62436⎞
⎛ 0 55884 − . − 1 ⎛ 0 09685 0⎞ ⎛ 0 0 5771 0⎞
.
.
.
W = =
2 ⎜ ⎝ − 0 62436 0 99693⎠ ⎟ ⎜ ⎝ 0 0⎠ ⎟ ⎜ ⎝ 0 3614 0⎠ ⎟
.
.
.
Therefore the vector proofs of bull 3 are:
⎛ ˆ a ⎞ ⎛ . 2 68225⎞ ⎛ 11 .9971⎞ ⎛ . 113343⎞ ⎛ 6 9235⎞ ⎛ 8 058⎞
.
.
31
⎜ ⎝ ˆ a ⎠ ⎟ = W 1 ⎜ ⎝ . 1 63375⎠ ⎟ + W 2 ⎜ ⎝ 0 ⎟ ⎠ = ⎜ ⎝ 0 6644⎠ ⎟ + ⎜ ⎝ 4 3358⎠ ⎟ = ⎜ ⎝ 5 000⎠ ⎟
.
.
.
32
The contribution from the DRP of bull 3 in country 1 accounts for over 85% of the
MACE proof in both countries, although the bull has no DRP in country 2. Thus,
with only 20 daughters, parental contribution was not very large, although in gen-
eral, parental contributions will be influenced by the heritability of the traits in both
countries and the genetic correlation between them.
When a bull has a proof only in country i and not in j, its proof in country j can
be obtained (Mrode and Swanson, 1999) as:
a = PA − (g /g )(aˆ − PA ) (5.21)
ˆ
j j ij ii i i
where g is the genetic variance in country i and g the genetic covariance between
ii ij
countries i and j. Therefore, if interest was only in calculating the proof of bull 3 in
country 2, it can be obtained from the above equation as:
a = 1.63375 − (12.839/20.5)(8.059 − 2.68225) = 5.001
ˆ
32
ˆ
Equation 5.21 can be derived from Eqn 5.18 as follows. The equation for a from
32
Eqn 5.18 is:
22
21
22
(g a )aˆ = g a (aˆ + 0.5(a ˆ + g ˆ ) + g a (a ˆ + 0.5(a ˆ + g ˆ )
bull 32 par sire2 mgs2 mgd2 par sire1 mgs1 mgd1
21
+ (g a )aˆ
bull 31
where a ˆ , a ˆ and g ˆ are the proofs for the sire, MGS and solution for the MGD
sirej mgsj mgdj
ii −1
bull par
in country j, respectively, and g are the inverse elements of G . Since a = 2a for
−1
bull 3, multiplying the above equation by (2a ) gives:
par
21
22
22
21
g aˆ = g (PA ) + g (PA ) − g aˆ
32 2 1 31
21
22
22
g a = g (PA ) − g (a − PA )
ˆ
ˆ
32 2 31 1
22
21
ˆ
a = PA − g /g (aˆ − PA )
32 2 31 1
a = PA − g /g (aˆ − PA )
ˆ
32 2 21 22 31 1
Thus the proof of a bull in country j is dependent on the parent average of the
bull in country j and the Mendelian sampling of the bull in the ith country.
Partitioning the proof of bull 2 with records in both countries and a
maternal grandson (bull 3) is as follows. The country deviations for bull 2 in
both countries are:
CD = y − m = 9.9717 − 7.268 = 2.7037
21 21 1
Multivariate Animal Models 93