Page 151 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

one fixed lactation curve was fitted for all cows and a vector (v) of actual daily fat
yield (kg) from days 4 to 310 can be obtained as:
310 nf
ˆ
v = Fb = å å f b ˆ j 2
= i 4 = j 1 ij
where F is a matrix of Legendre polynomials evaluated from 4 to 310 DIM, as
described in Appendix G. From the above equation, v , for instance, is:
38
38 [ 0 7071 - - ˆ =
=
v . 0 9525 0 6441 0 0176 - . 2 12 2001
.
0 6205]b
.
.
.
For the DIM in the example data set, v is:
(DIM) 4 38 72 106 140 174 208 242 276 310
v = [10.0835 12.2001 12.6254 12.2077 11.5679 11.0407 10.9156 11.1111 11.2500 10.8297]
A graph of the fixed lactation curve can be obtained by plotting the elements of v
against DIM.
The EBV for animals and solutions for permanent environmental effect obtained
by solving the MME are those for daily fat yield. To obtain EBV or solutions for pe
effects on the nth DIM, these solutions are multiplied by n. This is implicit from the
assumptions stated earlier of genetic correlations of unity among TD records. Thus
EBVs for 305 days, shown in the table of results above, were obtained by multiplying
the solutions for daily fat yield by 305.

PARTITIONING BREEDING VALUES AND SOLUTIONS FOR PERMANENT ENVIRONMENTAL EFFECTS
Similar to the repeatability model, EBVs of animals can be partitioned in terms of
contributions from various sources, using Eqn 3.8. The YD for an animal is now cal-
culated as the average of corrected TD records. The correction is for effects of HTD,
fixed regressions and pe. Thus for cow 6 with five TD records, YD is:
6
ˆ
ˆ
−1
YD = (Q′Q) Q′(y − X b − X b − pˆe)
6 6 1 1 2 2
ˆ
ˆ
withy − X b − X b − ˆ pe = y c
1
2
2
1
6
⎡10.4 ⎤ ⎡3.1893 ⎤ ⎡ 10..0835⎤ − ⎡ 1.68553⎤ ⎡ − 1.1875⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ − ⎥ ⎢ ⎥
⎢ 12.3 ⎥ ⎢ 3.3099 ⎥ ⎢ 12.2001 ⎥ ⎢ 1.6853 ⎥ ⎢ −11.5247 ⎥
⎢
⎥
⎥ ⎢
= ⎢13.2 − ⎢3.3897 ⎥ − 12.6254⎥ − − ⎢ 1.6853 = − 1.1298⎥
⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎢ 11.6 ⎥ ⎢ 0.6751 ⎥ ⎢ 12.2077 ⎥ ⎢ − 1.6853 ⎥ ⎢ 0.4025 ⎥
⎢ ⎣ 8.4 ⎥ ⎦ ⎢ ⎣ 0.0000 ⎥ ⎦ ⎢ ⎣ 11.5679 ⎥ ⎦ ⎢ ⎣ − 1.6853 ⎥ ⎢ − 1.4826 ⎥ ⎦
⎦ ⎣
and:
⎛ ⎡ − 1.1875⎤⎞
⎜ ⎢ ⎢ − 1.5247 ⎥⎟
⎥⎟
1
YD = (QQ − 1 ′( )y c = ⎜ ⎜ ′ q ⎢ − 1.1298⎥⎟ = − 4.9221/5 = − 0.9844
′ ) Q
6
5 ⎜ 6 ⎢ ⎥⎟
⎜ ⎢ 0.4025 ⎥⎟
⎥⎟
⎜ ⎝ ⎢ ⎣ − 1.48226 ⎠
⎦
Analysis of Longitudinal Data 135