Page 31 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

in the same herd and using the same parameters as in Example 1.6, predict the breed-
ing value of the bull calf for ADG and its accuracy.
From the parameters given:
2
p = p = p = s = 6400
11 22 33 y
2
1
1
p = cov(y , y ) = s a = (2752) = 1376
12 1 2 2 2
p = p = 1376
13 12
p = 0
23
2
g = s a = 2752
11
1
2
g = g = s a = 1376
12 13 2
The index equations are:
− 1
⎡ 1 b ⎤ ⎡ 6400 1376 1376⎤ ⎡2752 ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎥ ⎢
⎢ 2 b = 1376 6400 0000 ⎥ = 1376 ⎥
⎢
⎥ ⎢
⎢ ⎣ 3 b ⎦ ⎣ ⎣ 1376 0000 6400⎥ ⎦ ⎣ ⎢1376 ⎥ ⎦
Solutions to the above equations are b = 0.372, b = 0.135 and b = 0.135.
1 2 3
The index is:
I = 0.372(900 − m) + 0.135(800 − m) + 0.135(450 − m)
where m is the herd average. The accuracy is:
[
r= (0.372(2752) + 0.135(176) + 0.135(176))/2752] = 0 712
.
The high accuracy is due to the inclusion of information from both parents.
Using means of records from animal and relatives
Example 1.8
It is given that average protein yield for the first two lactations for a cow (y ) called
˜
1
˜
Zena is 230 kg and the mean protein yield of five other cows (y ), each with two lacta-
2
tions, is 300 kg. If all cows are all daughters of the same bull and no other relation-
ship exists among them, predict the breeding value of Zena, assuming a heritability of
0.25, a repeatability (t) of 0.5, standard deviation of 34 kg and herd average of 250 kg
for protein yield in the first two lactations.
From the given parameters:
2
2
2
2
g = s = h s = 0.25(34 ) = 289
11 a y
and:
1
2
g = covariance between half-sibs = (s a ) = (289) = 72.25
1
12 4 4
From calculations in Section 1.3.2:
⎛ (1 − t)⎞
y () =
p = var t + 2
11 1 ⎜ ⎝ n ⎠ ⎟ s y
2
= (0.5 + (1 − 0.5)/2)34 = 867
Genetic Evaluation with Different Sources of Records 15