Page 24 - Linear Models for the Prediction of Animal Breeding Values

Therefore:
1 2 2
b = s a /[t + (1 − t)/n]s y
4
Using the same calculations for obtaining b in Eqn 1.8:
b = n/(n + k)
The b value is half of the value of b in Eqn 1.8, thus the predicted breeding value of
a future daughter of the sire is equal to half the EBV of the sire. The performance of
a future daughter of the sire can be predicted as:
y = M + a ˆ
daugh.
where M is the management mean.
The accuracy of the predicted breeding value of the future daughter is:
y , ) / (var(
r a y = cov( a daugh. a) var(
y))
,
This could be expressed as:
1 2 2 1 h
a r ., y = 4 h s y = 4
2
2
daugh ⎛ (1 − 1 h )⎞ (1 − 1 h )
2
2
2 2 1 h + 4 2 1 h h + 4
⎝ n ⎠ n
h s y ⎜ 4 ⎟ s y 4
n
= 1
2
n+ k
which is equal to half of the accuracy of the predicted breeding value of the sire.
Reliability of the predicted breeding value equals 1 n/(n + k), which is one-quarter of
4
the reliability of the bull proof.
Example 1.3
Suppose the fat yield of 25 half-sib progeny of a bull averaged 250 kg in the first
lactation. Assuming a heritability of 0.30 and herd mean of 200 kg, predict the breed-
ing value of the bull for fat yield and its accuracy. Also predict the performance of a
future daughter of this bull for fat yield in this herd.
From Eqn 1.6:

a ˆ = b(250 − 200)
bull
with:
2
2
b = 2n/(n + (4 − h )/h ) = 2(25)/(25 + (4 − 0.3)/0.3) = 1.34
a ˆ = 1.34(250 − 200) = 67 kg
bull
r ~ = ( n / ( n k)) = é25 / 2 + (4 0 .3)
+
-
( 5
a,y ë / . ) ù =0 3 û . 0 82
The future performance of the daughter of the bull is:
y = (0.5)a + herd mean
bull
= 33.5 + 200 = 233.5 kg

8 Chapter 1