Page 22 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

Expected response to selection on the basis of mean of repeated records is:
R = 2 [ t + (1 − t) / n]
ay y
,
ih s

Example 1.2
Assume that a cow has a mean yield of 8000 kg of milk for first and second lacta-
tions. If the phenotypic standard deviation and heritability of milk yield in the first
two lactations are 600 kg and 0.30, respectively, and the correlation between first and
second lactation yields is 0.5, predict the breeding value of the cow for milk yield in
the first two lactations and its accuracy. Assume that the herd mean for first and
second lactations is 6000 kg.
From Eqn 1.7:
a ˆ = b(8000 − 6000)
cow
with:

b = 2(0.3)/(1 + (2 − 1)0.5)) = 0.4
Therefore:

a ˆ = 0.4(8000 − 6000) = 800 kg
cow
and:
r = 04. = 0.632
a,a ˆ

1.4 Breeding Value Prediction from Progeny Records

For traits where records can be obtained only on females, the prediction of breeding
values for sires is usually based on the mean of their progeny. This is typical of the
dairy cattle situation, where bulls are evaluated on the basis of their daughters. Let y ˜
i
be the mean of single records of n progeny of sire i and assume that the progeny are
only related through the sire (paternal half-sibs), and so the breeding value of sire i is:
ˆ
a = b(y˜ − m) (1.8)
i i
where:
˜
˜
b = cov(a, y)/var(y)
Now:
cov(a, y) = cov(a, 1 a + 1 a + Σe/n)
˜
2 s 2 d
where a is the sire breeding value and a represents the breeding value for the dams.
s d
Therefore:
˜ 2
s
cov(a, y) = 1cov(a, a ) = 1s a
2 2
Using the same principles in as in Section 1.3.2:
˜ 2
var(y) = [t + (1 − t)/n]s y
6 Chapter 1