Page 186 - Linear Models for the Prediction of Animal Breeding Values

2
2
2
predicted. From the genetic parameters in Example 10.1, s = s + s = 0.3 +
a
u
q
2
2
2
0.1 = 0.4 and s = 0.6; therefore, a = s /s = 0.6/0.4 = 1.5. The Z matrix in
e e a
Eqn 10.21 is now an identity matrix considering animals with records.
The matrix A below was calculated as the sum of A and A . The
a u(0.3/0.4) v(0.1/0.4)
matrices A and A have been calculated in Examples 10.2 and 10.4.
u v
é 1.000 0.000 0.500 0.800 0.705ù
ê ú
ê 0.000 1.000 0.500 0.200 0.295 ú
ê
A a = 0.5000 0.500 1.000 0.710 0.814ú
ê ú
ê 0.800 0.200 0.710 1.228 1.018 ú
ê ë 0.705 0.295 00.814 1.018 1.338 ú û
−1
The vector s for the ith animal needed to calculate A is shown in Table 10.2.
i a
−1
The matrix A calculated using Eqn 10.11 is:
a
é 2.2641 0.4854 - 0.4101 - 1.2080 - 0.1314ù
ê - ú
8
ê 0.4854 1.5007 1.0218 0.0030 0.0327 ú
ê
A -1 = - 0.4101 - 1.0218 2.7536 - 0.3673 - 0.9544ú
a
ê ú
ê - 1.2080 0.00030 - 0.3673 2.9811 - 1.4088 ú
ê ë ë - 0.1314 0.0327 - 0.9544 - 1.4088 2.4619 ú û
The MME (Eqn 10.23) for the example data is as follows:
⎡ 2.000 0.000 1.000 0.000 1.000 0.000 0.000⎤ ⎡ ˆ ⎤ ⎡ 15.3⎤
⎥
⎢ 1.000 ⎢ b 1 ⎥ ⎢ ⎥
⎢ 0.000 3.000 0.000 1.000 0.0000 1.000 ⎥ ˆ ⎥ ⎢ 17.5 ⎥
⎢ 1.000 0.000 4.396 0.728 − 0.615 − 1.812 − 0.197⎥ ⎢ b ⎢ 2 2 ⎥ ⎢ 6.8⎥
⎢ ⎥ ˆ a ⎢ ⎥
⎥
⎢ 0.000 1.0000 0.728 3.251 − 1.533 0.005 0.049 ⎢ ⎢ 1⎥ = ⎢ 4.5 ⎥
⎥
⎢ 1.000 0.000 − 0.615 − 1.533 5.130 −00.551 − 1.432 ⎥ ⎢ ˆ a 2 ⎥ ⎢ 8.5 ⎥
⎢ ⎥ ⎢ ˆ a ⎢ ⎥
⎢ 0.000 1.000 − 1.812 0.005 − 0.551 5.472 − 2.113 ⎥ ⎢ 3 ⎥ ⎥ ⎢ 6.0 ⎥
⎢ 0.000 1.0000 − 0.049 − − ⎥ ˆ a 4⎦ ⎢ ⎥
⎣
⎣ 0.197 1.432 2.113 4.693 ⎦ ⎣ 7..0 ⎦
Table 10.2. Vector (s ) with contributions from ancestors (animals
i
1 to 4) to animals 2 to 5, using the pedigree in Example 10.1.
Elements in s relating to animal
i
Animal 1 2 3 4

2 0.0000
3 0.5000 0.5000
4 0.5900 −0.0100 0.4200
5 0.0534 −0.0133 0.3877 0.5722

170 Chapter 10