Page 176 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

g = (1 − 0.9)g + (0.9)g = (0.1)0 + (0.9)1 = 0.9
3m,2m 2p,2m 2m,2m
g = (1 − 0.9)g + (0.9)g = (0.1)0 + (0.9)0 = 0
3m,3p 2p,3p 2m,3p
g = 1.0
3m,3m
10.4 An Alternative Approach for Calculating G
v
An alternative recursive method for the calculation of G and its inverse was presented
v
by Van Arendonk et al. (1994) using matrix notation. Their method accounts for
inbreeding and can be used to calculate a combined numerator relationship matrix
(A ) and its inverse. The matrix A = A + A , where A is the numerator relationship
a a u v u
matrix for animals for QTL not linked to the marker and A is the relationship matrix
v
for animals for MQTL linked to the marker. The inverse of A is useful for the direct
a
prediction of total additive genetic merit, i.e. additive genetic merit with information
from markers directly included.
The principles of their methodology are initially illustrated briefly using the calcula-
tion of the relationship matrix (A) among animals in the absence of marker information.
The representation of the rules for building A for animals 1 to i in matrix form is:
i
⎡ A i−1 A i−1 s i ⎤
A = ⎢ ⎥ (10.9)
i
i ⎣ ⎢s′ A i−1 a ii ⎦ ⎥
1
where s is the column vector of i − 1 elements containing two elements, , corre-
i 2
sponding to the sire or dam (if known) and zero elsewhere. A is the numerator
i−1
relationship matrix for animals 1 to (i − 1) and a is the diagonal element of A for
ii
animal i and is equal to 1 + F , where F is the inbreeding coefficient of the ith animal.
i i
Using the data in Example 10.1, the A matrix ignoring marker information is:
⎡ 1.000 0.000 0.500 0.750 0.625⎤
⎢ ⎥
⎢ 0.000 1.000 0.500 0.250 0.375 ⎥
⎢
A = 0.500 0.500 1.000 0.750 0.875⎥
0
⎢ ⎥
⎢ 0.750 0.250 0.750 1.250 1.000 ⎥
⎢ ⎣ 0.625 0.375 0..875 1.000 1.375 ⎥ ⎦
For animal 5, s′ = [0 0 0.5 0.5]; therefore, the column vector above the diagonal
5
for animal 5 (q ) in A using Eqn 10.9 can be calculated as q = A s .Thus the row
5 5 4 5
vector q′ = s′A = [0.625 0.375 0.875 1.00] and the diagonal element for animal 5,
5 5 4
a = 1 + 0.5(a ) = 1.375. Note also that given q , s can be computed as:
55 34 i i
s = A −1 q (10.10)
i i−1 i
This relationship will be used in subsequent sections when it is not possible to calcu-
late s directly.
i
Given A , for animal i − 1, Tier and Solkner (1993) demonstrated that the effect
−1
i−1
−1
of adding an additional row to A on the elements of A as:
⎡ A −1 ⎤ 0 − ⎡ss′ −s i ⎤
ii (
1
s
A −1 = ⎢ i−1 ⎥ + a − s′ A i− ) ⎢ i i ⎥ ⎥ (10.11)
i i 1 i −s′
⎣ ⎢ 0 0 ⎦ ⎥ ⎣ i 1 ⎦
160 Chapter 10