Page 191 - Linear Models for the Prediction of Animal Breeding Values 3rd Edition

Genotype at the two linked markers
Animal Sire Dam Marker 1 Marker 2

1 – – 11 22
2 – – 33 44
3 1 3 12 44
4 4 3 21 14
Assuming no double crossing over, the genetic parameters as in Example 10.2 and
letting p and q be equal to 0.8 and 0.2, respectively, predict the effects of the sex of
the calf, additive genetic effects (breeding values) not linked to the MQTL for animals
and additive genetic effects for the MQTL alleles of animals.
The alleles at the MQTL can be defined from the genotypes at the two linked
marker loci. Thus the paternal and maternal MQTL alleles for animal 1 will be v
s11
and v , respectively. Correspondingly, those for animal 4 will be v and v ,
s22 o21 o14
respectively. As in Example 10.3, a = 0.6./0.3 = 2 and a = 0.6/0.05 = 12. With the
1 2
assumption of no double crossing over, for calf 3, v = v (calf 2); therefore, the
o44 m44
row and column for v are deleted from G and the MME.
o44 v
The design matrix Z is an identity matrix of order four and W is:
⎡ 11 0 000 0⎤
0
⎢ 11 0 0000 ⎥
⎡ 11 00000⎤ ⎢ ⎥ ⎥
⎢ 00 11 000 ⎥ ⎢ ⎢ 0 011 0 0 0⎥ ⎥
W = ⎢ ⎥ and W W =′ ⎢ 00 1 2 1 0 0 ⎥
⎢ 000 11 00⎥
⎢ ⎥ ⎢ 000 11 00 ⎥
⎣ 00000 11 ⎦ ⎢ ⎥
⎢ 000 00 11 ⎥
⎢ ⎥
⎣ 000 00 11 ⎦
The covariance matrix G is:
v
é 1.000 0.000 0.000 0.000 0.200 0.800 0.040ù
ê ú
ê 0.000 1.000 0.000 0.0000 0.800 0.200 0.160 ú
ê 0.000 0.000 1.000 0.000 0.000 0.000 0.000ú
ê ú
G = 0.000 00.000 0.000 1.000 0.000 0.000 0.800
v ê ú
ê 0.200 0.800 0.000 0.000 1.000 0.3320 0.200 ú
ê ú
ê 0.800 0.200 0.000 0.000 0.320 1.000 0.064 ú
ê ú
0
ë 0.040 0.160 0.000 0.800 0.200 0.064 1.000 û
The calculation of G with elements g(i,j) for the first few animals is as follows.
v
For the first two animals, both parents are unknown; therefore, the diagonal element
of G for either the paternal or maternal allele is 1 for these animals. Calf 3 inherited
p
marker haplotype v from its sire; therefore, r = p in Eqn 10.7. Thus:
s12 o
g( ) = (1 − p)g( ) + pg( ) = q(1) + p(0) = q = 0.2
3p3p,1p1p 1p1p,1p1p 1m1m,1p1p
g( ) = (1 − p)g( ) + pg( ) = q(0) + p(1) = p = 0.8
3p3p,1m1m 1p1p,1m1m 1m1m,1m1m
g( ) = (1 − p)g( ) + pg( ) = q(0) + p(0) = 0
3p3p,2p2p 1p1p,2p2p 1m1m,2p2p
g( ) = (1 − p)g( ) + pg( ) = q(0) + p(0) = 0
3p3p,2m2m 1p1p,2m2m 1m1m,2m2m
Use of Genetic Markers in Breeding Value Prediction 175