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The infinitesimal animal model can be expanded by partitioning each animal's breeding value into genomic contributions from the sire and from the dam. The term genomic is used because reference is to the average relationship over the entire genome and not to one particular locus. The genomic relationship matrix, , was described by SMITH and ALLAIRE (1992) and the inverse of this matrix was derived by SCHAEFFER et al. (1993) although computations were too prohibitive for practical applications. Another attempt at a simple inverse was presented by ALI and FREEMAN (1994). However, using the algorithm of MEUWISSEN and LUO (1992), a very rapid method for determining the elements needed for the inverse of the genomic relationship matrix can be derived. Besides the inverse of the genomic relationship matrix, the same routines can provide the genomic relationships between any pair of individuals. The dominance genetic relationship between two individuals can be computed from the genomic relationships, even for inbred individuals.
With the ability to compute dominance relationships, the estimation of any number of non-additive genetic effects and variances, assuming a randomly mating population, can be accomplished using a simple animal model, and does not require the inverses of any of the non-additive genetic variance-covariance matrices. A description of this process is provided in this paper.
A genomic model separates the additive genetic merit of an
individual into contributions from the male parent and
from the female parent.
In matrix form, a genomic model is
The infinitesimal model assumes that there are an infinite number of loci affecting the trait, each with a small and nearly equal variance. The genomic relationship matrix can be constructed using fairly simple rules. Consider the following pedigrees.
Animal | Sire | Dam |
A | - | - |
B | - | - |
C | A | B |
D | A | C |
E | D | B |
Animal | Genome | Parent1 | Parent2 |
A | A1 | - | - |
A | A2 | - | - |
B | B1 | - | - |
B | B2 | - | - |
C | C1 | A1 | A2 |
C | C2 | B1 | B2 |
D | D1 | A1 | A2 |
D | D2 | C1 | C2 |
E | E1 | D1 | D2 |
E | E2 | B1 | B2 |
The genomic relationship matrix will have order equal to twice the number of animals. The diagonals of a genomic relationship matrix are all equal to 1. The completed matrix is shown below.
A | B | A | C | D | B | ||||||
A | B | C | D | E | |||||||
A1 | A2 | B1 | B2 | C1 | C2 | D1 | D2 | E1 | E2 | ||
A1 | 1 | 0 | 0 | 0 | .5 | 0 | .5 | .25 | .375 | 0 | |
A | |||||||||||
A2 | 0 | 1 | 0 | 0 | .5 | 0 | .5 | .25 | .375 | 0 | |
B1 | 0 | 0 | 1 | 0 | 0 | .5 | 0 | .25 | .125 | .5 | |
B | |||||||||||
B2 | 0 | 0 | 0 | 1 | 0 | .5 | 0 | .25 | .125 | .5 | |
C1 | .5 | .5 | 0 | 0 | 1 | 0 | .5 | .5 | .5 | 0 | |
C | |||||||||||
C2 | 0 | 0 | .5 | .5 | 0 | 1 | 0 | .5 | .25 | .5 | |
D1 | .5 | .5 | 0 | 0 | .5 | 0 | 1 | .25 | .625 | 0 | |
D | |||||||||||
D2 | .25 | .25 | .25 | .25 | .5 | .5 | .25 | 1 | .625 | .25 | |
E1 | .375 | .375 | .125 | .125 | .5 | .25 | .625 | .625 | 1 | .125 | |
E | |||||||||||
E2 | 0 | 0 | .5 | .5 | 0 | .5 | 0 | .25 | .125 | 1 |
Because the parents of A and B are unknown, then they are assumed to be randomly drawn from the large random mating population and assumed to have no genes identical by descent between them.
Let (A1,C1) indicate an element in the above table between the A1 male parent contribution of animal A and the C1 male parent contribution of animal C, then the value that goes into that location is
The dominance genetic relationship between animals X and
Y, in general, is given by
HENDERSON (1975) presented a fast method to invert the additive genetic relationship matrix for the case when animals were not inbred. QUAAS (1976) showed how to compute the inverse when animals were inbred. The fastest method of calculating the inbreeding coefficient was presented by MEUWISSEN and LUO (1992). A combination of the results of these papers leads to a rapid algorithm for inverting the genomic relationship matrix.
Any positive definite matrix may be partitioned as
Begin with C1, the parent genomes are A1 and A2. Form a table as follows:
Genome | t | B |
C1 | 1.0 | x |
A1 | 0.5 | 1 |
A2 | 0.5 | 1 |
The diagonal element for (C1,C1) in
is equal to
1, which is equal to
,
which is
For D2, the parent genomes are C1 and C2.
Genome | t | B |
D2 | 1.0 | x |
C1 | 0.5 | 0.5 |
C2 | 0.5 | 0.5 |
Genome | t | B |
D2 | 1.0 | x |
C1 | 0.5 | 0.5 |
C2 | 0.5 | 0.5 |
A1 | 0.25 | 1 |
A2 | 0.25 | 1 |
B1 | 0.25 | 1 |
B2 | 0.25 | 1 |
The next step would be to add the 'parents' of A1 and A2,
then B1 and B2, but these 'parents' are unknown, and so no
further additions to the table are made.
Now
as
So far, the diagonals of have been either 1 or .5. For E1, the parent genomes are D1 and D2. As the animals become younger, the length of these tables can become greater, and with n generations there can be up to 2^{n}+1 elements in the table.
Genome | t | B |
E1 | 1.0 | x |
D1 | 0.5 | 0.5 |
D2 | 0.5 | 0.5 |
A1 | 0.25 | 1 |
A2 | 0.25 | 1 |
C1 | 0.25 | 0.5 |
C2 | 0.25 | 0.5 |
A1 | 0.125 | 1 |
A2 | 0.125 | 1 |
B1 | 0.125 | 1 |
B2 | 0.125 | 1 |
Notice that A1 and A2 appear twice in the table, and their coefficients in must be added together before computing . The new table, after adding coefficents is
Genome | t | B |
E1 | 1.0 | x |
D1 | 0.5 | 0.5 |
D2 | 0.5 | 0.5 |
A1 | 0.375 | 1 |
A2 | 0.375 | 1 |
C1 | 0.25 | 0.5 |
C2 | 0.25 | 0.5 |
B1 | 0.125 | 1 |
B2 | 0.125 | 1 |
Then
Animal | Genome | Parent1 | Parent2 | |
A | A1 | - | - | 1 |
A | A2 | - | - | 1 |
B | B1 | - | - | 1 |
B | B2 | - | - | 1 |
C | C1 | A1 | A2 | 0.5 |
C | C2 | B1 | B2 | 0.5 |
D | D1 | A1 | A2 | 0.5 |
D | D2 | C1 | C2 | 0.5 |
E | E1 | D1 | D2 | 0.375 |
E | E2 | B1 | B2 | 0.5 |
The inverse of
is
i | p1 | p2 | |
i | b^{i} | 0.5b^{i} | 0.5b^{i} |
p1 | 0.5b^{i} | 0.25b^{i} | 0.25b^{i} |
p2 | 0.5b^{i} | 0.25b^{i} | 0.25b^{i} |
Applying these rules, then the complete inverse of for the example animals is
Given animals D and E from the previous pedigree and assuming that the diagonals of are computed for all animals, then dominance relationships can be easily computed using the same algorithm as for computing the diagonals of . Construct a table of the rows of for D1, D2, E1, and E2, for example, as follows:
D1 | D2 | E1 | E2 | B | |
A1 | .5 | .25 | .375 | 0 | 1 |
A2 | .5 | .25 | .375 | 0 | 1 |
B1 | 0 | .25 | .125 | .5 | 1 |
B2 | 0 | .25 | .125 | .5 | 1 |
C1 | 0 | .5 | .25 | 0 | .5 |
C2 | 0 | .5 | .25 | 0 | .5 |
D1 | 1 | 0 | .5 | 0 | .5 |
D2 | 0 | 1 | .5 | 0 | .5 |
E1 | 0 | 0 | 1 | 0 | .375 |
E2 | 0 | 0 | 0 | 1 | .5 |
The 2 by 2 block between animals D and E can be calculated by multiplying columns together times the diagonal of and summing down the table. For example,
and likewise,
Naturally, one does not need to explicitly make up the above table, and only ancestor parent genomes that are in common need to be involved. Computationally, this algorithm is very fast for one pair of animals, but with 10,000 animals there are over 50 million pairs of animals for which to compute these 2 by 2 blocks, and this could take a lot of time. A short cut would be to group animals by sire and dam pairs so that the above table would be the same for all of their progeny. There could be other strategies for reducing the number of calculations as well.
Once a 2 by 2 block is calculated then the additive and dominance relationships are easily computed. In the above example, a_{DE} = .75, and d_{DE}= .15625. Then the element for the additive by additive genetic effects would be aa_{DE} = (.75)^{2}, and the element for the additive by dominance genetic effects would be ad_{DE} = .1171875, and so on. Any number of locus interactions could be considered. All non-zero elements of these matrices need to be stored to a file in full-stored mode and sorted by columns within rows, to be used in the estimation of non-additive genetic effects as described in the next section.
HENDERSON (1984) described a general non-additive genetic
model as
where is the Hadamard product of two matrices, (i.e., element by element products). To simplify the description of the proposed algorithm, consider only and effects in the model. Extension to further non-additive genetic effects should become obvious.
In a non-inbred, random mating population in joint equilibrium, then where is the additive relationship matrix, and is the dominance relationship matrix. Much is known about and its inverse, but construction of and its inverse is not known in terms of easy methods to compute them. The same is true for all higher order epistatic genetic effects and the inverses of their variance-covariance matrices. The purpose of this paper is to show that the inverses of these matrices are not needed explicitly.
The mixed model equations, MME, are
Thus, the elements of and can be derived from the routines for the genomic relationship matrix, and elements of can be derived as usual. All epistatic effects can be written as a function of the additive genetic effects and the prior variance ratios.
The epistatic genetic covariance matrices should be stored in files (non-zero elements only) in full stored mode and sorted by columns within rows. The number of these files depends on the degree of non-additivity being considered in the model. Set the starting values for , , , etc., to null vectors to begin the iteration process. The iteration process involves a repetition of the following steps.
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Larry Schaeffer