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Relationships
L. R. Schaeffer, March 1999

Cockerham (1954) showed that under Hardy-Weinberg equilibrium and joint equilibrium, the total genetic covariance between animals X and Y is, for i+j>0,

\begin{displaymath}\sigma_{G_{XY}} = \sum_{i=0}^{m} \sum_{j=0}^{m}
(a_{XY})^{i} (d_{XY})^{j} \sigma^{2}_{ij} \end{displaymath}

where m is the level of interaction effects that one is willing to include, aXY is the additive genetic relationship between animals X and Y, dXY is the dominance genetic relationship between animals X and Y, and $\sigma^{2}_{ij}$ is the particular type of genetic variance. For example,

\begin{eqnarray*}\sigma^{2}_{10} & = & \mbox{additive genetic variance}, \\
\s...
...
\sigma^{2}_{02} & = & \mbox{dominance by dominance variance}.
\end{eqnarray*}


To illustrate, suppose that the following variances are non-zero,

\begin{eqnarray*}\sigma^{2}_{10} & = & 256, \\
\sigma^{2}_{01} & = & 32, \\
\sigma^{2}_{11} & = & 16,
\end{eqnarray*}


and suppose a full-sib situation where aXY=.5 and dXY=.25, then

\begin{eqnarray*}\sigma_{G_{XY}} & = & (.5)^{1}(.25)^{0} \sigma^{2}_{10}
+ (.5)...
..._{11} \\
& = & (.5)(256)+(.25)(32)+(.125)(16) \\
& = & 138.
\end{eqnarray*}


The task, therefore, is to compute the additive and dominance relationships between pairs of individuals. The additive relationship is the proportion of alleles that two animals have in common. Because a parent transmits a random sample half of its alleles to each offspring, the additive relationship between parent and offspring is usually .5. The relationship between parent and offspring can be greater than .5 if the parents are inbred.

The dominance relationship between two individuals reflects the probability that animals have the same pair of alleles in common. This is possible if the two animals have the same parents or the same set of grandparents, for example. Because many animals have different parents, most dominance relationships are zero or very small. In small populations of animals, the percentage of pairs of animals with dominance genetic relationships could be very large, and inbreeding could be increased.

Calculation of Relationships

Smith and Allaire(1986) described a method of computing dominance and additive relationships for inbred populations by the use of a genomic table of probabilities. To illustrate, suppose we have four individuals, A, B, C, and D, and that A and B are unrelated and are the parents of C. The parents of D are A and C. In a diploid species each animal has two genomes. Let A1 and A2 represent the two genomes of animal A, and similar notation for the other animals. Create a two-way table of animals and genomes within animals as shown in Table A. Each element of the table represents the probability that alleles are identical by descent. For example, the probability that A1 equals A1 is 1, but the probability that A1 equals A2 is zero if animal A is a randomly selected, non-inbred individual.

Table A.
Genomic Table Example
Starting Values

    A B C D
    A1 A2 B1 B2 C1 C2 D1 D2
  A1 1 0            
A                  
  A2 0 1            
  B1     1 0        
B                  
  B2     0 1        
  C1         1      
C                  
  C2           1    
  D1             1  
D                  
  D2               1

Begin Table A by putting ones on the diagonals, which are the probabilities that a genome is equal to itself. All diagonals of this table are equal to 1. Then for animals that are unrelated and non-inbred add zeros for the probabilities that their two genomes are equal. Because A and B are unrelated individuals, then

Pr(A1 = B1)=0, Pr(A1 = B2)=0,
Pr(A2 = B1)=0, Pr(A2 = B2)=0.
Now C is a progeny of A and B hence let C1 be the genome inherited from parent A and let C2 be the genome inherited from parent B, then
Pr(A1 = C1)= 1 if C1=A1, or
Pr(A1 = C1)= 0 if C1=A2.
Because C1 can be either A1 or A2 with equal probability, then
Pr(C1 = A1) = .5(1+0) = .5, and
Pr(C1 = A2) = .5.
Likewise,
Pr(C2 = A1)=0, and Pr(C2 = A2)=0
because C2 represents the genome from parent B.

Let XiYj represent an element from the genomic table, then the elements in the first two rows of Table A would be

A1C1 = .5 [ A1A1 + A1A2 ] = .5(1 + 0),
A1C2 = .5 [ A1B1 + A1B2 ] = .5(0 + 0),
A1D1 = .5 [ A1A1 + A1A2 ] = .5(1 + 0),
A1D2 = .5 [ A1C1 + A1C2 ] = .5(.5 + 0).

After completing a row of the table in the above manner, use the fact that the table is symmetric, so that C1A1 = A1C1, etc. The completed table is shown in Table B.

Table B.
Genomic Table Example
Completed.
    A B C D
    A1 A2 B1 B2 C1 C2 D1 D2
  A1 1 0 0 0 .5 0 .5 .25
A                  
  A2 0 1 0 0 .5 0 .5 .25
  B1 0 0 1 0 0 .5 0 .25
B                  
  B2 0 0 0 1 0 .5 0 .25
  C1 .5 .5 0 0 1 0 .5 .5
C                  
  C2 0 0 .5 .5 0 1 0 .5
  D1 .5 .5 0 0 .5 0 1 .25
D                  
  D2 .25 .25 .25 .25 .5 .5 .25 1

Additive and dominance genetic relationships may then be derived from the numbers in the genomic table. The additive genetic relationship between animals X and Y is given by the expression:

\begin{displaymath}a_{XY} = .5 ( \mbox{X1Y1} + \mbox{X1Y2} + \mbox{X2Y1} +
\mbox{X2Y2} ). \end{displaymath}

For example, the relationship between animals A and D is

aAD = .5 ( .5 + .25 + .5 + .25 ) = .5 ( 1.5 ) = .75.

The dominance relationship between animals A and D is given by

\begin{eqnarray*}d_{AD} & = & ( \mbox{(A1D1)(A2D2)} + \mbox{(A1D2)(A2D1)} ) \\
d_{AD} & = & ( (.5)(.25) + (.5)(.25) ) = .25.
\end{eqnarray*}


Note that the same formulas are used for the diagonal elements. For example,

\begin{eqnarray*}a_{DD} & = & .5 ( 1 + .25 + .25 + 1 ) = .5 ( 2.5 ) = 1.25, \\
d_{DD} & = & ( (1)(1) + (.25)(.25) ) = 1.0625.
\end{eqnarray*}


Therefore, the genomic table is very useful if dominance genetic effects are to be included in the animal model. However, computing dominance genetic relationships could be very cumbersome for large numbers of animals. Schaeffer et al. (1989) have shown a simple method for computing the inverse to the gametic relationship matrix (genomic table), and a possible application.

The Tabular Method

Additive genetic relationships among animals may be calculated more easily than through the genomic table using a similar recursive procedure which has been called the Tabular Method (attributable to Henderson and perhaps to Wright before him). To begin, make a list of all animals that have observations in your data, and for each of these determine their parents (called the sire and dam). An example list is shown below.

Animal Sire Dam
D A B
E A C
F E D
Now extend the list of animals to include the sires and dams as follows:
Animal Sire Dam
A - -
B - -
C - -
D A B
E A C
F E D
The list of animals should be arranged in chronological order so that no animal appears in the list before its parents. Note that the sire and dam of animals A, B, and C are assumed to be unknown, and consequently animals A, B, and C are assumed to be genetically unrelated. In some instances the parentage of animals may be traced for several generations, and for each animal the parentage should be traced to a common generation.

Using the completed list of animals and pedigrees, form a two-way table with n rows and columns, where n is the number of animals in the list, in this case n=6. Label the rows and columns with the corresponding animal identification and above each animal ID write the ID of its parents as shown below.

Table C.
Tabular Method Example
Starting Values.
  -,- -,- -,- A,B A,C E,D
  A B C D E F
A 1 0 0      
             
B 0 1 0      
             
C 0 0 1      
             
D            
             
E            
             
F            
             

For each animal whose parents were unknown a one was written on the diagonal of the table (i.e for animals A, B, and C), and zeros were written in the off-diagonals between these three animals, assuming they were unrelated. Let the elements of this table (refered to as matrix ${\bf A}$) be denoted as aij. Thus, by putting a 1 on the diagonals for animals with unknown parents, the additive genetic relationship of an animal with itself is one. The additive genetic relationship to animals without common parents or whose parents are unknown is assumed to be zero.

The next step is to compute relationships between animal A and animals D, E, and F. The relationship of any animal to another is equal to the average of the relationships of that animal with the parents of another animal. For example, the relationship between A and D is the average of the relationships between A and the parents of D, who are A and B. Thus,

aAD = .5 ( aAA + aAB ) = .5(1 + 0) = .5
aAE = .5 ( aAA + aAC ) = .5(1 + 0) = .5
aAF = .5 ( aAE + aAD ) = .5(.5 + .5) = .5
The relationship table, or ${\bf A}$ matrix, is symmetric, so that aAD = aDA, aAE = aEA, and aAF = aFA. Continue calculating the relationships for animals B and C to give Table D.

Table D.
Tabular Method Example
Partially Completed.
  -,- -,- -,- A,B A,C E,D
  A B C D E F
A 1 0 0 .5 .5 .5
             
B 0 1 0 .5 0 .25
             
C 0 0 1 0 .5 .25
             
D .5 .5 0      
             
E .5 0 .5      
             
F .5 .25 .25      
             

Next, compute the diagonal element for animal D. By definition this is one plus the inbreeding coefficient, i.e.

aDD = 1 + FD.
The inbreeding coefficient, FD is equal to one-half the additive genetic relationship between the parents of animal D, namely,
FD = .5 aAB = 0.
When parents are unknown, the inbreeding coefficient is taken to be zero assuming that the parents of the individual were unrelated. After computing the diagonal element for an animal, like D, then the remaining relationships to other animals in that row are calculated as they were before. The completed matrix is given in Table E. Note that only animal F is inbred in this example. The inbreeding coefficient is a measure of the percentage of loci in the genome of an animal that has become homogeneous, that is, the two alleles at a locus are the same. Sometimes these alleles may be lethal and therefore, inbreeding is generally avoided.

Table E.
Tabular Method Example
Completed Table.
  -,- -,- -,- A,B A,C E,D
  A B C D E F
A 1 0 0 .5 .5 .5
             
B 0 1 0 .5 0 .25
             
C 0 0 1 0 .5 .25
             
D .5 .5 0 1 .25 .625
             
E .5 0 .5 .25 1 .625
             
F .5 .25 .25 .625 .625 1.125
             

Generally, the matrix ${\bf A}$ is nonsingular, but if the matrix includes two animals that are identical twins, then the two rows and columns of ${\bf A}$ for these animals would be identical, and therefore, ${\bf A}$ would be singular. In this situation assume that the two twins are genetically equal and treat them as one animal (by giving them the same registration number or identification) (see Kennedy and Schaeffer, 1989).


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This LaTeX document is available as postscript or asAdobe PDF.

Larry Schaeffer
1999-02-26