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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,

where

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

and suppose a full-sib situation where

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

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. |

Pr(A1 = C1)= 1 if C1=A1, or |

Pr(A1 = C1)= 0 if C1=A2. |

Pr(C1 = A1) = .5(1+0) = .5, and |

Pr(C1 = A2) = .5. |

Pr(C2 = A1)=0, and | Pr(C2 = A2)=0 |

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.

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:

For example, the relationship between animals A and D is

The dominance relationship between animals A and D is given by

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

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 |

Animal | Sire | Dam |

A | - | - |

B | - | - |

C | - | - |

D | A | B |

E | A | C |

F | E | D |

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.

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 )
be denoted as *a*_{ij}. 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,

a_{AD} |
= .5 ( a_{AA} + a_{AB} ) |
= .5(1 + 0) = .5 |

a_{AE} |
= .5 ( a_{AA} + a_{AC} ) |
= .5(1 + 0) = .5 |

a_{AF} |
= .5 ( a_{AE} + a_{AD} ) |
= .5(.5 + .5) = .5 |

Table D.

Tabular Method Example

Partially Completed.

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.

a_{DD} = 1 + F_{D}. |

F_{D} = .5 a_{AB} = 0. |

Table E.

Tabular Method Example

Completed Table.

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 is nonsingular, but if the matrix includes two animals that are identical twins, then the two rows and columns of for these animals would be identical, and therefore, 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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