INBREEDING AND DIVERSITY – PART 2

Adding The Row ‘Emma’

Now that we have added a second row to form a column, a comment is in order that will greatly reduce your labor. Look ahead to the completed Table 3 e, for a moment. If you draw a diagonal line down the matrix from the cell Edmund-Edmund to the cell X-X, the numbers above that line will be the same as the numbers below that line. The diagonal is darkly shaded in the completed table in Table 3 e (step 4 of this Bob-Victoria breeding exercise). The shaded upper-right “triangle” in the completed table can be flipped around the diagonal axis to fill in the lower part of the table. The row for Edmund contains exactly the same entries as the column for Edmund. So to get started, we copy the entry from Edmund-Emma into Emma-Edmund, which is ‘0’. The rest of the entries follow as in step (1):

Emma – Tom = 0	Emma – X = 0/2 + 3/8	mma – Bob = 0/2 + 0/2 = 0	Emma – Annie = 0
Emma – Vin = 0/2 + ½	Emma – Jack = 0/2 = 0	Emma – Vic = ½ + ¼

Since Emma is “related to herself” (“has the same genes” is another way of saying this) by a factor of one, look across her row and see who else she is related to. To Vincent, it is 1 (herself as one of the parents) divided by 2 (since she is only one of the two parents.) In the same way, relationship to Victoria is calculated by dividing 1.5 (½ for Vincent’s R value and 1 for her own as Victoria’s dam) by 2, to give the ¾ you see in the table. Emma has no relation to Bob because she has no relation to his parents Jack or Annie (therefore 0/2 in each case).

And now, back to where we were, in the early stages of constructing that coefficient of inbreeding Table 3:

Table 3 c.

	Edmund	Emma	Tom	Annie	Tom Jack	Edm/Emma Vincent	Jacl/An Bob	Vin/Em Victoria	Bob/Vic X
Edmund	1	0	0	0	0	1/2	0	1/4	1/8
Emma	0	1	0	0	0	1/2	0	3/4	3/8

We continue to build our table. Remember that for convenience more than anything else, we put the oldest ones on the left, and X, the Bob-Victoria pup, on the right.

Rows ‘Tom’ through ‘Victoria’

The next six rows were filled in as outlined in steps (1) and (2) above. Note that the values in the triangle below the diagonal are the same as in the upper triangle, flipped around the diagonal. When we look at Victoria’s pedigree, though, we see something that requires special attention. We said earlier that we can use the tabular method to find inbreeding coefficients, and Victoria is inbred (on Emma).

Table 3 d.

	Edmund	Emma	Tom	Annie	Tom/ N/A Jack	Edm/Emma Vincent	Jacl/An Bob	Vin/Em Victoria	Bob/Vic X
Edmund	1	0	0	0	0	1/2	0	1/4	1/8
Emma	0	1	0	0	0	1/2	0	3/4	3/8
Tom	0	0	1	0	1/2	0	1/4	0	1/8
Annie	0	0	0	0	1	0	1/2	0	1/4
Jack	0	0	1/2	0	1	0	1/2	0	1/4
Vincent	1/2	1/2	0	0	0	1	0	3/4	3/8
Bob	0	0	1/4	1/2	1/2	0	1	0	1/2
Victoria	1/4	3/4	0	0	0	3/4	0	1 1/4	5/8

In the entry Victoria-Victoria, we see that the entry is 1 + ¼. Where did the ¼ come from and what does it mean? The 1 is Victoria’s relationship to herself in the absence of inbreeding. When an animal is inbred, or if you are not sure if an animal is inbred, you determine the coefficient of inbreeding from the table entry that corresponds to the relationship between its two parents. Victoria’s parents, Vincent and Emma, have a coefficient of relationship of ½. If we divide by 2, we get the ¼ in the table entry. Victoria is the most linebreed/inbred dog in this chart.

We can write the formula out more formally as: FAnyDog = ½ (RSire-Dam). To show that this works the way we assert it does, we’ll also find Bob’s coefficient of inbreeding: FBob = ½ (RVincent-Emma), and ½ of (0) is 0. Examination of Bob’s pedigree confirms that his coefficient of inbreeding is 0.

Row ‘X’

We are then left with only one row left to fill in, that belonging to X. If we fill out this last row as we have filled out all of the others, we will see that X is not inbred, despite the fact that his dam was. The fact that this is so may not come as much of a surprise because it is clear from the pedigree that Bob’s line is unrelated to Victoria’s.

Table 3 e.

	Edmund	Emma	Tom	Annie	Tom/ N/A Jack	Edm/Emma Vincent	Jack/An Bob	Vin/Em Victoria	Bob/Vic X
Edmund	1	0	0	0	0	1/2	0	1/4	1/8
Emma	0	1	0	0	0	1/2	0	3/4	3/8
Tom	0	0	1	0	1/2	0	1/4	0	1/8
Annie	0	0	0	0	1	0	1/2	0	1/4
Jack	0	0	1/2	0	1	0	1/2	0	1/4
Vincent	1/2	1/2	0	0	0	1	0	3/4	3/8
Bob	0	0	1/4	1/2	1/2	0	1	0	1/2
Victoria	1/4	3/4	0	0	0	3/4	0	1 1/4	5/8
X	1/8	3/8	1/8	1/4	1/4	3/8	1/2	5/8	1

We will now present a pair of brief examples to demonstrate two important ideas. The first point is that a table like the one above can be easily extended to answer “What-if…?” type questions about future matings. The second is that two inbred parents can produce offspring that are not inbred as long as the parents do not share a common ancestor. We will use the pedigree presented in Figure 4 for this example.

Figure 4. A Mating Between Bill and Victoria

Litter or Dog’s name (“Y”) here	Sire: Bill	Sire: Jack	Sire: Tom
			Dam: N/A
		Dam: Lisa	Sire: Tom
			Dam: N/A
	Dam: Victoria	Sire: Vincent	Sire: Edmund
			Dam:Emma
		Dam:Emma	Sire: N/A
			Dam: N/A

The sire of Y, Bill, is the product of a half-sib mating, while the dam, Victoria, is the product of a dam-son mating. The completed table of relationships and inbreeding coefficients is:

Table 4

	Edmund	Emma	Tom	Annie	Tom/ N/A Jack	Edm/Emma Vincent	Jacl/An Bob	Vin/Em Victoria	Bob/Vic X
Edmund	1	0	0	0	0	1/2	0	1/4	1/8
Emma	0	1	0	0	0	1/2	0	3/4	3/8
Tom	0	0	1	0	1/2	0	1/4	0	1/8
Annie	0	0	0	0	1	0	1/2	0	1/4
Jack	0	0	1/2	0	1	0	1/2	0	1/4
Vincent	1/2	1/2	0	0	0	1	0	3/4	3/8
Bob	0	0	1/4	1/2	1/2	0	1	0	1/2
Victoria	1/4	3/4	0	0	0	3/4	0	1 1/4	5/8
Y	1/8	3/8	1/4	5/16	5/16	3/8	9/16	5/8	1

Both parents of Y are inbred (FBill = 1/8 and FVictoria = ¼), but as they do not share a common ancestor.

Continued in PART 3

Fred Lanting

Fred Lanting is an internationally respected show judge, approved by many registries as an all-breed judge, has judged numerous countries’ Sieger Shows and Landesgruppen events, and has many years experience as one of only two SV breed judges in the US. He presents seminars and consults worldwide on such topics as Gait-&-Structure, HD and Other Orthopedic Disorders, and The GSD. He conducts annual non-profit sightseeing tours of Europe, centered on the Sieger Show (biggest breed show in the world) and BSP.

SIRIUS DOG

INBREEDING AND DIVERSITY – PART 2

Adding The Row ‘Emma’

Rows ‘Tom’ through ‘Victoria’

Row ‘X’

Books by Fred Lanting