| 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.
|
|
|
|
|
|
Jack |
Vincent |
Bob |
Victoria |
X |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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.
Table 3 d.
|
|
|
|
|
|
Jack |
Vincent |
Bob |
Victoria |
X |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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.
Table 3 e.
|
|
|
|
|
|
Jack |
Vincent |
Bob |
Victoria |
X |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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
|
|
|
|
|
|
Jack |
Vincent |
Bob |
Victoria |
X |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Both parents of Y are inbred (FBill = 1/8 and FVictoria = ¼), but as they do not share a common ancestor.
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 with SV. He presents seminars and consults worldwide on such topics as Gait-&-Structure, HD and Other Orthopedic Disorders, Anatomy, Training Techniques, and The GSD. Fred lives part of the year in Alabama, actively trains in schutzhund, and breeds for occasional litters. He invites all to join his annual non-profit Sieger Show and sightseeing tour.
All Things Canine -- consulting division, Willow
Wood Services Phone: 256-498-3319 Fax: 256-498-3311