New linebreeding feature - Page 1

I just noticed that there is a new feature in the linebreeding section of each pedigree.  It's the inbreeding coefficient calculation.  It looks awesome but I'm not sure how to interpret the results.  Can anyone summarize how to assess the percentages?
Judy

This was added today, good catch. I can provide some links but am on the road so will have to do it later.

Slight typo on PDB, it should be Coefficient not coefficiency!. Also the calculations are close to what I got after laborious work, which is a relief.
http://www.alpinek9forums.com/post39354.html#p39354
 

This is one of the better explanations I have found.

http://www.netpets.org/dogs/healthspa/demyst.html
 

Demystifying Inbreeding Coefficients

John Armstrong

While most breeders recognize that a mating between half-sibs or cousins represents inbreeding, the majority probably have no idea which represents the closer inbreeding. This is not helped by the non-standard definition of inbreeding in some books (e.g. Onstott's "Breeding Better Dogs").

The standard definition of inbreeding is that it is any scheme which results in the sire and the dam having common ancestors. This common heritage is expressed by a parameter called the inbreeding coefficient, first proposed by Sewell Wright in 1922. Designated F by Wright (but more commonly IC or IBC by breeders), it can theoretically range from 0 to 100%, and indicates the probability that the two alleles for any gene are identical by descent. Though the primary consequence of inbreeding is to increase homozygosity, the IC is not a direct measure of homozygosity because the two alleles may be the same for other reasons. Within a breed, some proportion of all the genes will be the homozygous because there was only one allele to start with. In that sense, the IC may be regarded as indicating what proportion of the remainder have been made homozygous by inbreeding.

The inbreeding coefficient is a function of the number and location of the common ancestors in a pedigree. It is not a function, except indirectly, of the inbreeding of the parents. Thus, one can mate two highly inbred individuals who share little common ancestry and produce a litter with a very low IC. (Because the potential number of ancestors doubles every generation, eventually you reach a point where the number of ancestors exceeds the number of individuals alive at that time. You are, therefore, bound to find some common ancestors if you go back far enough.) Conversely, it is possible to mate two closely related dogs, both of which have low ICs, and boost the IC substantially.

Calculating inbreeding coefficients

The accepted method is best illustrated by a simple example. Suppose we mate half-sibs, the common ancestor, Anson, being the father. Don is the son of Anson and Bess; Eva the daughter of Anson and Claire. Fred is one of their progeny.

 

To simplify, we don't show the ancestors that aren't shared:

Now we consider a gene for which Anson carries two different alleles, a1 and a2. Whichever one is passed to Don has a 50% probability of being passed to Fred. There is also a 50% probability that the same allele will be passed from Anson to Eva, and a 50% probability of it being passed from Eva to Fred, if Eva got it. When dealing with events that are contingent (this *and* that must happen), we multiply the probabilities - in this case 0.5 x 0.5 x 0.5 = 0.125 (12.5%). This final number is the probability that Fred will be homozygous for either a1 or a2 because of the common grandfather. If this were the only common ancestor, the inbreeding coefficient for Fred would be 12.5%.

In general, Wright's method is to determine the path from Fred to the common ancestor, Anson, and back again on the other side of the pedigree (Fred-Don-Anson-Eva-Fred), count the number of individuals in the path, excluding Fred (there are 3, Don-Anson-Eva) and then calculate 1/2 to the power n, where n is that number. So, in the present case, we have (1/2)³ or (1/2 x 1/2 x 1/2) = 1/8, which is 12.5% as we calculated above.

Now, suppose the common ancestor was one of the grandfathers of the parents (i.e. a great-grandfather of the litter). This adds an individual on each side of the pedigree, so that we will get a path of the type Fred-X-Don-Anson-Eva-Y-Fred, and the inbreeding on Anson will be (1/2)⁵ or 1/32 (3.125%).

Complications

If we had only a single common ancestor to deal with, life would be relatively simple. However, there are two complications to deal with. The first is that there will be more than one common ancestor. Let's consider the case of first cousins. In human populations such a pairing is prohibited in some societies but allowed in others. We have already calculated the inbreeding for a single shared grandparent. First cousins have two shared grandparents, and we simply add the inbreeding coefficient for each to get 6.25%.

The second complication is that the common ancestor may be inbred. If so, his or her inbreeding coefficient will have to be calculated. To account for this we have to multiply the inbreeding coefficient calculated for Fred by (1 + F_A), where F_A is the inbreeding coefficient calculated for Anson. For example, if Anson is the product of a mating of first cousins, the total inbreeding for Fred will be 0.125 x 1.0625 = 0.133 (13.3%) if there are no other shared ancestors in the pedigree.

Unfortunately, in the average pedigree, there are a large number of shared ancestors. Therefore, the total inbreeding for a dog cannot generally be calculated manually and appropriate software must be used (e.g. CompuPed). Calculating inbreeding for only the first few generations is not particularly useful. If there are more than one or two common ancestors in four or five generation pedigree, the inbreeding is probably already higher than desirable. Unfortunately, having none is no guarantee that common ancestors will not occur in abundance further back, and some pedigrees of this type still achieve moderately high inbreeding coefficients. Neither can be number of shared ancestors be used as a reliable guide, as the inbreeding coefficient is very sensitive to when and where they occur in a pedigree.

 

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http://en.wikipedia.org/wiki/Coefficient_of_relationship

http://highflyer.supanet.com/coefficient.htm

http://www.icelanddogs.com/inbreeding.html
 

Good spot Guddu.   Thomas z Dvorakoca sadu should be linebred to Dargo Ha-ja-Da with 3.125 (not 3.22)   Looking into this since the calculation path seems to be right (It can be tricky to debug pedigree's where wright's calculations jump back 7-10 times to the early 1900's)

> Thomas z Dvorakova sadu ( 1331083 ) is linebred and now we check the highest COI there
> Thomas z Dvorakova sadu ( 1331083 ) has a maximum base coefficiency of 03.13%

---> COI check - Dargo Ha-Ja-Da CS ( 463148 ) 

---> VD Dargo Ha-Ja-Da CS ( 463148 ) No linebreeding or missing pedigreedata - Addition is 0

> Thomas z Dvorakova sadu ( 1331083 ) is .03125 * (1+0)
Thomas z Dvorakova sadu ( 1331083 ) has a total coefficiency of: 03.1250

Found the problem.  A typo had me using the dogs coefficient as the addition (the (1+F_x) part)  instead of the ancestors.  So for Thomas vs. Dargo the calculations where doing  0.03125*(1+0.03125) instead of 0.03125*(1+0) which is what it should have been. 
That is fixed now.

True to Wright's formula the PDB system uses all available known ancestors.   For example the Coefficient for Ines von der Ehrenfeste  goes all the way to Horand von Grafrath :) 


Here is the path that the system is taking.   And remember it's doing full calculations on the fly (doesn't cache anything) every time..  So the Coefficient calculations are reading around 60-1000 pedigrees each time.

> COI check - Ines von der Ehrenfeste ( 1036 ) 

> V Ines von der Ehrenfeste ( 1036 ) is linebred and now we check the highest COI there
> V Ines von der Ehrenfeste ( 1036 ) has a maximum base coefficiency of 06.25%

---> COI check - Palme vom Wildsteiger Land ( 116 ) 

---> V Palme vom Wildsteiger Land ( 116 ) is linebred and now we check the highest COI there
---> V Palme vom Wildsteiger Land ( 116 ) has a maximum base coefficiency of 01.56%

------> COI check - Canto von der Wienerau ( 141 ) 

------> V1 Canto von der Wienerau ( 141 ) is linebred and now we check the highest COI there
------> V1 Canto von der Wienerau ( 141 ) has a maximum base coefficiency of 00.39%

---------> COI check - Hein vom Richterbach ( 247 ) 

---------> V, ROM (US) Hein vom Richterbach ( 247 ) is linebred and now we check the highest COI there
---------> V, ROM (US) Hein vom Richterbach ( 247 ) has a maximum base coefficiency of 00.39%

------------> COI check - Bodo von der Brahmenau ( 4981 ) 

------------> VA Bodo von der Brahmenau ( 4981 ) is linebred and now we check the highest COI there
------------> VA Bodo von der Brahmenau ( 4981 ) has a maximum base coefficiency of 00.20%

---------------> COI check - Jung Tell von der Kriminalpolizei ( 133370 ) 

---------------> SIEGER 1913 (NL) V Jung Tell von der Kriminalpolizei ( 133370 ) is linebred and now we check the highest COI there
---------------> SIEGER 1913 (NL) V Jung Tell von der Kriminalpolizei ( 133370 ) has a maximum base coefficiency of 03.13%

------------------> COI check - Graf Eberhard vom Hohen Esp ( 2018 ) 

------------------> V Graf Eberhard vom Hohen Esp ( 2018 ) is linebred and now we check the highest COI there
------------------> V Graf Eberhard vom Hohen Esp ( 2018 ) has a maximum base coefficiency of 25.00%

---------------------> COI check - Nelly II Eislingen (die Jüngere) ( 420562 ) 

---------------------> Nelly II Eislingen (die Jüngere) ( 420562 ) is linebred and now we check the highest COI there
---------------------> Nelly II Eislingen (die Jüngere) ( 420562 ) has a maximum base coefficiency of 12.50%

------------------------> COI check - Horand von Grafrath (Hektor Linksrhein) ( 1208 ) 

------------------------> V Horand von Grafrath (Hektor Linksrhein) ( 1208 ) No linebreeding or missing pedigreedata - Addition is 0

---------------------> Nelly II Eislingen (die Jüngere) ( 420562 ) is .125 * (1+0)

------------------> V Graf Eberhard vom Hohen Esp ( 2018 ) is .25 * (1+.125)

---------------> SIEGER 1913 (NL) V Jung Tell von der Kriminalpolizei ( 133370 ) is .03125 * (1+.28125)

------------> VA Bodo von der Brahmenau ( 4981 ) is .001953125 * (1+.0400390625)

---------> V, ROM (US) Hein vom Richterbach ( 247 ) is .00390625 * (1+.0020313262939453125)

------> V1 Canto von der Wienerau ( 141 ) is .00390625 * (1+.003914184868335723876953125)

---> V Palme vom Wildsteiger Land ( 116 ) is .015625 * (1+.00392153978464193642139434814453125)

> V Ines von der Ehrenfeste ( 1036 ) is .0625 * (1+.0156862740591350302565842866897583007813)
V Ines von der Ehrenfeste ( 1036 ) has a total coefficiency of: 06.3480

So the numbers are COI or COR ?? 

Wright's coefficient goes back as far back as the system can. (each time you look up the line breeding tab it's doing the calculations ) 

F_X=Σ[(¹/₂)^1+n1+n2_*(1+F_A)] <- This is the formula.   Where F_X is the Dog in question and F_A is the ancestor the system is checking.  So I check recursively each ancestor to see their COI > 0.   All the way to Horand if necessary.

So if a certain ancestor is line-bred and his ancestor also, the formula looks like this
F_X=Σ[(¹/₂)^1+n1+n2_*(1+F_A)]
FA=Σ[(¹/₂)^1+n1+n2_*(1+F_B)]
F_B=Σ[(¹/₂)^1+n1+n2_*(1+FC)]

And I need to calculate F_C before I can calculate F_B and (assuming) F_C is 0%, then I can finish calculating F_B, then I can calculate F_A and finally F_X 

Hardiman's is simpler.  There I check only 5 generations and for those ancestor's that occur > 1 times, I do the Hardimans calculation and check if that ancestor is line-bred and take his (the ancestor) COI and divide it by 2^generation .  So for an ancestor in the 5th generation it's his COI / 32.

I only do this once for Hardimans. i.e. I do not traverse as far as I can go.

Somebody take a look at this dog's pedigree .. 2-2 with a bunch of relatives already inbred themselves .. I would never do this and I am not sure the COR or COI reflect the danger in this type of breeding.  I can do a 2-3 with a sire and dam with somewhat unrelated pedigrees on the other half of the pedigree and no long history of inbreeding in their own individual pedigrees but inbreeding on inbreds that share the same ancestors on both sides of the pedigree makes me very nervous.  

http://www.pedigreedatabase.com/german_shepherd_dog/dog.html?id=1924558-benvillarosa-all-that-jazz

 

Thank you Oli, for a great tool.