Wright's Coefficient of Inbreeding

Introduction

Chromosomal genes have variations (alleles) which express individual characters. Humans inherit a copy of each gene from each parent, so have two alleles of each gene. (We ignore mitochondrial DNA -- inherited strictly from the mother -- and X and Y chromosomes -- which males inherit strictly from mother and father respectively.)

We will say an individual has an inbred gene if both copies are the same allele. Throughout this discussion we assume two individuals who are unrelated have different alleles.

In the following diagrams, we show a man marrying his half-sister and producing a daughter, Julia. Instead of four grandparents, Julia has two ordinary grandparents and one ``double grandparent.'' The distinct alleles of a gene are labelled in the diagrams: Sam has A & B alleles; Mary has C & D alleles in (a) but two C alleles in (b), and so on.

What is the probability Julia's gene will be inbred? Diagrams (a1) and (a2) are identical, just showing two different ways to depict the same pedigree. The three grandparents in all these diagrams are unrelated to each other, but in diagram (b) the double grandmother is, herself, already inbred.

(a1) (a2) (b) Sam Mary Jim Mary Sam Mary Jim Sam Mary Jim A.B C.D E.F C.D A.B === C.D === E.F A.B === C.C === E.F \ / \ / | | | | \ / \ / | | | | father mother father === mother father === mother \ / | | \ / | | ?.? ?.? ?.? Julia Julia Julia

The father inherits one allele from his mother, Mary, but has only a 50% chance of passing that allele to Julia. (The other 50% of the time he passes on Sam's gene.) Similarly the mother has a 50% chance of passing an allele from Mary to Julia. Julia has a 25% chance (50% * 50%) of getting both alleles from her double grandmother Mary.

Assuming Julia does inherit both of her alleles from Mary, in (a) she has equal chances to finish with C.C, C.D, D.C or D.D, while in (b) she will always then have C.C. Thus the net inbreeding probability for Julia is 0.125 in (a), and 0.250 in (b). This inbreeding probability is called the Wright Inbreeding Coefficient (``WIC'').

In (a), the double grandmother is not inbred (has WIC=0), while in (b) she is inbred (has WIC=1). If instead she had WIC = 0.1 (was inbred with 10% probability), then Julia's WIC would be .9*.125 + .1*.25 = .1375. More generally the WIC for the offspring of half-siblings is .125*(1+p) where p is the WIC for the double grandparent.

The discussion of probabilities may seem out of place: Why don't we just do a DNA test on Julia and see what actual alleles she has inherited? That would make sense if the human nucleus had but a single gene, but in fact there are many thousands, and Julia may inherit the grandfather's allele for some genes and the grandmother's allele for others. The Wright Coefficient thus estimates the proportion of genes that are inbred.

Many genes will have only a few distinct alleles possible, so an individual can end up with identical alleles even though his or her ancestors appear unrelated. But this fact does not change the utility of the Wright Inbreeding Coefficient, which is understood most easily with the simplified model of the diagrams above.

Calculating WIC with a single common ancestor

In diagram (a) or (b) draw a path from Mary to Julia and back to Mary which touches no individual twice except Mary herself. This can be done in only one way and produces a path of length L = 4. Julia's WIC is given by

WIC = 2^(1-L)*(1+p)

where p is Mary's WIC. Since L = 4 in the example, we end up with WIC = .125*(1+p), just as before.

In the diagram now, suppose that Jim is himself the son of Mary, by yet a third husband. Now there would be a second path from Mary to Julia back to Mary, with L = 5, and the net WIC would be the sum of the two WIC constituents,

WIC = (2^(1-4) + 2^(1-5)) *(1+p) = .1875*(1+p)

Finally suppose Sam is also Mary's son, by yet a fourth husband. This adds two more paths, of lengths 5 and 6 and the net WIC would be the sum of four contributing WIC's:

WIC = (2^(1-4) + 2^(1-5) + 2^(1-5) + 2^(1-6)) *(1+p) = .28125*(1+p)

Here are the four distinct paths for that final case. Note that no individual, except Mary herself, occurs twice in any path.

Mary - J's father - Julia - J's mother - Mary

Mary - Sam - J's father - Julia - J's mother - Mary

Mary - J's father - Julia - J's mother - Jim - Mary

Mary - Sam - J's father - Julia - J's mother - Jim - Mary

The reason the formula works is that it simply considers the way alleles can be inherited and sums the probability appropriately. The probability rule

prob(X or Y) = prob(X) + prob(Y)

only works when X and Y are exclusive events (that is prob(X and Y) = 0), but this condition holds here.

Calculating WIC with multiple common ancestors

Finally we are ready for the general formula. Again we simply add contributing WIC's. For example, if Julia's parents had two common ancestors, Mary and Malcolm, and neither Mary nor Malcolm is the other's ancestor, we can compute the WIC contribution from Malcolm to Julia as in the preceding section and simply add it to the contribution computed from Mary:

WIC_net = WIC_due_to_Malcolm + WIC_due_to_Mary

If Malcolm is Mary's ancestor (or vice versa) the simple addition will still work, as long as you follow the rule given above -- don't count paths from Malcolm to Julia to Malcolm in which Mary (or any individual) occurs twice.

Example 1

Let's practice the WIC calculation with a simple example. The first step is to prepare a pedigree of the target whose WIC is to be computed. One may go back just a few generations to keep the arithmetic simple, but of couse this may lead to a lower WIC being computed.

We'll look at part of the pedigree of Alphonso XII, a 19th-century King of Spain. (Throughout this page we assume that ``official genealogies'' are correct; Queen Isabella had many lovers and it is very doubtful that Francisco de Asis was the father of King Alphonso XII. Simlarly, King Charles IV was cuckolded and it is unlikely he fathered Francisco de Paula.)

Marie-Louise Charles IV of Parma ===== K of Spain (MLP) / | \ (CIV) / \ \ / \ \ Frances I of / \ Isabel ===== Two Sicilies / \ (IS) / \ (FI) / _____\________/ \ / / \ \ Francisco Louise of Ferdinand VII Cristina Maria de Paula ==== Two Sicil. K of Spain ==== of Two Sicil. (FDP) \ (L2S) (FVII) / (C2S) \ / Francisco Isabel de Asis ===== Q of Spain (FDA) | (QI) | Alphonso XII K of Spain

The next step is to prepare a list of the people who appear as ancestors of both the father, Francisco de Asis, and mother, Queen Isabel. Is this case the list would include Charles V and his wife Marie-Louise and Frances I and his wife Isabel. We need their WIC's as well; for now we assume them to be zero. Then we list the paths:

MLP -- FDP -- FDA -- Alphonso XII -- QI -- FVII -- MLP

MLP -- FDP -- FDA -- Alphonso XII -- QI -- C2S -- IS -- MLP

MLP -- IS -- L2S -- FDA -- Alphonso XII -- QI -- FVII -- MLP

CIV -- FDP -- FDA -- Alphonso XII -- QI -- FVII -- CIV

CIV -- FDP -- FDA -- Alphonso XII -- QI -- C2S -- IS -- CIV

CIV -- IS -- L2S -- FDA -- Alphonso XII -- QI -- FVII -- CIV

IS -- L2S -- FDA -- Alphonso XII -- QI -- C2S -- IS

FI -- L2S -- FDA -- Alphonso XII -- QI -- C2S -- FI

You should check that the eight paths shown are all correct and that no other paths have been overlooked. In summary, we have

For common ancestor MLP (Marie-Louise of Parma), three paths of lengths 6, 7, and 7.

For common ancestor CIV (King Charles VII), three paths of lengths 6, 7, and 7.

For common ancestor IS (Isabel), one path of length 6.

For common ancestor FI (Frances I), one path of length 6.

Since we assume (for now) that none of these common ancestors are themselves inbred, the computation becomes very simple. WIC = 4*2^(1-6) + 4*2^(1-7) = 0.1875

(When I do such calculations, I sometimes like to take advantage of the strange ``arithmetic rule'' 2k*{N} = k*{N-1}. With this rule, the ``sum'' of path lengths 4*{6} + 4*{7} can be simplified:

WIC = 4*{6} + 4*{7} = 4*{6} + 2*{6} = 6*{6} = 3*{5} = {4} + {5}

= 2^(1-4) + 2^(1-5) = 0.1875

Example 2, Henri de Rothschild

The most well-known examples of inbreeding are the dynasties of Egypt and the Hapsburg and Bourbon families, but the Rothschilds often married other Rothschilds. Here's a pedigree showing that Henri de Rothschild is descended from four of Mayer Amschel Rothschild's sons: the founders of the Bank branches in Austria, Italy, Britain and France.

Mayer Amschel Gutele Rothschild ======================= Schnapper / / \ \ / / \ \ / / \ Salomon / / \ (Austria) / / \ \ / / \ \ Calmann Nathan ======= Hannah Jacob \ (Italy) (Britain) / \ Cohen (France) === Betty | / \ | | / \ | Mayer ===== Luise Nathaniel ===== Charlotte \ / \ / \ / Laura T. ===== James E. | | Henri de Rothschild 1872 - 1947

Henri -> Laura -> Luise -> Nathan -> Nathaniel -> James -> Henri

Henri -> Laura -> Mayer -> Calmann -> M.A. -> Jacob -> Charlotte -> James -> Henri

Henri -> Laura -> Mayer -> Calmann -> M.A. -> Nathan -> Nathaniel -> James -> Henri

Henri -> Laura -> Luise -> Nathan -> M.A. -> Jacob -> Charlotte -> James -> Henri

Henri -> Laura -> Mayer -> Calmann -> M.A. -> Salomon -> Betty -> Charlotte -> James -> Henri

Henri -> Laura -> Luise -> Nathan -> M.A. -> Salomon -> Betty -> Charlotte -> James -> Henri

Henri's Cousin(?!) Alphonse

Another Rothschild, Alphonse, ends up with exactly the same 0.125 Wright's coefficient as Henri, though the details are quite different. Here is a chart with the relevant links. (With three women in the chart all named Charlotte, I've denoted them by their birth years: 1807, 1819, 1825.) How would you describe the cousin relationship(s) between Alphone and Henri? Solution: They are 2nd cousins 1x removed via Salomon and his wife, and also via Calmann and his wife. They are 2nd cousins via Jacob and Betty. They are double 2nd cousins and double 2nd cousins 1x removed via Nathan and his wife. Via Mayer they are 3rd cousins 9 ways, 3rd cousins 1x removed 11 ways, 4th cousins 2 ways. Add these up and see that if Henri married one of Alphonse's sisters (Charlotte or Valentine) their offspring would have consanguinity 35/256 -- slightly more than the 32/256 consang of Alphonse or Henri. (Of course the same result obtains if Alphonse married Henri's sister Louise.) I've also added, in red, links to Lionel Walter, 2nd Baron Rothschild (1868-1937). He and his siblings have inbreeding coefficient of 5/16, which may be the highest in the Rothschild family.

Example 3, Siblings marry

Suppose Adam and Eve are unrelated and not inbred; they mate to produce Seth and Azura, who mate in turn to produce Enosh and Noam, who mate in turn to produce Cainan and Mualeleth, who mate in turn to produce Mahalalel and his sister, What are the WIC's?

Enosh and Noam (generation 3) are the first inbred people in this lineage; WIC = .25 for the offspring of siblings. Cainan and his sister (generation 4) have WIC = 3/8 = .375.

For Mahalalel (generation 5) his common grandparents are themselves inbred; it turns out his WIC = .50. If the inbreeding continued in this fashion, WIC would be 0.594, 0.672, 0.734, 0.785, 0.826, 0.859, 0.886 for generations 6 - 12 respectively, eventually becoming asymptotic to 1.000.

Example 4, Single female

Suppose a black widow spider mates with the only male, kills him, but then gives birth only to a single male. She repeats this for many generations. The first two males mentioned have WIC=0, and the third has WIC=.25. For subsequent generations, WIC would be 0.375, 0.438, 0.469, 0.484, 0.492, 0.496, 0.498, 0.499, but would never exceed 0.500.

Example 5, Alphonso XII of Spain

We already did Alphonso XII above, but with an incomplete pedigree that ignored some important inbreeding paths. Let's do him again. He seems interesting to me because he finishes with WIC=0.250, exactly the same as for the offspring of siblings (when there is no other inbreeding). He has the highest WIC of anyone in my database, except for those who actually were the offspring of siblings.

From the following partial pedigree a WIC of .234 can be computed for Alphonso XII. Only two of Alphonso's great great grandparents are missing from the diagram; these are Philip's wife (who was Philip's own 1st cousin once removed) and Ferdinand I's wife (who was Ferdinand's own 2nd cousin once removed). These and other inbreedings bring the WIC up to .250.

If you want to try your hand at computing the WIC manually in this complicated pedigree, check your work (or check mine) with the relevant path lengths I've shown below.

Philip V Isabella K of Spain ===== Farnese / \ / \ Philip Charles III Marie-Amelie | K of Spain ====== of Saxony | / \ | / \ Marie-Louise Charles IV Ferdinand I of Parma ===== K of Spain of Sicily / | \ | / \ \ | / \ \ Frances I of / \ Isabel ===== Two Sicilies / \ / \ / _____\________/ \ / / \ \ Francisco Louise of Ferdinand VII Cristina Maria de Paula ==== Two Sicil. K of Spain ==== of Two Sicil. \ / \ / Francisco Isabel de Asis ===== Q of Spain | | Alphonso XII K of Spain

The relevant paths are summarized in this table: