A path of coefficient of relationship is defined as

$$\rho_{AO} = \left( \frac{1}{2}\right)^n \sqrt { \frac{1+f_A}{1+f_O}}$$

This SE post discusses this definition

From this, the coefficient of relatedness between two individuals $B$ and $C$ is defined as

$$R_{BC} = \sum \rho_{AB} \rho_{AC}$$

, where $A$ is (I think) the last common ancestor of $B$ and $C$. The sum is over all possible path coefficients.

Source: Wikipedia

I am a little bit confused, I don't really understand this formula. One reason is the following. Let's assume that $B$ has 2 paths and $C$ has 3 paths to go to $A$. If we denote $\rho_{AB,i}$, the path $i$ (or the $i^{th}$ if you prefer) between $A$ and $B$,

$$R_{BC} = \rho_{AB,i} \rho_{AC,i} + \rho_{AB,j} \rho_{AC,j} + \rho_{AB,k} \rho_{AC,k}$$

But the path $\rho_{AB,k}$ does not exist because there are only 2 paths between $A$ and $B$ (which are denoted $i$ and $j$). Also, there is no reason to count the path in a specific order and not another one. If we change the order we count the paths between $C$ and $A$, we can get:

$$R_{BC} = \rho_{AB,i} \rho_{AC,j} + \rho_{AB,j} \rho_{AC,i} + \rho_{AB,k} \rho_{AC,k}$$

, which is obviously different from the previous calculation of $R_{BC}$

Can you help me making sense of these formulas?

  • 1
    $\begingroup$ genetic-genealogy.co.uk/Toc115570135.html. The notation is confusing and the Wiki site is not helpful (imo). This site is better (many examples) and mentions Wright's original paper. $\endgroup$
    – daniel
    May 3, 2014 at 13:12
  • 1
    $\begingroup$ It would be more like: $\sum_i (\rho_{AB}\rho_{AC})_i.$ $\endgroup$
    – daniel
    May 3, 2014 at 13:59
  • 1
    $\begingroup$ Wright's 1922 paper. aipl.arsusda.gov/publish/other/wright1922.pdf $\endgroup$
    – daniel
    May 3, 2014 at 14:20
  • $\begingroup$ Your comments and links helped a lot! Thanks. [genetic-genealogy](genetic-genealogy.co.uk/Toc115570135.html) offers a perfect "for dummies introduction" I don't quite know how to answer my messy question though. You can give a try if you want and I'll check it. Otherwise I might just delete it. I still don't understand this and that though. $\endgroup$
    – Remi.b
    May 3, 2014 at 15:03

1 Answer 1


The expression $f = 2h-1,$ coefficient of inbreeding (COI) is a measure of homozygosity. Since in a randomly breeding herd (of cattle) we expect 50 per cent heterozygous and 50 per cent homozygous, the minimum for $2h-1 $ is 0. If the cattle's genome is purely homozygous (aa, AA) we have $2h-1 = 1.$ (h is total homozygosity--see Wright's paper, linked below.)

As we can see from this site, a basic calculation of relationship coefficient may not deal with f at all. For two descendants B, C, the quantities $f_B,f_C$ are an attempt to factor in existing relative homozygosity that the coefficient of relation (COR) would otherwise not capture. Likewise $f_A$ measures preexisting homozygosity in the common ancestor of B and C. If animal A is the product of siblings then the homozygosity could already be quite high and this would maybe have measurable effects on the vitality of a cross between B and C.

Now $f_A$ and $f_B$ vary between 0 and 1 so $s = (\frac{1+f_A}{1+f_B})^{1/2}$ varies between $(1/2)^{1/2}$ and $(2)^{1/2}.$ Note that the important thing here is relative homozygosity of an animal and its ancestor:

a. If the ancestor A and descendent B have the same known preexisting burden of heterozygosity we can disregard s (and f) entirely. Their COR will depend only on path distance.

b. The homozygosity of the ancestor A tends to increase COR while that of B tends to decrease COR.

The expression $r_{BC }= \sum p_{AB }\cdot p_{AC}$ is somewhat confusing. On page 334 of one of Wright's earlier papers we can get a clear idea of the use of the coefficient.

He defines $p_{BA} = (1/2)^n (\frac{1+f_A}{1+f_B})^{1/2}.$ If the path from C to A is $p_{CA} = (1/2)^m (\frac{1+f_A}{1+f_C})^{1/2}$ then the total path BAC would be

$$ p_{AB,AC} = (1/2)^{n+m}(\frac{1+f_A}{1+f_B})^{1/2}(\frac{1+f_A}{1+f_C})^{1/2} $$

$$= (1/2)^{n+m}\frac{1+f_A}{[(1+f_B)(1+f_C)]^{1/2}} $$

So the expression for the sum of such paths would be

$$r_{BC} = \sum_i p_{AB,AC} = \sum_i (1/2)^{k_i}\frac{1+f_{A_i}}{[(1+f_B)(1+f_C)]^{1/2}} $$

in which $A_i$ is the ith ancestor through whom B and C are related. The total path length for a given term is $(n_i+m_i)$ which I abbreviated $k_i.$

So a less confusing expression for $r_{BC}$ might be:

$$r_{BC }= \sum_i p_{A_iB}p_{A_iC}. $$

Note that the sum is over all common ancestors back to some arbitrary generation.

A good example of this is Figure 11 at link 1 above. It's really not ``dumbed down'' at all, it just leaves out the factor s, which for a carefully maintained herd may be close to 1 anyway.


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