Fine decomposition of the inbreeding and the coancestry coefficients by using the tabular method

Abstract

In this paper we describe a simple algorithm to decompose both inbreeding and coancestry coefficients. The decomposition is performed in pieces coming from each ancestor, including the founders and the Mendelian sampling terms of non-founders. The original algorithm presented here replaces the conventional tabular method formulae with an original set of recursive formulas. We illustrate the procedure with two small examples, including the pedigree of the bull Comet. The procedure was also tested with simulated pedigrees and it succeeded in analyzing the impact of successive bottlenecks into the average coancestry of the current cohort. Finally, we analyzed the origin of the current coancestry of the Asian Wild Horse population. The average coancestry of the current cohort was around 0.17 and the most relevant ancestors were the direct descendants of Woburn/7 and Mongolia/0, which converged in the Askania reintroduced population in Ukraine after surviving World War II.

Appendix

In this appendix, we will analyze the relationship between the conventional formula for self-coancestry (Eq. 2) and that one proposed in this paper, that is $\Phi_{ii} = \frac{1}{2}\left( {\Phi_{is} + \Phi_{id} } \right) + \delta_{ii} .$ We will distinguish tree cases depending on the knowledge about the parents.

Both parents are known

In this case, we will demonstrate that $\Phi_{ii} = \frac{1}{2}\left( {1 + \Phi_{sd} } \right)$ agrees with $\Phi_{ii} = \frac{1}{2}\left( {\Phi_{is} + \Phi_{id} } \right) + \frac{1}{8}\left( {1 - f_{s} } \right) + \frac{1}{8}\left( {1 - f_{d} } \right).$

Starting from the first equation

$$ 2\Phi_{ii} = 1 + \Phi_{sd} = 1 + \frac{1}{2}\Phi_{sd} + \frac{1}{2}\Phi_{sd} $$

and

$$ 2\Phi_{ii} = 1 - \frac{1}{2}\left( {\Phi_{ss} + \Phi_{dd} } \right) + \frac{1}{2}\left( {\Phi_{ss} + \Phi_{sd} } \right) + \frac{1}{2}\left( {\Phi_{dd} + \Phi_{sd} } \right) $$

From Eq. 1, the coancestry between the individual i and its parents is $\Phi_{is} = \frac{1}{2}\left( {\Phi_{ss} + \Phi_{sd} } \right)$ and $\Phi_{id} = \frac{1}{2}\left( {\Phi_{dd} + \Phi_{sd} } \right)$. Replacing them into 6 we have

$$ 2\Phi_{ii} = \frac{1}{2}\left( {1 - \Phi_{ss} } \right) + \frac{1}{2}\left( {1 - \Phi_{dd} } \right) + \Phi_{is} + \Phi_{id} $$

After the definition of inbreeding as coancestry between the parents, we can write the self-coancestry of s and d as $\Phi_{ss} = \frac{1}{2}\left( {1 + f_{s} } \right)$ and $\Phi_{dd} = \frac{1}{2}\left( {1 + f_{d} } \right)$ respectively and then the demonstration follows trivially by considering $1 - \Phi_{ss} = \frac{1}{2}\left( {1 - f_{s} } \right)$ and $1 - \Phi_{dd} = \frac{1}{2}\left( {1 - f_{d} } \right)$, namely

$$ 2\Phi_{ii} = \frac{1}{4}\left( {1 - f_{s} } \right) + \frac{1}{4}\left( {1 - f_{d} } \right) + \Phi_{is} + \Phi_{id} $$

and

$$ \Phi_{ii} = \frac{1}{8}\left( {1 - f_{s} } \right) + \frac{1}{8}\left( {1 - f_{d} } \right) + \frac{1}{2}\left( {\Phi_{is} + \Phi_{id} } \right) $$

One parent known

In this case, the self-coancestry is always 1/2 because of the lack of coancestry between the ancestors, but δ _ii will depend on the inbreeding of the known ancestor, as in the previous case. Now, $\Phi_{ii} = \frac{1}{2}\Phi_{ik} + \frac{1}{4} + \frac{1}{8}\left( {1 - f_{k} } \right),$ where k denotes the known parent, either s or d.

The demonstration is pretty similar to the previous one. Starting from equation $\Phi_{ii} = \frac{1}{2}\left( {1 + \Phi_{sd} } \right)$ with null Φ _sd

$$ 2\Phi_{ii} = 1 $$

and

$$ 2\Phi_{ii} = 1 - \frac{1}{2}\Phi_{kk} + \frac{1}{2}\Phi_{kk} $$

Then, applying Eq. 1 to the coancestry between the individual i and the known ancestor, $\Phi_{ik} = \frac{1}{2}\Phi_{kk} ,$

$$ 2\Phi_{ii} = \frac{1}{2} + \frac{1}{2}\left( {1 - \Phi_{kk} } \right) + \Phi_{ik} $$

And finally, applying the definition of the inbreeding as before

$$ 2\Phi_{ii} = \frac{1}{2} + \frac{1}{4}\left( {1 - f_{k} } \right) + \Phi_{ik} $$

and

$$ \Phi_{ii} = \frac{1}{4} + \frac{1}{8}\left( {1 - f_{k} } \right) + \frac{1}{2}\Phi_{ik} $$

completes the proof.

The third case is straightforward, that is, when both parents are unknown δ _ii = 1/2.