Modelling Inbreeding With Matrices
In the brother system problem, a male and a female mate, and from their direct descendants, two individuals of opposite sex are selected at random. Direct offspring of these two are mated and the process continues.The possible genotype for the parents 1st generation are AA, Aa and aa. The possible pairings are 1 - AA x AA, 2 - AA x Aa, 3 - Aa x Aa, 4 - Aa x aa, 5 - aa x aa, AA x aa. We can construct the table for the probabilities of the genotypes of Generation 2.
Generation 1\Generation 2 | AA x AA | AA x Aa | Aa x Aa | Aa x aa | aa x aa | AA x aa |
AA x AA | 1 | 0 | 0 | 0 | 0 | 0 |
AA x Aa | 1/4 | 1/2 | 1/4 | 0 | 0 | 0 |
Aa x Aa | 1/16 | 1/4 | 1/4 | 1/4 | 1/6 | 1/8 |
Aa x aa | 0 | 0 | 1/4 | 1/2 | 1/4 | 0 |
aa x aa | 0 | 0 | 0 | 0 | 1 | 0 |
AA x aa | 0 | 0 | 1 | 0 | 0 | 0 |
Generation 1\Generation 2 | AA x AA | aa x aa | AA x Aa | Aa x Aa | Aa x aa | AA x aa |
AA x AA | 1 | 0 | 0 | 0 | 0 | 0 |
aa x aa | 0 | 1 | 0 | 0 | 0 | 0 |
AA x Aa | 1/4 | 0 | 1/2 | 1/4 | 0 | 0 |
Aa x Aa | 1/16 | 1/16 | 1/4 | 1/4 | 1/4 | 1/8 |
Aa x aa | 0 | 1/4 | 0 | 1/4 | 1/2 | 0 |
AA x aa | 0 | 0 | 0 | 1 | 0 | 0 |
\[\left( \begin{array}{cc} I & O \\ S & Q \end{array} \right)\]
where\[I=\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right), \; O= \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right), \; S= \left( \begin{array}{cc} 1/4 & 0 \\ 1/16 & 1/16 \\ 0 & 1/4 \\ 0 & 0 \end{array} \right), \; Q= \left( \begin{array}{cccc} 1/2 & 1/4 & 0 & 0 \\ 1/4 & 1/4 & 1/4 & 1/8 \\ 0 & 1/4 & 1/2 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right)\]
Pick out the 4 x 4 matrix in the lower right.
\[Q= \left( \begin{array}{cccc} 1/2 & 1/4 & 0 & 0 \\ 1/4 & 1/4 & 1/4 & 1/8 \\ 0 & 1/4 & 1/2 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right)\]
The fundamental matrix is
\[F=(I-Q)^{-1}= \left( \begin{array}{cccc} 8/3 & 4/3 & 2/3 & 1/6 \\ 4/3 & 8/3 & 4/3 & 1/3 \\ 2/3 & 4/3 & 8/3 & 1/6 \\ 4/3 & 8/3 & 4/3 & 4/3 \end{array} \right)\]
.\[(I-Q)^{-1}S= \left( \begin{array}{cc} 1/4 & 1/4 \\ 1/2 & 1/2 \\ 1/4 & 1/4 \\ 1/2 & 1/2 \end{array} \right) \]
.These columns correspond to AA x AA and aa x aa, so after repeated inbreeding the genotype is either AA or aa.