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Background useful for this project comes from section 4.5 and chapter 5. (This project is based on an example taken from Applications of Linear Algebra by Rorres and Anton.)
Genotypes of Parents (father, mother) | ||||||
---|---|---|---|---|---|---|
(A,AA) | (A,Aa) | (A,aa) | (a,AA) | (a,Aa) | (a,aa) | |
A | 1 | 1/2 | 0 | 1 | 1/2 | 0 |
a | 0 | 1/2 | 1 | 0 | 1/2 | 1 |
AA | 1 | 1/2 | 0 | 0 | 0 | 0 |
Aa | 0 | 1/2 | 1 | 1 | 1/2 | 0 |
aa | 0 | 0 | 0 | 0 | 1/2 | 1 |
To investigate how these traits are propagated in a population, we consider a selective inbreeding program. We begin by selecting a male and a female from an initial population. From their offspring, we select a male and a female at random and mate them. We then repeat the process as many times as desired, selecting a male and a female at random from the offspring of each pair, and mating them.
To model this situation, we start by ordering the different pairs of genotypes in a mother-father pair, as shown in the table below.
Genotype Class |
Genotypes of pair (father,mother) |
1 | (A,AA) |
2 | (A,Aa) |
3 | (A,aa) |
4 | (a,AA) |
5 | (a,Aa) |
6 | (a,aa) |
Next, we let be a probability vector with six components, where component i is the probability that the mating pair chosen belongs to genotype class i. For example, consider a population where all of the males are of genotype A and all of the females are of genotype AA. Then a mating pair chosen at random would be of class 1 with probability 1, so the vector would have its first component equal to 1 and the other five components equal to zero. Recall that a probability vector must satisfy two conditions: each component must be non-negative () and the sum of the components must be 1. In our case, this means that the components must satisfy
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} = 1
To model our experiment, we let be the probability vector for the mating pair chosen at generation k. That is, the entry of is the probability that the mating pair chosen from generation k is of genotype class i. For example, suppose the initial pair belonged to genotype class 2. That means that the father is of type A and the mother is of type Aa. Then according to the table above, half the male offspring would be of genotype A and the other half would be of genotype a. The female offspring would be equally distributed between genotype AA and genotype Aa. Choosing a male and a female at random gives four different possible mating pairs: (A,AA), (A,Aa), (a,AA), and (a,Aa), each with probability 1/4.
In terms of our model, this example means that the initial vector would be given by
and the probability vector for the mating pair chosen from the first generation would be given byBy choosing different genotype classes for the initial probability vector and repeating this process, we could generate probability vectors for the mating pair chosen from the first generation for each of the six possible genotype classes of the first mating pair. However, this process this actually gives us a lot more. It gives us a matrix A which lets us compute from via the equation
where the columns of the matrix A are the six probability vectors whose computation was described above. If you actually go through the calculations, you will find that the matrix A is given byWilliam W. Farr