Subsections

# Modeling Inheritance of Genetic Traits

## Introduction

This project deals with a linear model describing the inheritance of genetic traits that are passed on from generation to generation. In particular we work with a model of what is called X-linked inheritance, in which the trait of interest depends on genes found only in the X chromosome. In humans, traits including color blindness, hereditary baldness, and muscular dystrophy are inherited in this way.

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.)

## Background

To keep things simple, we assume that the trait we wish to study is governed by a set of two genes, which we term A and a. These genes are present only on the X chromosome, and each chromosome will have one of the two genes. Since females have two X chromosomes, there are three different genotypes: AA, where both X chromosomes have the gene A, Aa, where one X chromosome has gene A and the other has gene a, and aa, where both chromosomes have the a gene. For males there are only two genotypes, A and a, since males have only one X chromosome. Male offspring receive one gene from the mother, with equal probability. That is if the mother's genotype is Aa, you would expect half the male offspring to be of genotype A and the other half to be of genotype a. If the mother's genotype is AA, all of the male offspring will be of genotype A. Females receive the one gene of the father, and one of the mother's two genes, with either being equally likely. For example, female offspring of a mating between a male of genotype A and a female of genotype Aa would be split equally between genotypes AA and Aa. All the possiblities are shown in the table below.

 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.

 GenotypeClass 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

x1 + x2 + x3 + x4 + x5 + x6 = 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 by

By 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 by

## Exercises

1.
Show that if is a probability vector and A is the matrix given above, then is also a probability vector.
2.
Verify that the fifth column in the matrix A is correct. That is, assume that the initial mating pair belongs to genotype class 5 and compute the probability vector for the mating pair to be chosen from the first generation.

3.
Assuming that there exists a matrix A such that

explain how the procedure described in the background can be used to compute the columns of A. (Hint - First, note that A is a linear transformation. Then consider the meaning of the standard basis vectors , in the context of our model.)

4.
Use the matrix A from the background section and perform simulations to determine what the long term probability vectors look like. Choose each of the standard basis vectors for in turn as initial conditions. That is, do a simulation for each possible choice of initial mating pair.

5.
Let a satisfy and consider an initial condition of the form

Perform simulations using this initial condition for a=1/4, a=1/2 and a=3/4. Describe your results.

6.
In a previous project, we found that the behavior of certain kinds of models is determined by a dominant eigenvalue. Use the Maple eigenvals command to determine the eigenvalues of the matrix A. (This matrix turns out to be one of those rare cases where you can determine the eigenvalues exactly.)

7.
In the previous exercise, you should have found that the number 1 appears twice as an eigenvalue of A. Use the Maple nullspace command to find a basis for the nullspace of A-I. Use this information to explain your results in exercise 5.