Application to Genetics





Living things inherit from their parents many of their physical characteristics. The genes of the parents determine these characteristics. The study of these genes is called Genetics; in other words genetics is the branch of biology that deals with heredity. In particular, population genetics is the branch of genetics that studies the genetic structure of a certain population and seeks to explain how transmission of genes changes from one generation to another. Genes govern the inheritance of traits like sex, color of the eyes, hair (for humans and animals), leaf shape and petal color (for plants).

There are several types of inheritance; one of particular interest for us is the autosomal type in which each heritable trait is assumed to be governed by a single gene. Typically, there are two different forms of genes denoted by A and a. Each individual in a population carries a pair of genes; the pairs are called the individualís genotype. This gives three possible genotypes for each inheritable trait: AA, Aa, and aa (aA is genetically the same as Aa).

Example in a certain animal population, an autosomal model of inheritance controls eye colouration. Genotypes AA and Aa have brown eyes, while genotype aa has blue eyes. The A gene is said to dominate the a gene. An animal is called dominant if it has AA genes, hybrid with Aa genes, and recessive with aa genes. This means that genotypes AA and Aa are indistinguishable in appearance.

Each offspring inherits one gene from each parent in a random manner. Given the genotypes of the parents, we can determine the probabilities of the genotype of the offspring. Suppose that, in this animal population, the initial distribution of genotypes is given by the vector


where the components represent the fraction ofanimals ofgenotypes AA, Aa, and aainitially. Let us consider a series of experiments in which we keep crossing offspring with dominant animals only. Thus we keep crossing AA, Aa, and aa with AA. We are interested in the probabilities of the offspring being AA, Aa, or aa in each of these cases.

        Consider the crossing of AA with AA. Since the offspring will have one gene from each parent, it will be of type AA. Thus the probabilities of AA, Aa, and aa resulting are 1, 0 and 0 respectively. All offspring have brown eyes.

        Next consider the crossing of Aa with AA. Taking one gene from each parent, we have the possibilities of AA, AA (taking A from the first parent and each A in turn from the second parent), aA, and aA (taking a from the first parent and each A in turn from the second parent). Thus the probabilities of AA, Aa, and aa, respectively, are and 0. All offspring again have brown eyes.

        Remains to consider the crossing of genotype aa with AA. there is only a possibility, namely aA. Thus the probabilities of AA, Aa, and aa are 0, 1 and 0 respectively.†† No offspring has blue eyes.

We conclude that crossing with genotype AA will produce offspring with brown eyes only. Our next step is to examine how the above fractions of initial genotypes will change from one generation to another. For this, we let Xn be the distribution vector of genotypes in the nth generation.By the above observations,†† the fractions of genotypes AA, Aa and aain the first generation can be expressed as 1.(1/3)+(1/2)(1/3)+0(1/3), 0(1/3)+(1/2)(1/3)+1.(1/3), 0.(1/3)+ 0.(1/3)+ 0.(1/3) respectively. In other words,X1=AX0, where

†††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††† †††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††††

is called the transition matrix.In general, Xn=AXn-1. Explicitly, we have:


Observe that the aa type disappears after the initial generation and that the Aa type becomes a smaller and smaller fraction of each successive generation. It is obvious that this sequence of vectors converges to the vector


in the long run.

Now try to create a similar model for crossing offspring with a hybrid animal. You will see that some offspring will have brown eyes and some blue eyes.

If you have already seen the application ``Markov chains'', then you would be able to recognize the vector X above as an eigenvector of the matrix A corresponding to the eigenvalue 1.