One way of measuring genotype frequency is from phenotype frequency. Consider the case of 3 blood groups A, AB and B determined by two alleles IA and IB at a single locus. In a random sample of 1000 humans, the A group occurred in 210, AB in 450 and B in 340 individuals.

The frequencies of the blood group phenotypes and their respective genotypes are obtained by dividing the number of individuals for each blood group by the total. Thus, the frequency of blood group B for instance would be 340/1000 = 0.34.

Another way of estimating genotype frequency is to first calculate gene frequency of genes A and B in the population. Assume that the above sample contains 210 AA, 450 AB and 340 BB individuals.

The gene frequency of A in the population is represented by the probability to find A allele at the AB locus and is exactly equivalent to the proportion of A alleles among all alleles at this locus in the sample or in the population (that is because we cannot determine the frequency of A in the whole population).

As each individual carries two alleles at the AB locus, the total number of alleles in the sample is 1000 x 2 = 2000. Out of these 210 + 210 + 450 = 870 are A. Therefore the frequency of the A allele is 870/2000 = 0.435. The number of B alleles is 450 + 340 + 340 = 1130, and the frequency of B allele is 1130/2000 = 0.565.

If we represent the gene frequency of A by p, then p represents a value between 0 and 1 (because the proportion of allele A must lie between 0 and 100 per cent). In our example p = 0.435. Similarly, if we symbolize the frequency of B by q, then q = 0.565. It may be noted that q = 1-p or 1- 0.435. Similarly p = 1 – q or 0.565. Thus p + q = 1.

For predicting genotype frequencies some assumptions have to be made, such as random mating in the population. That is to say, with respect to the trait of blood groups, an individual will mate with another without regard to whether the blood group of the mate is AA, AB or BB. Random mating implies random union of eggs and sperm, which in the example cited is the frequency of A and B alleles among the eggs and sperm (the gametes).

Now the probability for the allele A at the AB locus in the population is p, and this is also the probability for a randomly chosen gamete to carry allele A. Similarly the probability for a randomly chosen gemete to carry B is q. For producing individuals with genotype AA, an A sperm must fertilize an A egg.

This occurs with probability p x p =p2. An AB genotype results from fertilisation of A sperm with B egg or vice versa, and the probability is p x q or pq + qp = 2pq (that is pq for AB genotype and qp for BA genotype). The frequency of BB genotypes depends upon the chance fertilisation of a B sperm and B egg; this has probability q x q = q2.

Thus the frequencies of AA, AB and BB genotypes in the population should be expected to be p2, 2pq, and q2 respectively. If we substitute the values of p = 0.435 and q = 0.565 we can know that the frequency of AA = (0.435)2 = 0.189, AB = 2 x 0.435 x 0.565 = 0.246, and of BB = (0.565)2 = 0.319.

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