ADVERTISEMENTS:

Application of Hardy-Weinberg law in calculating Gene (Allele) frequencies in a population.

**The gene frequencies for the autosomal and sex-chromosomal allele can be determined by the help of Hardy-Weinberg law by the following method: **

**A. Calculation of Gene Frequencies of Autosomal Genes****: **

An autosomal gene locus may have codominant alleles, dominant and recessive alleles or multiple alleles. If one desires to determine the gene frequencies for each of these kinds of autosomal alleles in a given population, he has to adopt the different methods.

**(i) Calculation of Gene Frequencies for Codominant Alleles: **

ADVERTISEMENTS:

When codominant alleles are present in a two-allele system, each genotype has a characteristic phenotype. The numbers of each allele in both homozygous and heterozygous conditions may be counted in a sample of individuals from the population and expressed as a percentage of total number of alleles in a sample.

If the sample is representative of the entire population (containing proportionately same numbers as found in the entire population) then we can obtain an estimate of the allelic frequencies in the gene pool. If in a given sample of N individuals of which D are homozygous for one allele (A^{1} A^{1}), H are heterozygous (A^{1 }A^{2}), and R are homozygous for the allele (A^{2} A^{2}), then N D + H + R.

Since each of the N individuals are diploid at this locus, there are 2N alleles represented in the sample. Each A^{1} A^{1} genotype has two A^{1} alleles. Heterozygotes have only one A^{1} allele. Letting p represents the frequency of the A^{1} allele and q the frequency of the A^{2} allele, we have-

**Example: **

ADVERTISEMENTS:

The M-N blood type furnishes a useful example of a series of phenotypes due to a pair of codominant alleles. None of three possible phenotypes, M, MN and N, appears to have any selection value. The frequencies of the two alleles (viz., L^{M} and L^{N}) for a sample from a group of white Americans living in New York City, Boston, and Columbus, Ohio, can be calculated by the following ways-

**The sample of 6,129 Caucasian people includes the following three groups according to phenotypes and genotypes on M-N system:**

To calculate frequencies of the two codominant alleles, L^{M} and L^{N}, it should be kept in mind that these 6,129 persons possess a total of 6,129 x 2 = 12,258 genes. The number of L^{M} alleles, for example, is 1,787 + 1,787 + 3,039. Thus, calculation of the frequency of L^{M} and L^{N} alleles is worked out in this way.

Thus, the frequencies of the two codominant alleles in this sample are almost equal, and this is reflected in the close approximation to a 1: 2: 1 ratio, which is a simple monohybrid ratio for codominant alleles in Mendelian genetics.

Gene frequencies expressed as decimals may be used directly to state probabilities (a probability is a function that represents the likelihood of occurrence of any particular form of an event).

If we can assume this sample to be representative of the population, then there is a probability of 0.5395 that of the chromosomes bearing this pair of alleles, any one selected randomly will bear gene L^{M}, and 0.4605 that it will bear L^{N}.

**Let p represents genotypic frequency of L ^{M} allele and q represents frequency of L^{N} allele, then the frequencies of three genotypes to be expected in the population are as follows: **

**(ii) Calculation of Gene (Allele) Frequencies for Dominant and Recessive Autosomal Alleles:**

Calculation of the gene frequencies for alleles which exhibit dominance and recessive relationships requires a different approach from that used with codominant alleles. A dominant phenotype may have either of two genotypes, AA or Aa, but we have no way (other than by laboriously test- crossing each dominant phenotype) of distinguishing how many are homozygous or heterozygous in our sample.

The only phenotype whose genotype is known for certain is the recessive (aa). If the population is in equilibrium then we can obtain an estimate of q (the frequency of the recessive allele) from q^ (the frequency of the recessive genotype or phenotype).

**Example 1:**

ADVERTISEMENTS:

If 75% of a population was of the dominant phenotype (A-), then 25% would have recessive phenotype (aa). If the population is in equilibrium with respect to this gene locus, we expect q^{2} = frequency of aa.

Then q^{2} = 0.25, q = 0.5, p= 1 – q = 0.5.

**Example 2:**

An interesting pair of contrasting traits, which has been detected in human populations and has no known selective value is the ability or inability to taste the chemical phenylthiocarbamide (“PTC”, C_{7}H_{3}N_{2}S), also called phenylthiourea. This was reported by Fox in 1932, who found a similar situation for several other thiocarbamides.

ADVERTISEMENTS:

The test is a simple one which can easily be performed by any genetics class. The usual procedure is to impregnate filter paper with a dilute solution of PTC (about 0.5 to 1 gram per litre), allow it to dry, then place a bit of the treated paper on the tip of the tongue.

About 70 per cent of an American white population can taste the substance, generally as very bitter, rarely as sweetish. Although the physiological basis is unknown, tasting ability does depend on a completely dominant gene, which we will designate as T. Thus tasters are T-(TT or Tt) and nontasters are tt.

From a group of 146 genetics students who tested themselves for tasting ability, 105 were tasters and 41 were nontasters. From these results the frequencies of alleles T and t in the sample may be calculated readily.

The 41 (28 per cent of the sample) nontasters are persons of tt genotype, and in the Hardy-Weinberg theorem may be represented by q^{2} Therefore:

ADVERTISEMENTS:

q^{2} = 0.28 and q = VO.28 = 0.53 (frequency of t).

Since p + q = 1, p = 1 – q; p = 1 – 0.53 = 0.47 (frequency of T). The frequency of homozygous and heterozygous tasters may now be computed, using the expression p^{2} + 2pq + q^{2} = 1.

P^{2} = TT = (0.47)^{2} = 0.2209

2pq = Tt = 2(0.47 x 0.53) = 0.4983

q^{2} = tt = (0.53)^{2} = 0.2809/1.0000

By testing representative samples of different populations, the frequencies of T and t in those groups may similarly be calculated.

**(iii) Calculation of Gene (Allele) Frequencies for Autosomal Multiple Alleles:**

The binomial (p + q)^{2} = 1 applies when only two autosomal alleles occur at a given locus. For cases of multiple alleles we simply add more terms to the expression.

**Example: **

The four human blood types—A, B, AB, and O are determined by a series of three multiple alleles, L^{A} or I^{A}, L^{B} or I^{B}, and L^{0} or i, if we neglect the various subtypes.

**Hence, in a gene frequency analysis, we can here let: **

p = frequency of I^{A}

q = frequency of I^{B}

r = frequency of i

and p + q + r= 1

Thus, genotypes in a population under random mating will be given by (p + q + r)^{2}.

**In a sample of 23,787 persons from Rechester, New York, following frequencies for four blood types were recorded: **

The frequency of each allele may now be calculated from these data, remembering that we have let p, q and r represent the frequencies of genes I^{A}, I^{B}, i respectively.

**The value of r, that is, the frequency of gene i, is immediately evident from the figure given: **

r^{2} = 0.444, hence

r = √0.444 = 0.6663 (= frequency of i)

The sum of A and O phenotypes is given by (p + r)^{ 2} = 0.418 + 0.444 = 0.862; therefore,

p + r = √0.862 =0.9284

So p = (p + r)-r = 0.9284 – 0.6663 = 0.2621 (= frequency of I^{A}).

**Because p + q +r= 1, q = 1 – (p + r) = 1 -0.9284 = 0.0716 (= frequency of I ^{B}) we can now calculate genotypic frequencies as shown in the following table: **

**B. Calculation of Gene Frequencies for Sex-Linked Genes****: **

Alleles in the sex chromosomes may occur in a different frequency than those in autosomes because of the difficult arrangements of sex chromosomes in the two sexes. The same techniques with one small modification, may be used in treating sex-linked genes.

However, since human males or Drosophila males being heterogametic sexes contain only one X chromosome, they cannot reflect a binomial distribution for random combination of pairs of sex-linked genes as females. Equilibrium distribution of genotypes for a sex-linked trait, where p + q = I, is given by-

♂p + q

♀p^{2} + 2pq + q^{2}

**Example: **

In human population red-green colour blindness is a trait due to a sex-linked recessive, which we may designate r. About 8 per cent of males are colour-blind. This shows at once that q, the frequency of gene r, is 0.08 and p, the frequency of its normal allele, R is 0.92. Thus, the frequency of colour blind females is expected to be q^{2} = 0.0064.

This is about what is found. Sex-linked dominants may be handled in a similar fashion; in the case of normal colour vision, with the value of p – 0.92, the incidence of normal women is p^{2} + 2pq = 0.9936.