Hardy-Weinberg’s Law

 

Genetic Equilibrium

(Hardy-Weinberg’s Law)

(By Dr. Girish Chandra)

 

SUMMARY

 

            Fundamental ideas about population genetics were published independently in 1908 by G.H. Hardy in England and W. Weinberg in Germany. At that time it was thought that alleles of a gene would distribute in 3:1 ratio in F1 generation. But it is now known that heterozygotes appear in much higher frequencies in populations. Further study and treatment of gene frequencies was developed by R.A. Fischer (British), J.B.S. Haldane (British) and Sewall Wright (USA) in 1920-30.

 

            According to Hardy-Weinberg’s Law­—­­­ Relative frequencies of various kinds of genes in a large and randomly mating population tend to remain constant from generation to generation in the absence of mutation and natural selection.

 

            Hardy-Weinberg’s law describes a tendency of evolution to conserve gains of genetic changes and avoid too frequent changes in genotype. Therefore, the basic factors that cause evolution are: mutation, natural selection, non-random mating, small population and genetic drift.

            Gene frequency is the proportion of an allele in the gene pool as compared with other alleles at the same locus. It can be calculated by dividing the number of a particular gene by the total number of genes present in the population. For example if there are 100 individuals in a population, 40 of them being dominant MM, 40 heterozygous Mm and 20 recessive mm, the frequency of the dominant gene M, which is depicted by P, would be 40+20/100=0.6 and the frequency of the recessive gene m, denoted by q, would be 20+20/100=0.4. P+q should always be 1.0 and hence when frequency of one gene increases, that of the other must decrease.

            Genotype frequency is the total number of one kind of individuals in a population exhibiting similar characters (genotype) in respect to the locus. It can be determined by dividing the number of individuals having one kind of genotype by the total number of individuals in a population. Therefore, in the above example, the genotype frequency of the dominant homozygotes MM will be D/N=40/100=0.4, that of Mm will be H/N=40/100=0.4 and that of recessive mm will be r/N=20/100=0.2, where N=total number of individuals and D,H and r denote dominant, heterozygous and recessive respectively.

                       

The next generation will have the following composition:  0.25 MM  +  0.5 Mm  +  0.25 mm, which can also be written as:  1 MM  +  2 Mm  +  1 mm  OR  M2  +  2 Mm  +  m2 .  If P is the frequency of gene M and q is the frequency of gene m, then the above equation can be written as P2  + 2Pq  +  q2  =  (P+q)2. This is called Hardy-Weinberg’s equation, which can be used to find out genetic composition of a population and also to prove Hardy-Weinberg’s Law that gene frequencies remain unchanged from generation to generation, as illustrated in the following example.

 

            Example.  Let us presume there is population of rats, which has 50% brown (MM) and 50% white (mm) individuals. The gene frequencies will be P=0.5 and q=0.5 (P+q=1.0). Both gene frequencies and genotype frequencies are same here. Now let us substitute the figures in the Hardy-Weinberg’s equation as follows:

            P2 + 2Pq + q2  =  (0.5)2  +  2 . (0.5) (0.5)  +  (0.5)2  =   0.25  +  0.5  +  0.25 

            The genotype frequency in F1 generation will be—25% MM  +  50% Mm  +  25% mm but the gene frequencies will still remain as  P=0.5 and q=0.5, that is they remain unchanged even in the next generation.

            In this example dominant gene equals the recessive gene but in nature usually the recessive gene is far less in number. Will the gene frequencies remain unchanged even in such a case is illustrated by the following example.

             

Application of the Hardy-Weinberg’s equation: If we know the number of homozygous recessive animals in a population (which is easy because in these recessive gene has expressed), then we can find out the entire gene pool of the population or the entire genotype frequency can be found out using Hardy-Weinberg’s equation as in the following examples.

 

            Example.  If in a population of rats16% individuals are recessive white and rest are brown, what will be the composition of the population (i.e. find out the value of MM and Mm).

            If white individuals are 16% (mm), then m2 = 16 OR q2 = 0.16; q = square root of 0.16; q = 0.4.

            As we have found out the value of q, the value of P can be known by P=1.0-q; or 1.0-0.4 or P=0.6.

           Now to calculate the heterozygous Mm rats—2Pq = 2.(0.6)(0.4) = 0.48, which means 48% of rats were heterozygous brown (Mm), and since we know that 16% were white (mm), the rest of the population has to be MM or  homozygous brown (100-48+16) = 16%. The final composition of the population will be 36%MM + 48%Mm + 16%mm.

 

            Example.  Among 200 million Americans, 125,000 suffer from the heritable disease Cystic fibrosis, which is expressed only in recessive homozygous individuals. Carriers (heterozygotes) can be detected by excessive sodium in perspiration and in nail clippings. To find out the number of carriers in the population, calculate as follows:

            Sufferers (mm) = 125,000/200 million = 0.0625%;  q2=0.000625 (0.0625/100);  q=0.025.

            P=1-q (1.0-0.025) = 0.975.

            P2=(0.975)2 = 0.9506 OR 95.06% are homozygous dominant (MM) individuals.

            2Pq=2.(0.975 x 0.025) = 0.0488  OR  4.88% individuals are heterozygous (Mm) carriers of the disease.

            q2=(0.025)2 = 0.000625  OR  0.0625% individuals are homozygous recessive (mm) sufferers.

            The genotype frequency of the population reveals that one in every 20 Americans is a carrier of the disease.

              

            Example. In human population one in 20,000 is an albino, let us find out the genotype frequency.

            q2=1/20,000 = 0.00005;  q=square root of 0.00005 = 0.007

            P=(1.0-q) = (1.0-0.007) = 0.993;  P2=(0.993)2 = 0.986

            2Pq=2.(0.993 x 0.007) = 0.014

            Although one in 20,000 is an albino in human population, one in 70 is a carrier of the disease, which is 280 times as many as affected.

            Hardy-Weinberg’s equation is very useful in finding out the genotype frequencies in a population, which have not expressed and are hidden.