The Hardy-Weinberg law can be used under some circumstances to calculate genotype frequencies from allele frequences. Let A1 and A2 be two alleles at the same locus,
where the distribution of allele frequencies is the same in men and women, i.e.:
hommes (p,q) femmes (p,q)
these frequencies remain constant in successive generations.
Example: autosomal recessive inheritance with alleles A and a, and allele frequencies p and q:
Example: phenylketonuria (recessive autosomal), of which the deleterious gene has a frequency of 1/100 --> q = 1/100
Note that there are a lot of heterozygotes: 1/50, two hundred times more than there are individuals suffering from the condition.
For a rare disease, p is very little different from 1, and the frequency of the heterozygotes = 2q.
We use these equations implicitly, in formal genetics and in the genetics of pooled populations, usually without considering whether, and under what conditions, they are applicable.
The Hardy-Weinberg equilibium, which is also known as the panmictic equilibrium, was discovered at the beginning of the 20th century by several researchers, notably by Hardy, a mathematician and Weinberg, and physician.
The Hardy-Weinberg equilibrium is the central theoretical model in population genetics. The concept of equilibrium in the Hardy-Weinberg model is subject to the following hypotheses/conditions:
Under these circumstances, the genetic diversity of the population is maintained and must tend towards a stable equilibrium of the distribution of the genotype.
allele frequencies F(A):
In a population consisting of an infinite number of individuals (i.e. a very large population), which is panmictic (mariages occur randomly), and in the absence of mutation and selection, the frequency of the genotypes will be the development of (p+q)2, p and q being the allele frequencies.
The figure shows the correspondence between the allele frequency q of a and the genotype frequencies in the case of two alleles in a panmictic system. The highest frequency of heterozygotes, H, is then reached when p = q and H = 2pq = 0.50. In contrast, when one of the alleles is rare (i.e. q is very small), virtually all the subjects who have this allele are heterozygotes.
Let A be an autosomal gene that is found in a population in two allele forms, A1 and A2 (with the same frequencies in both sexes of course). As there is codominance, 3 genotypes can be distinguished. According to the hypotheses/conditions of Hardy-Weinberg (HW), the individuals of the n + 1 generation will be assumed to be the descendants of the random union of a male gamete and a femal gamete.
Consequently, if, by generation n, the probability of drawing an A1 allele is p, then that of producing an A1A1 zygote after fertilization is p x p = p2 and similarly for A2, that of producing an A2A2 zygote is q x q = q2. The probability of producing a heterozygote is pq + pq = 2pq. Finally, p2 + 2pq + q2 = (p+q)2 = 1
Change in HW across the generations (demonstration that the frequencies are invariable). In a population subject to HW, an equilibrium involving the distribution of the genotype frequencies is reached after a single reproductive cycle.
Is a population in the n generation
-> no change in allele frequencies:
The frequency of the (A1A1) genotype does not change between generation n and generation n+1 (same demonstration for the (A2A2 ) and (A1A2) genotypes).
The genotype structure no longer undergoes any further changes once the population reaches the Hardy Weinberg equilibrium.
In very many examples, the frequencies seen in natural populations are consistent with those predicted by the Hardy-Weinberg law.
The MN human blood groups.
Frequency of M = (1787 + 3039/2)/ 6129 = 0.540 = p
Frequency of N = (1303 + 3039/2)/6129 = 0.460 = q
Predicted proportion of MM = p2 = (0.540)2 = 0,2916
Predicted proportion of MN = 2pq = 2(0.540)(0.460) = 0.4968
Predicted proportion of NN = q2 = (0.460)2= 0.2116
Numbers predicted by Hardy-Weinberg :
In the present situation, there is no need to do c2 test to see that the actual numbers are not statistically different from those predicted.
3 alleles A1, A2, A3
with frequencies F(A1) = p, F(A2) = q, F(A3) = r
there will be 6 genotypes
A is dominant over a, which is recessive; in this case the genotypes (AA) and (Aa) cannot be distinguished within the population. Only the individuals with the phenotype [A], who number N1, will be distinguishable from the individuals with the phenotype [a], who number N2.
This is a method commonly used in human genetics to calculate the frequency of rare, recessive genes.
Frequencies of homozygotes and heterozygotes for rare recessive human genes.
Example: the ABO blood group system. Although the human (ABO) blood group system is often taken to be a simple example of polyallelism, it is in fact a relatively complex situation combining the codominance of A and B, the presence of a nul O allele and the dominance of A and B over O.
If we take
Bernsteins equation (1930) simplifies the calculations:
p = 1 - (F[B] + F[O])1/2
q = 1 - (F[A] + F[O])1/2
r = (F[O])1/2
then, if p+q+r # 1, correction by the deviation D = 1 - (p + q + r) --> p= p (1 + D/2)
q= q (1 + D/2)
r= (r + D/2) (1 + D/2)
p = 1 - (0.3660+0.1415)1/2 = 0.2876
q = 1 - (0.3660+0.4323)1/2 = 0.1065
r = = 0.6050
p+q+r = 0.9991 ... --> p= 0.2877, q= 0.1065, r= 0.6057.
frequency p and q in subjects XY; transmission to male descendants.
i.e. the frequency of the q allele, is qx in men, and qxx in women:
* calculation of the difference in allele frequencies between the 2 sexes:
qx(n) - qxx(n) = qxx(n-1) - (qxx(n-1))/2 - (qxx(n-1)) /2 = - 1/2 (qx(n-1) - qxx(n-1))
--> qx(n) - qxx(n) = (- 1/2)n (qx(0) - qxx(0)) : tends towards zero in 8 to 10 generations
* mean frequency q:
1/3 of the X chromosomes belong to men, 2/3 to women: q = 1/3 qx(n) + 2/3 qxx(n)
the mean frequency is invariable (develop q1 into q0 ...... --> q1 = q0)
at equilibrium, q(e) est : qx(e) = qxx(e) = q(e)
* exercise: For generation G0, consisting of 100% of normal men and 100% of color-blind women, calculate the frequencies of the gene up to G6.
For a sex-linked locus, the Hardy Weinberg equilibrium is reached asymptotically after 8-10 generations, whereas it is reached after 1 generation for an autosomal locus.
Kalmes R, Huret JL
Atlas of Genetics and Cytogenetics in Oncology and Haematology 2001-02-01
Online version: http://atlasgeneticsoncology.org/teaching/30076/hardy-weinberg-model