I | THE INTUITIVE APPROACH |
II | THE HARDY-WEINBERG EQUILIBRIUM |
II-1 | FOR AN AUTOSOMAL, DIALLELE, CO-DOMINANT GENE |
EXERCISE | |
III | THE HW LAW |
III-1 | DEMONSTRATION OF THE LAW |
III-2 | EXERCISES |
III-3 | CONSEQUENCES OF THE LAW |
III-3.1 | WHAT IS THE ALLELE FREQUECY IN THE n+ 1 GENERATION? |
III-3.2 | WHAT IS THE GENOTYPE FREQUENCY IN THE n+ 1 GENERATION? |
III-3.3 | EXAMPLE |
IV | EXTENSION OF HW TO OTHER GENE SITUATIONS |
IV-1 | TO AN AUTOSOMAL, TRIALLELE, CO-DOMINANT GENE |
IV-2 | TO AN AUTOSOMAL, DIALLELE, NON CO-DOMINANT GENE |
IV-3 | TO AN AUTOSOMAL, TRIALLELE, NON CO-DOMINANT GENE |
IV-3.1 | BERNSTEIN's EQUATION |
IV-4 | TO A HETEROSOMAL (= gonosomic) GENE |
IV-4.1 | Y CHROMOSOME |
IV-4.2 | X CHROMOSOME |
V | SUMMARY- CONSEQUENCES OF HW's LAW |
*
I- THE INTUITIVE APPROACH
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,
p is the frequency of allele A1 | 0 =< p =< 1 | |
q is the frequency of allele A2 | 0 =< q =< 1 and p + q = 1 |
--> frequency of the genotypes: : | AA = p^{2} | and the phenotypes [ ]: | [A] = p^{2} + 2pq |
Aa = 2pq | [a] = q^{2} | ||
aa = q^{2} |
Example : phenylketonuria (recessive autosomal), of which the deleterious gene has a frequency of 1/100:
--> q = 1/100
therefore, the frequency of this disease is q^{2} = 1/10 000,
and the frequency of heterozygotes is 2pq = 2 x 99/100 x 1/100 = 2/100;
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 EQUILIBRIUM
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:
II-1. FOR AN AUTOSOMAL, DIALLELE, CO-DOMINANT GENE (Alleles A1 and A2)
Let:
Génotypes | A1A1 | A1A2 | A2A2 | ||
Number of subjects | DN | HN | RN | (total number N) | |
Frequencies F(G) | D | H | R | with (D+H+R) = 1 |
allele frequencies F(A) : | de A1 | D + H/2 = p | |
de A2 | R + H/2 = q | with p+q=1 |
NOTES
proba of drawing A1 = | drawing A1A1: : D x 1 then drawing A1 into A1A1 |
or | drawing A1A2: H x 1/2 then drawing A1 amongst A1A2 |
sum: | -> -> P(A1) = D + H/2 |
EXERCISE
let | les phénotypes | [A1] | [A1A2] | [A2] | ||
les génotypes | A1A1 | A1A2 | A2A2 | |||
number of subjects | 167 | 280 | 109 | total N : 556 |
calculate the following frequencies: F(P: phenotypes), F(G: genotypes), F(A: alleles), F (gametes) : F(A) = F(gam), because there is 1 allele (of each gene) per gamete In addition, here F(p) = F(G), because these are co-dominant alleles
F(P) = F(G) | 167/556 | 280/556 | 109/556 | |
Where : | D=0.300 | H=0.504 | R=0.196 | confirm: S(D,H,R)=2 |
confirm:S(p,q)=1
F(A) = F(gam.) | p = D+H/2 = (167+280/2)/ 556 or 0.300+0.504/2 = 0.552 |
q = R+H/2 = (109+280/2)/ 556 or 0.196 + 0.504/2 = 0.448 | |
III- LOI DE HW
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.
FIG.1
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.
III-1. DEMONSTRATION OF THE LAW
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 = p^{2} and similarly for A2, that of producing an A2A2 zygote is q x q = q^{2}. The probability of producing a heterozygote is pq + pq = 2pq. Finally, p^{2} + 2pq + q^{2} = (p+q)^{2} = 1
A1A1 | A1A2 | A2A2 | |
D = p^{2} | H=2pq | R = q2 | <- seulement sous HW |
Table of gametes
A1 | A2 | |
(p) | (q) | |
_________ | __________ | |
A1 (p) | A1A1 (p^{2}) | A1A2 (pq) |
A2 (q) | A1A2 (pq) | A2A2 (q^{2}) |
III-2. EXERCICES
if H = 0-> | p = D + H/2 = 0.5 | -> | D = 0.5 | H = 0 | R = 0.5 |
if H = 1-> | D = R = 0 | -> | D = 0 | H = 1 | R = 0 |
AA | AB | BB | |
1787 | 3039 | 1303 | N=6129 |
DN | HN | RN |
Confirmation: :
c2 = |
S (0i - Ci)^{2} Ci | = |
(1787 - 1787.2)^{2} 1787.2 |
+ (3039 - 3044.9)^{2} 3044.9 |
+ (1303 - 1296.9)^{2} 1296.9 |
= NS |
III-3. CONSEQUENCES OF THE LAW
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
III-3.1. WHAT WILL THE ALLELE FREQUENCY BE IN THE n+1 GENERATION?
A1A1 | A1A2 | A2A2 | |
n | p^{2} | 2pq | q^{2} |
n + 1 | F(A1) = D + H/2 = p^{2} +1/2 (2pq) = p (p+q) = p |
F(A2) = R + H/2 = q^{2} +1/2 (2pq) = q (p+q) = q |
-> no change in allele frequencies:
III-3.2. WHAT WILL THE GENOTYPE FREQUENCY BE IN THE n+1 GENERATION ?
male | p2 | 2pq | q^{2} | |||
female | A1A1 | A1A2 | A2A2 | |||
p2 | A1A1 | A1A1 | A1A1 | no A1A1 | ||
2pq | A1A2 | 1/2A1A1 | 1/4A1A1 | no A1A1 | Generation n+1 | |
q2 | A2A2 | no A1A1 | no A1A1 | no A1A1 |
Frequency of (A1A1) in the generation n+1 | = (p^{2})^{2} + 1/2 (2 pq.p^{2}) + 1/2 (p^{2}.2pq) + 1/4 (2pq)^{2} |
= p^{4} + p^{3}q + p^{3}q + p^{2}q^{2} = p^{2} (p^{2} + 2pq + q^{2}) = p^{2} |
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.
III-3.3. EXAMPLE
The MN human blood groups.
Group | MM | MN | NN | |
Number: | 1787 | 3039 | 1303 | Total, N = 6129 |
IV- EXTENSION OF HW TO OTHER GENE SITUATIONS
IV-1.TO AN AUTOSOMAL, TRIALLELE, CO-DOMINANT GENE
3 alleles A1, A2, A3 |
with frequencies F(A1) = p, F(A2) = q, F(A3) = r |
A1A1 | A1A2 | A1A3 | A2A2 | A2A3 | A3A3 | |
Genotype frequencies according to HW | p^{2} | 2pq | 2pr | q^{2} | 2qr | r^{2} |
p | q | r | |||
A1 | A2 | A3 | |||
p | A1 | p^{2} | pq | pr | |
q | A2 | pq | q^{2} | qr | |
r | A3 | pr | qr | r^{2} |
IV-2. TO AN AUTOSOMAL, DIALLELE, NON CO-DOMINANT GENE
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.
Genotypes | AA | Aa | aa |
Phenotypes | [A] | [a] | |
Number | N1 | N2 | N |
Frequency of genotype | 1-q^{2} | q^{2} | |
and the frequency of the allele a = F(a) =(q^{2})1/2 = (N2/(N1 + N2))^{1/2}
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.
Gene
Gene | Incidence in population q^{2} | Frequency of allele q | Frequency of heterozygotes 2pq |
Albinism | 1/22 500 | 1/150 | 1/75 |
Phenylketonuria | 1/10 000 | 1/100 | 1/50 |
Mucopolysaccharidosis | 11/90 000 | 1/300 | 1/150 |
IV-3. TO AN AUTOSOMAL, TRIALLELE, NON CO-DOMINANT GENE
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
rdiffering genotype and phenotype frequencies are found by applying the Hardy-Weinberg law. .
Phenotype | Genotype | Genotype frequency | Phenotype frequency |
[A] | (AA) | p^{2} | |
(AO) | 2pr | p^{2}+2pr | |
[B] | (BB) | q^{2} | |
(BO) | 2qr | q^{2}+2qr | |
[AB] | (AB) | 2pq | 2pq |
[O] | (OO) | r^{2} | r^{2} |
Using::
IV-3.1. BERNSTEIN's EQUATION (1930)
Bernstein's 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)
Example :
Group | A | B | O | AB |
Number | 9123 | 2987 | 7725 | 1269 |
Frequency | 0.4323 | 0.1415 | 0.3660 | 0.601 |
p0= 1 % h0.3660+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.
IV-4. TO A HETEROSOMAL (= gonosomic) GENE
IV-4.1. Y CHROMOSOME: frequency p and q in subjects XY; transmission to male descendants.
IV-4.2. X CHROMOSOME:
Female | XA1XA1 | XA1XA2 | XA2XA2 |
p^{2} | 2pq | q^{2} |
M&ale | XA1/Y | XA2/Y |
p | q |
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:
q_{x}^{(n)} - q_{xx}^{(n)} = q_{xx}^{(n-1)} - (q_{xx}^{(n-1)})/2 - (q_{xx}^{(n-1)}) /2 = - 1/2 (q_{x}^{(n-1)} - q_{xx}^{(n-1)})
--> q_{x}^{(n)} - q_{xx}^{(n)} = (- 1/2)^{n} (q_{x}^{(0)} - q_{xx}^{(0)}) : tends towards zero in 8 to 10 generations
* mean frequency q: :
* 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:
answer
G0: X | XDXD |
G0 : qx(0) = 0.00 | qxx(0) = 1.00 |
G1 : qx(1) = 1.00 | qxx(1) = 0.50 |
G2 : qx(2) = 0.50 | qxx(2) = 0.75 |
G3 : qx(3) = 0.75 | qxx(3) = 0.63 |
G4 : qx(4) = 0.63 | qxx(4) = 0.69 |
G5 : qx(5) = 0.69 | qxx(5) = 0.66 |
G6 : qx(6) = 0.66 | qxx(6) = 0.60 |
FIG.2
Therefore:
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.
V- CONSEQUENCES OF THE HW LAW
Contributor(s) |
Written | 2001-02 | Robert Kalmes, Jean-Loup Huret |
Genetics, Dept Medical Information, UMR 8125 CNRS, University of Poitiers, CHU Poitiers Hospital, F-86021 Poitiers, France (JLH) |
© Atlas of Genetics and Cytogenetics in Oncology and Haematology | indexed on : Mon Sep 18 16:46:58 CEST 2017 |
For comments and suggestions or contributions, please contact us