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Hardy-Weinberg model

ITHE INTUITIVE APPROACH
IITHE HARDY-WEINBERG EQUILIBRIUM
II-1FOR AN AUTOSOMAL, DIALLELE, CO-DOMINANT GENE
EXERCISE
IIITHE HW LAW
III-1DEMONSTRATION OF THE LAW
III-2EXERCISES
III-3CONSEQUENCES OF THE LAW
III-3.1WHAT IS THE ALLELE FREQUECY IN THE n+ 1 GENERATION?
III-3.2WHAT IS THE GENOTYPE FREQUENCY IN THE n+ 1 GENERATION?
III-3.3EXAMPLE
IVEXTENSION OF HW TO OTHER GENE SITUATIONS
IV-1TO AN AUTOSOMAL, TRIALLELE, CO-DOMINANT GENE
IV-2TO AN AUTOSOMAL, DIALLELE, NON CO-DOMINANT GENE
IV-3TO AN AUTOSOMAL, TRIALLELE, NON CO-DOMINANT GENE
IV-3.1BERNSTEIN's EQUATION
IV-4TO A HETEROSOMAL (= gonosomic) GENE
IV-4.1Y CHROMOSOME
IV-4.2X CHROMOSOME
VSUMMARY- 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 A10 ≤ p ≤ 1
q is the frequency of allele A20 ≤ q ≤ 1 and p + q = 1
 where the distribution of allele frequencies is the same in men and women, i.e.:
hommes (p,q)     femmes (p,q) if they procreate :    (p + q)2 = p2 + 2pq + q2 = 1
where:
p2 = frequency of the A1 A1 genotype <-- HOMOZYGOTE
2pq = frequency of the A1 A2 genotype   <-- HETEROZYGOTE
q2 = frequency of the A2 A2 genotyp  <-- HOMOZYGOTE
these frequencies remain constant in successive generations.
 
Example : autosomal recessive inheritance with alleles A and a, and allele frequencies p and q:
--> frequency of the genotypes: : AA = p2and the phenotypes [ ]: [A] = p2 + 2pq
Aa = 2pq [a] = q2
aa = q2

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 q2 = 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:

  1. The population is panmictic (couples form randomly (panmixia), and their gametes encounter each other randomly (pangamy))
  2. The population is "infinite" (very large: to minimize differences due to sampling).
  3. There must be no selection, mutation, migration (no allele loss /gain).
  4. Successive generations are discrete (no crosses between different generations).

Under these circumstances, the genetic diversity of the population is maintained and must tend towards a stable equilibrium of the distribution of the genotype.

II-1. FOR AN AUTOSOMAL, DIALLELE, CO-DOMINANT GENE (Alleles A1 and A2)
Let:

  • The frequencies of genotypes F(G) be called D, H, and R with 0 ≤ [D,H,R] = < 1 and D + H + R = 1
  • The frequencies of alleles F(A) be called p, and q with 0 ≤ [p,q] ≤ 1 and p+q = 1
    Génotypes A1A1A1A2A2A2
    Number of subjectsDNHNRN (total number N)
    Frequencies F(G)DHR with (D+H+R) = 1

    allele frequencies F(A) :de A1D + H/2 = p
    de A2 R + H/2 = qwith p+q=1

    NOTES

  • The genotype frequencies F(G) can always be used to calculate the allele frequencies F(A)
  • F(A) contains less information than F(G)
  • if p = 0: allele is lost; if p = 1: allele is fixed..

  • first demonstration that p = D + H/2, by counting the alleles:
  • second demonstration, by calculating the probabilities:
    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
    similarly for A2 ...;

    EXERCISE
    let les phénotypes[A1][A1A2][A2]
    les génotypesA1A1A1A2A2A2
    number of subjects167280109    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)167/556280/556109/556
    Where :D=0.300H=0.504R=0.196 confirm: Σ (D,H,R)=2

    confirm:Σ (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 = 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

    A1A1A1A2A2A2
    D = p2H=2pqR = q2<- seulement sous HW

    Table of gametes
    A1A2
    (p)(q)
    ___________________
    A1 (p)A1A1 (p2)A1A2 (pq)
    A2 (q)A1A2 (pq)A2A2 (q2)

  • (The allele frequencies can only be used to calculate the genotype frequencies if they are subject to HW)
  • the allele frequencies remain the same from one generation to another
  • the genotype frequencies remain the same from one generation to another

    III-2. EXERCICES

  • exercise : Show that, in the absence of panmixia, two populations with similar allele frequencies can have different genotype frequencies (by doing this, you show that there is a loss of information between genotype and allele frequencies):
    example : for p = q = 0,5 answer
    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

  • exercise : calculation of the genotype and allele frequencies, calculation of the numbers predicted by HW (theoretical numbers of individuals), and confirmation that we are indeed in a situation subject to HW :
    AAABBB
    178730391303N=6129
    DNHNRN
     answer:
    F(A) = (1787 + 3039/2) / 6129 = 0.54 = p
    F(B) = (1303 + 3039/2) / 6129 = 0.46 = q … and Σ (p,q)=1

    genotype frequencies predicted by HW genotype frequencies predicted by HW
    AA : p2 = (0.54)2 = 0.2916
    AB = 2pq = 2x 0.54 x 0.46 = 0.4968
    BB : q2 = (0.46)2 = 0.2116

    Numbers predicted by HW
    AA : p2N = 0.2916 x 6129 = 1787.2
    AB : 2pqN = 0.4968 x 6129 = 3044.9
    BB : q2N = 0.2116 x 6129 = 1296.9

    Confirmation: :
    χ2 = Σ (0i - Ci)2
    Ci    		 = (1787 - 1787.2)2
    1787.2    		 + (3039 - 3044.9)2
    3044.9     		+ (1303 - 1296.9)2
    1296.9    		 = NS
    --> we are in a situation subject to HW

     

    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?

    A1A1A1A2A2A2
    np22pqq2

    n + 1F(A1) = D + H/2 = p2 +1/2 (2pq) = p (p+q) = p
    F(A2) = R + H/2 = q2 +1/2 (2pq) = q (p+q) = q

    -> no change in allele frequencies:

    in the n generation, we have p and q
    in the n+1 generation, we have p and q

    III-3.2. WHAT WILL THE GENOTYPE FREQUENCY BE IN THE n+1 GENERATION ?

    malep22pqq2
    femaleA1A1A1A2A2A2

    p2    		 A1A1A1A1A1A1no A1A1
    2pq A1A2 1/2A1A11/4A1A1no A1A1Generation n+1
    q2 A2A2 no A1A1no A1A1 no A1A1
    Frequency of (A1A1) in the generation n+1 = (p2)2 + 1/2 (2 pq.p2) + 1/2 (p2.2pq) + 1/4 (2pq)2
    = p4 + p3q + p3q + p2q2 = p2 (p2 + 2pq + q2) = p2

    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.

    GroupMMMNNN
    Number:178730391303Total, N = 6129

    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 :
    for MM = p2N = 0,2916 x 6129 = 1787.2
    for MN = 2pqN = 0,4968 x 6129 = 3044.9
    for NN = q2N = 0,2116 x 6129 = 1296.9
    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.

     

    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

    there will be 6 genotypes

    A1A1A1A2A1A3A2A2A2A3A3A3
    Genotype frequencies
    according to HW     		p2 2pq 2prq22qr r2
    pqr
    A1A2A3
    p A1p2pqpr
    q A2pqq2qr
    r A3prqrr2

    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.

    GenotypesAAAaaa
    Phenotypes [A] [a]
    Number N1 N2N
    Frequency of genotype1-q2 q2
    with q2 = N2/N = N2 / (N1 + N2)

    and the frequency of the allele a = F(a) =(q2)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

    GeneIncidence in
    population q2     		Frequency
    of allele q    		Frequency of
    heterozygotes 2pq
    Albinism1/22 5001/1501/75
    Phenylketonuria1/10 0001/1001/50
    Mucopolysaccharidosis11/90 0001/3001/150

    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

    p to designate the frequency of allele A
    q to designate the frequency of allele B

    (p + q + r = 1)

    rdiffering genotype and phenotype frequencies are found by applying the Hardy-Weinberg law. .

    PhenotypeGenotypeGenotype frequency Phenotype frequency
    [A] (AA) p2
    (AO)2prp2+2pr
    [B](BB)q2
    (BO)2qrq2+2qr
    [AB](AB) 2pq2pq
    [O](OO)r2r2

    Using::

    p2 +2pr +r2 = (p+r)2
    q2 +2qr +r2 = (q+r)2     where F[A] + F[O] = (p+r)2
    F[B] + F[O] = (q+r)2 et F[O] = r2

     

    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 :

    GroupABOAB
    Number9123298777251269
    Frequency 0.43230.14150.36600.601

    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.

    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:
    FemaleXA1XA1XA1XA2XA2XA2
    p2 2pq q2
    M&ale XA1/YXA2/Y
    p q

    i.e. the frequency of the q allele, is qx in men, and qxx in women:

  • the X chromosome of the boys (in generation n) has been transmitted from the mothers (generation n-1) --> qx(n) = qxx(n-1) qx(n) = qxx(n-1)
  • the X chromosome carrying the q allele in the daughters has:
  • 1/2 chance of coming from their father,
  • 1/2 chance of coming from their mother
    1/2 chance de venir du père --> qxx(n) = ( qx(n-1) + qxx(n-1))/2
  • the frequency of the allele in men = the frequency in women in the previous generation
  • the frequency of the allele in women = mean of the frequencies in the 2 sexes in the previous generation.

    * 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:
    answer
    G0: XNYXDXD
    G0 : qx(0) = 0.00qxx(0) = 1.00
    G1 : qx(1) = 1.00qxx(1) = 0.50
    G2 : qx(2) = 0.50qxx(2) = 0.75
    G3 : qx(3) = 0.75qxx(3) = 0.63
    G4 : qx(4) = 0.63qxx(4) = 0.69
    G5 : qx(5) = 0.69qxx(5) = 0.66
    G6 : qx(6) = 0.66qxx(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

  • Regardless of whether we are in a situation subject to HW or not, the genotype frequencies (D, H, R) can be used to calculate the allele frequencies (p,q), from : p = D + H/2, q = R + H/2.
  • Whereas, if and only if we are subject to HW, the genotype frequencies can be calculated from the allele frequencies, from D = p2, H = 2pq, R = q2.
  • The dominance relationships between alleles have no effect on the change in allele frequencies (although they do affect how difficult the exercises are!)
  • The allele frequencies remain stable over time; and so do the genotype frequencies.
  • The random mendelian segregation of the chromosomes preserves the genetic variability of populations.
  • Since "evolution" is defined as a change in allele frequencies, an ideal diploid population would not evolve.
  • It is only violations of the properties of an ideal population that allow the evolutionary process to take place.

  • The practical approach to a problem is always the same:
    1. The Numbers Observed --> the (Observed) Genotype Frequencies;
    2. Calculate the Allele Frequencies: p=D/2 + S Hi/2 , q = ...
    3. If we are subject to HW (hypothetically), then D=p2, H= 2pq, etc ... : we calculate the Theoretical Genotype Frequencies according to HW.
    4. The Calculated Genotype Frequencies --> the Calculated Numbers;
    5. Comparison of Observed Numbers - Calculated Numbers: : c2 = S (Oi - Ci)2/Ci
    6. If c2 is significant: we are not in accordance with HW; this
      ---> Consanguinity?
      ---> Selection?
      ---> Mutations ?

    Contributor(s)

    Written2001-02Robert 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 : Sat Dec 5 19:11:08 CET 2020

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