Hardy-Weinburg Equilibrium

Remember Punnett Analysis?

hybrid hybrid
Aa X Aa
A a
A A A A a
a A a a a
offspring
genotype count phenotype prob
AA 1 type A 25%
Aa 2 hybrid 50%
aa 1 type a 25%

For a population breeding problem,
we must add probabilities to the Punnett Analysis
. . -where probability = [relative] frequency

population
P-1
A
p = 0.7		 a
q = 0.3
A
p = 0.7		 A A
0.7*0.7 = 0.49		 A a
0.7*0.3 = 0.21
a
q = 0.3		 A a
0.3*0.7 = 0.21		 a a
0.3*0.3 = 0.09
offspring
F-1
genotype count phenotype prob
AA 0.49 type A p2 = 0.49
Aa 2*0.21 hybrid 2pq = 0.42
aa 0.09 type a q2 = 0.09

  • or.. with dominance
    offspring
    F-1
    genotype count phenotype prob
    AA 0.91 dominant p2+2pq = 0.91
    Aa
    aa 0.09 recessive q2 = 0.09


  • and.. the F-1 gene pool is:
    alleles
    genotype number A a
    AA 49 98 0
    Aa 42 42 42
    aa 9 0 18
    subtotal - 140 60
    total 100 200
    p 0.70
    q 0.30

  • which is exactly the gene pool with which we started..

    QED (quod erat demonstratum):
  • The allele frequencies remain constant from generation to generation!
    . . and the F-n generation is:
    . . . . . {AA, Aa, aa} = {p2, 2pq, q2},
    . . . . . where p = probability of allele 'A,'
    . . . . . and q = (1 - p) = probability of allele 'a.'
  • assuming, "of course"...
    . . 1) mating is random,
    . . 2) there is no selection,
    . . 3) there is no mutation,
    . . 4) there is no immigration nor emmigration, and
    . . 5) the population is arbitrarily large (infinite)

  • However, in the Real World, these five assumptions are not met;
    . .. so the task becomes estimating which of the five
    . . . has the most influence on the observed results.

    Normally, we would use Chi-square to determine whether or not the Hardy-Weinburg predicts the observed results, where significant Chi-square means that at least one assumption is not met.
    Other statistical tests may provide a means to identify the critical assumptions.


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    © 2004 Prof. LaFrance, Ancilla College