﻿ Genetics Evolution

# 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

 populationP-1 Ap = 0.7 aq = 0.3 Ap = 0.7 A A0.7*0.7 = 0.49 A a0.7*0.3 = 0.21 aq = 0.3 A a0.3*0.7 = 0.21 a a0.3*0.3 = 0.09 offspringF-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.