Hardy-Weinberg Principle
Background Principle
• Gene and allele frequencies in a
randomly-breeding population
remain constant from generation
to generation unless specific
disturbing influences are
introduced
• They are considered to be in
equilibrium
Disturbing influences
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Non-random mating
Mutations
Selection
Limited population size
Random genetic drift
Gene flow
• Hardy Weinberg equilibrium is
impossible in nature
• Deviation from the HardyWeinberg equilibrium denotes
the evolution of a species
HW Equation is derived using a
punnett square
Hardy-Weinberg Equilibrium
Equation
• Used to discover the probable
genotype frequencies in a population
and to track their changes from one
generation to another
p2 + 2pq + q2 = 1
p = frequency of the dominant allele
q = frequency of the recessive allele
• P2 = the predicted frequency of
homozygous dominant individuals
in a population
• 2pq = predicted frequency of
heterozygous individuals in a
population
• q2 = the predicted frequency of
homozygous recessive individuals
in a population
Observation of phenotypes
• It is usually only possible to
know the frequency of
homozygous recessive
individuals
• Individuals showing the
dominant trait could either be
homozygous or heterozygous
ALBINISM: A SAMPLE HARDYWEINBERG PROBLEM
• Albinism is a rare
genetically inherited trait
that is only expressed in
the phenotype of
homozygous recessive
individuals (aa)
• The average human
frequency of albinism in
North America is only about 1
in 20,000.
Referring back to the Hardy-Weinberg
equation (p² + 2pq + q² = 1), the
frequency of homozygous recessive
individuals (aa) in a population is
q². Therefore, in North America the
following must be true for albinism:
q² = 1/20,000 = .00005
• By taking the square root of
both sides of this equation,
we get: q = .007
• In other words, the
frequency of the recessive
albinism allele (a) is .00707
• Knowing one of the two
variables (q) in the HardyWeinberg equation, it is
easy to solve for the other
(p).
p=1-q
p = 1 - .007
p = .993
The frequency of the
dominant, normal allele (A)
is, therefore, .99293 or about
99 in 100.
• The next step is to plug the
frequencies of p and q into the
Hardy-Weinberg equation:
p² + 2pq + q² = 1
(.993)² + 2 (.993)(.007) + (.007)² = 1
.986 + .014 + .00005 = 1