HARDY WEINBERG: POPULATION GENETICS
Using mathematical approaches to calculate
changes in allele frequencies…this is evidence
of evolution.
Hardy-Weinberg equilibrium

Hypothetical, non-evolving population
 preserves

allele frequencies
natural populations rarely in H-W equilibrium
 useful
model to measure if forces are acting on a
population

measuring evolutionary change
G.H. Hardy
mathematician
W. Weinberg
physician
Evolution of populations

Evolution = change in allele frequencies in a
population

hypothetical: what conditions would cause allele
frequencies to not change?
1.
2.
3.
4.
5.
very large population size (no genetic drift)
no migration (no gene flow in or out)
no mutation (no genetic change)
random mating (no sexual selection)
no natural selection (everyone is equally fit)
H-W occurs ONLY in non-evolving populations!
Populations & gene pools

Concepts
a
population is a localized group of interbreeding
individuals
 gene pool is collection of alleles in the population

remember difference between alleles & genes!
 allele
frequency is how common is that allele in the
population

how many A vs. a in whole population
H-W formulas

Alleles:
p+q=1
B

b
p2 + 2pq + q2 = 1
Individuals:
BB
BB
Bb
Bb
bb
bb
Origin of the Equation

Assuming that a trait is recessive
or dominant


Allele pairs AA, Aa, aa would
exist in a population
Male
Gametes
a(q)
Female
gametes
A(p)
AA
p2
Aa
pq
Female
Gametes
a(q)
Aa
pq
aa
q2
p+q=1




Male
Gametes
A(p)
The probability that an individual
would contribute an A is called p
The probability that an individual
would contribute an a is called q
Because only A and a are present
in the population the probability
that an individual would donate
one or the other is 100%
p2 + 2pq + q2
Hardy-Weinberg theorem

Counting Alleles
 assume
2 alleles = B, b
 frequency of dominant allele (B) = p
 frequency of recessive allele (b) = q

Frequencies are usually
written as decimals!
frequencies must add to 1 (100%), so:
p+q=1
BB
Bb
bb
Hardy-Weinberg theorem

Counting Individuals
of homozygous dominant: p x p = p2
 frequency of homozygous recessive: q x q = q2
 frequency of heterozygotes: (p x q) + (q x p) = 2pq
 frequency

frequencies of all individuals must add to 1 (100%), so:
p2 + 2pq + q2 = 1
BB
Bb
bb
Practice Problem:


In a population of 100 cats, there are 16 white
ones. White fur is recessive to black.
What are the frequencies of the genotypes?
Use Hardy-Weinberg equation!
q2 (bb): 16/100 = .16
q (b): √.16 = 0.4
p (B): 1 - 0.4 = 0.6
p2=.36
BB
2pq=.48
Bb
q2=.16
bb
MustWhat
assume
are population
the genotype
is in
frequencies?
H-W equilibrium!
Answers:
p2=.36
Assuming
H-W equilibrium:
Expected data
Observed data
How do you
explain the data?
2pq=.48
q2=.16
BB
Bb
bb
p2=.20
=.74
BB
2pq=.64
2pq=.10
Bb
q2=.16
bb
Tips for Solving HW Problems:




Solve for q first.
Then solve for p.
Don’t assume you can just solve for p2 if only given
dominant phenotypic frequency.
READ carefully!!! HW Math is fun 
Homework:


Answer all the questions on the “Hardy Weinburg
Practice Problems” handout
Complete the following prelab questions:
 List
the conditions for a Hardy-Weinberg Population.
 Write a hypothesis for expected allele outcome for
Case 1.
 Write down the formula of how to determine the allele
frequencies.
 Write a hypothesis for expected allele outcome for
Case 2.
 Write a hypothesis for expected allele outcome for
Case 3.