HOW TO:
HARDY - WEINBERG
WHAT IS HARDY-WEINBERG?
Hardy-Weinberg says
that the frequency of
alleles and genotypes
remain constant in a
population generation
after generation when
certain conditions are
met.
WHAT ARE THE CONDITIONS?
No Mutations
No Gene Flow
Random Mating
No Genetic Drift
No Selection
*note that these conditions are rarely met*
HARDY-WEINBERG EQUATION
p+q=1
p= frequency of dominant allele
q= frequency of recessive allele
EXAMPLE 1
If 60 people out of 100 can roll their tongues, and tongue
rolling is dominant, then what is the frequency of the
recessive alleles?
60/100 = .60 or 60% can roll tongues
p+q=1
.60 + q = 1
-.60
-.60
q= .40 or 40% cannot roll their tongues
YOUR TURN
If 23 out of 100 people do not have a hitchhikers
thumb, and possessing a hitchhikers thumb is
recessive, then what is the frequency of the
dominant allele?
Click here for the answer.
But…
we know that we can have a
mixture of dominant and
recessive alleles (heterozygous)
HARDY-WEINBERG EQUATION
2
p
+ 2pq +
2
=
q 1
or
(p x p) + (2 x p x q) + (q x q) = 1
2
p
2
+ 2pq +
2
=
q 1
𝑝 =frequency of homozygous dominant genotypes
2
𝑞 =frequency of homozygous recessive genotypes
2pq = frequency of heterozygous genotypes
EXAMPLE 2
 You have sampled a population in which you know
that the percentage of the homozygous recessive
genotype (aa) is 36%. Using that 36%, calculate the
following:
a) The frequency of the "aa" genotype.
b) The frequency of the "a" allele.
c) The frequency of the "A" allele.
d) The frequencies of the genotypes "AA" and "Aa."
a)
The frequency of the "aa" genotype. Answer: 36%, as given in the
problem itself.
b)
The frequency of the "a" allele. Answer:The frequency of aa is 36%,
which means that q2 = 0.36, by definition. If q2 = 0.36, then q = 0.6,
again by definition. Since q equals the frequency of the a allele, then
the frequency is 60%.
c)
The frequency of the "A" allele. Answer: Since q = 0.6, and p + q = 1,
then p = 0.4; the frequency of A is by definition equal to p, so the
answer is 40%.
d)
The frequencies of the genotypes "AA" and "Aa." Answer:The
frequency of AA is equal to p2, and the frequency of Aa is equal to
2pq. So, using the information above, the frequency of AA is 16% (i.e.
p2 is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).
YOUR TURN
 You have sampled a population in which you know
that the percentage of the homozygous dominant
genotype (AA) is 25%. Using that 25%, calculate the
following:
a) The frequency of the “AA" genotype.
b) The frequency of the "a" allele.
c) The frequency of the "A" allele.
d) The frequencies of the genotypes “aa" and "Aa."
Click here for the answer.
.77 or 77%
.25 (given in the problem)
b) .5 (take the square root of .25 and subtract from 1)
c) .5 (take the square root of .25)
d) .25, .50 (your number from C and D into the HardyWeinberg equation)
a)