PASS SESSION 6 - DETERMINING ALLELE FREQUENCIES USING HARDY-WEINBURG
EQUATION
BACKGROUND
HARDY-WEINBERG EQUATION: p + q = 1 (allele frequencies)
To conceptualize The Hardy-Weinberg Equilibrium Model, it is easiest to think of it as the Mendelian
genetics of breeding populations. Using a Punnett Square, it is possible to calculate genotype frequencies
or the percent of a certain genotype (dominant homozygotes, heterozygote, recessive homozygotes) in a
breeding population if the frequencies are known in the same way that we calculate Mendelian ratios if the
genotypes of both parents are known. The study of genetics of breeding populations is called population
genetics and the Hardy-Weinberg Equilibrium Model is central to understanding and doing research in
population genetics.
Table 1: Punnett square for Hardy-Weinburg equilibrium
Males
A (p)
a (q)
Females
A (p)
a (q)
2
AA (p ) Aa (pq)
Aa (pq) aa (q2)
The final three possible genotypic frequencies in the offspring become:
Determining allele frequencies is not very difficult. One method simply involves counting identifiable
genotypes. For example, in the human MN blood-group system (this is similar to the ABO system) there are
three identifiable genotypes and three identifiable phenotypes because the MN system is a codominant
system. If two alleles are codominant, then both will be expressed when both are present. One allele is not
dominant and the other is not recessive. In a codominant system, the heterozygote is distinguishable from
the dominant homozygote. The three identifiable phenotypes are M, MN, and N. Because these are
identifiable phenotypes, it is not difficult to determine genotypes. Phenotype M has the genotype MM or two
copies of the M allele; Phenotype MN has the genotype MN because both alleles are expressed; and
Phenotype N has the genotype NN, or two copies of the N allele.
Example 1:
PHENOTYPE
M
MN
N
NUMBER OF
INDIVIDUALS
GENOTYPE
600
MM
300
MN
100
NN
1000 individuals
NUMBER OF ALLELES
M
N
______
______
______
______
______
______
______ alleles ______ alleles
By doing a census in this hypothetical population of 1000, it was determined that 600 individuals were
blood type M, 300 were blood type MN, and 100 were blood type N. Using this information, allele
frequencies, or the percent of alleles at this locus that are M and the percent that are N, can be determined.
Remember that each individual has two alleles at each locus, so for every 1000 individuals there will be
2000 alleles (1000 X 2 = 2000).
Example 1 cont.: Using this information, it is then a simple matter to determine the frequency of the M
allele and the frequency of the N allele in this population.
The frequency of the M allele in this population equals _______(total number) which equals ___ (proportion
i.e. value between 0.0 and 1.0) or_____(percent). The total number of M alleles is _______(total number)
and the total number of M and N alleles is ______(total number). The frequency of the N allele in this
population equals _______(total number) which equals ____ (proportion )____(percent). Notice that the
frequency of M plus the frequency of N equals 1.0 or 100 percent. It is impossible to have more than 100
percent of the alleles at a given gene locus.
Example 2:
NUMBER OF
PHENOTYPE
M
MN
N
INDIVIDUALS
75
75
50
200 individuals
NUMBER OF ALLELES
GENOTYPE
M
N
MM
______
______
MN
______
______
NN
______
______
______alleles ______ alleles
The frequency of M in this population equals ________(total number) which equals ______(proportion )
____(percent). The frequency of the N allele equals __________(total number) which equals ____(
proportion)____(percent).
Example 3:
In a hypothetical population, a genetic anthropologist has gone in and determined the following genotypes
and genotype frequencies for earlobe attachment. From this data, this scientist wants to determine allele
frequencies at this locus for the dominant and recessive alleles. Unattached earlobes (E) are the dominant
form of the trait while attached earlobes (e) are the recessive form of the trait. At this particular gene locus,
the following genotype distribution was determined:
(1) dominant homozygote (EE) = 0. 16 (or 16 percent of the individuals are dominant homozygotes for this
trait and have unattached earlobes) (2) heterozygote (Ee = 0.48 (or 48 percent of the individuals in this
population is heterozygous at this locus and have unattached earlobes) (3) recessive homozygote (ee) =
0.36 (or 36 percent of the individuals in this population are recessive homozygotes and have attached
earlobes)
The Hardy-Weinberg Equilibrium Formula: p + q = 1 (allele frequencies)
If we square both sides: (p + q) 2 =12 therefore = (p + q) 2 =1
If we expand the formula: p 2+ 2pq + q2= 1 (genotype frequencies)
p always equals the frequency of the dominant allele.
q always equals the frequency of the recessive allele.
To determine allele frequencies:
p = p 2 + 1/2(2pq)
q = q2 + 1/2(2pq)
The first step in determining the frequency of the E and e alleles in this population is to place the genotype
frequencies in the expanded Hardy-Weinberg Formula:
p 2 + 2pq + q2 = 1.0
___________ = 1.0
Example 3 cont. : The second step is to use the formulas above for determining p and q. Remember,
because E is the dominant allele it equals p while e equals q because it is the recessive allele.
p = p 2+ 1/2(2pq)
= ____________
= ____________
= ____________
q = q2 + 1/2(2pq)
= ____________
= ____________
= ____________
The frequency of the dominant allele equals _____(proportion)_____percent of the alleles at this locus are
dominant. The frequency of the recessive allele equals _____(proportion)_____percent of the alleles at this
locus are recessive. Together, they add up to 100 percent of the alleles at this locus: ____+____
(proportions) = 1.0 or ___+___= 100 percent. It is important to remember that these are alleles, not
individuals.
Example 4:
In another population, another genetic anthropologist examines the same trait as in Example 3. This
scientist is also interested in determining the allele frequencies for both alleles at this locus. Initial research
has determined the following genotype distribution:
EE
p2 +
0.25 +
Ee
ee
2pq + q 2
= 1.0
0.50 + 0.25 = 1.0
p= p2 + 1/2(2pq)
= __________
= __________
= __________
q = q2 + 1/2(2pq)
= __________
= __________
= __________
In this population, _____percent of the alleles are dominant (E) and _____percent of the alleles are
recessive (e).
It is also possible to determine both genotype and phenotype frequencies from allele frequencies.
Example 5:
In another population, other researchers interested in the attachment of earlobes has determined that the
frequency of E equals 0.80 (or that 80 percent of the alleles are dominant) and that the frequency of e
equals 0.20 (or 20 percent of the alleles are recessive). Their goal is to determine genotype and phenotype
distributions in this population.
The first step here is to substitute the above data into the expanded Hardy-Weinberg formula. Remember,
p = E = 0.80 and q = e = 0.20.
p2 + 2pq + q2 = 1.0
(_____)2 + 2(______)+ (___)2 = 1.0
_____+ ____+ _____= 1.0
Example 5 cont. : In this population, ____ percent of the individuals are dominant homozygotes,
____percent are heterozygotes, and ______percent are recessive homozygotes. In order to determine
phenotype frequencies, since this system is a case of simple dominance and recessiveness, the dominant
phenotype equals ____+ ____= ____(proportion) or ____percent of this population will exhibit the dominant
phenotype or unattached earlobes. The recessive phenotype equals the frequency of the recessive
homozygote. In this case ____, or ____percent of this population will exhibit the recessive trait, or attached
earlobes.
Example 6:
In yet another population, it was determined that the frequency of the E allele is 90 percent and the
frequency of the e allele equals 10 percent. The researcher here was also interested in determining
genotype and phenotype frequencies with respect to this locus in this population.
p2 + 2pq + q2
(_____) 2+2(____x ____) +(____)2 = ______
_____+ _____+ ____= 1.0
EE =_____
Ee = _____
ee = _____
Under the Hardy-Weinberg Equilibrium Model these genotype frequencies are what we would expect in the
next generation. If the allele and genotype frequencies change from one generation to the next then
evolution has taken place!
The dominant phenotype = ____+ ____= ____(proportion) or ____percent of this population has
unattached earlobes. The recessive phenotype = ______(proportion) or ______percent of this population
has attached earlobes.
In order to determine if evolution with respect to this trait has occurred in this population, the scientists
calculated allele frequencies from the genotype frequencies of the next generation :
EE
Ee
ee
p 2+ 2pq + q2 = 1.0
______ +
______ +
______
= 1.0
p = _____+ 1/2(_____)
= _____+ _____
= _____
q = ______+ 1/2(_____)
= ______+ ______
= ______
The allele frequencies for these alleles did not change so evolution has not taken place with respect to this
gene locus.
This activity adapted from:
Mathematics across the curriculum (n.d.) Population Genetics: The theory and practice of the HardyWeinburg equation [online]. University of Nevada. [Course notes in the topic ANTH102]. Available:
http://www.unr.edu/mathcenter/mac/disciplines/anth/102/102.genetics.exercise.html. Accessed 30/8/07.