THE HARDY-WEINBERG EQUILIBRIUM
MODEL: TWO EXAMPLES
EXAMPLE #1 TAY-SACHS DISEASE
Place: New York State
 Population: 10,000,000
 Character of the disease: caused by the
recessive allele and the recessive homozygous
genotype is 100% fatal.
 Alleles: T, t

TT
9,697,700
Tt
tt
300,000
2,300
Step #1 Divide the number of people in each genotype by the total
population in order to determine the frequency of each
genotype in the population.
TT = 9,697,700/10,000,000 = .96977
Tt = 300,000/10,000,000 = .03
tt = 2,300/10,000,000 = .00023
These are your observed genotype frequencies.
Step #2 Calculate the allele frequencies.
To calculate the frequency of the T allele, you
have to count the number of T alleles in each
genotype. Remember, everybody has two
alleles.
T = 2 x 9, 697,700 = 19,395,400 + 300,000
= 19,695,400
This has to be divided by the total number of
alleles in the population = 20,000,000.
19,695,400/20,000,000 = .98477 = p


The same for the t allele:
t = 2 x 2300 = 4,600 + 300,000 = 304,600
304,600 / 20,000,000 = .01523 = q
Both allele frequencies should add up to 1. If not,
then there has been a mistake in the math
somewhere.
Step number 3: use the allele frequencies in
the Hardy-Weinberg Equilibrium model. This will
give you genotype frequencies that are in
equilibrium.
p2 + 2(pq) + q2=1
TT
Tt
tt
(.98477)2 + 2(.98477 x .01523) + (.01523)2= 1
.96977
.029996
.00023
These are the expected genotype frequencies.
They do not represent the genotype frequencies
of a real population.

Step #4: Compare the observed (real) genotype
frequencies with the expected genotype
frequencies (projection) to see if the observed
genotype frequencies are in equilibrium.
TT
Tt
tt
observed: _ .96977 _.03
_ .00023
expected:
.96977 .029996 .00023
.0
.000004 0
Result: The original population is in equilibrium.
There are no forces affecting this population.

EXAMPLE #2: SICKLE CELL ANEMIA
Place: Nigeria.
 Population: 12,387.
 Character of the disease: Caused by the
deleterious HbS allele. This allele is codominant
with the HbA allele.

HbA HbA
HbA HbS
HbS HbS
9365
2993
29
Step #1 Calculate the observed genotype frequencies.
9365/12,387
=.7560
2993/12,387
29/12,387
=.2416
=.0023
Step #2: Calculate the allele frequencies.
HbA = (2 x 9,365) + 2,993/(12,387x2)
21,723/24,774 = .8768 = p
HbS = (2 x 29) + 2,993/(12,387 x 2)
3051/24,774 = .1232 = q

Step #3: Place the values for p and q into the
Hardy-Weinberg Equilibrium model to calculate
the expected genotype frequencies.
(.8768)2 + 2(.8768 x .1232) + (.1232)2 = 1
.7688
+
.2160
+ .0152 = 1
Step #4: Compare the observed genotype
frequencies with the expected genotype
frequencies to determine whether or not the
observed population is in equilibrium.
Observed: _ .7560
_.2416
_.0023
Expected: .7688
.2160
.0152
.0128* .0256
.0129*
Conclusion: The genotypes of the observed
population are in disequilibrium as there is a
significant difference between the observed and
expected values.
* Only absolute differences are relevant.
DISTINGUISHING THE IMPACT OF SELECTION
Since the population is in disequilibrium, looking
at how the genotypes of the population are
affected is warranted. First we want to measure
the survival efficiency of the three genotypes.
To do this, we divide the observed genotype
frequencies by the expected genotype
frequencies.
Observed:
Expected:
Survival efficiency:
HbAHbA
HbAHbS
HbSHbS
.7560
.7688
.2416
.2160
.0023
.0152
9833
1.185
.1513
.
If a survival efficiency statistic is higher than
1, then the people who have this genotype
are doing rather well in the given
environment. The reverse is true for values
lower than 1. The heterozygotes are doing the
best.
RELATIVE FITNESS
We can gain some additional insights into
selection by adjudging the fitness of the
genotypes relative to each other. We
accomplish this by making the most successful
genotype the standard of fitness, and then by
dividing all three genotypes by this value.
HbA HbA
HbAHbS
HbSHbS
.9833
1.185
= .83
1.185
1.185
=1
.1513
1.185
=.13
The homozygote genotype HbA HbA has 83% of
the fitness of the heterozygote genotype, while
the homozygote HbSHbS genotype has 13% of
the fitness of the heterozygote. Selection is
acting against both homozygotes.