Introduction to Hardy Weinberg equation and examples of its usage in biology

1. Hardy Weinberg
Equilibrium & Equation

2. Hardy-Weinberg Equilibrium
• A model to explain a stable allele and
genotype frequency for a population
– A population in Hardy-Weinberg equilibrium
would indicate a population which is NOT evolving
or being acted upon by natural selection
– only applies to sexually reproducing diploid
organisms
– 5 conditions must be met for a population to
remain in Hardy-Weinberg equilibrium
2

3. Hardy-Weinberg Equilibrium
• The 5 conditions
– There is no mutation, i.e. no new alleles
– There is no selection among genotypes, i.e.
individuals with different genotypes have equal
probabilities of survival and rates of reproduction
– There is no gene flow, i.e. no migration into or out
of a population
– The population size is infinite
– Mating is random, i.e. individuals do not choose
mates with particular genotypes preferentially
3

4. Using the Hardy-Weinberg Equation
• Used to determine the “expected” allele
frequency in an evolving population
• Deviating from Hardy-Weinberg
equilibrium indicates the population is
evolving
• for a gene with 2 alleles in equilibrium
p + q = 1
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5. Hardy-Weinberg Equilibrium
• Allele frequency in a population for gene with
2 alleles
• 1 = p + q
• Calculate frequency of genotypes
– (p + q)2 = 1
– p2 + 2pq + q2 = 1
• 3 alleles (p + q + r)2 = 1
5

6. Hardy-Weinberg Equilibrium
• The average human frequency of albinism (a
recessive trait) in North America is only about
1 in 20,000
• Therefore, q² = 1/20,000 = .00005 and q ≈ .007
• So dominant wild-type allele frequency
• p = 1 - q
• p = 1 - .007
• p = .993
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7. Hardy-Weinberg Equilibrium
• Plugging in allele frequencies
p² + 2pq + q² = 1
(.993)² + 2 (.993)(.007) + (.007)² = 1
.986 + .014 + .00005 = 1
• p² = predicted frequency of homozygous dominant
individuals = .986 = 98.6%
• 2pq = predicted frequency of heterozygous individuals
(carriers) = .014 = 1.4%
• q² = predicted frequency of homozygous recessive
individuals (the albinos) = .00005 = .005% 7

8. Hardy-Weinberg Equilibrium
• Useful for calculating “expected” gene frequencies
because not meeting even one of the conditions for
many genes the allele frequencies remaining stable is
unlikely
– Therefore one can make a null hypothesis that the allele
frequency is stable, calculate the expected using Hardy-
Weinberg and then find that the null hypothesis is rejected
– Then the fun begins…which of the conditions is
responsible for NOT maintaining allele frequencies?
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9. In a population of humans in a village in central Africa, doctors took
blood samples from 200 adolescent boys to study sickle-cell anemia,
which is a recessively inherited disease caused by a mutation in a
single gene coding for hemoglobin production. In the sample, 116 boys
were found to be carriers (heterozygote) for sickle cell anemia, and 75
were homozygous for the normal-hemoglobin allele. Is there evidence
for evolution in the sickle-cell anemia gene in this population?
Using H-W Equation to Detect
Evolution Forces Acting On a
Population

10. General Rule for Estimating Allele Frequencies
from Genotype Frequencies:
Genotypes: AA Aa aa
Frequency: p2 2pq q2
 Frequency of the A allele:
p = (p2 + ½ (2pq))/n
 Frequency of the a allele:
q = (q2 + ½ (2pq))/n

11. In a population of humans in a village in central Africa, doctors took
blood samples from 200 adolescent boys to study sickle-cell anemia,
which is a recessively inherited disease caused by a mutation in a
single gene coding for hemoglobin production. In the sample, 116 boys
were found to be carriers (heterozygote) for sickle cell anemia, and 75
were homozygous for the normal-hemoglobin allele. Is there evidence
for evolution in the sickle-cell anemia gene in this population?
Using “S” as the wild-type allele and “s” as the sickle cell allele
75 are SS
116 are Ss
116 + 75 = 191 therefore 200-190 = 9 are ss
[the only phenotype not mentioned in the paragraph above]
allele frequency of s is [9 + (116/2)]/200 = 0.33
allele frequency of S is [75 + (116/2)]/200 = 0.67

12. In a population of humans in a village in central Africa, doctors took
blood samples from 200 adolescent boys to study sickle-cell anemia,
which is a recessively inherited disease caused by a mutation in a
single gene coding for hemoglobin production. In the sample, 116 boys
were found to be carriers (heterozygote) for sickle cell anemia, and 75
were homozygous for the normal-hemoglobin allele. Is there evidence
for evolution in the sickle-cell anemia gene in this population?
So, plugging in the H-W equation to get the expected
Homozygous sickle cell p2 = (0.33)2 = 0.109 of population so
0.109 * 200 = 21.8
Heterozygotes 2pq = 2 (0.33)*(0.67) = 0.442 of population so
0.442 * 200 = 88.4
Homozygous wild-type q2 = (0.67)2 = 0.449 of population so
0.449 * 200 = 89.8

13. χ2 analysis
Observed Expected (O-E)2/E
homozygous sickle cell 9 21.8 7.5
heterozygotes 116 88.4 8.6
homozygous wild-type 75 89.8 2.5
χ2 18.6
[3 genotypes -1] and an additional -1 because you had to calculate allele
frequencies based on genotype data so d.f. =1
Probability that the values would be observed by chance alone if the
alleles were in Hardy Weinberg equilibrium is less than 0.001* so you
would reject the hypothesis and therefore state there is some
evolutionary force at work here [likely natural selection for
heterozygotes]
* χ2 = 18.6 therefore p << 0.01
*