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Chapter 23 – Part 1
Hardy Weinberg Equilibrium
Context
 Evolution appears to be the typical situation
 Sometimes evolution is not occurring
 Hardy-Weinberg Equilibrium
 Conditions that specify if evolution is NOT happening
 Equilibrium = no evolution
 More of a theoretical case than a real test
 But affords comparison to evolution
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Definitions
 Population Genetics – study of how populations
change genetically over time
 Population – Group of same species individuals that
can and do interbreed successfully
 Gene pool – all of the alleles at all loci in all the
members of a population
 Fixed – only one allele exists for a locus in the
population
 More fixed alleles = lower species diversity
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Hardy-Weinberg
 If gene pools are NOT Evolving, can use Hardy-
Weinberg
 Describes a population that is NOT evolving
 Allelic frequencies will remain constant
 Gene frequencies remain constant throughout the
generations
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Conditions for Hardy-Weinberg
1. No mutations
2. Random mating
3. No natural selections
4. Large population size
5. No gene flow
 Immigration or emigration
 Genes coming into or leaving the population
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Math in Biology?
p = Frequency of the dominant allele = f(A)
q = Frequency of the recessive allele = f(a)
There are only 2 alleles (dominant + recessive), so
p+q=1
square both sides:
(p + q)2 = 12
expand the binomial: p2 + 2pq + q2 = 1
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Hardy-Weinberg Math (Page 2)
p2 = pp = f(AA) = Frequency of homozygous dominant
2pq = f(Aa) = frequency of heterozygote
 However, remember from Punnett Square, there were 2
heterozygotes for the F1 generation; So
 2pq = f(Aa)
q2 = qq = f(aa) = frequency of homozygous recessive
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In Summary
 Summary:
 p2 + 2pq + q2 = 1
 f(AA) + f(Aa) + f(aa) = 1
 f(Homo. Dom.) + f(Hetero.) + f(Homo. Recess.) = 1
 Are there any other genotypes?
 Makes sense?
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Problem (Part 1)
1. A trait has two characters, dominant (A) and recessive
(a). In a population of 500 individuals in HardyWeinberg equilibrium, 25% display the recessive
phenotype (aa).
a) What is the frequency of the dominant allele in
the population?
b) What is the frequency of the recessive allele in
this population?
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Answer
 f(aa) = 0.25 OR q2 = 0.25
 q = √(0.25) = 0.5
OR f(a) = 0.5
 So the frequency of the recessive allele = 0.5
From before: p + q = 1 AND q = 0.5, so
p = 0.5 = f(A)
 So the frequency of the dominant allele = 0.5
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Part 2 of Problem
 What are the frequencies of
a) homozygous dominant genotype?
b) Heterozygous genotype?
c) homozygous recessive genotype?
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Part 2 of Problem (Answer)
 What are the frequencies of
a) homozygous dominant genotype?
Since p = 0.5, p2 = (0.5)2 = 0.25
b) Heterozygous genotype?
Since p = 0.5 & q = 0.5,
2pq = 2(0.5)(0.5) = 0.5
c) homozygous recessive genotype?
Yep, it was given, but if we must: q = 0.5, q2 = (0.5)2 = 0.25
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Part 3 of Problem
 How many individuals have the
a) homozygous dominant genotype?
b) Heterozygous genotype?
c) homozygous recessive genotype?
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Part 3 of Problem (Answer)
 How many individuals have the
a) homozygous dominant genotype?
500 members & p2 = 0.25,
(0.25)(500) = 125
b) Heterozygous genotype?
500 members & p = 0.5, q = 0.5 2pq = 0.5
So 500 (0.5) = 250 heterozygotes
c) homozygous recessive genotype?
500 members & q2 = 0.25,
(0.25)(500) = 125
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Part 4 of Problem
 How many individuals have the
a) Dominant phenotype?
b) Recessive phenotype?
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Part 4 of Problem (Answer)
 How many individuals have the
a) Dominant phenotype?
= (Homo. Dom.) + (Hetero.) = 125 + 250 = 375
b) Recessive phenotype?
= Homo. Recess. = 125
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