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Hardy-Weinberg Equilibrium
Lecture 5
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The Hardy-Weinberg Equilibrium
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Hardy-Weinberg Theorem
Hardy-Weinberg
original proportions of
genotypes in a population
will remain constant from
generation to generation
-
- Sexual reproduction
(meiosis and fertilization)
alone will not change
allelic (genotypic)
proportions.
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Assumptions of the H-W Theorem
1.Large population size
-small populations can have chance fluctuations
in allele frequencies (e.g., fire, storm, floods).
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Assumptions of the H-W Theorem
2. No migration
- immigrants can change the frequency of an
allele by bringing in new alleles to a population.
3. No net mutations
-if alleles change from one to another, this will
change the frequency of those alleles
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Assumptions of the H-W Theorem
4. Random mating
- if certain traits are more desirable, then
individuals with those traits will be selected and
this will not allow for random mixing of alleles.
5. No natural selection
- if some individuals survive and reproduce at a
higher rate than others, then their offspring will
carry those genes and the frequency will change
for the next generation.
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Hardy-Weinberg Theorem (2 alleles at 1 locus)
Allele freq.
f(A) = p
f(a) = q
p + q = 1
Sum of all alleles = 100%
Gametes
A (p)
a(q)
A(p)
AA
(pp)
Aa
(pq)
a(q)
aA
(qp)
aa
(qq)
Genotypic freq.
f(AA) = p2 Dominant homozygous
f(Aa) = 2 pq Heterozgous
f(aa) = q2 Recessive homozygous
p2 + 2 pq + q2 = (p+q)2 = 1
Sum of all genotypes = 100%
AA = p*p
= p2
Aa = pq + qp = 2pq
aa = q*q
= q2
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Hardy-Weinberg Principle
The gene pool of a non-evolving population remains constant
over multiple generations; i.e., the allele frequency does not
change over generations of time.
The Hardy-Weinberg Equation: binomial expansion
(p+q)2 = p2 + 2pq + q2 = 1.0
p2 = frequency of AA homozygous genotype;
2pq = frequency of Aa plus aA heterozygous genotypes;
q2 = frequency of aa homozygous genotype
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Assumptions of the H-W Equilibrium
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Assumptions of the H-W Equilibrium
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Assumptions of the H-W Equilibrium
for a population
with genotypes:
calculate:
Genotype frequencies
100 GG
160 Gg
140 gg
Phenotype frequencies
Allele frequencies
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Assumptions of the H-W Theorem
for a population
with genotypes:
100 GG
160 Gg
calculate:
Genotype frequencies
260
100/400 = 0.25 GG
0.65
160/400 = 0.40 Gg
140/400 = 0.35 gg
Phenotype frequencies
260/400 = 0.65 green
140/400 = 0.35 brown
140 gg
Allele frequencies
2*100 + 160/800 = 0.45 G
2*140 + 160/800 = 0.55 g
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Assumptions of the H-W Theorem
another way to calculate
allele frequencies:
Genotype frequencies
100 GG
160 Gg
0.25 GG
0.40 Gg
0.35 gg
G 0.25
G 0.40/2 = 0.20
g 0.40/2 = 0.20
g 0.35
Allele frequencies
140 gg
360/800 = 0.45 G
440/800 = 0.55 g
OR [0.25 + (0.40)/2] = 0.45
[0.35 + (0.40)/2] = 0.65
p (G) = D + H/2
q (g) = R + H/2
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Hardy-Weinberg Equilibrium
Population of cats
n=100
16 white and 84 black
aa = white
A_ = black
Can we figure out the allelic frequencies of individuals AA and Aa?
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Hardy-Weinberg Equilibrium
p2 + 2pq + q2
and
p+q = 1 (always two alleles)
16 cats white = 16 (aa) then (q2 = 16/100 = 0.16)
This we know we can see and count!!!!!
If p + q = 1 then we can calculate p from q2
q = square root of q2 = q
√0.16
q = 0.4
p + q = 1 then p = 0.6 (0.6 +0.4 = 1)
P2 = 0.36
All we need now are those that are heterozygous (2pq)
(2 x 0.6 x 0.4)=0.48
0.36 + 0.48 + 0.16 = 1
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Hardy-Weinberg Equilibrium
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Example use of H-W theorem
1000-head sheep flock. No selection for color. Closed
to outside breeding.
910 white (B_); 90 black (bb)
Start with known: f(black) = f(bb) = 0.09 =q2
q  q  .09  0.3  f (b)
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Then, p = 1 – q = 0.7 = f(B)
f(BB) = D = p2 = (0.7)2 = 0.49
f(Bb) = H = 2pq = 2 * 0.7 * 0.3 = 0.42
f(bb) = R= q2 = (0.3)2 = 0.09
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In summary:
Allele freq.
f(B) = p = 0.7 (est.)
f(b) = q = 0.3 (est.)
Genotypic freq.
f(BB) = p2 = 0.49 (est.)
f(Bb) = 2pq = 0.42 (est.)
f(bb) = q2 = 0.09 (actual)
Phenotypic freq.
f(white) = 0.91 (actual)
f(black) = 0.09 (actual)
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