1. 2 hardy-weinberg equilibrium lecture 5 3 the hardy-weinberg equilibrium
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Hardy-Weinberg Equilibrium
Lecture 5
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The Hardy-Weinberg EquilibriumThe 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 TheoremAssumptions 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 TheoremAssumptions 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 TheoremAssumptions 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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Allele freq.f(A) = pf(a) = qp + q = 1Sum of all alleles = 100%
Genotypic freq.f(AA) = p2 Dominant homozygousf(Aa) = 2 pq Heterozgousf(aa) = q2 Recessive homozygousp2 + 2 pq + q2 = (p+q)2 = 1Sum of all genotypes = 100%
Gametes A (p) a(q)
A(p) AA (pp)
Aa(pq)
a(q) aA (qp)
aa (qq)
AA = p*p = p2
Aa = pq + qp = 2pqaa = q*q = q2
Hardy-Weinberg Theorem (2 alleles at 1 locus)
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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
Hardy-Weinberg Principle
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Assumptions of the H-W Assumptions of the H-W EquilibriumEquilibrium
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Assumptions of the H-W Assumptions of the H-W EquilibriumEquilibrium
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for a populationwith genotypes:
100 GG
160 Gg
140 gg
Genotype frequencies
Phenotype frequencies
Allele frequencies
calculate:
Assumptions of the H-W Assumptions of the H-W EquilibriumEquilibrium
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for a populationwith genotypes:
100 GG
160 Gg
140 gg
Genotype frequencies
Phenotype frequencies
Allele frequencies
100/400 = 0.25 GG160/400 = 0.40 Gg140/400 = 0.35 gg
260/400 = 0.65 green140/400 = 0.35 brown
2*100 + 160/800 = 0.45 G2*140 + 160/800 = 0.55 g
0.65260
calculate:
Assumptions of the H-W TheoremAssumptions of the H-W Theorem
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another way to calculateallele frequencies:
100 GG
160 Gg
140 gg
Genotype frequencies
Allele frequencies
0.25 GG
0.40 Gg
0.35 gg
360/800 = 0.45 G440/800 = 0.55 g
OR [0.25 + (0.40)/2] = 0.45 [0.35 + (0.40)/2] = 0.65
G
g
Gg
0.250.40/2 = 0.200.40/2 = 0.200.35
p (G) = D + H/2q (g) = R + H/2
Assumptions of the H-W TheoremAssumptions of the H-W Theorem
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Hardy-Weinberg Equilibrium
Population of cats n=10016 white and 84 blackaa = whiteA_ = black
Can we figure out the allelic frequencies of individuals AA and Aa?
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p2 + 2pq + q2
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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 = 0.4p + q = 1 then p = 0.6 (0.6 +0.4 = 1)P2 = 0.36All 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
Hardy-Weinberg Equilibrium
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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
Then, p = 1 – q = 0.7 = f(B)
f(BB) = D = p2 = (0.7)2 = 0.49f(Bb) = H = 2pq = 2 * 0.7 * 0.3 = 0.42f(bb) = R= q2 = (0.3)2 = 0.09
)(3.009.2 bfqq
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In summary:
Allele freq.f(B) = p = 0.7 (est.)f(b) = q = 0.3 (est.)
Phenotypic freq.f(white) = 0.91 (actual)f(black) = 0.09 (actual)
Genotypic freq.f(BB) = p2 = 0.49 (est.)f(Bb) = 2pq = 0.42 (est.)f(bb) = q2 = 0.09 (actual)