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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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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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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for a populationwith genotypes:

100 GG

160 Gg

140 gg

Genotype frequencies

Phenotype frequencies

Allele frequencies

calculate:


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for a populationwith genotypes:

100 GG

160 Gg

140 gg


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:


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another way to calculateallele frequencies:

100 GG

160 Gg

140 gg


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


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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


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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)

1. 2 hardy-weinberg equilibrium lecture 5 3 the hardy-weinberg equilibrium

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