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BIOE 109 Summer 2009 Lecture 5- Part I Hardy- Weinberg Equilibrium.

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Presentation on theme: "BIOE 109 Summer 2009 Lecture 5- Part I Hardy- Weinberg Equilibrium."— Presentation transcript:

1 BIOE 109 Summer 2009 Lecture 5- Part I Hardy- Weinberg Equilibrium

2 The Hardy-Weinberg-Castle Equilibrium

3 Godfrey Hardy Wilhelm Weinberg William Castle

4 Conclusions of the Hardy-Weinberg principle

5 Conclusions of the Hardy-Weinberg principle 1. Allele frequencies will not change from generation to generation.

6 Conclusions of the Hardy-Weinberg principle 1. Allele frequencies will not change from generation to generation. 2. Genotype proportions determined by the “square law”.

7 Conclusions of the Hardy-Weinberg principle 1. Allele frequencies will not change from generation to generation. 2. Genotype proportions determined by the “square law”. for two alleles = (p + q) 2 = p 2 + 2pq + q 2

8 Conclusions of the Hardy-Weinberg principle 1. Allele frequencies will not change from generation to generation. 2. Genotype proportions determined by the “square law”. for two alleles = (p + q) 2 = p 2 + 2pq + q 2 for three alleles (p + q + r) 2 = p 2 + q 2 + r 2 + 2pq + 2pr +2qr

9 Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies

10 Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A 1 = 0.80, A 2 = 0.20 A 1 A 1 = 0.64, A 1 A 2 = 0.32, A 2 A 2 = 0.04

11 Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A 1 = 0.80, A 2 = 0.20 A 1 A 1 = 0.64, A 1 A 2 = 0.32, A 2 A 2 = 0.04 A 1 = 0.50, A 2 = 0.50 A 1 A 1 = 0.25, A 1 A 2 = 0.50, A 2 A 2 = 0.25

12 Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A 1 = 0.80, A 2 = 0.20 A 1 A 1 = 0.64, A 1 A 2 = 0.32, A 2 A 2 = 0.04 A 1 = 0.50, A 2 = 0.50 A 1 A 1 = 0.25, A 1 A 2 = 0.50, A 2 A 2 = 0.25 A 1 = 0.10, A 2 = 0.90 A 1 A 1 = 0.01, A 1 A 2 = 0.18, A 2 A 2 = 0.81

13 Assumptions of Hardy-Weinberg equilibrium

14 1. Mating is random

15 Assumptions of Hardy-Weinberg equilibrium 1. Mating is random… but some traits experience positive assortative mating

16 Assumptions of Hardy-Weinberg equilibrium 1. Mating is random 2. Population size is infinite (i.e., no genetic drift)

17 Assumptions of Hardy-Weinberg equilibrium 1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration

18 Assumptions of Hardy-Weinberg equilibrium 1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration 4. No mutation

19 Assumptions of Hardy-Weinberg equilibrium 1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration 4. No mutation 5. No selection

20 Hardy-Weinberg principle: A null model 1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration 4. No mutation 5. No selection The Hardy-Weinberg equilibrium principle thus specifies conditions under which the population will NOT evolve. In other words, H-W principle identifies the set of events that can cause evolution in real world.

21 Does Hardy-Weinberg equilibrium ever exist in nature?

22 Example: Atlantic cod (Gadus morhua) in Nova Scotia

23 Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia as a juvenile…

24 Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia … and as an adult

25 Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia a sample of 364 fish were scored for a single nucleotide polymorphism (SNP)

26 Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia a sample of 364 fish were scored for a single nucleotide polymorphism (SNP) A 1 A 1 = 109 A 1 A 2 = 182 A 2 A 2 = 73

27 Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia a sample of 364 fish were scored for a single nucleotide polymorphism (SNP) A 1 A 1 = 109 A 1 A 2 = 182 A 2 A 2 = 73 Question: Is this population in Hardy-Weinberg equilibrium?

28 Testing for Hardy-Weinberg equilibrium

29 Step 1: Estimate genotype frequencies

30 Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Step 2: Estimate allele frequencies

31 Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Step 2: Estimate allele frequencies Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium

32 Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Step 2: Estimate allele frequencies Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium Step 4: Compare observed and expected numbers of genotypes  2 =  (Obs. – Exp.)2 Exp.

33 A simple model of directional selection

34 Persistent selection changes allele frequencies over generations (Obvious) Conclusion: Natural selection can cause rapid evolutionary change!

35 A simple model of directional selection consider a single locus with two alleles A and a

36 A simple model of directional selection consider a single locus with two alleles A and a let p = frequency of A allele

37 A simple model of directional selection consider a single locus with two alleles A and a let p = frequency of A allele let q = frequency of a allele

38 A simple model of directional selection consider a single locus with two alleles A and a let p = frequency of A allele let q = frequency of a allele relative fitnesses are: AAAaaa w11 w12 w22

39 A simple model of directional selection consider a single locus with two alleles A and a let p = frequency of A allele let q = frequency of a allele relative fitnesses are: AAAaaa w 11 w 12 w 22 it is also possible to determine relative fitness of the A and a alleles:

40 A simple model of directional selection consider a single locus with two alleles A and a let p = frequency of A allele let q = frequency of a allele relative fitnesses are: AAAaaa w11 w12 w22 it is also possible to determine relative fitness of the A and a alleles: let w 1 = fitness of the A allele

41 A simple model of directional selection consider a single locus with two alleles A and a let p = frequency of A allele let q = frequency of a allele relative fitnesses are: AAAaaa w11 w12 w22 it is also possible to determine relative fitnesses of the A and a alleles: let w 1 = fitness of the A allele let w 2 = fitness of the a allele

42 The fitness of the A allele = w 1 = pw 11 + qw 12

43 The fitness of the a allele = w 2 = qw 22 + pw 12

44 Directional selection let p = frequency of A allele let q = frequency of a allele relative fitness of different genotypes are: AAAaaa w 11 w 12 w 22 it is also possible to determine relative fitness of the A and a alleles: The fitness of the A allele = w1 = pw11 + qw12 The fitness of the a allele = w2 = qw22 + pw12 Mean population fitness = w = pw1 + qw2


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