Mutation and genetic variation
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1 Mutation and genetic variation
2 Thymidine dimer
3
4 Point mutations 1. transitions (e.g., A G, C T) 2. transversions (e.g., T A, C G)
5 Models of Point mutation dynamics α A G α A G α α α α β β β β C T α Jukes-Cantor Model C T α Kimura - 2 parameter Model
6 Genetic code
7 Mutation hotspots Cytochrome b
8
9 Molecular clock Emile Zuckerkandl and Linus Pauling, 1965
10
11 Synonymous substitutions v.s. Non-synonymous substitutions Ka/Ks is an indicator of selection
12 General classes of mutations Point mutations Copy-number mutations Chromosomal mutations Genome mutations
13 Chromosomal inversions lock blocks of genes together
14 Changes in chromosome number are common Robertsonian fusions and fissions are common and can have major effects on speciation in mammals, chromosome numbers range from N = 3 to N = 42. in insects, the range is from N = 1(some ants) to N = 220 (a butterfly) karyotypes can evolve rapidly!
15 Muntiacus reevesi; N = 23 Muntiacus muntjac; N = 4
16 Genome mutations polyploidization events cause the entire genome to be duplicated. polyploidy has played a major role in the evolution of plants. ancient polyploidization events have also occurred in most animal lineages.
17 Generation of a tetraploid
18 Where do new genes come from?
19 Where do new genes come from? An example: the antifreeze glycoprotein (AFGP) gene in the Antarctic fish, Dissostichus mawsoni Convergent evolution of an AFGP gene in the arctic cod, Boreogadus saida
20 Hardy-Weinberg Equilibrium
21 The Hardy-Weinberg Equilibrium Godfrey Hardy Wilhelm Weinberg
22 Gregor Mendel Udny Yule Reginald Punnett & William Bateson ( ) ( ) William Castle
23 Reginald Punnett Godfrey Hardy
24 The Hardy-Weinberg-Castle Equilibrium Godfrey Hardy Wilhelm Weinberg William Castle
25 The Hardy-Weinberg Equilibrium consider a single locus with two alleles and three genotypes exist:,, let p = frequency of allele let q = frequency of allele since only two alleles present, p + q = 1 Question: If mating occurs at random in the population, what will the frequencies of and be in the next generation?
26 What are the probabilities of matings at the gamete level?
27 What are the probabilities of matings at the gamete level? Egg Sperm Zygote Probability x p x p = p 2
28 What are the probabilities of matings at the gamete level? Egg Sperm Zygote Probability x p x p = p 2 x p x q = pq
29 What are the probabilities of matings at the gamete level? Egg Sperm Zygote Probability x p x p = p 2 x p x q = pq x q x p = qp
30 What are the probabilities of matings at the gamete level? Egg Sperm Zygote Probability x p x p = p 2 x p x q = pq x q x p = qp 2pq
31 What are the probabilities of matings at the gamete level? Egg Sperm Zygote Probability x p x p = p 2 x p x q = pq x q x p = qp 2pq x q x q = q 2
32 Therefore, zygotes produced in proportions: Genotype: Frequency: p 2 2pq q 2
33 Therefore, zygotes produced in proportions: Genotype: Frequency: p 2 2pq q 2 what are the allele frequencies?
34 What are the allele frequencies?
35 What are the allele frequencies? Frequency of = p 2 + ½ (2pq)
36 What are the allele frequencies? Frequency of = p 2 + ½ (2pq) = p 2 + pq
37 What are the allele frequencies? Frequency of = p 2 + ½ (2pq) = p 2 + pq = p(p + q)
38 What are the allele frequencies? Frequency of = p 2 + ½ (2pq) = p 2 + pq = p(p + q) = p
39 What are the allele frequencies? Frequency of Frequency of = p 2 + ½ (2pq) = p 2 + pq = p(p + q) = p = q 2 + ½ (2pq)
40 What are the allele frequencies? Frequency of Frequency of = p 2 + ½ (2pq) = p 2 + pq = p(p + q) = p = q 2 + ½ (2pq) = q 2 + pq
41 What are the allele frequencies? Frequency of Frequency of = p 2 + ½ (2pq) = p 2 + pq = p(p + q) = p = q 2 + ½ (2pq) = q 2 + pq = q(q + p)
42 What are the allele frequencies? Frequency of Frequency of = p 2 + ½ (2pq) = p 2 + pq = p(p + q) = p = q 2 + ½ (2pq) = q 2 + pq = q(q + p) = q
43 What are the allele frequencies? Frequency of Frequency of = p 2 + ½ (2pq) = p 2 + pq = p(p + q) = p = q 2 + ½ (2pq) = q 2 + pq = q(q + p) = q ALLELE FREQUENCIES DID NOT CHANGE!!
44 Conclusions of the Hardy-Weinberg principle
45 Conclusions of the Hardy-Weinberg principle 1. Allele frequencies will not change from generation to generation.
46 Conclusions of the Hardy-Weinberg principle 1. Allele frequencies will not change from generation to generation. 2. Genotype proportions determined by the square law.
47
48 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
49 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
50 Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies
51 Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies = 0.80, = 0.20 = 0.64, = 0.32, = 0.04
52 Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies = 0.80, = 0.20 = 0.64, = 0.32, = 0.04 = 0.50, = 0.50 = 0.25, = 0.50, = 0.25
53 Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies = 0.80, = 0.20 = 0.64, = 0.32, = 0.04 = 0.50, = 0.50 = 0.25, = 0.50, = 0.25 = 0.10, = 0.90 = 0.01, = 0.18, = 0.81
54 Assumptions of Hardy-Weinberg equilibrium
55 Assumptions of Hardy-Weinberg equilibrium 1. Mating is random
56 Assumptions of Hardy-Weinberg equilibrium 1. Mating is random but some traits experience positive assortative mating
57 Assumptions of Hardy-Weinberg equilibrium 1. Mating is random 2. Population size is infinite (i.e., no genetic drift)
58 Assumptions of Hardy-Weinberg equilibrium 1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration
59 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
60 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
61 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 The Hardy-Weinberg equilibrium principle thus predicts that no evolution will occur unless one (or more) of these assumptions are violated!
62 Does Hardy-Weinberg equilibrium ever exist in nature?
63 Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia
64 Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia as a juvenile
65 Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia and as an adult
66 Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia
67 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)
68 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) = 109 = 182 = 73
69 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) = 109 = 182 = 73 Question: Is this population in Hardy-Weinberg equilibrium?
70 Testing for Hardy-Weinberg equilibrium
71 Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies
72 Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of = 109/364 =
73 Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of = 109/364 = Frequency of = 182/364 =
74 Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of = 109/364 = Frequency of = 182/364 = Frequency of = 73/364 =
75 Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies
76 Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of = p = Freq ( ) + ½ Freq ( )
77 Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of = p = Freq ( ) + ½ Freq ( ) = ½ (0.5000)
78 Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of = p = Freq ( ) + ½ Freq ( ) = ½ (0.5000) =
79 Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of = p = Freq ( ) + ½ Freq ( ) = ½ (0.5000) = Frequency of = q = Freq ( ) + ½ Freq ( )
80 Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of = p = Freq ( ) + ½ Freq ( ) = ½ (0.5000) = Frequency of = q = Freq ( ) + ½ Freq ( ) = ½ (0.5000)
81 Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of = p = Freq ( ) + ½ Freq ( ) = ½ (0.5000) = Frequency of = q = Freq ( ) + ½ Freq ( ) = ½ (0.5000) =
82 Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of = p = Freq ( ) + ½ Freq ( ) = ½ (0.5000) = Frequency of = q = Freq ( ) + ½ Freq ( ) = ½ (0.5000) = Check that p + q = = 1
83 Testing for Hardy-Weinberg equilibrium Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium
84 Testing for Hardy-Weinberg equilibrium Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium Expected No. of = p 2 x N
85 Testing for Hardy-Weinberg equilibrium Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium Expected No. of = p 2 x N = (0.5495) 2 x 364
86 Testing for Hardy-Weinberg equilibrium Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium Expected No. of = p 2 x N = (0.5495) 2 x 364 = 109.9
87 Testing for Hardy-Weinberg equilibrium Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium Expected No. of = p 2 x N = (0.5495) 2 x 364 = Expected No. of = 2pq x N
88 Testing for Hardy-Weinberg equilibrium Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium Expected No. of = p 2 x N = (0.5495) 2 x 364 = Expected No. of = 2pq x N = 2(0.5495)(0.4505) x 364
89 Testing for Hardy-Weinberg equilibrium Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium Expected No. of = p 2 x N = (0.5495) 2 x 364 = Expected No. of = 2pq x N = 2(0.5495)(0.4505) x 364 = 180.2
90 Testing for Hardy-Weinberg equilibrium Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium Expected No. of = q 2 x N
91 Testing for Hardy-Weinberg equilibrium Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium Expected No. of = q 2 x N = (0.4595) 2 x 364
92 Testing for Hardy-Weinberg equilibrium Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium Expected No. of = q 2 x N = (0.4595) 2 x 364 = 73.9
93 Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes
94 Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
95 Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
96 Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
97 Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected
98 Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected χ 2 = Σ (Obs. Exp.) 2 Exp.
99 Testing for Hardy-Weinberg equilibrium Step 4: Compare observed and expected numbers of genotypes Genotype Observed Expected χ 2 = Σ (Obs. Exp.) 2 = Exp.
100 Suppose we sampled a mixed population
101 Suppose we sampled a mixed population Population A p = = 1.0 q = = 0 100%
102 Suppose we sampled a mixed population Population A Population B p = = 1.0 p = = 0 q = = 0 q = = % 100%
103 Suppose we sampled a mixed population Population A Population B p = = 1.0 p = = 0 q = = 0 q = = % 100% æ å Mixed population
104 Suppose we sampled a mixed population Population A Population B p = = 1.0 p = = 0 q = = 0 q = = % 100% æ å Mixed population 50% from population A (all ) 50% from population B (all )
105 Suppose we sampled a mixed population 50% from population A (all ) 50% from population B (all ) Sample 1000 individuals
106 Suppose we sampled a mixed population 50% from population A (all ) 50% from population B (all ) Sample 1000 individuals Observed = 500 = 0 = 500
107 Suppose we sampled a mixed population 50% from population A (all ) 50% from population B (all ) Sample 1000 individuals Observed Expected = 500 = 250 = 0 = 500 = 500 = 250
108 Sampling a mixed population generates a deficiency of heterozygotes This is called a Wahlund effect
109 Deafness in Tristan da Cunha observed AA = 1228 Aa = 352 aa = 253 Total: 1833 Inbreeding: Example p = 1228/ /2*1833 = = q = 253/ /2*1833 = = 0.234
110 Deafness in Tristan da Cunha observed AA = 1228 Aa = 352 aa = 253 Total: 1833 Inbreeding: Example p = 1228/ /2*1833 = = q = 253/ /2*1833 = = n*p 2 = (0.766) 2 * 1833 = n * 2pq = 2(0.766 * 0.234) * 1833 = n * q 2 = (0.234) 2 * 1833 =
111 Inbreeding: Example Deafness in Tristan da Cunha observed AA = 1228 Aa = 352 aa = 253 Total: 1833 expected p = 1228/ /2*1833 = = q = 253/ /2*1833 = = 0.234
112 Inbreeding: Example Deafness in Tristan da Cunha observed expected AA = Aa = aa = 253 (13.8%) (5.4%) Total: 1833 p = 1228/ /2*1833 = = q = 253/ /2*1833 = = 0.234
113 Deafness in Tristan da Cunha Inbreeding: Example observed expected AA = Aa = aa = 253 (13.8%) (5.4%) Total: 1833 p = 1228/ (352/2)/1833 = = q = 253/ (352/2)/1833 = = p = / (657.10/2)/1833 = q = / (657.10/2)/1833 = 0.234
114 A simple model of directional selection
115 A simple model of directional selection consider a single locus with two alleles and
116 A simple model of directional selection consider a single locus with two alleles and let p = frequency of allele
117 A simple model of directional selection consider a single locus with two alleles and let p = frequency of allele let q = frequency of allele
118 A simple model of directional selection consider a single locus with two alleles and let p = frequency of allele let q = frequency of allele relative fitnesses are: w 11 w 12 w 22
119 A simple model of directional selection consider a single locus with two alleles and let p = frequency of allele let q = frequency of allele relative fitnesses are: w 11 w 12 w 22 it is also possible to determine relative fitnesses of the and alleles:
120 A simple model of directional selection consider a single locus with two alleles and let p = frequency of allele let q = frequency of allele relative fitnesses are: w 11 w 12 w 22 it is also possible to determine relative fitnesses of the and alleles: let w 1 = fitness of the allele
121 A simple model of directional selection consider a single locus with two alleles and let p = frequency of allele let q = frequency of allele relative fitnesses are: w 11 w 12 w 22 it is also possible to determine relative fitnesses of the and alleles: let w 1 = fitness of the allele let w 2 = fitness of the allele
122 What is the fitness of an allele? Consider fitness from the gamete s perspective:
123 What is the fitness of an allele? Consider fitness from the gamete s perspective:
124 What is the fitness of an allele? Consider fitness from the gamete s perspective:
125 What is the fitness of an allele? Consider fitness from the gamete s perspective: realized fitness w 11
126 What is the fitness of an allele? Consider fitness from the gamete s perspective: realized fitness w 11 This route will occur with a probability p, since p is the frequency of the allele
127 What is the fitness of an allele? Consider fitness from the gamete s perspective: p realize fitness w 11 This route will occur with a probability p, since p is the frequency of the allele
128 What is the fitness of an allele? Consider fitness from the gamete s perspective: p realized fitness w 11
129 What is the fitness of an allele? Consider fitness from the gamete s perspective: p realized fitness w 11 realized fitness w 12
130 What is the fitness of an allele? Consider fitness from the gamete s perspective: p q realized fitness w 11 realized fitness w 12 This route will occur with a probability q, since q is the frequency of the allele
131 What is the fitness of an allele? Consider fitness from the gamete s perspective: p q realized fitness w 11 realized fitness w 12 Therefore, w 1 = pw 11 + qw 12
132 What is the fitness of an allele? Consider fitness from the gamete s perspective: p q realized fitness w 1 w 1 realized fitness w 12 Therefore, w 1 = pw 11 + qw 12 (this is equivalent to a weighted average of the two routes)
133 What is the fitness of an allele? Similarly for the A2 allele: q p realized fitness w 22 realized fitness w 12 Therefore, w 2 = qw 22 + pw 12
134 The fitness of the A1 allele = w1 = pw11 + qw12
135 The fitness of the allele = w 1 = pw 11 + qw 12 The fitness of the allele = w 2 = qw 22 + pw 12
136 The fitness of the allele = w 1 = pw 11 + qw 12 The fitness of the allele = w 2 = qw 22 + pw 12 One final fitness to define.
137 The fitness of the allele = w 1 = pw 11 + qw 12 The fitness of the allele = w 2 = qw 22 + pw 12 One final fitness to define. Mean population fitness
138 The fitness of the allele = w 1 = pw 11 + qw 12 The fitness of the allele = w 2 = qw 22 + pw 12 One final fitness to define. Mean population fitness = w = pw 1 + qw 2
139 The fitness of the allele = w 1 = pw 11 + qw 12 The fitness of the allele = w 2 = qw 22 + pw 12 One final fitness to define. Mean population fitness = w = pw 1 + qw 2 (This too is a weighted average of the two allelic fitnesses.)
140 Let p = frequency of allele in the next generation
141 Let p = frequency of allele in the next generation p = pw 1 /(pw 1 + qw 2 )
142 Let p = frequency of allele in the next generation p = pw 1 /(pw 1 + qw 2 ) p = p(w 1 /w)
143 Let p = frequency of allele in the next generation p = pw 1 /(pw 1 + qw 2 ) p = p(w 1 /w) Let q = frequency of allele in the next generation
144 Let p = frequency of allele in the next generation p = pw 1 /(pw 1 + qw 2 ) p = p(w 1 /w) Let q = frequency of allele in the next generation q = qw 2 /(pw 1 + qw 2 )
145 Let p = frequency of allele in the next generation p = pw 1 /(pw 1 + qw 2 ) p = p(w 1 /w) Let q = frequency of allele in the next generation q = qw 2 /(pw 1 + qw 2 ) q = q (w 2 /w)
146 Let p = frequency of allele in the next generation p = pw 1 /(pw 1 + qw 2 ) p = p(w 1 /w) Let q = frequency of allele in the next generation q = qw 2 /(pw 1 + qw 2 ) q = q (w 2 /w)
147 An example of directional selection
148 An example of directional selection Let p = q = 0.5
149 An example of directional selection Let p = q = 0.5 Genotype:
150 An example of directional selection Let p = q = 0.5 Genotype: Fitness:
151 An example of directional selection Let p = q = 0.5 Genotype: Fitness: w 1 = pw 11 + qw 22 = 0.975
152 An example of directional selection Let p = q = 0.5 Genotype: Fitness: w 1 = pw 11 + qw 22 = w 2 = qw 22 + pw 11 = 0.925
153 An example of directional selection Let p = q = 0.5 Genotype: Fitness: w 1 = pw 11 + qw 12 = w 2 = qw 22 + pw 12 = w = pw 1 + qw 2 = 0.950
154 An example of directional selection Let p = q = 0.5 Genotype: Fitness: w 1 = pw 11 + qw 22 = w 2 = qw 22 + pw 11 = w = pw 1 + qw 2 = p = p(w 1 /w) = 0.513
155 An example of directional selection Let p = q = 0.5 Genotype: Fitness: w 1 = pw 11 + qw 22 = w 2 = qw 22 + pw 11 = w = pw 1 + qw 2 = p = p(w 1 /w) = q = q(w 2 /w) = 0.487
156 An example of directional selection Let p = q = 0.5 Genotype: Fitness: w 1 = pw 11 + qw 22 = w 2 = qw 22 + pw 11 = w = pw 1 + qw 2 = p = p(w 1 /w) = q = q(w 2 /w) = In ~150 generations the allele will be fixed
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