Mutation and genetic variation Thymidine dimer Point mutations 1. transitions (e.g., A → G, C → T) 2. transversions (e.g., T → A, C → G) Models of Point mutation dynamics A α G A α G α α α α β β β β C α T C α T Jukes-Cantor Kimura - 2 parameter Model Model Genetic code Mutation hotspots Cytochrome b Molecular clock Emile Zuckerkandl and Linus Pauling, 1965 Synonymous substitutions v.s. Non-synonymous substitutions Ka/Ks is an indicator of selection General classes of mutations Point mutations “Copy-number” mutations Chromosomal mutations Genome mutations Chromosomal inversions lock blocks of genes together 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! Muntiacus reevesi; N = 23 Muntiacus muntjac; N = 4 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. Generation of a tetraploid Where do new genes come from? 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 Hardy-Weinberg Equilibrium The Hardy-Weinberg Equilibrium Godfrey Hardy Wilhelm Weinberg Gregor Mendel Udny Yule Reginald Punnett & William Bateson (1875-1967) (1861-1926) William Castle Reginald Punnett Godfrey Hardy The Hardy-Weinberg-Castle Equilibrium Godfrey Hardy Wilhelm Weinberg William Castle The Hardy-Weinberg Equilibrium consider a single locus with two alleles A1 and A2 • three genotypes exist: A1A1, A1A2, A2A2 • let p = frequency of A1 allele • let q = frequency of A2 allele • since only two alleles present, p + q = 1 Question: If mating occurs at random in the population, what will the frequencies of A1 and A2 be in the next generation? What are the probabilities of matings at the gamete level? What are the probabilities of matings at the gamete level? Egg Sperm Zygote Probability 2 A1 x A1 → A1A1 p x p = p What are the probabilities of matings at the gamete level? Egg Sperm Zygote Probability 2 A1 x A1 → A1A1 p x p = p A1 x A2 → A1A2 p x q = pq What are the probabilities of matings at the gamete level? Egg Sperm Zygote Probability 2 A1 x A1 → A1A1 p x p = p A1 x A2 → A1A2 p x q = pq A2 x A1 → A2A1 q x p = qp What are the probabilities of matings at the gamete level? Egg Sperm Zygote Probability 2 A1 x A1 → A1A1 p x p = p A1 x A2 → A1A2 p x q = pq 2pq A2 x A1 → A2A1 q x p = qp What are the probabilities of matings at the gamete level? Egg Sperm Zygote Probability 2 A1 x A1 → A1A1 p x p = p A1 x A2 → A1A2 p x q = pq 2pq A2 x A1 → A2A1 q x p = qp 2 A2 x A2 → A2A2 q x q = q Therefore, zygotes produced in proportions: Genotype: A1A1 A1A2 A2A2 2 2 Frequency: p 2pq q Therefore, zygotes produced in proportions: Genotype: A1A1 A1A2 A2A2 2 2 Frequency: p 2pq q what are the allele frequencies? What are the allele frequencies? What are the allele frequencies? 2 Frequency of A1 = p + ½ (2pq) What are the allele frequencies? 2 Frequency of A1 = p + ½ (2pq) 2 = p + pq What are the allele frequencies? 2 Frequency of A1 = p + ½ (2pq) 2 = p + pq = p(p + q) What are the allele frequencies? 2 Frequency of A1 = p + ½ (2pq) 2 = p + pq = p(p + q) = p What are the allele frequencies? 2 Frequency of A1 = p + ½ (2pq) 2 = p + pq = p(p + q) = p 2 Frequency of A2 = q + ½ (2pq) What are the allele frequencies? 2 Frequency of A1 = p + ½ (2pq) 2 = p + pq = p(p + q) = p 2 Frequency of A2 = q + ½ (2pq) 2 = q + pq What are the allele frequencies? 2 Frequency of A1 = p + ½ (2pq) 2 = p + pq = p(p + q) = p 2 Frequency of A2 = q + ½ (2pq) 2 = q + pq = q(q + p) What are the allele frequencies? 2 Frequency of A1 = p + ½ (2pq) 2 = p + pq = p(p + q) = p 2 Frequency of A2 = q + ½ (2pq) 2 = q + pq = q(q + p) = q What are the allele frequencies? 2 Frequency of A1 = p + ½ (2pq) 2 = p + pq = p(p + q) = p 2 Frequency of A2 = q + ½ (2pq) 2 = q + pq = q(q + p) = q ALLELE FREQUENCIES DID NOT CHANGE!! Conclusions of the Hardy-Weinberg principle Conclusions of the Hardy-Weinberg principle 1. Allele frequencies will not change from generation to generation. Conclusions of the Hardy-Weinberg principle 1. Allele frequencies will not change from generation to generation. 2. Genotype proportions determined by the “square law”. Conclusions of the Hardy-Weinberg principle 1. Allele frequencies will not change from generation to generation. 2. Genotype proportions determined by the “square law”. 2 2 2 • for two alleles = (p + q) = p + 2pq + q Conclusions of the Hardy-Weinberg principle 1. Allele frequencies will not change from generation to generation. 2. Genotype proportions determined by the “square law”. 2 2 2 • for two alleles = (p + q) = p + 2pq + q 2 2 2 2 • for three alleles (p + q + r) = p + q + r + 2pq + 2pr +2qr Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A1 = 0.80, A2 = 0.20 A1A1 = 0.64, A1A2 = 0.32, A2A2 = 0.04 Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A1 = 0.80, A2 = 0.20 A1A1 = 0.64, A1A2 = 0.32, A2A2 = 0.04 A1 = 0.50, A2 = 0.50 A1A1 = 0.25, A1A2 = 0.50, A2A2 = 0.25 Conclusions of the Hardy-Weinberg principle 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies Allele frequencies Genotype frequencies A1 = 0.80, A2 = 0.20 A1A1 = 0.64, A1A2 = 0.32, A2A2 = 0.04 A1 = 0.50, A2 = 0.50 A1A1 = 0.25, A1A2 = 0.50, A2A2 = 0.25 A1 = 0.10, A2 = 0.90 A1A1 = 0.01, A1A2 = 0.18, A2A2 = 0.81 Assumptions of Hardy-Weinberg equilibrium Assumptions of Hardy-Weinberg equilibrium 1. Mating is random Assumptions of Hardy-Weinberg equilibrium 1. Mating is random… but some traits experience positive assortative mating Assumptions of Hardy-Weinberg equilibrium 1. Mating is random 2. Population size is infinite (i.e., no genetic drift) Assumptions of Hardy-Weinberg equilibrium 1. Mating is random 2. Population size is infinite (i.e., no genetic drift) 3. No migration 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 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 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! Does Hardy-Weinberg equilibrium ever exist in nature? Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia as a juvenile… Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia … and as an adult Does Hardy-Weinberg equilibrium ever exist in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia 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) 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) A1A1 = 109 A1A2 = 182 A2A2 = 73 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) A1A1 = 109 A1A2 = 182 A2A2 = 73 Question: Is this population in Hardy-Weinberg equilibrium? Testing for Hardy-Weinberg equilibrium Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995 Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995 Frequency of A1A2 = 182/364 = 0.5000 Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies Frequency of A1A1 = 109/364 = 0.2995 Frequency of A1A2 = 182/364 = 0.5000 Frequency of A2A2 = 73/364 = 0.2005 Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000) Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000) = 0.5495 Testing for Hardy-Weinberg equilibrium Step 2: Estimate allele frequencies Frequency of A1 = p = Freq (A1A1) + ½ Freq (A1A2) = 0.2995 + ½ (0.5000) = 0.5495 Frequency of A2 = q = Freq (A2A2) + ½ Freq (A1A2) Testing for