Population Genetics

Review: 1859: Darwin and the birth of modern biology (explaining why living things are as they are) – Heritable Traits and Environment  Evolution

Review: 1859: Darwin and the birth of modern biology (explaining why living things are as they are) – Heritable Traits and Environment  Evolution Mendel: Heredity works by the transmission of particles (genes) that influence the expression of traits

Review: 1859: Darwin and the birth of modern biology (explaining why living things are as they are) – Heritable Traits and Environment  Evolution Mendel: Heredity works by the transmission of particles (genes) that influence the expression of traits Avery, McCarty, and MacLeod: Genes are DNA

Review: 1859: Darwin and the birth of modern biology (explaining why living things are as they are) – Heritable Traits and Environment  Evolution Mendel: Heredity works by the transmission of particles (genes) that influence the expression of traits Avery, McCarty, and MacLeod: Genes are DNA Watson and Crick: Here’s the structure of DNA

Review: 1859: Darwin and the birth of modern biology (explaining why living things are as they are) – Heritable Traits and Environment  Evolution Mendel: Heredity works by the transmission of particles (genes) that influence the expression of traits Avery, McCarty, and MacLeod: Genes are DNA Watson and Crick: Here’s the structure of DNA Modern Genetics: Here’s how DNA influences the expression of traits from molecule to phenotype throughout development

Review: 1859: Darwin and the birth of modern biology (explaining why living things are as they are) – Heritable Traits and Environment  Evolution Mendel: Heredity works by the transmission of particles (genes) that influence the expression of traits How does evolution work at a genetic level? Population Genetics and the Modern Synthesis Avery, McCarty, and MacLeod: Genes are DNA Watson and Crick: Here’s the structure of DNA Modern Genetics: Here’s how DNA influences the expression of traits from molecule to phenotype throughout development

Review: 1859: Darwin and the birth of modern biology (explaining why living things are as they are) – Heritable Traits and Environment  Evolution Mendel: Heredity works by the transmission of particles (genes) that influence the expression of traits How does evolution work at a genetic level? Population Genetics and the Modern Synthesis Avery, McCarty, and MacLeod: Genes are DNA Watson and Crick: Here’s the structure of DNA Modern Genetics: Here’s how DNA influences the expression of traits from molecule to phenotype throughout development How can we describe the patterns of evolutionary change through DNA analyses? Evolutionary Genetics

The Modern Synthesis The Darwinian Naturalists The Mutationists Ernst Mayr Selection is the only mechanism that can explain adaptations; mutations are random and cannot explain the non-random ‘fit’ of organisms to their environment T. H. Morgan R. Goldschmidt The discontinuous variation between species can only be explained by the discontinuous variation we see expressed as a function of new mutations; the probabilistic nature of selection is too weak to cause the evolutionary change we see in the fossil record

The Modern Synthesis Sewall Wright Random chance was an important source of change in small populations J. B. S. Haldane Developed mathematical models of population genetics with Fisher and Wright R. A. Fisher Multiple genes can produce continuous variation, and selection can act on this variation and cause change in a population Theodosius Dobzhansky Described genetic differences between natural populations; described evolution as a change in allele frequencies.

Population Genetics I. Basic Principles

Population Genetics I. Basic Principles A. Definitions: - Population: a group of interbreeding organisms that share a common gene pool; spatiotemporally and genetically defined - Gene Pool: sum total of alleles held by individuals in a population - Gene/Allele Frequency: % of genes at a locus of a particular allele - Gene Array: % of all alleles at a locus: must sum to 1. - Genotypic Frequency: % of individuals with a particular genotype - Genotypic Array: % of all genotypes for loci considered = 1.

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: 1. Determining the Gene and Genotypic Array: AA Aa aa Individuals 60 80 (200)

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: 1. Determining the Gene and Genotypic Array: AA Aa aa Individuals 60 80 (200) Genotypic Array 60/200 = 0.30 80/200 = .40 = 1

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: 1. Determining the Gene and Genotypic Array: AA Aa aa Individuals 60 80 (200) Genotypic Array 60/200 = 0.30 80/200 = .40 = 1 ''A' alleles 120 200/400 = 0.5

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: 1. Determining the Gene and Genotypic Array: AA Aa aa Individuals 60 80 (200) Genotypic Array 60/200 = 0.30 80/200 = .40 = 1 ''A' alleles 120 200/400 = 0.5 'a' alleles

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: 1. Determining the Gene and Genotypic Array 2. Short Cut Method: - Determining the Gene Array from the Genotypic Array a. f(A) = f(AA) + f(Aa)/2 = .30 + .4/2 = .30 + .2 = .50 b. f(a) = f(aa) + f(Aa)/2 = .30 + .4/2 = .30 + .2 = .50 KEY: The Gene Array CAN ALWAYS be computed from the genotypic array; the process just counts alleles instead of genotypes. No assumptions are made when you do this.

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: 1. If a population acts in a completely probabilistic manner, then: - we could calculate genotypic arrays from gene arrays - the gene and genotypic arrays would equilibrate in one generation

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: 1. If a population acts in a completely probabilistic manner, then: - we could calculate genotypic arrays from gene arrays - the gene and genotypic arrays would equilibrate in one generation 2. But for a population to do this, then the following assumptions must be met (Collectively called Panmixia = total mixing) - Infinitely large (no deviation due to sampling error) - Random mating (to meet the basic tenet of random mixing) - No selection, migration, or mutation (gene frequencies must not change)

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: Sources of Variation Agents of Change Mutation N.S. Recombination Drift - crossing over Migration - independent assortment Mutation Non-random Mating VARIATION So, if NO AGENTS are acting on a population, then it will be in equilibrium and WON'T change.

Non-Random Mating does not alter gene frequencies, but it DOES alter genotypic frequencies and thus, the genetic structure of the population – causing a deviation from HWE expectations.

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: 3. PROOF: - Given a population with p + q = 1. - If mating is random, then the AA, Aa and aa zygotes will be formed at p2 + 2pq + q2 - They will grow up and contribute genes to the next generation: - All of the gametes produced by AA individuals will be A, and they will be produced at a frequency of p2 - 1/2 of the gametes of Aa will be A, and thus this would be 1/2 (2pq) = pq - So, the frequency of A gametes in the “gamete/gene pool” will be p2 + pq = p(p + q) = p(1) = p - Likewise for the 'a' allele (remains at frequency of q). - Not matter what the gene frequencies, if panmixia occurs then the population will reach an equilibrium after one generation of random mating...and will NOT change (no evolution)

Initial genotypic freq. Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: AA Aa aa Initial genotypic freq. 0.4 0.2 1.0 Gene freq. Genotypes, F1 Gene Freq's Genotypes, F2

Initial genotypic freq. Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: AA Aa aa Initial genotypic freq. 0.4 0.2 1.0 Gene freq. f(A) = p = .4 + .4/2 = 0.6 f(a) = q = .2 + .4/2 = 0.4 Genotypes, F1 Gene Freq's Genotypes, F2

Initial genotypic freq. Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: AA Aa aa Initial genotypic freq. 0.4 0.2 1.0 Gene freq. f(A) = p = .4 + .4/2 = 0.6 f(a) = q = .2 + .4/2 = 0.4 Genotypes, F1 p2 = .36 2pq = .48 q2 = .16 = 1.00 Gene Freq's Genotypes, F2

Initial genotypic freq. Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: AA Aa aa Initial genotypic freq. 0.4 0.2 1.0 Gene freq. f(A) = p = .4 + .4/2 = 0.6 f(a) = q = .2 + .4/2 = 0.4 Genotypes, F1 p2 = .36 2pq = .48 q2 = .16 = 1.00 Gene Freq's f(A) = p = .36 + .48/2 = 0.6 f(a) = q = .16 + .48/2 = 0.4 Genotypes, F2

Initial genotypic freq. Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: AA Aa aa Initial genotypic freq. 0.4 0.2 1.0 Gene freq. f(A) = p = .4 + .4/2 = 0.6 f(a) = q = .2 + .4/2 = 0.4 Genotypes, F1 p2 = .36 2pq = .48 q2 = .16 = 1.00 Gene Freq's f(A) = p = .36 + .48/2 = 0.6 f(a) = q = .16 + .48/2 = 0.4 Genotypes, F2 .36 .48 .16 1.00

Initial genotypic freq. PROOF: Consider a gene pool of p + q = 1 Panmixia AA Aa aa Initial genotypic freq. p2 2pq q2 1.0 These zygotes grow up to reproductive age and produce gametes… at what frequencies? A Gametes a Gametes ALL A, f = p2 ½ A; f = pq ½ a; f = pq ALL a, f = q2 So, the gene pool of gametes from which the new generation of zygotes will be made has frequencies: Gamete Frequ. F(A) = p2 + pq = p(p+q) =p(1) = p F(a) = q2 + pq = q(p+q) =q(1) = q Genotypes, F2

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility 1. If no real populations can explicitly meet these assumptions, how can the model be useful?

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility 1. If no real populations can explicitly meet these assumptions, how can the model be useful? It is useful for creating an expected model that real populations can be compared against to see if a population is EVOLVING.

F(A) = 0.756 + ½(0.242) = 0.756 + 0.121 = 0.877 F(S) = 0.002 + ½(0.242) = 0.002 + 0.121 = 0.123

Wrong… F(A) = 0.756 + ½(0.242) = 0.756 + 0.121 = 0.877 F(S) = 0.002 + ½(0.242) = 0.002 + 0.121 = 0.123 Calculate HWE expectations (book values incorrect): (0.877)(0.877)(12,387) = 9527.2 2(0.877)(0.121)(12,387) = 2672.4 (0.123)(0.123)(12,387) = 187.4

Genotype Obs. Exp. (O-E)2/E AA 9365 9527.2 2.76 AS 2993 2672.4 38.48 SS 29 187.4 133.89 X2 = sum = 175.13

Genotype Obs. Exp. (O-E)2/E AA 9365 9527.2 2.76 AS 2993 2672.4 38.48 SS 29 187.4 133.89 X2 = sum = 175.13 NOT HWE

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility 1. If no real populations can explicitly meet these assumptions, how can the model be useful? It is useful for creating an expected model that real populations can be compared against to see if a population is EVOLVING. 2. Also, If HWCE is assumed and the frequency of homozygous recessives can be measured, then the number of heterozygous carriers can be estimated.

For example: If you observe that the frequency of individuals expressing a homozygous recessive condition is 1/100, then f(aa) = 0.01 Then, if you ASSUME HWE, you can calculate f(a) = q = √(0.01) = 0.1 If you assume a two-locus model, then and f(A) = p = (1 – q) = 1 – 0.1 = 0.9 And then, the frequency of heterozygous ‘carriers’ = f(Aa) = 2(0.1)(0.9) = 0.18

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility E. Extensions

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility E. Extensions 1. 2 alleles in diploids: (p + q)2 = p2 + 2pq + q2

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility E. Extensions 1. 2 alleles in diploids: (p + q)2 = p2 + 2pq + q2 2. More than 2 alleles (p + q + r) 2 = p2 + 2pq + q2 + 2pr + 2qr + r2

Population Genetics I. Basic Principles A. Definitions: B. Basic computations: C. Hardy-Weinberg Equilibrium: D. Utility E. Extensions 1. 2 alleles in diploids: (p + q)2 = p2 + 2pq + q2 2. More than 2 alleles (p + q + r) 2 = p2 + 2pq + q2 + 2pr + 2qr + r2 3. Tetraploidy: (p + q)4 = p4 + 4p3q + 6p2q2 + 4pq3 + q4 (Pascal's triangle for constants...)

Population Genetics I. Basic Principles II. X-linked Genes

Population Genetics I. Basic Principles II. X-linked Genes A. Issue

Population Genetics I. Basic Principles II. X-linked Genes A. Issue - Females (or the heterogametic sex) are diploid, but males are only haploid for sex linked genes. A A A

Population Genetics I. Basic Principles II. X-linked Genes A. Issue Females (or the heterogametic sex) are diploid, but males are only haploid for sex linked genes. As a consequence, Females will carry 2/3 of these genes in a population, and males will only carry 1/3. A A A

Population Genetics I. Basic Principles II. X-linked Genes A. Issue Females (or the heterogametic sex) are diploid, but males are only haploid for sex linked genes. As a consequence, Females will carry 2/3 of these genes in a population, and males will only carry 1/3. So, the equilibrium will occur when: p(eq) = 2/3p(f) + 1/3p(m) A A A

Population Genetics I. Basic Principles II. X-linked Genes A. Issue Females (or the heterogametic sex) are diploid, but males are only haploid for sex linked genes. As a consequence, Females will carry 2/3 of these genes in a population, and males will only carry 1/3. So, the equilibrium will occur when: p(eq) = 2/3p(f) + 1/3p(m) - This won’t occur in one generation, either. A A A

Population Genetics I. Basic Principles II. X-linked Genes A. Issue B. Example 1. Calculating Gene Frequencies in next generation: p(f)1 = ½[p(f)+p(m)] Think about it. Daughters are formed by an X from the mother and an X from the father. So, the frequency in daughters will be AVERAGE of the frequencies in the previous generation of mothers and fathers.

Population Genetics I. Basic Principles II. X-linked Genes A. Issue B. Example 1. Calculating Gene Frequencies in next generation: p(f)1 = ½[p(f)+p(m)] Think about it. Daughters are formed by an X from the mother and an X from the father. So, the frequency in daughters will be AVERAGE of the frequencies in the previous generation of mothers and fathers. p(m)1 = p(f) Males get all their X chromosomes from their mother, so the frequency in males will equal the frequency in females in the preceding generation.

Population Genetics I. Basic Principles II. X-linked Genes A. Issue B. Example 2. Change over time: - Consider this population: f(A)m = 0, and f(A)f = 1.0.

Population Genetics I. Basic Principles II. X-linked Genes A. Issue B. Example 2. Change over time: - Consider this population: f(A)m = 0, and f(A)f = 1.0. - In f1: p(m) = 1.0, p(f) = 0.5

Population Genetics I. Basic Principles II. X-linked Genes A. Issue B. Example 2. Change over time: - Consider this population: f(A)m = 0, and f(A)f = 1.0. - In f1: p(m) = 1.0, p(f) = 0.5 - In f2: p(m) = 0.5, p(f) = 0.75

Population Genetics I. Basic Principles II. X-linked Genes A. Issue B. Example 2. Change over time: - Consider this population: f(A)m = 0, and f(A)f = 1.0. - In f1: p(m) = 1.0, p(f) = 0.5 - In f2: p(m) = 0.5, p(f) = 0.75 - In f3: p(m) = 0.75, p(f) = 0.625

Population Genetics I. Basic Principles II. X-linked Genes A. Issue B. Example 2. Change over time: - Consider this population: f(A)m = 0, and f(A)f = 1.0. - In f1: p(m) = 1.0, p(f) = 0.5 - In f2: p(m) = 0.5, p(f) = 0.75 - In f3: p(m) = 0.75, p(f) = 0.625 There is convergence on an equilibrium = p = 0.66 p(eq) = 2/3p(f) + 1/3p(m)