1 GENERAL BIOLOGY LAB 1 (BSC1010L) Lab #11: Population Genetics ______________________________________________________________________________ OBJECTIVES: • To gain a general understanding about the field of population genetics and how it can be used to study evolution and population dynamics. • Understand the concepts of evolution, fitness, natural selection, genetic drift and mutations. • To simulate Hardy-Weinberg equilibrium conditions. ______________________________________________________________________________ INTRODUCTION: A population is a group of individuals (plant or animal) of one species that occupy a defined geographical area and share genes through interbreeding. Within a large population, new, genetically distinct subpopulations can arise through isolation by distance (IBD). In this mechanism, as the subpopulations become geographically isolated, genetic differentiation between groups in the general population increases (Jensen, Bohonak and Kelley, 2005). These groups, more commonly referred to as local populations or demes (See Fig. 1), consist of members that are far likelier to breed with each other than with the remainder of the population. As such, their gene pool differs significantly from that of the general population and with continued isolation, demes may eventually evolve into new species. Figure 1. Local populations (demes) present within a large population2 Demes also arise from other mechanisms including geographical, ecological, temporal and/or behavioral isolation (Fig. 2). Figure 2. Mechanisms of isolation Population genetics, which emerged as separate branch of genetics in the early 1900s, is a direct extension of Mendel’s laws of inheritance, Darwin’s ideas of natural selection, and the concepts of molecular genetics. The field of population genetics focuses on the population to which an individual belongs rather than on the individual. Within any given population, every individual has its own set of alleles; for diploid organisms, there are 2 alleles for each gene, one from the mother and one from the father. Collectively, every individual’s set of alleles comprises the population’s gene pool. The role of a population geneticist is to study the allelic and3 genotypic (Formulas 1 and 2, respectively) variation present within a population’s gene pool and to assess how this variation changes from one generation to the next. Example: In a population of 100 students, 64 are PTC tasters with genotype TT, 32 are PTC tasters with genotype Tt and the last 4 are non-tasters with genotype tt. a. What is the allelic frequency t? Allelic frequency of t = 2 [t allele in recessive genotype (tt)] + t allele in heterozygous genotype (Tt) Total # of alleles for the PTC gene in the population* *2 (# of alleles in homozygous dominant condition) + 2 (# of alleles in heterozygous condition) + 2 (# of alleles in homozygous recessive condition) = 2[4] + 32 = 40 = 0.2 or 20% 2[64] + 2[32] + 2[4] 200 b. What is the genotypic frequency of tt? Genotypic frequency of tt = # off tt individuals Total # of individuals in the population = 4 = 4 = 0.04 or 4% 64 + 32 + 4 100 1. Allelic frequency = # of copies of an allele in a population Total # of all alleles for that gene in a population 2. Genotypic Frequency = # of individuals with a particular genotype in a population Total # of all individuals in a population4 In general, populations are dynamic units that change from one generation to the next. To be able to predict how a gene pool changes in response to fluctuations in size, geographic location and/or genetic composition, population geneticists have developed mathematical models that quantify these parameters. The most recognized of these is the Hardy-Weinberg (HW) equation, formulated by G. Hardy and W. Weinberg. This equation relates allele and genotype frequencies in a population and indicates the proportion of each allele combination that should exist within a population. In this formula, p2 = the frequency of a homozygous dominant genotype (e.g. BB) q2 = the frequency of a recessive genotype (e.g. bb) 2pq = the frequency of a heterozygote genotype (e.g. Bb) The HW equation predicts equilibrium, i.e., the allelic and genotypic frequencies remain constant over the course of many generations, if the following five assumptions are met: (1) Large population size, (2) Random mating, (3) No mutation, (4) No migration, (5) No natural selection. In reality, no population ever satisfies HW equilibrium completely. Example 1: If in a population of 100 cats, 84 carry a dominant allele for black coat (B) and 16 carry the recessive allele for white coat (b), then the frequency of the black phenotype is 0.84 and of the white phenotype is 0.16. a. Using the HW equation, calculate the frequencies of alleles B and b. frequency of white (bb) cats = 16/100 = 0.16 q2 = 0.16 therefore q = √0.16 = 0.4 since p + q =1, then p = 1 – q therefore, p =1- 0.4 = 0.6 (p + q)2 = 1 or p2 + 2pq + q2 = 15 b. Using the HW equation, calculate the frequencies of the BB and Bb genotypes. From part a, we know that p = 0.6 and q = 0.4 therefore, the frequency of BB cats is p2 = (0.6)2 = 0.36 and the frequency of Bb cats = 2pq = 2(0.6)(0.4) = 0.48 To check: since p2 + 2pq + q2 = 1, then 0.36 + 0.48 +0.16 = 1 Example 2: In a population of fruit flies, the genotypes of individuals present are: 50 RR, 20 Rr and 30 rr where R = red eyes and r = white eyes. Assuming the population is in Hardy-Weinberg equilibrium, the proportion of each genotype would be determined as follows: a. Using Formula 1, calculate the frequency of each allele, in this case R and r. Frequency of r allele = 2[30] + 20 = 80 = 0.4 2[50] + 2[20] + 2[30] 200 Therefore, q, the frequency of the recessive allele, equals 0.4. b. Since q is known, the p+q = 1 is equation is used to determine p, the frequency of the dominant allele. since p + q =1, then p = 1 – q therefore, p = 1 – 0.4 = 0.6 c. Now using the HW equation, we can calculate the proportion of RR, Rr and rr individuals in the population.6 From part a and b, we know that p = 0.6 and q = 0.4 therefore, the frequency of RR individuals is p2 = (0.6)2 = 0.36 the