October 12, 2009 Bioe 109 Fall 2009 Lecture 8 Microevolution 1 - selection The Hardy-Weinberg-Castle Equilibrium - consider a single locus with two alleles A 1 and A 2. - three genotypes are thus possible: A 1 A 1, A 1 A 2, A 2 A 2. - let p = frequency of A 1 allele - let q = frequency of A 2 allele - since only two alleles present, p + q = 1 Question: If mating occurs at random in the population, what will the frequencies of A 1 and A 2 be in the next generation? - what does random mating mean? - very simply, that matings take place in the population independently of the genotype of an individual. - if mating is random the frequencies of matings among different genotypes are determined simply by their frequencies. - random mating can be described among genotypes or, more simply, among the gametes produced by these individuals. - what are the probabilities of such matings at the gamete level? - these will occur at proportions determined by the frequencies of the A 1 and A 2 alleles in the population: Egg Sperm Zygote Probability A 1 x A 1 A 1 A 1 p x p = p 2 A 1 x A 2 A 1 A 2 p x q = pq A 2 x A 1 A 2 A 1 q x p = qp = 2pq A 2 x A 2 A 2 A 2 q x q = q 2 - therefore, zygotes are produced in the following proportions: A 1 A 1 A 1 A 2 A 2 A 2 p 2 2pq q 2
- what are the allele frequencies in the next generation? 1. Frequency of A 1 = p 2 + ½ (2pq) = p 2 + pq = p(p + q) = p 2. Frequency of A 2 = q 2 + ½ (2pq) = q 2 + pq = q(q + p) = q NOTE THAT THE FREQUENCIES OF THE ALLELES DO NOT CHANGE There are three conclusions to be drawn from the H-W principle: 1. Allele frequencies will not change from generation to generation. 2. Genotype proportions will occur according to 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 3. Hardy-Weinberg equilibrium occurs independently of allelic frequencies - proportions of genotypes will shift in accordance with those predicted by the square law. Allele frequencies Genotype frequencies p = 0.80, q = 0.20 A 1 A 1 = 0.64; A 1 A 2 = 0.32; A 2 A 2 = 0.04 p = 0.50, q = 0.50 A 1 A 1 = 0.25; A 1 A 2 = 0.50; A 2 A 2 = 0.25 p = 0.20, q = 0.80 A 1 A 1 = 0.04; A 1 A 2 = 0.32; A 2 A 2 = 0.64 Assumptions of Hardy-Weinberg Equilibrium: 1. Random mating 2. Infinite population size (i.e., no genetic drift) 3. No migration (i.e., no gene flow) 4. No mutation 5. No selection - The H-W principle thus states that allele frequencies in populations will not change unless some evolutionary process is acting to cause a change in allele frequency. - therefore, this principle thus predicts that no evolution will occur unless one of the above assumptions is violated.
- therefore, the study of microevolution is largely concerned with understanding the conditions under which the above assumptions are violated. - in other words, it involves determining the relative importance of random drift, migration, mutation, and natural selection in affecting the frequency of genetic polymorphism in natural populations. - given that many (most?) of these assumptions are likely to violated, an important question to ask is whether Hardy-Weinberg equilibrium ever exists in nature? Example: Atlantic cod (Gadus morhua) in Nova Scotia - a sample of 364 fish scored for a DNA SNP (single nucleotide polymorphism) - the following genotypes were observed. A 1 A 1 = 109 A 1 A 2 = 182 A 2 A 2 = 73 364 Question: Is this population in Hardy-Weinberg equilibrium? Testing for Hardy-Weinberg equilibrium Step 1: Estimate genotype frequencies. Frequency of A 1 A 1 = 109/364 = 0.2995 Frequency of A 1 A 2 = 182/364 = 0.5000 Frequency of A 2 A 2 = 73/364 = 0.2005 Step 2: Estimate allele frequencies. Frequency of A 1 = p = Freq (A 1 A 1 ) + ½ Freq (A 1 A 2 ) = 0.2995 + ½ (0.5000) = 0.5495 Frequency of A 2 = q = Freq (A 2 A 2 ) + ½ Freq (A 1 A 2 ) = 0.2005 + ½ (0.5000) = 0.4505 Check that p + q = 0.5495 + 0.4505 = 1 Step 3: Estimate expected genotype frequencies under the assumption of H-W equilibrium. - here, we are testing if genotypes are present in proportions consistent with the square law.
Expected No. of A 1 A 1 = p 2 x N = (0.5495) 2 x 364 = 109.9 Expected No. of A 1 A 2 = 2pq x N = 2(0.5495)(0.4505) x 364 = 180.2 Expected No. of A 2 A 2 = q 2 x N = (0.4505) 2 x 364 = 73.9 Step 4: Compare observed and expected numbers of genotypes. Genotype Observed Expected A 1 A 1 109 109.9 A 1 A 2 182 180.2 A 2 A 2 73 73.9 - we can test if the proportions differ significantly by performing a Chi-square test (see Box 6.5 on pages 192-193): χ 2 = Σ (Obs. Exp.) 2 = 0.036 Exp. - the observed Chi-square is not significant. - although there is an excellent agreement between observed proportions and those expected under the assumption of Hardy-Weinberg equilibrium. - does this mean that one, or more, assumptions are in fact being violated. - if they are, then their effect is so slight as to not cause any appreciable perturbation. - suppose we were to sample a mixed population. Population A Population B p = A 1 = 1.0 p = A 1 = 0 q = A 2 = 0 q = A 2 = 1.0 100 % A 1 A 1 100 % A 2 A 2 Mixed population 50% from population A (all A 1 A 1 ) 50% from population B (all A 2 A 2 )
Sample 1000 individuals Observed Expected A 1 A 1 = 500 A 1 A 1 = 250 A 1 A 2 = 0 A 1 A 2 = 500 A 2 A 2 = 500 A 2 A 2 = 250 - here, there is a marked departure from Hardy-Weinberg due to the violation of the random mating assumption. - a deficiency of heterozygotes always occurs when a mixed population is sampled this is called a Wahlund effect. A simple model of directional selection - consider a simple two allele case where p = A 1 and q = A 2. - let the relative fitnesses of the three genotypes be: Genotype A 1 A 1 A 1 A 2 A 2 A 2 Rel. fitness w 11 w 12 w 22 - there are three equations needed to predict the changes in frequencies of alleles in a population caused by directional selection. - the first two estimate the fitness of the A 1 and A 2 alleles: w 1 = p w 11 + q w 12 w 2 = q w 22 + p w 12 - the mean fitness of the A 1 allele is thus a weighted average of its fitnesses in the two genotypes in which it can occur (i.e., A 1 A 1 and A 1 A 2 ). - the third equation gives the mean fitness of the entire population: wbar = pw 1 + qw 2 - designating the frequency of allele A 1 in the next generation as p p = pw 1 / (pw 1 + qw 2 ) = p (w 1 /wbar) similarly, q = qw 2 / (pw 1 + qw 2 ) = q (w 2 /wbar)
- these equations tell us that changes in the frequencies of alleles A and a over one generation are proportional to the difference between the allele s fitness and the mean population fitness An example: suppose frequencies of alleles A 1 and A 2 are p = q = 0.50. Genotype A 1 A 1 A 1 A 2 A 2 A 2 Rel. fitness w 11 w 12 w 22 1 0.95 0.90 - we can intuitively see the outcome - the small a allele will be eliminated from the population. - how rapidly will this occur? w 1 = p(w 11 ) + q(w 12 ) = 0.5(1.0) + 0.5(0.95) = 0.975 w 2 = q(w 22 ) + p(w 12 ) = 0.5(0.9) + 0.5(0.95) = 0.925 wbar = p(w 1 ) + q(w 2 ) = 0.5(0.975) + 0.5(0.925) = 0.950 thus, p = p(w 1 /wbar) = 0.5(0.975/0.95) = 0.513 q = q(w2/wbar) = 0.5(0.925/0.95) = 0.487 Conclusion: the frequency of the A 1 allele increased by only 0.013 even though the strength of selection is moderately strong. - even though this is a small change, it will take only 150 generations for the A 1 allele to be fixed (i.e., reach a frequency of 1).