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A Brief Look at Population Genetics and the Quantitative Study of Natural Selection Population Genetics • Concerned with genotype and gene frequencies • Vast numbers of loci; human genome may contain 100,000 loci • Easiest to consider models of evolution at one locus Consider a population of eight individuals • Aa, AA, aa, aa, AA, Aa, AA, Aa • To find the genotype frequencies, simply count the number of individuals with each genotype • So, frequency of AA = 3/8 = 0.375, Aa = 3/8 = 0.375, and aa = 2/8 = 0.25 What about gene frequencies? A a A A a a a a A A A a A A A a Each genotype contains two genes, so there are a total of 16 genes per locus in a population of eight individuals. The frequency of A is 9/16 or 0.5625, while the frequency of a is 7/16, or 0.4375 If we identify: P = genotype frequency of AA Q = genotype frequency of Aa R = genotype frequency of aa Then p = gene frequency of A = P + ½ Q and q = gene frequency of a = R + ½ Q …. and p + q = 1. An elementary population genetics model has four steps. In the absence of selection, genotype frequencies go to the Hardy-Weinberg equilibrium. p2 + 2pq +q2 = 1 Hardy-Weinberg frequencies of genotypes AA, Aa, and aa in relation to the frequency of the gene a. We can test, by observation, whether genotypes in a population are in H-W equilibrium. We don’t really expect to see genotypes in natural populations in H-W equilibrium, mainly because of natural selection. A simple model of selection can concentrate on how genotypic frequencies are modified between birth and the adult reproductive stage. Start with a simple situation….. Natural selection operating on a single genetic locus, at which there are two alleles, A and a, with A dominant. Suppose that the three possible genotypes have the following relative chances of survival from birth to the adult stage: Genotype AA, Aa Chance of survival 1 Aa 1-s “s” is a number between 0 and 1, and is called the selection coefficient. With the fitnesses given, selection will act to eliminate the a gene and fix the A allele. To “fix” a gene means to carry its frequency up to one. How rapidly will the population change through time? We need an expression of the gene frequency of A (p') in one generation in terms of its frequency in the previous generation (p). The difference between the two: Δp = p‘ – p Is the change in gene frequency between two successive generations. Look at algebraic example: Genotype AA Aa Aa frequency p2 2pq q2 fitness 1 1 1–s 2pq q2(1 – s) Birth Adult relative freq. p2 frequency p2/1 – sq2 2pq/1 – sq2 q2(1 – s)/1 – sq2 And a numerical example AA Aa aa Birth Total Number 1 18 81 Frequency 0.01 0.18 0.81 Fitness 1 1 0.9 Number 1 18 73 Frequency 1/92 18/92 73/92 100 Adult 92 So, how are p’ and p2 related? Recall that the frequency of A at any time is equal to the frequency of AA plus half the frequency of Aa. So: p 2 pq p p 2 2 1 sq 1 sq Since the denominator is less than one, p΄ is greater than p. Selection is increasing the frequency of A. What is Δp? p p p p 2 1 sq p p spq 2 1 sq 2 spq 2 1 sq 2 We can use this to calculate the change in gene frequency between successive generations for any selection coefficient, s, and any gene frequencies. Similarly, if we know the gene frequencies for two successive generations we can rearrange the equation to solve for the selection coefficient: p s pq 2 So, how rapidly can selection change gene frequencies? Apply the model to Biston betularia If we can come up with estimates of the frequencies of the different color forms at two different times, we can use them to estimate the gene frequencies, and substitute those into our equation to get an estimate of s. A disadvantageous gene in mutation-selection balance has a frequency of m/s, where m is the mutation rate and s is its selective advantage. J.B.S. Haldane guessed that m = 10-6 and s = 0.1 for the melanic C gene. This would give the gene a frequency of 10-5 in 1848 when it was first seen. By 1898, the frequency of the light-colored genotype was 110% in polluted areas, give it a frequency of about 0.2, and the C gene a frequency of about 0.8 So, what selection coefficient would generate an increase in the frequency of the C allele from 10-5 to 0.8 in 50 generations? If our equation is iterated 50 times, we find that a selection coefficient of about 0.33 will lead to the observed change in gene frequencies. In other words, the peppered moths would have about 2/3 the survival rate of the melanic moths. Theoretical changes in gene frequencies in the evolution of melanism in the peppered moth We can get a second estimate of fitness by looking at the survivorship of different genotypes in mark-recapture studies