SP08-PopulationGeneticsHandout[1] - 1B Trenham POPULATION...

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1B Trenham Spring 2008 POPULATION GENETICS The five conditions required for Hardy-Weinberg equilibrium are: 1. Large population size 2. Mating must be at random with respect to the phenotypic trait 3. No migration 4. No net change in allele frequencies as a result of mutation 5. No advantage to individuals carrying particular alleles at a locus If these conditions are satisfied, then there should be no change in genotype or allele frequencies across generations. So, if we have information about allele or genotype frequencies for one generation, we can determine the expected genotypic frequencies for future generations under Hardy-Weinberg equilibrium. We compare expected frequencies (assuming equilibrium) with observed frequencies--if they match we assume the 5 conditions are satisfied, and if they don't match, then we can determine which condition is violated. There are several different ways to use information about genotype and allele frequencies, depending on what you know at the beginning and what you want to find out. I. How to estimate genotype frequencies if you know allele frequencies: IA . If Hardy-Weinberg assumptions are valid for the population Example: Begin with a population of 1000 breeding adult budgerigars, in which there are two alleles at a feather color locus. In the parental generation, the frequency of the y allele = q = 0.2, and the frequency of the Y allele = p = 0.8. Check : Allele frequencies at a given locus must add up to 1.00 ( this is a math check ; it does not say anything about equilibrium). Here, 0.2 + 0.8 = 1.0. 1. Suppose we want to determine the expected frequency of yy offspring or YY offspring in the next generation: The chances that an egg drawn at random from this population has the y allele = q. Similarly, the chances that a sperm drawn at random has a y allele = q. The chances that an offspring will obtain two y alleles = q x q = q 2 . In this population, q = .2, so the likelihood of getting a yy genotype = q 2 = .04 Similarly, the chances of getting two Y alleles, one from each parent = p x p = p 2 . In this population: .8 x .8 = 0.64 2. Suppose we want to know: What is the expected frequency of a heterozygote ? There are two ways this could happen: (i). Egg = y and sperm = Y : q x p = qp = .8 x .2 = .16 OR (ii). Egg = Y and sperm = y : p x q = pq = .2 x .8 = .16 -1-
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Hence, the likelihood of getting a heterozygote is qp + pq—we add the two ways together to account for both of them. Notice that qp + pq is the same thing as 2pq. In this example, 2pq = .16 + .16 = .32. If Hardy-Weinberg conditions are satisfied, the expected genotype and phenotype frequencies in the offspring generation should be: Genotypes Phenotypes q 2 = .04 4% homozygous recessives blue p 2 = .64 64% homozygous dominants green 2pq = .32 32% heterozygotes green Check ...the genotype frequencies at a given locus must add up to 1.00. Here, .04 + .64 + .32 = 1.00.
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SP08-PopulationGeneticsHandout[1] - 1B Trenham POPULATION...

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