EXAM 3 2008 Key

EXAM 3 2008 Key - Biology 471 Evolutionary Mechanisms Name_...

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Biology 471 – Evolutionary Mechanisms Name__________________________ EXAM 3- April 3, 2008 1. Name TWO requirements for a population to be in Hardy Weinberg Equilibrium. (2 points) No selection. No mutation. No migration. Random Mating. Infinite Population Size 2. Define frequency dependent selection. (3 points) Selection in which the fitness of a genotype depends on whether it is rare or common. 3. Describe ONE example of frequency dependent selection. (3 points) In evening primrose, flower phenotypes are either pin (long stamen, short pistils) or thrum (short stamen, long pistils). Pins do not pollinate pins, and thrums do not pollinate thrums. In that case, the rare phenotype has an advantage because many more pollinators are available to it. Any plausible example was acceptable if you explained it. 4. Which of the following two schools of thought predicts the existence of more polymorphism in natural populations? Circle the correct answer. (3 points) The Classical School The Neutral Theory 5. In class we talked about an experiment in which Buri monitored evolution in numerous replicate populations of Drosophila, all of which had been founded by 16 flies that were heterozygous at the brown eye locus (in other words, all 16 flies were genotype Bb). These populations were maintained at a size of 16 flies over many generations so that Buri could monitor the effect of Genetic Drift on allele frequency. a. If he had continued to monitor evolution until one of the two alleles had achieved fixation in every population, what proportion of the populations would have fixed the B allele (i.e. achieved frequency(B) = 1). (2 points) 0.5 b. How would your answer in (a) change if Buri had founded his populations with 15 BB flies and 1 bb fly? (2 points) 15/16 = 0.937 1
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Biology 471 – Evolutionary Mechanisms Name__________________________ 6. Three alleles exist at the self-incompatibility locus in Evening Primrose, and allele frequencies are found to be f(A1) = .1, f(A2) = .4, f(A3) = .5. a. Assuming random mating, what is the expected frequency of homozygotes? (I want the proportion of all homozygotes combined). (4 points) (.1)x(.1) + (.4)x(.4) + (.5)x(.5) = 0.01 + 0.16 + 0.25 = 0.42 b. Since this is a self-incompatibility locus, is it reasonable to assume random mating? Circle the correct answer. (2 points) YES NO c. What is the real expected frequency of homozygotes at a self-incompatibility locus? (2 points)
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EXAM 3 2008 Key - Biology 471 Evolutionary Mechanisms Name_...

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