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Unformatted text preview: BSCI205: Genetic Drift Practice Problems 09/23/2010 1. In how many generations will the expected heterozygosity be 5% of the initial value in populations of (a) size 10 and in populations of (b) size 100? Answer: (a) , (b) , . → . . → →. . →. →. →. 1 BSCI205: Genetic Drift Practice Problems 09/23/2010 2. What is the effective population size Ne when Nf = 5 and Nm = 1 for a diploid organism? Answer: → → . 2 BSCI205: Genetic Drift Practice Problems 09/23/2010 3. 20 and 40 founders from a species of snail colonized the Greek islands Paros and Naxos, respectively. The allele frequency A for a two-allele locus (A, a) was 0.30 for the Paros founders and 0.40 for the Naxos founders and both sets of founding populations were in Hardy-Weinberg equilibrium. What is the heterozygosity in each island after 20 generations of random mating? Answer: Paros Island: → ≅. → . . → Naxos Island: → ≅. → . . → 3 BSCI205: Genetic Drift Practice Problems 09/23/2010 4. A population of 100 individuals contains 100 A alleles and 100 a alleles at a single locus and is in Hardy-Weinberg Equilibrium. If no mutation occurs and the three genotypes are selectively neutral, what would you expect the genotype frequencies to be over the long term, such as 10,000 generations? Answer: → → . → ≅ If heterozygosity is 0, then the frequency of either A or a is 0. Thus, we expect either 100% A or 100% a (there is an equal chance of each getting fixed in the population, because they both started at equal frequency). 4 BSCI205: Genetic Drift Practice Problems 09/23/2010 5. If the neutral mutation rate is μ=10-8 at a locus, what is the rate of neutral evolution at that locus if the population size is (a) 100 individuals? (b) 1,000 individuals? Answer: The rate of neutral evolution equals the mutation rate and is not influenced by population size. Thus, the answer to both (a) and (b) is k=μ=10-8. 5 BSCI205: Genetic Drift Practice Problems 09/23/2010 6. (FROM 2009 EXAM) A species is divided in three populations A, B, and C. Population A contains 500 individuals and has a mutation rate of 10-5, population B contains 15,000 individuals and has a mutation rate of 10-6, and population C contains 20 individuals and has a mutation rate of 10-4. For each population calculate the (a) number of new alleles created by mutation in each generation, (b) the chance of a new allele drifting to fixation, and (c) the number of new alleles that are created by mutation in each generation and are destined to drift to fixation. (d) What is the neutral rate of evolution for each population? Answer: (a) The number of new alleles created by mutation is given by 2Nμ population A: 2x500x10-5=10-2 population B: 2x15,000x10-6=3x10-2 population C: 2x20x10-4=4x10-3 (b) The chance that a new allele will drift to fixation is given by 1/2N population A: 1/(2x500)= 10-3 population B: 1/(2x15,000)=3.33x10-5 population C: 1/(2x20)=0.025 (c) The number of new alleles destined to drift to fixation is given by the mutation rate μ population A: 10-5 population B: 10-6 population C: 10-4 (d) The neutral rate of evolution for each population is equal to the mutation rate (same answer as in section (c)). 6 ...
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This note was uploaded on 10/24/2010 for the course BSCI 205 taught by Professor Rokas during the Spring '10 term at Vanderbilt.
- Spring '10