Part+IIproblem+set+answers

Part+IIproblem+set+answers - Part II. Questions form Old...

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Part II. Questions form Old Exams and Questions on Population Genetics 1. 0.18, no. 2. e. You could calculate this if you knew the population was at HW equilibrium. 3. c 4. B. Write it out as a 3 x 3 Prout square and you will see. 5. e 6. b 7. d 8. c 9. a: gene flow; b. fewer alleles, c: bottlenecks, d: drift, e: non-random mating. The key word is join (as opposed to create—founder effects). 10. c 11. b. The key is that positive assortative mating (inbreeding) creates a dearth of heterozygotes. 12. a. 0.62; b. no 13. e. There are two conditions under which we can get genotypes from phenotypes. 1) knowing a popn is at HW (and we know this one is not, given the behavior of the pollinators); 2) incomplete dominance so that heterozygotes advertise their genotype through their phenotype. 14. There is no good phenotypic definition of evolution: phenotypes changing through time can be because of underlying genotypic change (evolution) or environmental change and the expression of a non-changing genotype. 15. The population (communities and ecosystems do not share a uniform set of alleles
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This note was uploaded on 10/08/2009 for the course BIS 2 taught by Professor Schwartzandkeen during the Spring '09 term at UC Davis.

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