{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}


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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online