Midterm Key66

Midterm Key66 - NAME_ POPULATION GENETICS AND...

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NAME______________________________________________________ POPULATION GENETICS AND MICROEVOLUTIONARY THEORY MIDTERM EXAMINATION WRITE YOUR NAME ON EVERY PAGE! The following table may be useful in answering one or more of the questions. Use it as needed: Chi-Square Test Statistic Value Probability df = 1 df = 2 df = 3 df=4 0.10 2.706 4.605 6.251 7.779 0.05 3.841 5.991 7.815 9.488 0.01 6.635 9.210 11.345 13.277 1. Briefly define (5 points each): a) Evolution A change through time of the frequencies of alleles or allele combinations in the gene pool. b) Infinite sites model Every mutation is at a different nucleotide site from all previous mutations. c) AMOVA A partitioning of genetic variation within and among populations that is similar to f st but that uses a molecule genetic distance to compare paired homologous sequences rather than heterozygosity.
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NAME______________________________________________________ 2. A captive population of an endangered species is surveyed at an autosomal locus with two alleles, yielding the following results: Genotype AA Aa aa Total Number 30 60 10 100 a. (5 points) Test the null hypothesis that this population has Hardy-Weinberg genotype frequencies. b. Genotype AA Aa aa Sum: Numbers 30 60 10 100 Genotype Freq. 0.3000 0.6000 0.1000 1 Alleles A a Allele Freq. 0.600 0.400 1 1 pt H-W. Freq. 0.3600 0.4800 0.1600 1 Exp. 36.000 48.000 16.000 100 1 pt (o-e)2/e 1 3 2.25 6.25 0.012419332 Df=1, 1 pt. chi-square, 1 pt p-value<.05, 1 pt
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NAME______________________________________________________ b. (4 points) An analysis of the pedigrees of the individuals in the genetic survey reveals that their average inbreeding coefficient is 0.271. Is this result consistent with your answer to part a), and why or why not? It is consistent. Even though part a) shows a significant excess of heterozygotes, and hence f < 0, since this inbreeding coefficient is calculated from pedigree data, it must be the average F , and reflects more of the role of genetic drift and not system of mating (just like in the example of Speke’s gazelles). c. (6 points) The captive population was established 3 generations ago.
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This note was uploaded on 12/14/2009 for the course CHEM CHEM 140B taught by Professor Burkhart during the Spring '08 term at UCSD.

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Midterm Key66 - NAME_ POPULATION GENETICS AND...

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