Molecular Evolution, course # Final Exam, May 3, 2006
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1 Molecular Evolution, course #27615 Final Exam, May 3, 2006 This exam includes a total of 12 problems on 7 pages (including this cover page). The maximum number of points obtainable is 150, and at least 85 points are required to pass. Note that different problems are worth different amounts of points. Problems should be answered in the space provided on these pages. The duration of the exam is 4 hours (13:00-17:00). All non-living resources may be used (books, internet including exercise web-pages, notes, etc. You should, however, not contact your uncle who is a professor of evolutionary biology at Harvard). You can work on your own, or in the groups you worked in during the computer exercises. Please do not discuss the test outside your group. Members of the group must sign below: Name Student ID number Signature 1
2 Table of chi-square critical values Significance level DF
3 1: (5 points) You are performing an experiment on a population of haploid, asexual organisms. Assume that the organisms exhibit exponential growth with discrete, non-overlapping generations, and that genetic drift can be ignored. At a certain locus you observe two alleles: Frequent and Rare. The absolute fitness of both alleles is R = 5 offspring/generation. At the start of the experiment the Rare allele is present in 7% of the population. What is the frequency of the Rare allele after 10 generations? 2: (10 points) Below we have listed four widely used nucleotide substitution models and a list of four PAUP model specifications. Use lines to connect the model names with the corresponding PAUP specifications: (A) Jukes and Cantor model (JC) (B) General Time Reversible model (GTR) (C) Felsenstein 81 (F81) (D) Kimura 2 parameter model (K2P) (1) lset nst=1 basefreq=empirical (2) lset nst=2 basefreq=equal tratio=estimate (3) lset nst=1 basefreq=equal (4) lset nst=6 basefreq=empirical rmatrix=estimate 3: (5 points) You have a strictly bifurcating, unrooted tree with 37 tips. In how many different positions can you place a root? 4: (15 points) You are using parsimony to reconstruct the phylogeny of a set of 9 sequences. Each of your sequences is 1 nucleotide long (!). What is the length of the two trees below? Which tree is the best one according to the parsimony criterion? 3
4 5: (15 points) Below is an alignment of two 45 bp long DNA sequences: Seq_1: AGCTGGCATTGCGTTATTATCATTGCATGCATCTTGTCATATGGC Seq_2: AGCAGAGTCTCCGCTGTTATCGATGTTTCCATCTTGCCTTACAGT What is the (uncorrected) distance between these two sequences? The above result does not take multiple substitutions into account. Assuming the Jukes and Cantor model of evolution, compute the estimated, corrected distance between the sequences: Now, assume the Kimura 2-parameter model of evolution, and again compute the estimated, corrected distance between the sequences: 6: (5 points) Under the so-called HKY85 model of evolution, the likelihood of a tree will depend on where the root is placed 1) True 2) False 7: (5 points) The likelihood of a model is (select one): 1) The probability of the data given the model P(Data Model) 2) The probability of the model given the data P(Model Data) 3) The probability of the model before any evidence has been collected P(Model) 4) The total probability of the data P(Data) 8: (5 points) Branch and bound searching of tree-space is guaranteed to always find the best possible tree: 1) True 2) False Heuristic searching of tree-space is guaranteed to always find the best possible tree: 1) True 2) False 4
5 9: (15 points) As a consequence of your graduate studies at the University of Copenhagen you have become interested in the pubic louse - Pthirus pubis. You therefore want to construct a maximum likelihood tree for a set of sequences from 35 species of lice. Your first step is to decide which model to use. Unfortunately your supervisor has run out of funding and you only have access to some extremely cheap Swedish software that is able to fit just two different models to your data: the Kimura 2-parameter model (K2P) and the general time reversible model (GTR). Below we have listed the number of free parameters in the different models (K) along with the maximal log likelihood (lnl) of the fitted models. Use likelihood ratio testing to decide which model is best (use a significance level of 5%). Model K lnl K2P GTR : (10 points) Below we have listed the names of five different methods for reconstructing phylogenies as well as five very brief descriptions of the methods. Use lines to connect the method names with the corresponding descriptions: (A) Maximum likelihood (B) Least squares (1) The ultimate goal of this method is to find the full posterior probability distribution over trees and other parameters (2) The best tree is the one implying fewest mutations (C) Parsimony (D) Bayesian (E) Minimum evolution (3) For a given tree topology, the best branch lengths are those having the smallest deviation between observed and patristic distances; the best tree is the shortest one. (4) The best tree, and the best branch lengths for a given tree, are those having the smallest deviation between observed and patristic distances. (5) The best tree is the one giving the highest probability of the alignment having evolved from some common ancestor. 5
6 11: (35 points) You are investigating an outbreak of Sundbus elsinorii virus. This virus attacks only Latvian chipmunks and is characteristic in having a genome that is merely 2 nucleotides long (its fascinating life-cycle involves heavy use of programmed frame-shifting during translation of the encoded poly-protein). From a population of chipmunks living in cages at the Latvia University of Agriculture (Latvijas Lauksaimniecibas Universitate) you have obtained 6 fullgenome sequences that are known to be related as shown on the unrooted tree below. Note that two of the sequences are placed at the ancestral nodes. All branches have the same length t 1. Indicated below the tree is the time-reversible substitution-probability matrix corresponding to this branch length (specifically, the matrix is of the F81 type). The equilibrium frequencies of the nucleotides are: π A =0.2, π C =0.3, π G =0.4, π T =0.1. Compute the probability of the data given the information provided here. (Tip: perform the computation one nucleotide position at a time and combine the results). to A C G T A From C t 1 G T
7 12: (25 points) During a ski trip in Norway with 5 friends from your water polo team, you suddenly remember that your supervisor asked you to finish a Bayesian analysis of a set of mitochondrial sequences (obtained from a population of Latvian chipmunks) just before you left. Regrettably, you forgot to do this. However, you fortunately brought a printout of the full data set (in fasta-format) with you to Norway. Since there is no electricity (not to mention a computer) in the ski hut you decide to do the analysis by hand and then send the result to your supervisor by ordinary surface mail. To simplify the job somewhat, you decide to use the Kimura 2 parameter substitution model in a discretized version where only 6 possible values of the transition/transversion ratio are considered: T=1.0, T=2.0, T=3.0, T=4.0, T=5.0, and T=6.0. Based on your previous experience with mitochondrial sequences you have specified the following prior probabilities for the six possible parameter values: Prior probabilities: P(T=1.0) = 0.05 P(T=2.0) = 0.10 P(T=3.0) = 0.25 P(T=4.0) = 0.40 P(T=5.0) = 0.15 P(T=6.0) = 0.05 For each of these possible values you have furthermore computed the probability of the data given that particular parameter value (the likelihood): Likelihoods: P(D T=1.0) = 0.01 P(D T=2.0) = 0.05 P(D T=3.0) = 0.15 P(D T=4.0) = 0.23 P(D T=5.0) = 0.78 P(D T=6.0) = 0.15 Find the posterior probability of the six possible parameter values: P(T=1.0 D) = P(T=2.0 D) = P(T=3.0 D) = P(T=4.0 D) = P(T=5.0 D) = P(T=6.0 D) = 7
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