Biology 3300- Phylogenies

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BIOL 3300
Theresa Burg

Biology 3300 Feb.28 , 2014 UPGMA  unweighted pairwise group method with arithmetic average 1) Estimate all pairwise distances - A transition and transversion will not occur at the same rate - Can weigh more heavily if there is a transversion - Have individuals who are closely related (A and B at 0.08) - Have individuals who are not as closely related (C and D at 0.80) 2) Take smallest distance d AB = 0.08 , connect dOA = dOB = 0.04 3) Next smallest distance d DE= 0.12 4) Now we need to connect OTU (B) that is already in a clade d CB = 0.17 - d = (d AB+ dBC /2 = 0.18 5) Finally connect clades ABC and DE - d = avg d ADd ,BD ,CD , AE,aBE d CE Problem with UPGMA is it assumes all rates are equal - What can influence the rate of evolution? o Selection o Generation time o DNA vs. RNA, double stranded vs. single stranded o Amount of offspring/ population size o Type of reproduction (asexual vs. sexual) o BMR; metabolic rate Principle of Parsimony - The simplest explanation is almost always the right one o Smallest number of changes is probably the right tree o Alternative methods  Maximum likelihood • Probability approach, what is the probability of observing data under different models of evolution? • Computationally VERY SLOW  Bayesian inferences • Creates a 3D space • Local vs. global peaks • Problem is you can get stuck in the local peaks rather than finding the global peaks • The global peak is the highest peak and is the best tree • Some limitations depending on how you set up your program - Parsimony score: number of character changes (mutations) along the evolutionary tree o The lowest score will be considered the most parsimonious tree and the best - Most p
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