1 / 12

Tree Reconstruction

Tree Reconstruction. Basic Principles of Phylogenetics Distance Parsimony Compatibility Inconsistency Likelihood. Central Principles of Phylogeny Reconstruction. s1. s1. s1. s3. s3. s3. s2. s2. s2. s4. s4. s4. 0. 1. 2. Parsimony. Total Weight: 3. 0. 0.

Download Presentation

Tree Reconstruction

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Tree Reconstruction Basic Principles of Phylogenetics Distance Parsimony Compatibility Inconsistency Likelihood

  2. Central Principles of Phylogeny Reconstruction s1 s1 s1 s3 s3 s3 s2 s2 s2 s4 s4 s4 0 1 2 Parsimony Total Weight: 3 0 0 0.6 1 3 2 3 2 0 0.7 1.5 Distance 0.4 0.3 L=3.1*10-7 Parameter estimates Likelihood TTCAGT TCCAGT GCCAAT GCCAAT

  3. Molecular clock a b c d e a - 22 10 22 22 b 7 - 22 16 14 c 7 8 - 22 22 d 12 13 9 - 16 e 13 14 10 13 - No Molecular clock c 11 a d 3 2 6 8 2 1 7 5 7 4 e b d a c e b a 7 7 c b e b 8 14 From Distance to Phylogenies What is the relationship of a, b, c, d & e?

  4. UGPMA Unweighted Group Pairs Method using Arithmetic Averages A B C D E A 1715 2147 3091 2326 B 2991 3399 2058 C 2795 3943 D 4289 E UGPMA can fail: 857 ? A B 1096 AB C D E AB 2529 3245 2192 C 2795 3943 D 4289 E 857 C A B E A and B are siblings, but A and C are closest 1347 ABE C D ABE 3027 3593 C 2795 D A 1096 857 B A D E C B 1655 Siblings will have [d(A,?)+d(B,?)-d(A,B)]/2 maximal. 1347 1096 ABE CD ABE 3310 CD 857 A D E B C From Molecular Systematics p486

  5. Assignment to internal nodes: The simple way. A G T C ? ? ? ? ? ? C C C A What is the cheapest assignment of nucleotides to internal nodes, given some (symmetric) distance function d(N1,N2)?? If there are k leaves, there are k-2 internal nodes and 4k-2 possible assignments of nucleotides. For k=22, this is more than 1012.

  6. 3 5 4 6 13 11 9 7 15 17 14 10 12 16 8 2 1 5S RNA Alignment & Phylogeny Hein, 1990 Transitions 2, transversions 5 Total weight 843. 10 tatt-ctggtgtcccaggcgtagaggaaccacaccgatccatctcgaacttggtggtgaaactctgccgcggt--aaccaatact-cg-gg-gggggccct-gcggaaaaatagctcgatgccagga--ta 17 t--t-ctggtgtcccaggcgtagaggaaccacaccaatccatcccgaacttggtggtgaaactctgctgcggt--ga-cgatact-tg-gg-gggagcccg-atggaaaaatagctcgatgccagga--t- 9 t--t-ctggtgtctcaggcgtggaggaaccacaccaatccatcccgaacttggtggtgaaactctattgcggt--ga-cgatactgta-gg-ggaagcccg-atggaaaaatagctcgacgccagga--t- 14 t----ctggtggccatggcgtagaggaaacaccccatcccataccgaactcggcagttaagctctgctgcgcc--ga-tggtact-tg-gg-gggagcccg-ctgggaaaataggacgctgccag-a--t- 3 t----ctggtgatgatggcggaggggacacacccgttcccataccgaacacggccgttaagccctccagcgcc--aa-tggtact-tgctc-cgcagggag-ccgggagagtaggacgtcgccag-g--c- 11 t----ctggtggcgatggcgaagaggacacacccgttcccataccgaacacggcagttaagctctccagcgcc--ga-tggtact-tg-gg-ggcagtccg-ctgggagagtaggacgctgccag-g--c- 4 t----ctggtggcgatagcgagaaggtcacacccgttcccataccgaacacggaagttaagcttctcagcgcc--ga-tggtagt-ta-gg-ggctgtccc-ctgtgagagtaggacgctgccag-g--c- 15 g----cctgcggccatagcaccgtgaaagcaccccatcccat-ccgaactcggcagttaagcacggttgcgcccaga-tagtact-tg-ggtgggagaccgcctgggaaacctggatgctgcaag-c--t- 8 g----cctacggccatcccaccctggtaacgcccgatctcgt-ctgatctcggaagctaagcagggtcgggcctggt-tagtact-tg-gatgggagacctcctgggaataccgggtgctgtagg-ct-t- 12 g----cctacggccataccaccctgaaagcaccccatcccgt-ccgatctgggaagttaagcagggttgagcccagt-tagtact-tg-gatgggagaccgcctgggaatcctgggtgctgtagg-c--t- 7 g----cttacgaccatatcacgttgaatgcacgccatcccgt-ccgatctggcaagttaagcaacgttgagtccagt-tagtact-tg-gatcggagacggcctgggaatcctggatgttgtaag-c--t- 16 g----cctacggccatagcaccctgaaagcaccccatcccgt-ccgatctgggaagttaagcagggttgcgcccagt-tagtact-tg-ggtgggagaccgcctgggaatcctgggtgctgtagg-c--t- 1 a----tccacggccataggactctgaaagcactgcatcccgt-ccgatctgcaaagttaaccagagtaccgcccagt-tagtacc-ac-ggtgggggaccacgcgggaatcctgggtgctgt-gg-t--t- 18 a----tccacggccataggactctgaaagcaccgcatcccgt-ccgatctgcgaagttaaacagagtaccgcccagt-tagtacc-ac-ggtgggggaccacatgggaatcctgggtgctgt-gg-t--t- 2 a----tccacggccataggactgtgaaagcaccgcatcccgt-ctgatctgcgcagttaaacacagtgccgcctagt-tagtacc-at-ggtgggggaccacatgggaatcctgggtgctgt-gg-t--t- 5 g---tggtgcggtcataccagcgctaatgcaccggatcccat-cagaactccgcagttaagcgcgcttgggccagaa-cagtact-gg-gatgggtgacctcccgggaagtcctggtgccgcacc-c--c- 13 g----ggtgcggtcataccagcgttaatgcaccggatcccat-cagaactccgcagttaagcgcgcttgggccagcc-tagtact-ag-gatgggtgacctcctgggaagtcctgatgctgcacc-c--t- 6 g----ggtgcgatcataccagcgttaatgcaccggatcccat-cagaactccgcagttaagcgcgcttgggttggag-tagtact-ag-gatgggtgacctcctgggaagtcctaatattgcacc-c-tt-

  7. Cost of a history - minimizing over internal states d(C,G) +wC(left subtree) A CGT A CGT A C G T

  8. Cost of a history – leaves (initialisation). A C G T Initialisation: leaves Cost(N)= 0 if N is at leaf, otherwise infinity G A Empty Cost 0 Empty Cost 0

  9. Compatibility and Branch Popping Definition: Two columns can be placed on the same tree – each explained by 1 mutation. A GCACGTGCAGTTAGGA B GCACGTGCAGTTAGGA C TCTCGTGCAGTTAGGA D TCTCATGCAATTAGGA E TCTCATGCAATTATGA F TCTCATGCAATTATGA ABC This is equivalent to: In the two columns only 3 or the 4 possible character pairs are observed EFG Multistate Definition: The number of mutations needed to explain a pair of columns is the sum of the mutations needed to explain the individual columns ABC A GCACGTGCAGTTAGGA B GCACGTGCAGTTAGGA C TCTCGTGCAGTTAGGA D TCTCATGCAATTAGGA E TCTCATGCAATTATGA F TCTCATGCAATTATGA E EF For imperfect data: Find the maximal compatible set of characters and then branch-pop A GCACGTGCAGTTAGGA B GCACGTGCAGTTAGGA C TCTCGTGCAGTTAGGA D TCTCATGCAATTAGGA E TCTCATGCAATTATGA F TCTCATGCAATTATGA C AB 1 2 3 4 5 6 1 +? ? ? ? ? 2 +? ? ? ? 3 +? ? ? 4 +? ? 5 +? 6 + D EF

  10. The Felsenstein Zone Felsenstein-Cavendar (1979) Reconstructed Tree True Tree s1 s4 s1 s2 s3 s4 s2 s3 Patterns:(16 only 8 shown) 0 1 0 0 00 0 0 0 0 1 0 01 0 1 0 0 0 1 01 1 0 0 0 0 0 10 1 1

  11. Hadamard Conjugation & binary characters on a tree Closely related to inclusion-exclusion principle and Sieve Methods • 1 • 1 -1 H1= Hk-1 Hk-1 Hk-1 -Hk-1 Hk= Branch lengths – s, Bipartition lengths - q q=Hs From branch lengths to bipartitions Inconsistency in presence of a Clock: D D A A E E C C B B s=H-1 q From bipartition to lengths True Tree with Clock More Likely Tree Felsenstein (2004) Inferring Phylogenies p 118

  12. Bootstrapping Felsenstein (1985) 1 500 2 ATCTGTAGTCT ATCTGTAGTCT ATCTGTAGTCT ATCTGTAGTCT ?????????? ?????????? ?????????? ?????????? ?????????? ?????????? ?????????? ?????????? 2 3 2 3 2 3 4 4 1 1 4 1 ATCTGTAGTCT ATCTGTAGTCT ATCTGTAGTCT ATCTGTAGTCT 10230101201

More Related