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Trees & Topologies Chapter 3, Part 1

Trees & Topologies Chapter 3, Part 1. Terminology. Equivalence Classes – specific separation of a set of genes into disjoint sets covering the whole set of genes Jump Process – describes which pair of genes coalesce at each coalescence event

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Trees & Topologies Chapter 3, Part 1

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  1. Trees & TopologiesChapter 3, Part 1

  2. Terminology • Equivalence Classes – specific separation of a set of genes into disjoint sets covering the whole set of genes • Jump Process – describes which pair of genes coalesce at each coalescence event • Waiting Time Process – the waiting time to the next coalescent event when there are k genes left COMP 790-Trees & Topologies

  3. Coalescent Tree COMP 790-Trees & Topologies

  4. Coalescent vs. Phylogenic Trees • Phylogenetic tree: branch length = #of mutations • Coalescent tree: branch length = time to coalescence (coalescent time x 2N generations x generation time) • Expected number of mutations = /2 Coalescent time Four representations of a coalescent tree Rooted Phylogenetic Tree COMP 790-Trees & Topologies

  5. Counting Trees & Topologies (Ck) # of coalescent topologies with k leaves (Bk) # of binary unrooted tree topologies with k leaves COMP 790-Trees & Topologies

  6. Recursion Illustrated Basic recursion for the number of unrooted tree topologies as a function of leaves COMP 790-Trees & Topologies

  7. Recurrence Intuition COMP 790-Trees & Topologies

  8. Gene Trees • Graph that shows the ancestral relationship between genes. • Assume infinite sites model to build gene trees. (Ch. 5 discusses what happens without this assumption) • Not a coalescent tree. • Clusters genes according to their type and mutation pattern. COMP 790-Trees & Topologies

  9. Example Gene Tree Data set with five sequences and four segregating sites with relative positions. Built up, starting with first site, and continually adding more sites to the tree. COMP 790-Trees & Topologies

  10. Building Gene Trees • Determine if data passes 4-gamete test. If not, there cannot be a gene tree. • If each column is a binary number, sort the numbers in decreasing order, with largest binary number in column one. • Add each sequence with all its characters one at a time. The characters of a sequence to be added is a specific row, which is read right to left. The sequence is placed by tracing from the leaves towards the root. It has its own edges until the prefix is encountered where it coincides with the last added character. • Root is labeled with an open circle. It can be removed to form an unrooted tree. COMP 790-Trees & Topologies

  11. Example Given the following table, build a gene tree. • Determine if data passes 4-gamete test. If not, there cannot be a gene tree. • If each column is a binary number, sort the numbers in decreasing order, with largest binary number in column one. • Add each sequence with all its characters one at a time. The characters of a sequence to be added is a specific row, which is read right to left. The sequence is placed by tracing from the leaves towards the root. It has its own edges until the prefix is encountered where it coincides with the last added character. • Root is labeled with an open circle. It can be removed to form an unrooted tree. COMP 790-Trees & Topologies

  12. Nested Subsamples • Assume a sample A, is taken of size n, and within that sample a subsample B, of size m is taken, m  n. • Process describing the number of ancestors starts out in (m,n) and jumps to either (m,n-1) or (m-1,n-1) COMP 790-Trees & Topologies

  13. More nested subsamples • Probability that the MRCA of B is also the MRCA of A • Special case: A is the whole population (n  , or n = 2N, and 2N is large) COMP 790-Trees & Topologies

  14. More nested subsamples Remember: time until whole population has found a MRCA is 2 (in coalescent units) and the time until a sample of size two has found a MRCA is 1. COMP 790-Trees & Topologies

  15. Hanging Subtrees COMP 790-Trees & Topologies

  16. Unbalanced Trees • Probability that the basal split into two lineages at the root of the tree results in the labeled, unordered partition (i, n-i), i = 1,2,…,n/2 • In large samples, unbalanced trees are unlikely. COMP 790-Trees & Topologies

  17. Neanderthal Example • Nordborg(1998) studied the tree of a combined sample of 986 human mitochondrial sequences and 1 Neanderthal sequence. • Assuming random mating: 2 /(986 *985) = 2 * 10-6 • Nordborg pointed out that a large part of the human sample had found a common ancestor during the time the sequence Neanderthal lived (30,000-100,000 years ago) • For example, if there were 5 ancestors to present human sample 30,000 years ago, the probability is 2 /(5*4) = 10%. • Does not provide strong evidence against interbreeding between Neanderthals and humans. COMP 790-Trees & Topologies

  18. Next Time • More Trees & Topologies • A single lineage • Disjoint subsamples • A sample partitioned by a mutation • The probability of going from n ancestors to k ancestors. COMP 790-Trees & Topologies

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