Trees & Topologies Chapter 3, Part 1. Terminology Equivalence Classes – specific separation of a set of genes into disjoint sets covering the whole set.

Trees & TopologiesChapter 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

• Waiting Time Process – the waiting time to the next coalescent event when there are k genes left

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Coalescent Tree

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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

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Rooted Phylogenetic Tree

Four representations of a coalescent tree

Counting Trees & Topologies

(Ck) # of coalescent topologies with k leaves

(Bk) # of binary unrooted tree topologies with k leaves

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Recursion Illustrated

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Basic recursion for the number of unrooted tree topologies as a function of leaves

Recurrence Intuition

K 2 3 4 5 6 7 8 10 15 20

Bk 1 1 3 15 105 945 10395 2027025 7.9x1012 2.2x1020

Ck 1 3 18 180 2700 56700 1587600 2571912000 7.0x1018 5.6x1029

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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.

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Example Gene Tree

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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.

Building Gene Trees

1. Determine if data passes 4-gamete test. If not, there cannot be a gene tree.

2. If each column is a binary number, sort the numbers in decreasing order, with largest binary number in column one.

3. 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.

4. Root is labeled with an open circle. It can be removed to form an unrooted tree.

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Example

Given the following table, build a gene tree.1. Determine if data passes 4-gamete test. If not, there cannot be a gene tree.2. If each column is a binary number, sort the numbers in decreasing order, with

largest binary number in column one.3. 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.

4. Root is labeled with an open circle. It can be removed to form an unrooted tree.

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A B C D

1. 0 0 1 0

2. 0 0 0 1

3. 1 0 0 0

4. 0 0 0 1

5. 1 1 0 0

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)

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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)

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More nested subsamples

M 1 2 3 5 9 19 29

P (A = B) 0 / 2 (no info) 1/3 1/2 2/3 = 0.67 4/5 = 0.80 9/10 = 0.90 14/15 = 0.9333

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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.

Hanging Subtrees

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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.

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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.

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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.

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