Transcript Coalescent Tree

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
• 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
Four representations of a coalescent tree
Rooted Phylogenetic Tree
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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
Basic recursion for the number of unrooted tree topologies as a function of leaves
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Recurrence Intuition
K
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Bk 1
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105
945
10395
2027025
7.9x1012 2.2x1020
Ck 1
3
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180
2700
56700 1587600
2571912000 7.0x1018 5.6x1029
1E+18
1E+16
1E+14
1E+12
1E+10
B
100000000
C
1000000
10000
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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
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.
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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.
2.
3.
4.
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.
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B
C
D
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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
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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
Probability (A = B)
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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.
1
0.8
0.6
0.4
0.2
0
M
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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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